16_NCM10extSB_ans Page 612 Thursday, May 18, 2006 11:01 AMAnswers Chapter 1 10 Any number greater than Start up 11 56 b −6 f −10 j −32 1 a 3 e −8 i 3 2 a 3 --7 b 3 a 7400 e 0.535 4 a 48% 3 --5 6 a e 7 a d g j 8 a d g 9 a d g 10 a 11 a 12 a d 7 -----25 d −48 h 9 l −2 c 25 g 0 k 2 c 17 -----40 9 -----16 d $8.70 b $15 f 1:1 1 : 30 1 : 10 3:2 7:5 18 : 13 4:7 9:8 5:6 5:6 7:5 b 4:3:6 3:8 12 : 35 21 -----25 1 -----40 d e 1 --2- -----e 2 21 50 27 : 40 3:1 4:5 1:4 d 2 : 15 17 : 5 : 13 9 : 80 9 : 50 000 11 -----4 13 b − ----- e − 5--2 a 2 1--- b 1 1--- 1 1--9 2 4--7 7 3 a 100 --------3 2 f 1 --7 d − --18 21 -----4 c −4 1--- d -1 2--- 1 2--3 − 3--2 4 b e 4 -----11 = 3 c 15 -----18 c = −1 1--2 3 --5 d 4 --3 e 8 -----11 c 6 1--- d 4 4--- 9 e 2 ----- f 1 5--- -----g 2 23 1 h 6 ----- 20 15 -----28 f 3 --13 1 --2 17 -----20 5 24 10 2 50 2 5 9 -----10 c 1 --3- 19 - e d 1 ----- g 14 -----45 h 2 1--- i 4 21 7 10 612 0.12 0.2 1.8 6.63 0.663 0.663 d h l d h l 0.36 8.8 0.028 0.663 663 0.0663 c 4 8--9 g 39 --------100 b 1.075, terminating c 0.32, terminating d 0.7˙ 2˙ , recurring 5 --8 17 -----50 b 25 d 9 1--3 -----h 2 16 37 3 - d c 5 ----- e 4 --5 41 --------200 h i 3 --5- j 7 -----20 c 0.666 d 10 1 --------400 b 0.33 1 --------250 8 13 -----40 e 2 --9 4 a 7 --9 b 38 -----99 c 875 --------999 d 22 -----45 e 17 --------198 g 35 --------198 h 2 --------225 i 541 --------990 -----j 1 37 f 5 -----48 -----j 2 19 13 -----20 f 0.55 1 -----11 99 Skillbank 1B 2 a e i 4 a e i 0.5 90 0.08 16 1.6 0.016 b f j b f j 90 0.7 3.5 160 16 160 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.2/ 5.3 c g k c g k 8 8 4 1.6 160 0.16 d h l d h l 0.9 30 0.3 1.6 0.16 0.0016 g 1 --3 h 3 --8 d 1 --------200 e 4 i 1 1--2 j 11 --------200 b 0.08 e 0.0004 c 0.042 3 a 32.5% b 1.5% c 66 --2- % e 180% f 325% h 65% i 60% 87 1--2 % g 133 1--- % 3 4 5 6 b 0.1 c 636.0 d 3.765 f 2.47 g 8.00 h 244.00 j 24.371 k 20 l 3.092 b 12.6 c 36.843 e 245.6615 f 7.37 h 54.87 i 2.9861 b 19.71 c 1 600 000 e 0.0079 f 1.00 h 0.31 i 8 000 000 k 4 l 81 3 c 2 d 4 e 2 4 h 5 i 2 b 2.72 c 0.04729 e 3 360 000 f 0.002 7 43 500 6091.2 cents b $60.90 i 2 ii 4 10 B 11 A 3 a 0.79 e 22 1--- d 6 5 7 f − ----- c g k c g k 56.37 19 0.7 3.90 0.6154 8.1 56 000 30 000 3.01 0.0520 3 b 2 g 4000 0.94 25 3 2 - j 8 --h 17 1--- i 8 ----- 3 --8 1 a e i 2 a d g 3 a d g j 4 a f 5 a d 6 1 8 a c 9 B 6 - g f 3 ----- 5 b b e 12 3--- 2 c 2 2--7 3 9 a 1 --5 g 7 --6- 8 a 6 --2f 6 b 2.4 0.027 0.012 6630 663 66.3 j 0.85, terminating b 3 1--- 7 a b f j b f j Exercise 1-02 2 a 7 6 a 1 ----- 4 3.5 0.8 0.24 66.3 6.63 6630 33 --------400 7 -----25 2 a 0.37 d 1.15 d e 1.375, terminating f 0.3˙, recurring g 4.6, terminating h 0.46˙ , recurring i 5.6˙ , recurring 10 5 a 40 16 $180 Exercise 1-03 7 f − ----- 29, 30, 31, 32, 33, 34 33, 34, 35 13, 14, 15, 16, 17 49, 50, 51, 52, 53, 54, 55 12 -----15 f 7 Skillbank 1A 2 a e i 4 a e i 20 20 1 a 0.7˙ , recurring 5 --3 4 a b c d b d 15 -----8 f 2 e c 8 Exercise 1-04 7 9 - b 3 ------ c 1 a 1 ----- 3 Exercise 1-01 1 a and less than 12 2 --4- times 5 $24 c 14 kg d 32 L $28 b 2:1 c 1:2 e 13 : 12 f 5:1 h 1:2 i 1 000 000 : 1 b 3:8 c e 3:2 f h 4:3 b 21 : 16 c e 25 : 1 f h 10 : 3 1:5 c 20 : 1 b 6:3:8 c b 3:4 c e 16 : 5 f 15 -----20 11 e 8.45% 20 31 -----40 (to the nearest second) 14 15 15 4 2--- cans 3 9 - c b 1 ----- such as 213 2 ----minutes or 2 minutes 11 seconds b 158 c 0.3 d 0.0301 f 0.0072 b 66 2--- % c 175% d 10.8% 5 a 16 ------ , 20 7 8 10 12 j 68% a $24 d $5100 a $50.40 d 1.827 cm a $101.20 d $921.38 a $1406 $37.50 $20.05 a 58.3˙ % 19 20 22 23 $161 c 36 kg $35.70 f $2.98 $1500 c 13.5 kg 97.2 kg f $1940.13 $20.91 c 90.2 kg 12.9 L f 132 m b $1258 9 $70 157.50 11 $831.25 b 8% c 76% e 33.3˙ % h 16.6˙ % d 30% g 7.5% 13 28.6% 16 40.6% 18 a 600 d 666.6˙ b e b e b e 3 14 30% 17 14.8% b 900 e 545.4˙ 5˙ f 30% i 7.1% 15 10% c 500 f 68 4--7 g 53.3˙ h 32.6˙ i 102 6--7 a $75 b $750 $16.32 21 19 443 a $4347 b $24 153 $275 24 80 25 $500 26 $560 Exercise 1-05 1 a c e g i 2 C 4 a c e g i k 5 a c e g i 6 a b 7 a d 8 a c e 5.6 × 103 7 × 10−2 7.128 × 102 2.78 × 10−4 9 × 108 b d f h 7.2 × 107 1 × 10−9 4 × 103 5 × 10−1 3 D 37 000 b 0.0987 0.000 000 8 d 15 760 000 0.3 f 80 700 0.000 046 1 h 1 280 000 0.030 61 j 99 100 000 000 0.001 l 0.000 000 021 1.21 × 1013 b 1.44 × 10−15 1.76 × 1017 d 2.37 × 10−4 4.19 × 10−9 f 2.92 × 1015 8.23 × 1023 h 1.21 × 1012 1.96 × 10−5 j 9.13 × 10− 4 Maximum 2.49 × 108 km, minimum 2.06 × 108 km 2.275 × 108 km 1.5 × 104 b 7 × 106 c 3.5 × 105 7.61 × 107 1.9 × 10−3 kg b 1.66 × 10−24 g 1 × 10− 6 m d 1.66 242 × 108 km2 3 × 108 f 2.817 × 10−15 m 16_NCM10extSB_ans Page 613 Thursday, May 18, 2006 11:01 AM Exercise 1-06 1 a d g j m p 2 a d g j m 3 a c e g i 4 a e i m q 5:6 b 2 : 11 e 9:8 h 3:2 k 50 : 3 n 9:4 1:2 b 1 : 10 e 1:8 h 15 : 1 k 1 : 50 000 n 3:2:7 7 : 12 : 4 8:1:6 15 : 4 : 10 12 : 7 : 10 48 b 48 72 f 50 9 j 55 4.8 n 28 55 r 20 7:6 16 : 11 1 : 12 5:6 32 : 15 c f i l o 9 : 13 21 : 55 10 : 1 50 : 1 20 : 1 2:3 c 3 : 40 1:7 f 1 : 10 8:1 i 15 : 1 500 : 3 l 3:8 9:4 o 1 : 300 b 6:1:4 d 3:4:2 f 9 : 8 : 12 : 12 h 4:7:5 c g k o 5 625 39 37.5 d h l p 7 36 4 175 1080 a 32 years $45 000 a 8 cups d 12 cups 6 a 12 -----25 b 38 : 18 = 19 : 9 4 $146.70 b 9 cups c 10 cups = 48% b 60 white, 40 red 7 a 43 : 1 : 6 b 2% c 45 kg plastic, 270 kg glass d 8.3% 8 a 1000 b i 444.44 ii 750 9 972 tickets 10 84 STD calls 11 75 students 12 1.24 kg of nickel and 2.17 kg of copper 13 96 kg 14 30 m3 gravel, 22.5 m3 sand, 7.5 m3 cement 15 a i 12 : 12 = 1 : 1 ii 18 : 6 = 3 : 1 b 18 g Exercise 1-08 1 a 1 : 500 b 1 : 100 c 2000 : 1 d 1 : 50 000 e 1 : 10 f 1:5 g 250 : 1 h 1000 : 1 i 1 : 200 000 j 800 : 1 k 1:8 l 1 : 120 000 m 500 : 1 n 1 : 937.5 o 1 : 12 500 2 a 1.5 m b 2.4 m c 1.15 m d 2.75 m e 1.9 m 3 C 4 825 m 5 a 12 m b 4m 6 a 3.5 mm b 2.65 mm c 2.25 mm d 1.4 mm e 0.4 mm 7 A 8 D 9 B 10 1 : 2 000 000 11 1 : 25 000 000 12 a i 20 km ii 37.5 km b 9.7 km c 41 km d 2.5 km 13 a 1 : 20 000 000 b 4000 km 14 a 1 : 100 b 80 cm 15 a 1 : 1000 b 1 : 200 Exercise 1-09 1 a 90 km/h c 13 km/L e 18 mm/h 3 4 5 6 7 8 9 10 11 12 13 v 27 727.92 peso vi d $172 e i $1.31 ii $2.38 iv $5.79 v $819.67 Exercise 1-07 1 2 3 5 2 g $34/m h 1250 parts/h i $8.25/bottle j 25 km/day k 42 words/min l 20 g/m2 m $0.90/min n 6.67 m/s o 333.33 mL/s o 11 km/L a $22.50/hour b $787.50 c 120 hours a $11.50/kg b $69 c 3.478 kg Sydney a 2000 L/h b 4000 L a 25 min b 4 h 35 min a 65 km b 162.5 km c 487.5 km 85 km/h a 12 hours b 70.9 km/h a 250 g b 1500 kg a i 255 km ii 495 km b 5.5 hours c midnight a 1 909 090.9 km2 b 930 769 231 a $2360 NZ b $A4594 c i $76.18 US ii $129 F iii 28 154 yen iv 29 820 baht b $36/h d 80c/kg f 76 cents/call 610 000 iii $2.34 vi $12.43 14 L/100 km 11 L/100 km 12.25 L/100 km 12.5 L/100 km 1 a Teacher to check b i 20 100 2 a i 1 iv b i 1 --2 ii 1 --2 iii iii ii 1 1--- iii 2 2 2 1--2 v 4 1--2 + 2 --3 = 1 --1- + 2 --3 + 2 --3 + 1 --3 + 1 --3 1 --2 + 1 --3 + 5 --6 = 7 --1- (1 + N) vi 49 1--2 2 + 1 --4 + 1 --4 + 2 --4 + 2 --4 + 3 --4 =3 + 3 --4 + 1 --5 +… 2 22 --12 iv N ---2 ii 500 500 v 2475 3 3, 37, 111 4 a 3 b 5 c 7 f 13 g 15 h 17 5 2519 6 a i 15.5% increase ii 4% decrease iii 8% decrease iv 17.2% decrease d 9 i 19 e 11 17 16 2--- % Exercise 1-10 1 a c e g Power plus 3 b d f h 7.5 L/100 km 8 L/100 km 12.8 L/100 km 10 L/100 km w w y x x 18 a ---- , ---- , -- , -- , ---x y x y w b Depends on the values assigned to the pronumerals, e.g. if x = 3, y = 2, w = 1 i 4 2--- L/100 km 3 2 a 500 km b 50 km c 150 km d 25 km e 47.9 km f 8.3 km 3 a 67.6 L b 39.52 L c 152.88 L d 6.864 L e 0.26 L f 78 L 4 a 61.5 L b $67.59 5 a 1 birth/1000 people b 2.5 births/1000 people c 13 births/1000 people d 1.2 deaths/1000 people e 34 births/1000 people f 4.5 deaths/1000 people 6 a 240 000 b 130 000 c 5.5 per 1000 7 John uses 324 L, Ken uses 144 L Exercise 1-11 1 a 1167 m/min b 5000 m/min c 22.2 m/s d 26.4 m/s e 30.56 m/s f 1000 m/min g 2400 m/min h 1.8 km/h i 1.5 km/h j 252 km/h k 7.2 km/h l 36 km/h 2 a 50 g/cm b 125 kg/h c 80 mL/g d 0.04 g/m2 e 8.64 kg/day f 54 L/h g 11.52 t/day h 0.000 010 8 s/mm3 i 2 g/cm2 j 1.25 beats/s k 8 m/mL l 300 mL/s 3 a 4500 m b 9 km/h 4 454.55 km 5 66.67 m 6 2480 seconds 7 a 15 m b 20 m c 27.5 m 8 15 per 1000 9 Wonder Gal 10 240 km/h then 3 --2 2--- ; if x = 6, y = 2, w = 1, then 6 --2 1 2 --1 9 10M2 10 a Timmy d --------720 b h Chapter 1 Review 1 1 a − ----- 3 b 2 ----- 3 1--8 -----− 25 32 15 e i 1 2--3 2 a $200 10 f c 9 -----35 19 d 1 ----- g −3 1--4 h 13 1--- 20 3 j 3 2--- k 1 3--- l 3 3--- b 60 kg c 90 d 5 8 3 a 23.75 b 2.303 d −0.8516 e 12.43 4 a 38 920 b 39 000 5 9 461 000 000 000 km 6 a 0.2˙ 7˙ , recurring 4 17 --------120 c 0.6 f 4.859 c 40 000 b 0.575, terminating c 1.5˙ 71428˙ , recurring d 0.95, terminating e 3.6˙ , recurring 7 a 0.84 8 a d 9 a 10 a c 11 a c 12 a c 13 a 21 -----31 b $24.96 b 6.12 m e 62.5 b 1.27 × 105 7 × 10−6 0.000 000 43 0.000 005 8.3 × 1011 5.1 × 1012 2:5 b c 5 --8 $52.80 c 12 kg $132.60 f $10.50 $84.80 c $432.32 b 7.01 × 10−2 d 3.7 × 103 b 87 530 b 2.0 × 1013 d 1.1 × 10−10 4:3 c 1 : 10 ANSWERS 613 16_NCM10extSB_ans Page 614 Thursday, May 18, 2006 11:01 AM d g 14 a d 15 a b 9:2 e 3:8 2:1 h 1:2 6:4:3 b 7:2:3 5:9:6 e 3:8:6 4 : 5 = 12 : 15 = 32 : 40 2 --3 = 12 -----18 = f i c f 1:3 4:9 1:6:2 1:2:4 Exercise 2-03 14 -----21 16 a 180 b 30 c 48 d $16 250 e i 240 : 360 ii 100 : 200 : 300 iii 120 : 60 : 240 : 180 17 a 1 : 400 b 20 : 1 c 1 : 250 d 1 : 10 000 000 18 a 93.33 km b 126.67 km c 133.33 km d 266.67 km 19 a i 36 km/h ii $5.98/kg b i 36.60 ii 40 220 yen iii 15.25 iv $18.65 20 a i 12 L/100 km ii 17.2 L/100 km b i 10.5 births/1000 people ii 6.4 deaths/1000 people 21 a 1.5 kg/min b 0.3456 kL/day c 12.2 L/100 km d 10.53 km/L e 33.33 m/s f 18 km/h Chapter 2 115 b 15.7 c 37.7 d 69.9 201 f 76.1 g 103 h 905 48.5 1681 mm2 b 1650 mm2 1248 mm2 d 4750 mm2 2310 mm2 f 1800 mm2 182 mm2 h 770 mm2 680 mm2 y = 29 b k = 29 d = 19 33 cm, 85 cm2 b 88 cm, 616 cm2 396 cm, 12 469 cm2 581 cm, 26 880 cm2 4.913 m; 1.508 m2 38.562 cm; 88.357 cm2 12.727 m; 8.727 m2 105 m3 b 480 m3 c 640 m3 1 a 282 m2 b 204 m2 c 165 m2 2 a 1036 cm2 b 1020 mm2 c 204 m2 d 390 cm2 e 672 cm2 f 5672 mm2 3 a 57 m2 b 22 m c 269 m2 d 226 m2 e 20 m2 f 40 m2 4 a 657.1 cm2 b 2858.8 cm2 c 2714.2 cm2 d 2440.5 cm2 e 150.3 cm2 5 a 26.14 m2 b 28 m2 6 35 026 cm2 Exercise 2-02 a a a a a 275 m2 166.4 m2 843 cm2 432 cm2 144 m2 b b b b b 614 7 8 9 10 11 12 d a a a a a a b c 3 --5 b 30.16 cm c 4.8 cm 8 cm e 193 cm2 17.3 cm b 16.5 cm 7.1 cm b 5.6 cm c 360 cm2 24.4 cm b 573.1 cm2 3.99 cm b 27.9 cm c 27.6 cm 6.9 cm b 85.3 cm c 85 cm r = 6.31, l = 11.2, h = 9.25 r = 10.9, l = 12.9, h = 6.90 r = 13.1, l = 17.0, h = 10.9 1 a 2827.43 mm2 b 380.13 m2 c 366.44 cm2 2 a 432π m2 b 192π cm2 c 768π mm2 3 a 628 m2 b 314 m2 c 236 m2 d 1257 m2 4 5.1 × 108 km2 5 a 716.3 m2 b 72 L 6 a 6.91 cm b 7.98 cm c 9.77 cm 7 a Teacher to check b i 3800 cm2 ii 8.5 m2 c 6.5 cm Exercise 2-05 1 a d g j m cm2 857.7 5670.9 cm2 282.7 cm2 3769.9 cm2 4805.8 cm2 b e h k n 564 mm2 3456 mm2 1592 cm2 2150 cm2 150 m2 c c c c c 87.4 cm2 743.1 cm2 3116 cm2 173 cm2 138 m2 cm2 412.3 343.0 cm2 652.9 cm2 1148.8 cm2 3908.4 cm2 c f i l o cm2 1042.0 628.3 cm2 537.4 cm2 3095.3 cm2 328 cm2 1 a 64.8 m3 b 1534.5 m3 c 13.6 m3 d 135.7 m3 e 107.5 m3 f 146.3 m3 2 a 1539.4 m3 b 14 431.7 cm3 c 226.4 m3 d 4.7 m3 3 a 23 562 cm3 b 4825 cm3 c 5027 cm3 d 1989 cm3 e 536 cm3 f 12 900 cm3 g 33 117 cm3 h 2639 cm3 i 794 cm3 4 a 20 slices b 138 cm2 c 0.3 m2 5 Triangular prism 36.4 m2, half cylinder 28.8 m2. The triangular tent has the greater surface area. 6 2073.5 cm2 or 0.21 m2 7 a 1508.6 cm3 b 75.4 cm3 8 1028.00 cm3 9 a 29 040 cm3 b 24 L N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.2/ 5.3 12 10.7 cm 15 3375 cm3 149.3 cm3 b 240 cm3 120 cm3 d 336 m3 1200 cm3 f 106.7 m3 950.7 cm3 b 41.1 m3 2400 mm3 d 840 cm3 14 842.7 mm3 f 21.9 m3 7883.3 cm3 h 8064 mm3 i 24 cm ii 3200 cm3 iii 3200 mL b i 20 m ii 19 200 m3 iii 19 200 kL c i 40 mm ii 4320 mm3 iii 4.32 mL d i 60 mm ii 28 160 mm3 iii 28.16 mL e i 7.7 m ii 133.056 m3 iii 133.056 kL f i 84 cm ii 564 480 cm3 iii 56448 mL or 56.448 L a Volume 343 cm3, capacity 343 mL b Volume 400 cm3, capacity 0.4 L c Volume 1152 cm3, capacity 1152 mL d Volume 6100 cm3, capacity 6100 mL e Volume 6048 cm3, capacity 6048 mL f Volume 2500 cm3, capacity 2500 mL a 33.75 m3 b 18.98 tonnes b 0.412 m3 a 946 729 m3 27 m 8 24 cm2 9 19 mm 4.9 cm 1 a c e 2 a c e g 3 a 4 5 6 7 10 Exercise 2-08 209 m3 b 872 cm3 c 1272 mm3 616 cm3 e 393 cm3 f 2545 mm3 12 566.4 cm3 b 3141.6 cm3 18 849.6 cm3 d 4712.4 cm3 Perpendicular height 6.9 cm, volume 115.6 cm3 b Perpendicular height 27.2 m, volume 13 786.1 m3 c Perpendicular height 12.4 cm, volume 378.6 cm3 d Perpendicular height 3.5 m, volume 2.3 m3 e Perpendicular height 244.6 m, volume 296 103.1 m3 f Perpendicular height 71.9 cm, volume 129 674.2 cm3 a i Teacher to check. ii 40 mm iii 3392.9 mm3 b i Teacher to check. ii 12 cm iii 314.2 cm3 c i Teacher to check. ii 3 m iii 8.0 m3 d i Teacher to check. ii 72 mm iii 33 250.6 mm3 a 9 cm b 4 cm c 15 cm d 2 cm a 5.7 m b 9.7 cm c 9.2 cm d 8.5 m 2 cm 8 5.23 mm 1 a d 2 a c 3 a Exercise 2-06 Exercise 2-01 1 2 3 4 5 6 a c 0.0238 kL 10 4948 cm3 11 9.0 m2 13 36 mm 14 3.2 m2 16 64 m2 Exercise 2-07 1 a 101 cm2 b 628 cm2 c 2419 cm2 2 a 392.7 mm2 b 62.8 m2 c 192.4 cm2 3 a 90π b 224π c 577π 4 a 942 cm2 b 393 cm2 c 1017 cm2 d 402 cm2 5 5525 cm2 Exercise 2-04 Start up 1 a e i 2 a c e g i 3 a c 4 a c d 5 a b c 6 a 6 44 436 m2 7 a 1344 mm2 b 180 cm2 c 343.4 m2 8 a 42 m b A = 735 m2 c 35 m d 28 m 4 5 6 7 16_NCM10extSB_ans Page 615 Thursday, May 18, 2006 11:01 AM Exercise 2-09 Power plus mm3 1 a 14 137 b c 660 cm3 d e 1072 cm3 f 2 a 33 510.3 cm3 b c 179.6 cm3 d e 356.8 cm3 f 3 a 4 mm b 7 cm 4 a 11 500 mm3 b c 409 000 mm3 d e 38.8 m3 f 5 1.1 × 1012 km3 m3 697 3619 m3 8579 mm3 97.7 m3 0.9 m3 4.1 cm3 c 3m 5150 cm3 1.07 m3 51 000 mm3 Exercise 2-10 1 a cylinder and cone b 59 m3 c 59 kL 2 a Volume 31 416 cm3, capacity 31.416 L b Volume 616 cm3, capacity 0.616 L c Volume 264 cm3, capacity 0.264 L 3 a Volume 1950 cm3, capacity 1.95 L b Volume 22 988 cm3, capacity 22.988 L c Volume 1309 cm3, capacity 1.309 L d Volume 455 cm3, capacity 0.455 L e Volume 25 656 cm3, capacity 25.656 L f Volume 1527 cm3, capacity 1.527 L 4 a Tank A 28.27 m3, Tank B 56.55 m3 b 28.27 kL 5 a 12 balls b 60 balls c 31 416 cm3 d 48% 6 a 1963 cm3 b 0.55 cm3/s b 210 kL c $205.80 7 a 250 m3 8 1.447 × 1015 km3 7 a 2 b H ---4 −4 Chapter review 5236 cm2 b 277.6 m2 104.3 m2 d 14 294.2 cm2 5871.2 cm2 f 4427.8 cm2 960 cm2 b 7776 cm2 c 1356 cm2 704 mm2 b 3270 mm2 2490 mm2 452 m2 b 681 m2 c 5890 m2 3318 cm2 b 1728 cm2 5429 cm2 d 1078 cm2 16 416 cm2 f 3016 cm2 11 cm3 b 20 160 cm3 10 472 cm3 183 m3 b $21.96 323 m3 b 540 cm3 c 348 cm3 i 1340.4 m3 ii 10 262.5 m3 iii 31.7 m3 b i 10.7 cm ii 6.9 cm 10 a i V = 180 000 mm3, C = 180 mL ii V = 5.75 m3, C = 5.75 kL b i 12 mm ii 3 m 11 a 360 498 mm3 b 145 125 mm3 c 455 cm3 d 1152 m3 e 3054 cm3 f 18 096 m3 12 a 250 cm3 b i Increases 27 times 1 a c e 2 a 3 a c 4 a 5 a c e 6 a c 7 a 8 a 9 a ii Decreases by 1 --4 7 a b 8 a b c 9 a b c 1 ------3 2 of original length 10 a 162 cm3 b 154 mL c 25 : 49 d 363.41 cm2 e Increases 9 times f Decreases by 27 -----64 11 a V1 = 72 cm3 b V2 = 4V1 d V4 = 9V1 e V5 = 12 350% increase −2 4 a 5 a 17 , 5, 1 4--7 , π --2 3 --4 V1 c V3 = 2V1 , Chapter 3 Start up m2 + 10m + 21 b k2 + 7k + 10 y2 − 3y − 4 d w2 + 4w − 21 f 6d2 + 11d + 3 n2 − 5n + 6 4 − 17p − 15p2 h 3a2 + 17af + 10f 2 3x2 − 7vx + 2v2 j ac + ad + bc + bd 10e2 + 15eg + 6e + 9g 6h2 − 7h − 5 i a2 + 2ab + b2 ii a2 − 2ab + b2 i x2 + 8x + 16 ii y2 − 6y + 9 iii 4k2 + 4k + 1 iv 9m2 − 24m + 16 v 9k2 − 12kf + 4f 2 vi a2 + 4ab + 4b2 3 a a2 − b2 b i d2 − 9 ii 9a2 − 16 iii 16w2 − y2 iv 25h2 − 9e2 v 1 − 9y2 vi 81d2 − 16w2 4 a 26 b 5−2 c 1010 d 70 e 3−5 1 a c e g i k l 2 a b 19 ------ f 157 g 4 6 h 74 i 32 5 4 and 5 6 a 1.93 b 1.78 c 3.66 d 4.33 e 207.06 f 7.37 g 5623.41 h 26.75 Exercise 3-01 b 0.7 f 2.6 0 187% 2 --59- 1 c R h R mR 2 d R i I n I b π, 2 7--9 b 2 10 , 3 , 3 e I j R o I 11 20 , 2.6˙ 6 b i 57 mm ii Approximately 1.4 (= 56 ÷ 40) 7 a Teacher to check. b i Construct a right-angled triangle with the two shorter sides equal to 10 units. The hypotenuse is 10 2 units. ii Construct a right-angled triangle with the two shorter sides equal to a units. Exercise 3-02 1 a 2 e 0.09 b 5 f 28 c 27 g 45 d 250 h 50 2 a 2 2 b 3 3 c 2 6 d 3 6 e 9 3 f 3 5 g 4 3 h 10 2 i 4 6 j 3 7 k 12 2 l 6 3 m5 3 n 7 3 o 4 2 p 11 2 q 9 2 r 7 5 s 5 5 t 16 2 u 3 35 v 10 7 w4 7 x 3 11 d 56 2 1 a −1.8 e −2.5 −1 26 3 a 25 2 ii Length of side is divided by 2 1 : 11.56 1 : 39.304 i 4 : 25 ii 8 : 125 i 1:4 ii 1 : 8 i 49 : 16 ii 343 : 64 i 3:2 ii 9 : 4 i 5:3 ii 125 : 27 i Increased 8 times ii −3 2 a R b I f R g R k I l R 3 A, C, D π --2 74% y 4 10 Exercise 2-11 1 a 1 : 16 b 9 : 16 c 25 : 4 d 4 : 49 2 a 9:1 b 9 : 25 c 81 : 25 d 4:9 3 a 3:5 b 1 : 10 c 8:5 d 4:9 4 a 7:6 b 7:9 5 x = 3.5 6 a 54 cm2 b 44.1 cm2 c 7.5 cm d 154 cm e i Area is quadrupled (×4) 4 -----11 − 12 − 3 15 −1 --45 1, 2, 3, 4, 5, 6 Teacher to check. d −3.5 h 1.9 e c 12 3 10 f 3 g 7 ------3 h 6 6 i 18 17 j 5 5 ---------2 k 3 2 l 3 3 n 15 3 o 14 17 q 2 5 r 6 5 s 4 5 t 6 2 u 12 10 v 5 10 w3 6 x 49 7 m 40 10 p 13 ---------3 y 3 30 4 a F b F c T d T e T f F Exercise 3-03 b − 35 c 4 3 d 6 e −5 2 f 45 g 15 30 h −10 21 i 140 j −30 2 k 36 l − 60 2 1 a 14 m −112 n 24 6 o 80 q −396 r 160 5 s 216 2 t −96 6 u 36 5 v − 60 10 w 252 3 x 72 14 b − 3 c 3 6 p 90 6 2 a c 0.4 g 1.6 b 6 2 5 d 2 2 e − ------7- or − 1--- f 21 g 1 2 ANSWERS 2 7 h 8 615 4 16_NCM10extSB_ans Page 616 Thursday, May 18, 2006 11:01 AM i 5 3 j 5 2 l 2 6 n − ------24 or o 1 q 4 r −21 2 t 2 u b 4 6 3 a 2 2 -----45 2 --3 30 c e 14 3 f 2 Exercise 3-04 b −5 2 c 7 5– 7 d 4 5 e 0 f g 8 15 h 0 i 2 3 j 10 5 k 6 10 l −14 3 1 a 7 7 f g 1 ------5w or 3−1h−4 1 ----x2 b 9 + 2 14 c 21 − 4 5 d d 49 + 12 10 e 77 + 30 6 f 179 − 20 7 g 110 + 60 2 h 35 + 20 3 1 --3 h−4 j 73 − 12 35 c 67 d −57 g 2 h −35 5 a 88 − 30 7 b −10 + 21 2 9 -----m6 i 25 --------9d 2 6 a 5 f 10 k 2 125 --------k6 j w4 -------------m 8 x 12 u 1 v 1 g 29 h 92 − 12 5 7 a 3 i −6 j − 4 30 b 13 10 + 3 2 k 4 l −12 + 2 15 c −9 3 + 5 2 d 7 15 + 8 2 m −9 n 30 + 4 15 e −3 7 − 2 5 f 2 6 −8 3 2 a 5 5 −8 i −6 7 3 a C b D j −3 5 c B d A e C b 3 3 c −2 5 d − 7 e 5 6 f 7 5 g − 10 h 8 11 j 8 3 5 a 11 3 c −6 3 14 ---------6 m 2– 2 ---------------2 b − 5 o 35 – 5 ------------------5 d 30 3 p 22 – 11 ---------------------11 k 3 2 l 9 2 f 41 2 g 5 6 h 29 2 i −15 3 j 0 k 6 2 +2 3 l 12 3 + 3 6 m3 2 −6 5 n −10 2 o 4 6 p −5 2 q 17 5 − 10 7 r −10 10 − 7 3 t −4 a v −13 n + 10 k b 6 3 − 4 15 ------------5 l b 1 --4 f 8v e 14 3 g 73 t h c -----8 1 5 b y3 6 5– 3 -----------------------3 e 1 ------4 x 1 --6 g d 5 --6 h 1 --2 2 --5 c 6 5 w 2 d 32 3 p 5 f 1 ------------8 4 a3 g k 243 l 1 --------128 n 1000 o 1 --------216 s 5 -----32 t 1 --4 6 a 1.7 e 0.046 d 1 --4 f d4 9e x3 3m2n 4g2 6c9 5g4 3 a 1 ----42 c f i l o r u k15 12k9 2a4 l14m13 1 36t5 wt3 x 5 p2 --------3 c −1 g 15 b 8 f 3 c 1 --8 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.2/ 5.3 d d −2 h 24 1 ----53 e 1 ----27 2 -----25 c 8 j 625 4 --9 5 -----24 1 --2 h 10 7 k 3 1 --2 m 27 e b 1 ---------------5 64a 3 i 4 -----35 1 ----65 b 1 ---------4 3 d h 256 p b -1 h (5y4) 3 g 3 --5 16 -----25 3 f x− 5 f 100 000 o c - e 27 n 4:3 2 --3 1 c y2 1 4 1 -----25 m 9:7 b 2x f 4 15w a d k 9 : 20 17 -----40 e p− 3 1 --8 j 5:9 r l 4:5 c - h3 k5 -1 f 74 2 60 3 a d4 4 a 2 -1 c 8k 2 3 2 3 i 5:9 1 --4 h a -7 5 --7 2 e 5p 3 1 3 g a2 c d 2 1--- -1 5 a 125 4 --5 x 24 c 4 -1 - Skillbank 3 2 --3 w1 b d t t 8 b (2d) 2 5 d 11 ---------11 g 16 10 + 54 21 − 2 3 c 28 6 + 21 + 8 2 + 2 3 616 or 2 − e 109 + 10 77 10 − 6 5 − 3 2 g (9ab) 2 a 10 ---------8 −1 w fh3 55 i 24 − 3 6 f 72 − 23 6 35 ---------5 −27e6 8 2 −4 h 5 5 − 75 10 5 2+ 6 ------------------------4 v y 2 a e g 42 − 8 7 d 20 + n b e h k n q t f − 4 11 + b 7+2 7 − 2 ------2 y8 1 4g3 20p3q7 48b17 3 5w2 e 12 + 6 6 2 a 10 + 77 ---------2 1 a d g j m p s d 3 10 − 15 h Exercise 3-07 6 c −2 21 + 7 g k =1− or d 4 2 3 2 ---------4 5 6 ---------6 s 4 Exercise 3-08 c 1+2 7 -------------------2 3 a 10 j r 27 16 d (15w) 3 s Exercise 3-05 15 − f 5 ------20 l 1 f 4 ----- 2 4 5 ---------5 r 1+ q x − 2 b 5 7 ---------7 9 64n 3 -----------27d 6 e 2 j 3 o 3 3 -1 2 3– 2 -----------------------2 e u 8 y 1 a 1 -----16 1 a 36 3 q 2 a s 30 5 − 15 3 2a 7 e i e 5 7 w9 d − 3 ------3 k 16c 10 ------------a4 h 7 --1- d 1 i 6 n 1 b 3 1--- e 2 1--- Exercise 3-06 1 a 4 a 6 2 i 6 2 4 -1 h − 6 13 + 11 7 g 13 11 − 3 e 83 5 ----------a2 b3 b 5 c 8 g 3 h 25 l 150 m 25 f 85 − 3 77 d 73 + 40 3 g w ---4 d 27 f q c 29 + 5 35 e 2p−3 -----c 2 10 4 p 1 --1- 10 or 15 -----n3 j c (4k)−2 h –4 ------3 b 6 1--- e 4d 2 --------y5 i b 5y−4 5 a 3 i 159 − 30 6 4 a 5 b 22 e 1 f 166 3 ----g4 h 4 a 12−1 3 a 8 − 2 15 2 s 12 i − 15 − 1 35 j 45 − 5 21 m 2 14 − 1--4 p 10 d h −16 − k 4 3 b 0.23 f 0.30 c 2.4 g 15 d 0.0098 h 2.3 c g k o d h l p Exercise 3-09 1 a y12 e a15 i p19 m2 2a -----3c b f j n k9 m8 c11 81e2 m d4 5k5 4mn6 w9 x3 12q10 t8w7 r 2x3 s 2d11 t 16a6b7 u 2c2h 2 a 9 e 1 b 50 f 64 c 16 g 24 d 2 h 81 i 2 j 2 k −27 l q -1 3 a 82 e y 4 a 4 3 1 ----23 -1 b (2m) 4 f (9k) b 3 5 1 --4 -1 -1 c 20 3 g c 1 --2 d k5 7 4 8 - h (5b) 3 c 1 ----k2 since 1 --------------------7– 2 7+ 2 by 1 ⎛⎝ -------------------7+ 2 = 1⎞ ⎠ has been multiplied .16_NCM10extSB_ans Page 617 Thursday.288 L c 154.824 b 64 : 125 c 1.7 cm3 c 512 cm3 28 a i 1 : 5. terminating c 0.9 b 0. recurring b 0.18 × 1015 15 20.25 14 a 7.5˙ 4˙ .4 m2 b 192.22 23 a 22.2 L b 8.89 m 31 10 cm 32 a 10 cm b 400 cm3 33 a 6 b 12 c 9 d 1 37 a k 4k6 b b e h 13 a 145.43 m2 b 868 m2 c 1555.574 × 10−3 8 a 75 km/h b 70c/kg c $7.0 cm3 21 a 580 cm2 b 7.7 × 107 b 6 × 10−6 c 3.064 3:2 1:4 1 : 180 d 12 4 7 -----55 3 32m 5 1 e --------4 45 b 39 a −2 2 c 2 3--- b y2 d 27 p 3 g 14.8 m2 c 303 m2 24 a 18 473 mm3 b 2413 mm3 c 1105 mm3 25 a 800 mm2 b 8.65/m d 85 words/min 9 a 0.82 b 1.56 m d −12 5 d 23 − 8 7 e 77 + 10 6 7 a b 5 Powerplus 2 a f 8 7 4 Mixed Revision 1 1 --9 b i 2 3 b 10 000 e 4 343 27k 6 e 2 1--i 9 a d 7a 2 ----h5 f e 81y8 d 4mn 1 -------------( 5g ) 2 or 30 ---------2 j 1 b 7 3 ---------3 e 11 ---------55 f 2 2 ---------3 h 6 i 2 35 ------------5 l 6 + 15 -------------------3 k 44 a 5 − 3 14 – 7 --------------------------21 10 c 2 5 + 1 --2 (9 − 10 b 5 d 7 − 2 10 2 e 5 − 2 10 45 c f 2 10 − 7 5) cm2 Chapter 4 Start up 1 a y=5 b x = −3 c m = −24 d a=7 e y = 2 --3- f x=1 b x −13 c x > −8 e y < −10 f m 10 4 2 a y < 50 d x < −1 1--2 3 a c e g i 4 a c e 5 a c e 6 a c e 7 a c e g + 13a + 30 x2 + 9x + 20 k2 + 2k − 15 25y2 + 30y + 9 a2 + 14a + 51 (4 − m)(4 + m) 2y(7 − y) 5(x − 8)(x + 8) (k + 1)(k + 4) (m − 8)(m + 7) (w − 7)(w − 3) (y + 2)2 (n − 6)2 (2w − 3)2 (3a + 1)(a + 3) (2y − 5)(3y + 8) (5v + 3)(v − 7) (3h − 4)(5h − 1) a2 ANSWERS b − 3y + 1 d y2 − y − 6 f m2 − 4m + 4 h 9a2 − 24a + 26 2y2 b d f b d f b d f b d f h (d − 11)(d + 11) 5p(2p + 5) 2(3w − 5)(3w + 5) (y − 8)(y − 2) (u + 13)(u − 5) (x − 6)(x + 4) (d − 3)2 (p + 9)2 (8q + 5)2 (5x + 2)(x − 3) (3t − 1)(5t + 4) (2y + 5)(4y + 7) (4p − 3)(3p + 5) 617 .9 cm3 b 8143.56 L 11 a $12 520 b $28 170 12 a 3.9 cm c 4904 cm2 26 a 1122 cm2 b 61 m2 27 a 320 cm3 b 746. recurring 10 a 95.0019.63 m2 20 a 15 927.335. terminating d 0. terminating e 0.96 m c 7 cm 22 a 46 m3 b $49. 7+ 2 --------------------5 ii Yes b 2 15 + 3 ----------------------17 d 8–3 6 -------------------2 3 a 1189 mm × 841 mm b 3 2 ---------2 1 ------2 2 ------2 c 10 − 2 13 + 5 5 ----------------------11 4 s= D ------3 or s = 3D -----------3 5 6 (2 + 2 3 ) m 7 4 --3 = 10 3 ------------3 Chapter 3 Review 1 a I f R b R g R 2 a 6 2 c R h R d I i N e R j I b 7 2 c 5 11 d 8 2 e 15 6 f 14 7 g 48 2 h 15 5 i 28 3 j k 6 m 4 11 50 -----3 l 2 2 5 n 6 o 6 2 p 3 6 3 a 21 b 2 10 c 4 3 d 55 e 2 6 f g 4 6 j 3 2 ---------2 h 1 k i 2 --3 4 a −2 3 14 6 ------6 b 5 2 c 7 5 d 3 7 −3 2 e 2+6 5 f 3 7 b 5 2 −7 5 5 a 13 2 c 14 2 + 17 3 d 32 5 − 9 7 e 38 2 − 24 3 f 8 11 6 a −12 + 9 2 b c 7 35 − 27 8 a 3 10 ------------10 10 − 10 5 h 70 + 38 5 b 2 2+1 -------------------3 c 3 ------4 5 6 ---------6 30m3 m 16 a 65 -----99 d y c b 3 c 3 --8 c c f i 34 a 3 2 g 14 7 35 a 2 6 b $A204.6% d 60 2 f 43 g 9 d c w 38 a 3 2 6 a 673.64˙ .6 m b 2.85.06 × 1020 g 15 b 2 5 c −6 2 f 13 2 h −2 3 i −18 10 h 36 a 118 d 12 7 5 c −10 14 e − 2 f 2 ------2 i b y7 5 --6 b (3m5) 2 f 5m3 1 -1 d 2y 3 1 --8 3 3 ---------2 c 3k7m2 e 6 5 - c 2w 4 f 24 5 10 2 ( 4n ) 5 b 15 5 d 7 2 e 0 f 32 5 − 6 3 g −14 6 h 9 3 − 6 10 b Rational e Rational c Rational f Irrational 9 41 a 1 ----- b 2 5--- c 12 16 6 b 24 2 − 12 42 a 5 c 4 7 + 10 d −7 − 7 2 e 33 + 4 14 f 54 + 22 10 g 8 − 2 15 i 5 h 163 + 8 30 j 259 k −20 6 43 a l −18x − 6 wx 5 ------5 d 3 5 g 990 4 --3 h c i −9 y 40 a Irrational d Neither 313 c 1 -------- e 2 5 b 1 ---------------------3125 c 5 c 9 3 2. 2006 11:01 AM d 3 ----a5 1 -----------25g 2 e p3 g 1 -----------d2 f 3 h --------x4 i 93 j 25 -----m4 k 9 ----h3 l 5 a 1 5 b d 10 a 32 1 d -------- ----------y 12 1 --4 c 2 d 1 -----15 f 4 g 4 h 1 -----36 j 1 1--- k 32 l 1 --------216 -----n 4 17 o b 2 4 --5 2 m 3 3--8 27 1 -----32 1 20 2 a 17. May 18.86 m d i 3.0 3 a 3:7 d 3:4 g 12 : 1 4 40 320 km/h p 27 5 a 6 2--- 1 a Yes.76 ii 1 : 13.375 : 1 ii 45 cm 29 14 mm 30 4.0010 7 a 1.72 5:3 10 : 3 4:1 3 1 --2 17 a 2145 mm3 b 8181 mm3 c 7069 mm3 18 a 6362 mm3 b 90 mm3 c 30 708 cm3 19 a 1294. Chris is 14. 215. y = 3 f x = 5.50 12 a 410.5 y=8 x = −10 m=6 y = −7 a = −1 4 --3 k a= y = 15 k = 57 x = 31 x = −29 b e h k 5 a m = 2 --2- x=8 a = −2 y = −2 y=5 x=1 m = 1.4 km/h iii 180 km/h b 30. 216 8 26 Scott is 11 and his mother is 34 85 10-cent coins. y = 4 .9 m2 1 a y= g R=± 5 −3 k w −3 Exercise 4-04 i y = -14 --12 1--2 j a −1 31. 62.1 m 21.8 cm c 12 cm. $18 18 mm.54 Exercise 4-01 1 a d g 2 a d g 3 a d g y=2 a=1 w=6 y = 10 a = −3 y = −1 x = 16 y=8 m = −8 j a= 4 a d g j b e h b e h b e h 11 -----13 a=5 a = −5 x = 7.2/ 5. 50°.0 cm2 e r= 13 b b b b c 325 km c a= 5 6 -----11 7 15.50. 24 b length = 23. 117 20-cent coins 6 The son is 3 and the father is 27.5 y=4 x = 11 y = −5 l a= a=9 n = −35 y = 46 x = 24 b x=4 5 c f i c f i c f i c f i l 9 -----10 m=7 y = −7 m = 18 m = 10 c w= −3 ----4 d x=3 e y=3 f a= −8 3--5 g p = 9 2--- h y=3 i y= −5 j w = 10 k w = 50 l w = 9 3--- 3 n y = 60 11 o a = 1 ----- 6 a x= 7 -----15 b y = 15 4--- c m = 19 10 2--9 10 3--5 d x= e x= 9 h a = 7 ----- g a = -41 j a= 14 5 5 ----14 k w = 11 f y= l y= Exercise 4-02 1 2 3 5 6 7 9 10 11 12 13 15 2 k–m ------------p c y= 5 – 2x ---------------d d y= P–8 ------------k e y= 5m ------3 f y= K–D --------------M g y= 4d – 5 ---------------8 i y= 20m ---------3 20m – 9 -------------------3 w–5 ------------x l y = kx2 my=± a2 – p --------------m n y = T 2c − k o y= d h= 2A -----------a+b f s= v2 – u2 ----------------2a h l= A – πr 2 ------------------πr i n= s + 360 -----------------180 j T= 1 2 D ---S 12 15 3 5 6 4 0 2 160 2600 180 46 40 44 b f j n r v 70 900 770 26 8 135 c 240 g 220 k 18 o 18 s 104 w 600 d h l p t x 900 300 34 12 24 80 j m 3 --2- k d 2 --3- l m −2 --1- b y −8 e p −6 h k −12 c k −11 f p −4 i y4 k x 12 l h −18 n m 29 q p −4 o h −9 r y −6 Exercise 4-03 −6 −4 −2 0 1 --2 1 a 52 b 17 2 a i 36 km/h ii 86. y = 3 e x = 1.5˙ m/s 3 43 4 a I = 300 b P = 781. 13 cm 61. 2006 11:01 AM i (4d + 5)2 8 a 5. May 18.8 cm 105. 63 Vatha is 22. y = 4 b x = 1.6 7 8 9 10 b 13.2 cm.9 m 55.3 618 0 1 2 3 −2 −1 0 1 2 −4 −3 −2 −1 0 1 −10 −9 −8 −7 −6 −5 −4 0 2 4 6 8 10 −1 0 1 2 3 4 −2 e a1 −2 −2 --14 −3 −10 −8 −2 −1 0 2 −6 −4 2 3 4 5 6 7 −8 h y 20 5 10 15 20 25 1 2 3 4 5 i a3 c k 0 2 3 −4 −3 4 −5 1--2 −7 −6 5 a m −10 1 −2 3 1--4 2 i d −12 −10 −8 −6 −4 −2 0 2 −3 4 g a5 0 5 m w −11 p x −8 4 a w −1 h m f w −10 0 2 g k −4 −12 −10 −8 −1 0 i m 3 --1- Skillbank 4A 2 a e i m q u −2 h a6 4 d m −8 e a −4 d x4 9 −2 −6 g a4 f x −4 2 f w0 −5 c y −4 −6 1 c x −7 b x5 0 0 5 −4 1 a x9 6 −1 b y −2 Exercise 4-05 3 2 e a 3 --2- 1 --2 d r 5 5n – d ---------------n 2 ( s – ut ) ---------------------t2 A+ ------------------π 1 c y -8 j t −2 2--- b b = ± c2 – a2 πr 2 0 b m6 3 a x3 d m0 g x −3 I ------Pn 3 −2 −8 2 a x1 5 k y=± 0 $7.0 137. y = −1 c x = 7. 12 cm.84 kg 4. 25°. 36 mm.3 1 a x = 3. 21 cm d 3. 213. 5 and 17 Exercise 4-06 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. y = 3 d x = 2.5˙ km/h d 1 1--.hours 11 a $97.5 a = −2. 120 students 2c + k --------------a j y = ± t2 – d2 3V ------4π −3 −10 b y= h y= −1 l a −6 5–x -----------2 − 3 or y = −2 −4 b 620 km b 73.3 cm3 80 km/h 2 a r= ----6 m a= 10 6--7 a a a a 1 --3 −5 b y −2 --12 d x 11 -----3 6 a 22. width = 5. 214.16_NCM10extSB_ans Page 618 Thursday. 36 mm 57 mm. 105° 14 72 L 8 teachers.25 5 a 27°C b 0°C c 100°C d 39°C 6 a 11.2 b 9 c 17. 23. 19 mm 4 29 cm. y = 3 x = 5. y = −1 3 Exercise 4-09 1 a a = 2.40 or −0. 10 2 − 5 or 1+2 5 1–2 5 --------------------.or 2 c u= − 1--8 or −5 1 --2 2 d m= 1 --7 or 1 e p = −4 or 7 g t = --3.or − 1--- x2 + 2x + 1 = (x + 1)2 p2 − 6p + 9 = (p − 3)2 m2 − 8m + 16 = (m − 4)2 k2 + 4k + 4 = (k + 2)2 e y2 + 7y + e Exercise 4-12 1 a m = −7 or −3 c y = −5 or 3 e t = −7 or 0 1 a b c d 3 2 1 --2 l a = −2 or 1 --3 o 4 a c e g i b = −2 ± 2 x = −0.16_NCM10extSB_ans Page 619 Thursday. Exercise 4-08 1 a x = −2.5 g x = 3.or 5 f e = −1 or 5 h d = − --7.35 j w = ±7. c and f 5 a x = 1 or x = −3 2--3 c m = −5 or m = 6 b a = 0 or a = −5 2 --3 d k= or k = −1 1--3 e p = −7. y = −3 d n = 1 1--. 2006 11:01 AM 2 a c 3 a c e x = 2.47 e = 1.40 h = 1. y = 1 --2 1 --2 x = 10. y = 2 x = 2.76 140. y = −3 x = 1.6 1. p = −2 e x = 3. y = −2 m = −1. w = 5 x = 6..29 d m = ±1.13 24. y = 2 x = 4. y = 3 a c e g i k b a = 3. y = 2 e x= 2 a c e g 3 --4 . 1 2 e y= 5 --. d = −2 Skillbank 4B 3 a e i 6 a d g j 13 b 2460 c 0. y = − 4 h = −2.80 h k = ±9.3 2.or −1 --1- m = 0 or −2 f = −5 or 5 x = 3 or −3 t = −3 or 6 n=7 p = 4 or 6 d = 6 or −3 k = −9 or 3 c = −5 or 3 y=2 m=1 7 2 --.47 p = −2 or 1. y = -3 b x = 8.or 3 3 49 -----4 2 + 3 3 −2 – 3 3 -. c = −3 b e = 4. y = −5 c x = 2. y = 8 b x = 1.6 h 2460 l 246 140. y = 2 x = -13. y = 1 x = 4.14 or −1. y = 1 x = 3. -------------------2 2 3 2 6 + 82 -------------------2 .6 e 140.. y = 6 f p = 1. y = −3 x = 3. y = −3 i x = 6. −1 − d 1+ 1 --2 7 5 2. y = 2 1--3 x = 5.)2 2 1 2 --.076 14 076 k 14 076 c f i l d 24. y = 2 b x = −3. y = −5 2 i c = 1 1--. y = 7 a = 2. y = − 4 h x = 3.33 n = 1 or −2..) 2 = (w − 3 3 --5 l f= --57 o e= 1 --4 h h2 − 5h + 2 4 9 --4 f w2 − 3w + g x2 + x + = (y + 2 + 42 − 2 – 42 -.76 1407.88 or −5. y = 1 x = −2. g = 4 x = 2. q = 0 x = 4. y = 1 x = 3.88 Exercise 4-14 1 a x = 36.or 3 4 h a = 2 --12 j q = −4 or 1 --13 or 1 l c = −4 or 3 4 a x = −2 1--.76 140 760 h 14. 5 3–4 2 -------------------4 3 a h = −1 ± 6 b r=1± 2 c m = −3 ± 7 d w=2± 3 f x= 7 ± 61 -------------------2 ± 21 ----------------------2 h c= 9 ± 73 -------------------2 ± 17 ----------------------2 j y= 3 ± 17 -------------------2 k x= 3± 5 ----------------2 l e= − 5 ± 17 ----------------------2 mk= −7 ± 61 ----------------------2 n u= 1 ± 21 -------------------2 e a=5+ g p= i f= 30 −1 −5 5 or −2 --1- ----5 a+ k 2+ 3 a k = −3 or − 2--5 5 2 --.. y = −2 h x = 5..94 e k = ±0. -----------------------g −----------------------- h n = − 1--. c = 3 a = 3.07 k a = ±9. --------------------i −-------------------3 3 3 + 71 -------------------2 l .49 i y = ±0.72 or −1.5 or p = 4. y = 3 i x = -8.58 f x = ±7. g = − 4 c x = 1. y = 3 f x = 1. y = 6 j x = 3 --1. May 18. y = 1 c h = −2.5 b d f h m = 4.or --1- i h = ±5 j f = 0 or 2 k w= 5 8 1 --6 49 -----16 = (a + 7 --. w = 6. y = 2 The lines are parallel.)2 4 j v2 − --53 v+ 25 -----36 = (v − 5 2 --.) 6 2 a −3 + or 3 7 . 4600 children 4 videos 8 11 Supreme pizzas Jenni is 19 years old Adam is 12 years old a Pie = $2. y = 10 d x = 1.) 2 = (x + 2 j x = − --1.39 w = 1.80 b 115 × 50-cent coins Exercise 4-13 x = ±20 y = ±1 t = ±4 w = ±10 y = ±15 w = ±4 a = ±4 d = ±12 c f i l o x = ± --1- n w = ±5 2 p m = ±6 q y = ±1 r p = ±3 s k = ±2 t y = ±10 u x = ±9 2 a m = +2 b a = ±9 c m = ±5. 4 a. 1 6 −1 ----. 6 – 82 ------------------2 2 + 7 −2 – 7 -. 2 5 g a= 1 ± 17 -------------------4 h m= i c= 3 ± 41 -------------------8 j n= c k= ANSWERS d p= −2 ----3 .20 3 Because the square of a positive number or a negative number is always a positive number. c = −8 b d f h j l a = 1. y = 2 3--4 x = 1.27 or −2.1− + 1 --2 5.58 g k = ±9. y = 3 x = 1.27 or −0. e = 1 2 2 a x = 6. n = 1 x = 1.5 f x = 21 or x = −19 2 --3 g m = −2 or m = 14 h y = or y = 2 i x = −32 or x = 33 g w = 0 or i a= 1 --2 k c= 5 --2 or 2 a c e g i k m o q s u 3 2 1 --2 1 --2 y = 0 or 3 p = 4 or −4 g = −1 or −2 u = −2 or −24 w = −11 or 6 k = 3 or 4 y = 5 or −3 a = 3 or −2 r = 11 or −3 d = −6 b g = −1 or −1 1--- −2 d t = −2 1--.46 j 13 k 130 14 076 b 1407.or −1 5 --6 e m= or g x= 1 --3 i u= −4 k w= −1 2--5 2 −2 ----3 2 or 3 f y = −1 1--.5− c −1 + 10 .13 or −0. p = 3 e x = 3. y = 4 Exercise 4-11 b x = 5. y = 2 g x = 5..6 1407.) 2 7 --2 j n h = −1 or b d f h j l n p r t ----3 = (h − i a2 + h or 1 c d = −1 or 25 -----4 3 --. y = 2 x = 1. y = 1 x = 9. −3 b n = −2 1--. y = −10 f w = −1.24 l y = ±6. 3 – 71 ------------------2 5. -----------------------f −------------------------ b d = 3 or 7 d k = 0 or 3 f p = 0 or 3 2 --3 m c = − 1--. y = 9 x = 5.2− 3+4 2 -------------------4 . y = 2 d x = 0. w = 5 h x = −3. y = 4 x = 2. 38 laser 10 400 adults. y = 1 x = −2. y = 6 x = 4. y = −1 f x = −2. y = −3 b x = 0.20 b Hotdog = $1.77 y = 1. −2 f x = −3 ± 7 ± 22 ----------------------3 −1 3 ± 21 -------------------6 619 . −3 − b 5+ 5.6 f 1300 g 1. y = 8 f x= b d f h 2 --3 . so they do not intersect. y = −7 p = 2. c = 2 d x = − 4.12 g = 0.4076 Exercise 4-10 1 3 4 6 7 9 10 11 12 m = ±12 k = ±13 x = ±2 k = ±6 b e h k m k = ±1 Exercise 4-07 1 a c 2 a c e g 3 b 1 a d g j 200 women 2 280 children b 110 adults 210 children 5 22 inkjet. y = 2 d x = 24.or 3 b t = −2 1--. 62. y = 11 b x = −4.09. y = −11 b x = 3.33. y = 29 b x = 0.or 2 3 c y= 26 ±3 3 ----------------------2 −3 16 a m = −5.48. 1 or 2 solutions . y = 4 b x = 0.11 2. −2.7 d m = 1. 0. 7.2/ 5.. −2 1 --3 b x= 6 Intersects twice e −4.81 l h= 9 ± 17 -------------------4 n w= 1± 7 ----------------3 p n= 2± 7 ----------------3 b d f h j l n i no solution d −2 ± 11 1 a x= 1. y = 7 1--e x = −2.77. y = 6 9 a x = −3.4 s 6 a $9200 b $5550 c 471 (truncate) or c y = ±2 19 a x = 4.15 2 ± 2 10 ----------------------3 n 2 a −1. corresponding angles are not equal. -0.2 d x = ±0.69 4.5 m × 13 m c 17 cm × 12 cm d x=6 2 a 24 and 25 b 42 and 44 c 15 and 27 3 a 1275 b 3240 c 1965 d 36 e 44 4 15 and 17 5 a 1800 m b 1080 m c 19 s (approximately) d 13.83.5 14 a m = −2 or 2 b x = 0 or 10 c p = ±6 d m = -6 or 8 15 a y = -4 ± 1 --2 b m = ± 2 or ±2 Chapter 5 f 1 a 58 m × 38 m b 6.39. −1. co-interior angles have a sum of 180°. y = −2 c x = 9.8 or −2. 9 2. corresponding angles) p = 81 (corresponding angles) w = 99 (angles on a straight lines.26 Exercise 4-18 1 a x = 3.18.2 or 4. y = 14 d x = −1. y = 11 f x = −2. y = −4 d k = −3. y = 1 11 a m = 5. y = 5 and x = 1 1--.9 or ±1. 1.83 4. Yes.5 d d x = ± k2 – 2 y 3 Teacher to check. y = 20 g x = 7.2 4 a b c m = 63 (vertically opposite angles) x = 51 (angles on a straight line) y = 57 (angles at a point) k = 318 (angles at a point) w = 89 (vertically opposite angles) a = 19 (angles at a point) d = 127 (alternate angles) 4m = 88 (corresponding angles) m = 22 2h + 66 = 180 (co-interior angles) 2h = 114 h = 57 a = 72 (alternate angles.9 or 2. y = 0 and x = 3. y = −22 and x = 5. q = 3 b 20.4 x = −0.5 or 0.2 c p=0 i q 4 --5 5 m = 8 1--3 y = --57 −1 ----2 f m = −1. alternate angles) h = 46 (angle sum of a triangle) m + 35 = 124 (exterior angle of a triangle) m = 89 10h = 180 (angle sum of a triangle) h = 18 5y + 137 + 83 = 360 (angle sum of a quadrilateral) 5y = 140 y = 28 3k + 105 = 180 (co-interior angles) 3k = 75 k = 25 a + 27 + 197 + 34 = 360° (angle sum of a quadrilateral) a = 102 No. − b or a = c b y −5 Exercise 4-17 16a2p2 1 a b c d e f 2 a b 2 0 8 −2 −1 x −3 −2 −1 0 1 y 7 6 5 4 3 x −3 −2 −1 0 1 y 4 6 8 10 12 e m = − 5--.0 17 a 32 and 34 3 ± 41 -------------------4 Power plus Exercise 4-15 1 a −5.62 1.5 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. y = 12 c x = 1.28 1..31. 93. 1 --3 Chapter 4 review f no solution m 1 --1. y = 0 and x = 4. y = 9 and x = 2.0 or 4.5 b x = −5. −1. Jane is 16 5 a i 160 mm3 ii 300 m2 iii 1. 0. y = 39 h no solution 620 d e 2 A – bh -------------------h 1 2 3 4 5 −5 3 a b −4 −3 −2 −1 0 −1 0 1 2 3 c y −1 c −2 d d m −2 --15 −3 4a2t2 . −3.. −0. y = 6 7 a y5 Exercise 4-16 − 4--5 2A ------h 6 a= ----9 b m = 1 --1- l −7 −1 c m= 11 k 3 --1. −1. w = −10 12 a 900 children b 17 DVDs and 13 videos 13 a y = ±11 b m = ±5 c m = 6 or −2 d y = 1. −1. −1.16_NCM10extSB_ans Page 620 Thursday.89.8 or ±1. y = −20 and x = −3.27. y = 32 c x = 4.54 c k = 2..or ±3 2 2 .13 3.30. −2 j 7 1 a w = −3 d m = 2. corresponding angles) k = 104 (vertically opposite angles) m = 76 (angles on a straight line.4 b 14 m × 22 m Start up 3 a p 3.5 b y = −2 2 a m = 4 3--- 1 b w = 13 ----- d -3 ± 11 ---------------------2 4 ± 46 -------------------3 o 5±3 5 -------------------4 ± 33 ----------------------4 r no solution 4 a 92.8 or −2.41. −9 j 29 6± 6 ----------------3 b y = −1. -2. 1 1--3 3 a x = ±2 or ±3 1 . alternate angles (or corresponding angles) are not equal.3 f x = −3 or 5 b x=3± d k= e f ∴ x = −2. 94.894 or 1.41 3. 2.27 0.5 or -4. c = −9. 0.65 0.35 2. 5 b x = 1.or ±2 c m = ± -----3 2 b x = ±2 or ±2 2 d y = ± 1--. May 18.5 cm b 120 m 1 a +7 c 11 + 6h + h2 e u2 − 3u + 4 b b 2at − d 3k2 − 6k − 4 f k2 − 4k + 5 1 2--5 c − 1--2 . 2006 11:01 AM k d = -2 ± ± 6 -------------------5 mg= −1 o v= −2 2 a c e g i k m o 5 ± 14 ----------------------5 8. y = 1 d x = −1. 2 2 3 11 -----13 c x = −8 c 5± h 3 ± 37 -------------------2 18 a y = 2 x 10 4 p = 4.65. Exercise 5-01 1 a y = 50 (angle sum of an isosceles triangle) b m = 108 (angle sum of an isosceles triangle) c h = 120 (exterior angle of an equilateral triangle) d x = 45 (angle sum of a right-angled isosceles triangle) e a = 135 (exterior angle of a right-angled isosceles triangle) f 5d = 180 (angle sum of an isosceles triangle) d = 36 g y + 70 + 70 = 180 (angle sum of an isosceles triangle) y = 40 . y = 26 and x = −2.or ±2 e k = ± -----5 4 a c 5 a c e y = ±3 b m = ±4 no solutions m = 1 or 2 b m = −1 or 10 x = ±2. 95 b Grace is 13. −1 1--2 e −1 1--. −5 −1 2 0. y = 18 and x = 5 1 --3 5 .73 g 2. y = 3 10 a x = 2. No.31.85.09 3.48 8. −7 ± 97 ----------------------6 e y = 2. 0. 5 Let ∠XYP = x°.8% n 78.3% 84% 5. and vertically opposite angles) f w = 24 (alternate angles and angle sum of an isosceles triangle) a = 102 (angle sum of straight line or alternate angles) g ∠DXA = 90° (diagonals of a rhombus meet at right angles) ∴ p = 56 (angle sum of ΔDXA) h h = 70 (corresponding angles. AE || BD) ∴ ∠CBD = ∠AEB But ∠CDB = ∠CBD = ∠EAB ∴ ∠AEB = ∠EAB ∴ ΔABE is an isosceles triangle (two equal angles) 13 ∠LMN = ∠LNM (equal angles of isosceles ΔLMN) ∠KLN = ∠LMN + ∠LNM (exterior angle of triangle) = 2 × ∠LMN ∠KLP = 1--2 × (2 × ∠KMN) (LP bisects ∠KLN) ∠KLP = ∠LMN ∴ LP || MN (corresponding angles are equal) 14 Let ∠B = ∠A = x° (equal angles of isosceles ΔABC) ∠ACB = 180 − 2x° (angle sum of ΔABC) ∠DCE = 180 − 2x° (vertically opposite angles) ∠D = ∠E = 1 --2 [180 − (180 − 2x)] (angle sum of isosceles ΔDCE) ∠D = ∠E = x° ∠B = ∠D ∴ AB || DE (alternate angles are equal) 15 ∠YUX = ∠UYX = ∠UXY = 60° (angles in equilateral ΔUXY) ∠UXW = 120° (angles on a straight line) ∠XWU + ∠WUX + 120° = 180° (angle sum of ΔWXU) ∠XUW = ∠XWU = 30° (ΔWXU is isosceles) ∴ ∠WUY = ∠XUW + ∠YUX = 30° + 60° = 90° 16 ∠WTP = ∠P and ∠YTQ = ∠Q (alternate angles. HK) ∴ x = 72 − 36 = 36 ∴ y = 180 − x − x (equal angles in isosceles triangle) y = 108° e g°= 180 − 67 − 33 = 80° (angle sum of ΔABD) ΔDAB and ΔDEC are similar triangles. x°= 180 − 54 = 126° (angle in straight line) g°= 126 + 20 = 146° (exterior angle of ΔSTR) 12 ∠CDB = ∠CBD (equal angles of isosceles ΔCDB) ∠EAB = ∠CBD (corresponding angles. BF || CE) ∠BFD = ∠DEC (corresponding angles.16_NCM10extSB_ans Page 621 Wednesday. DE.° d 174° a 18 b 24 c 8 a 60° b 30° a 6 b 24 c 45 24 7 d 45 e 12 c 12° d 8 e 15 Skillbank 5A 2 a e i m 4 a d g j m 70% 75% 55% 90% 25% 60% 59% 70% 91% b f j n 66% c 45% 75% g 80% 35% k 30% 45% o 52% b 68% e 10% h 70. CB || DE) ∠BEC = 42° (equal angles in isosceles ΔBCE) ∠ABE = ∠BCE + ∠BEC (exterior angle of ΔBCE) = 84° ∴ m = 84 11 a y° = 60 (angle in equilateral triangle) x = 120 (co-interior angle in a rhombus) b ∠ACB = 60° (angle in equilateral triangle) x + x = 60 (exterior angle in isosceles triangle) ∴ x = 30 ∠B = 60° (angle in equivalent triangle) ∴ y = 90 (angle sum of ΔABD) c 2x + 40 = 180 (angle sum of isosceles triangle) ∴ x = 70 y + y = 40 (exterior angle of isosceles triangle) ∴ y = 20 d ∠HJK = ∠JHK = 36 (equal angles in isosceles triangle) ∠IJK = 72° (co-interior angles on parallel lines IJ.∠ SP SQ S in common. a T b T c F d F e T f F g T h T i F j F k T l T mF n T ∠LMK = ∠LKM = 45° (angle sum of a right-angled isosceles triangle) ∠PMN = 135° (angles on a straight line) 2x + 135 = 180 (angle sum of isosceles triangle MNP) ∴ x = 22. 2006 9:02 AM 2 3 4 5 6 7 8 h 2p = 130 (exterior angle of an isosceles triangle) p = 65 i h = 84 (exterior angle of an isosceles triangle) a y = 127 (rhombus.-----------= SR .2% k 42.4% .∠D DA DB in common ∴ x = 67° f similar triangle ST.-------------= DC.5% 31. BE||CD) and ∠EBD = ∠AEB (alternate angles.25% ANSWERS d 88% h 80% l 80% c f i l o 621 17% 33. ∠TWP = y° ∠PYW = x° (YP bisects ∠XWY) and ∠PWY = y° (WP bisects ∠TWY) 2x + 2y = 180 (co-interior angles. AE || BD) ∠EBD = ∠CDB (alternate angles. WY || PQ) ∴ Angle sum ΔPQT = ∠P + ∠PTQ + ∠Q = ∠WTP + ∠PTQ + ∠YTQ (angles on a straight line) = 180° 17 ∠BAD + ∠DAH + ∠ BAC + ∠CAF = 180° (angles on a straight line) But ∠BAD = ∠DAH (AD bisects ∠HAB) and ∠BAC = ∠CAF (AC bisects ∠FAB) 2 ∠BAD + 2 ∠BAC = 180° ∠BAD + ∠BAC = 90° ∴ ∠CAD = 90° Exercise 5-02 1 a d 2 a 3 a 3240° b 3960° e 28 b 12 c 108° b 4 a 168° 5 6 7 8 1440° c 5040° f 7 d 40 e 140° c b 172° 2340° 1800° 14 f 9 150° c 128 --4. co-interior angles) b k = 127 (opposite angles of a parallelogram are equal) c w = 35 (angles in a rectangle) d h = 109 (co-interior angles in a trapezium) e f = 25 (angle sum of a right-angled triangle in a rhombus) f a = 46 (opposite angles of a parallelogram are equal. YX || WT) ∴ x + y = 90 But x° + y° + ∠YPW = 180° (angle sum of ΔYPW) 90° + ∠YPW = 180° ∴ ∠YPW = 90° ∠DBF = ∠DCE (corresponding angles. angles on a straight line) a t = 70 (angle sum of an isosceles triangle) b m = 58 (alternate angles) y = 61 (angle sum of an isosceles triangle) c d + 74 = 120 (exterior angle equals sum of interior opposite angles) d = 46 d m = 115 (corresponding angles) k = 115 (vertically opposite angles) w = 65 (co-interior angles) h = 65 (corresponding angles) e c = 94 (angle sum of isosceles triangle. May 24. angle sum of a triangle) i w = 93 (isosceles triangle. and angle sum of a quadrilateral) a q = 70 (alternate angles) p = 70 (angle sum of a straight line) w = 40 (angle sum of ΔXYZ) b ΔXYZ is isosceles since p = q = 70. BF || CE) But ∠DBF = ∠BFD (equal angles of isosceles ΔBDF) ∴ ∠DCE = ∠DEC ∴ ΔCDE is an isosceles triangle (two angles equal) 9 ∠NKL + 93° = 147° (exterior angle of ΔNKL) ∠NKL = 54 ∠NKH = 54 (NK bisects ∠NHK) ∠HKL = 108° ∠KHL + 108° = 147 (exterior angle of ΔHKL) ∴ ∠KHL = 39° 10 a ∠CED = ∠CDE = 42° (equal angles in isosceles ΔCDE) ∠DCE = 96 (angle sum of ΔCDE) ∴ k = 96 b ∠BCE = ∠CED = 42° (alternate angles. ∴ ΔBCY ≡ ΔDCY (SAS) iv BY = DY (matching sides of congruent triangles) Exercise 5-04 1 a TX = TY (given) XW = WY (given) TW is common. 6 a ∠XDE = ∠XFG (alternate angles.2/ 5.3 3 a ∠ABD = ∠CDB (alternate angles. w = 25 d a = 33. ∠UXY = ∠WXY (YX bisects ∠UXW) ∴ ΔUXY ≡ ΔWXY (SAS) AB = CB (equal sides of a square) AY = CX (given) ∠A = ∠C = 90° (angles in a square) ∴ ΔABY ≡ ΔCBX (SAS) LM = NP (given) LP = NM (given) PM is common. ∴ ΔADB ≡ ΔACB (SSS) ii ∠DAC = ∠CAB (matching angles of congruent triangles) ∴ AB bisects ∠DAC e i ∠HEF = ∠GFE (given) EH = FG (given) EF is common. centre B) AB is common. May 18.16_NCM10extSB_ans Page 622 Thursday. ∴ ΔVWX ≡ ΔXYV (AAS) c ∠V = ∠X (matching angles in congruent triangles WXY and YVW) ∠W = ∠Y (matching angles in congruent triangles VWX and XYV) ∴ Opposite angles of a parallelogram are equal. DE || FG) DE = FG (opposite sides of a parallelogram are equal) ∴ ΔDEX ≡ ΔFGX (AAS) b DX = FX and EX = GX (matching sides of congruent triangles) ∴ Diagonals of a parallelogram bisect each other (X is the midpoint of both diagonals). AB ||| CD) BD is common. SAS i Yes. RHS Yes. AD || CB) ∠ABD = ∠CDB (alternate angles. PM bisects ∠NPL and ∠NML. Similarly. ∴ ΔABC ≡ ΔADC (SSS) ii ∴ ∠BCA = ∠DCA (matching angles of congruent triangles) iii Since ∠BCA = ∠BCY and ∠DCA = ∠DCY. ∴ Angles of a rhombus bisected by the diagonals. ∴ ΔLMN ≡ ΔLPN (SSS) b MN = ML (equal sides of a rhombus) PN = LP (equal sides of a rhombus) PM is common. ∴ ΔABD ≡ ΔCDB (AAS) b AB = CD (matching sides of congruent triangles) and AD = CB ∴ Opposite sides of a parallelogram are equal. ∴ ΔHEF ≡ ΔGFE (SAS) ii ∠EHF = ∠FGE (matching angles of congruent triangles) f i OT = ON (equal radii) OL = OM (equal radii) LT = MN (given) ∴ ΔLOT ≡ ΔMON (SSS) ii ∠LOT = ∠MON (matching angles of congruent triangles) g i OC = OE (equal radii) OD is common. BX = CX (AX bisects BC) ∴ ΔAXB ≡ ΔAXC (SSS) b ∴ ∠AXB = ∠AXC (matching angles of congruent triangles) . AAS d = 31 b k = 25 y = 12. proved in ii) CB = CD (given) CY is common. ∠ODC = ∠ODE = 90° (OD⊥CE) ∴ ΔOCD ≡ ΔOED (RHS) ii CD = ED (matching sides of congruent triangles) ∴ OD bisects CE h i AB = AD (given) CB = CD (given) AC is common. ∠HNF = ∠GMF = 90° (HN ⊥ FG. AAS h Yes. WX || VY) ∠WVX = ∠YXV (alternate angles. ∠BCY = ∠DCY (∠BCA = ∠DCA. ∴ ΔAYB ≡ ΔCYB (SSS) ii ∠A = ∠C (matching angles of congruent triangles) c ∴ ∠A = ∠B = ∠C But ∠A + ∠B + ∠C = 180° ∴ ∠A = ∠B = ∠C = 60° 5 a ∠XWY = ∠VYW (alternate angles. N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. ∴ ΔPMN ≡ ΔPML (SSS) c ∠PNL = ∠MNL and ∠PLN = ∠MLN (matching angles of isosceles triangles LMN and NPL) ∴ LN bisects ∠PNM and ∠PLM. p = 9 p = 109 f k = 11 AB = CB (given) EB = DB (given) ∠ABE = ∠CBD (vertically opposite angles) ∴ ΔABE ≡ ΔCBD (SAS) QW = PT (given) TW is common. GM ⊥ FH) ∴ ΔFHN ≡ ΔFGM (AAS) i PQ = PR (ΔPQR is isosceles) QA = RB (given) ∠Q = ∠R (equal angles of an isosceles triangle) ∴ ΔPQA ≡ ΔPRB (SAS) ii PA = PB (matching sides of congruent triangles) ∴ ΔPAB is isosceles (two sides of the triangle are equal) i TP = XP (given) AP = CP (given) ∠TPA = ∠XPC (vertically opposite angles) ∴ ΔTAP ≡ ΔXCP (SAS) ii ∠T = ∠X (matching angles of congruent triangles) ∴ TA || XC (alternate angles proved equal) i ∠ADB = ∠CBD (alternate angles. AD || CB) BD is common. SSS e No f Yes. DE || FG) ∠XED = ∠XGF (alternate angles. ∴ ΔWXT ≡ ΔXYT (RHS) b ∴ WT = YT (matching sides of congruent triangles) 8 a AB = AC (given) AX is common. ∴ ΔWXY ≡ ΔYVW (AAS) b ∠WXV = ∠YVX (alternate angles. by considering matching angles in ΔPMN and ΔMLP. centre A) BD = BC (equal radii of circle. WX || VY) ∠XYW = ∠VWY (alternate angles. SAS c No Yes. ∴ ΔABD ≡ ΔCDB (AAS) ii ∴ AD = CB and AB = CD (matching sides of congruent triangles) 622 d i AD = AC (equal radii of circle. DH || FG) CH = EG (given) ∴ ΔCDH ≡ ΔEFG (AAS) UX = WX (given) XY is common. WV || XY) WY is common. 7 a XW = XY (given) ∠XTW = ∠XTY = 90° (XT⊥WY) XT is common. WV || XY) XV is common. CH || EG) ∠HDC = ∠GFE (corresponding angles. ∠QTW = ∠PWT = 90° (given) ∴ ΔQTW ≡ ΔPWT (RHS) ∠HCD = ∠GEF (corresponding angles. 4 a i AC = BC (equal sides of an equilateral triangle) AX = BX (given) CX is common. ∴ ΔLMP ≡ ΔNPM (SSS) OA = OC (equal radii of small circle) OB = OD (equal radii of large circle) ∠AOB = ∠COD (vertically opposite angles) ∴ ΔAOB ≡ ΔCOD (SAS) FH = FG (given) ∠F is common. AB || CD) ∠ADB = ∠CBD (alternate angles. 2006 11:01 AM Exercise 5-03 1 a d g 2 a c e 3 a b c d e f g h 4 a b c No b Yes. ∴ ΔAXC ≡ ΔBXC (SSS) ii ∠A = ∠B (matching angles of congruent triangles) b i AB = CB (equal sides of an equilateral triangle) AY = CY (given) BY is common. ∴ ΔTXW ≡ ΔTYW (SSS) b ∠X = ∠Y (matching angles of congruent triangles) 2 a PL = ML (equal sides of a rhombus) PN = MN (equal sides of a rhombus) NL is common. ∠MLT = ∠PLT (LN bisects ∠MLP (proved in a) ΔLMT ≡ ΔLPT (SAS) MT = PT (matching sides of congruent triangles) ∴ Diagonal MP is bisected. ∠CFH = ∠EDH and ∠CDH = ∠EFH (matching angles of congruent triangles) CF || DE and CD || FE (alternate angles equal) ∴ CDEF is a rhombus (opposite sides parallel and all sides equal). ∴ ΔPXT ≡ ΔRXT (AAS) b TP = TR (matching sides of congruent triangles) ∴ Sides opposite the equal angles are equal. and diagonals bisect each other. TW = TV = TY = TX Then ΔTWV ≡ ΔTXY (SAS). Also ΔWXV ≡ ΔYVX ≡ ΔWVY ≡ ΔWXY (AAS) ∠W = ∠X = ∠Y = ∠V (matching angles of congruent triangles) Since the angle sum of WXYV = 360° ∠W = ∠X = ∠Y = ∠V = 90°. b c d e f Exercise 5-05 1 a ∠A = ∠C and ∠B = ∠D Now ∠A + ∠C + ∠B + ∠D = 360° (angle sum of a quadrilateral) ∴ 2∠A + 2∠B = 360° (∠C = ∠A. 11 a In ΔLMN and ΔLPN. c Prove ΔFBC ≡ ΔDBC (SAS) FB = DB (matching sides of congruent triangles) ∴ Diagonal DF is bisected at B. since the sides of PQRT are equal. Since ∠M = 90°. ∠FGM = ∠HLM (matching angles of congruent triangles FGM and HLM) FG || HL Opposite sides are parallel ∴ FGHL is a parallelogram. AD || EC) AD = EC (sides of a rhombus) ∴ ΔDAE ≡ ΔCEB (SAS) ii ED = BC (matching sides in congruent triangles above) AE = DC and AE = EB ∴ DC = EB ∴ BCDE is a parallelogram because opposite pairs of sides are equal i AP = CR (given) ∠A = ∠C (opposite angles of a parallelogram) AS = CQ (given) ∴ ΔAPS ≡ ΔCRQ (SAS) ∴ PS = RQ PB = RD (given AP = CR and opposite sides of a parallelogram) ∠B = ∠D (opposite angles of a parallelogram) ANSWERS 623 . 2006 11:01 AM c ∠AXB + ∠AXC = 180° (angles on a line) ∠AXB = ∠AXC = 90° ∴ AX ⊥ BC 9 a ∠P = ∠R (given) ∠TXP = ∠TXR = 90° (TX⊥PR) TX is common. d i Matching angles of congruent triangles CBF and EBF. In ΔLMF and ΔGMH. GL = GH = LK = KH (opposite sides of a rectangle are equal) ∴ GHKL is a square (all sides are equal. The diagonals bisect each other at right angles. ∴ ΔLMN ≡ ΔLPN (SSS) i ∴ ∠LMN = ∠LPN (matching angles of congruent triangles) ii ∠MLN = ∠PLN and ∠MNL = ∠PNL (matching angles of congruent triangles) ∴ ∠MLP and ∠MNP are bisected by the diagonal LN. Since ∠A + ∠D = 180° and ∠A + ∠B = 180° AB || DC and AD || BC (co-interior angles have a sum of 180°) ABCD is a parallelogram with right angles ∴ ABCD is a rectangle. Similarly. However. In ΔLMP and ΔNPM. ∴ ABCD is a parallelogram. ∠D = ∠B) ∴ ∠A + ∠B = 180° Note: A pair of co-interior angles have a sum of 180° ∴ AD || BC Also. PQRT is a rhombus. Join the diagonal PR. since TN = PM) ∠M = ∠N = ∠P = ∠T = 90° (angle sum of a quadrilateral and matching angles of congruent triangles) ∴ MNPT is a square. May 18. ΔPQR ≡ ΔRTP (SSS) ∠QPR = ∠TRP and ∠QRP = ∠TPR (matching angles of congruent triangles) PQ || TR and PT || QR (alternate angles proved equal) ∴ PQRT is a parallelogram. In ΔPQR and ΔRTP. ∴ WXYV is a rectangle. LM = GM (given) FM = HM (given) ∠LMF = ∠GMH (vertically opposite angles) ΔLMF ≡ ΔGMH (SAS) ∠FLM = ∠HGM (matching angles of congruent triangles) ∴ FL || GH (alternate angles proved equal). If GL = GH. so MNPT is a rhombus. GHKL is a rectangle (proved in part g). LM || NP) ΔLMP ≡ ΔNPM (SAS) ∠LPM = ∠NMP (matching angles of congruent triangles) LP || NM (alternate angles proved equal) ∴ LMNP is a parallelogram (opposite sides are parallel). Since the sides are equal. ii ∠CBF = ∠EBF But ∠CBF + ∠EBF = 180° (angles on a straight line) ∠CBF = ∠EFB = 90° ∴ FB ⊥ CE and FD ⊥ CE e The diagonals of a rhombus bisect each other at right angles. Draw the diagonal PM. Also. ∠LTM = ∠LTP (matching angles of congruent triangles) and ∠LTM + ∠LTP = 180° (angles on a straight line) ∠LTM = ∠LTP = 90° LT ⊥ MP and LN ⊥ MP ∴ Diagonal MP is bisected at right angles by diagonal LN. Since the angles of the quadrilateral are right angles. LM = NP (given) PM is common. TWNE is a rhombus (proved in part d). TWME is a square (a square is a rhombus with a right angle). b LM = LP (given) LT is common. i BX = DY (given) ∠B = ∠D (opposite angles of a parallelogram) AB = CD (opposite sides of a parallelogram) ∴ ΔABX ≡ ΔCDY (SAS) ii AX = CY ∴ XC = AY as BX = YD and BC = AD (opposite sides of a parallelogram) ∴ AXCY is a parallelogram as pairs of opposite sides are equal i AE = EB (given) ∠DAE = ∠CEB (corresponding angles. LM = LP (given) NM = NP (given) LN is common. and ΔTVY ≡ ΔTWX (SAS) ∠VWT = ∠XYT and ∠TVY = ∠TWX (matching angles of congruent triangles) g h i j 2 a b c WV || XY and VY || WX (alternate angles equal) ∴ WXYV is a parallelogram.16_NCM10extSB_ans Page 623 Thursday. ∠D = ∠B) ∠A + ∠D = 180° AB || DC (co-interior angles have a sum of 180°) Opposite sides are parallel. Since WY = XV. ∠CFB = ∠EFB (diagonals of a rhombus bisect the angles) ∴ ΔCBF ≡ ΔEBF (SAS) b CB = EB (matching sides of congruent triangles) ∴ Diagonal CE is bisected at B. ∠LMP = ∠NPM (alternate angles. ∴ MN = NP = PT = MT ΔMNT ≡ ΔNPT ≡ ΔPTM ≡ ΔMNP (SSS. ΔFHC ≡ ΔFHE ≡ ΔDHE ≡ ΔDHC (SAS) FC = FE = DE = DC (matching sides of congruent triangles) Also. all angles are 90°). 10 a FC = FE (equal sides of a rhombus) FB is common. PQ = RT (given) QR = TP (given) PR is common. from ∠A + ∠B + ∠C + ∠D = 360° 2∠A + 2∠D = 360° (∠C = ∠A. 5 --8 b A and B. AD || BE) ∠B = ∠D (opposite angles of a parallelogram) ∴ ΔEAB || ΔAFD (AA) iii ∠ECF = ∠ADF (alternate angles. and DZ = ZG = FX = EX. e SQ and PR are the diagonals. ∴ ΔAXB ≡ ΔAYB (SSS) ∴ ∠XAC = ∠YAC (matching angles of congruent triangles) b XA = YA (equal radii) AC is common.53˙ h 16.25 c i ∠F = ∠E (given) ∠FCB = ∠ECD (vertically opposite angles) ∴ ΔBCF ||| ΔDCE (AA) ii x = 7. ∠PLM = ∠PTW = 90° (given) ∴ ΔPLM ||| ΔPTW (AA) d KX -------TX GX -------YX ∴ = = KX -------TX 7.4 f i ∠E is common.4. ∠TFR = ∠TPD = 90° (given) ∴ ΔTFR ||| ΔTPD (AA) ii TR = 14. PO = RO (equal radii small circle) SO = QO (equal radii large circle) PR ⊥ SQ (given) ∴ PQRS is a rhombus because the diagonals bisect each other at right angles. AD || EC) ∠EFC = ∠AFD (vertically opposite angles) ∴ ΔEFC ||| ΔAFD (AA) iv AB = 14.. Also A and D. ∴ ii WXYZ is a rhombus (all sides equal).86 3 --4 2 a BC --------DC b i ∠F = ∠C (alternate angles) ∠G = ∠D (alternate angles) ∴ ΔCDE ||| ΔFGE ii k = 11. ∠ECF = ∠EBA (corresponding angles. 2. ∴ ΔGMK ||| ΔLHK (AA) f ∠Y is common. FC || AB) ∴ ΔEFC ||| ΔEAB (AA) ii ∠AEB = ∠FAD (alternate angles. WP = 2 13 .625 b No f Yes c Yes d Yes b 6.5 4 a A and C. PR -------PN = PQ --------PM = 1 --2 ∴ PR = RN 4 a ∠A is common ∠ADC = ∠ACB = 90° (given) ΔADC ||| ΔACB (AA) AC -------AB = AD -------AC (matching sides are in the same ratio) ∴ AC2 = AB × AD b ∠B is common.5 ˙ ------.6˙ or 6 2--3 d 10 1 f 5. centre B) AB is common.0˙ 9˙ or 5 ----11 3 a 18 d 7 --17 g 10. WZ = WX = YX = ZY (by Pythagoras’ theorem) ∴ i WXYZ is a parallelogram (opposite sides equal). ∴ ΔEFG ||| ΔHFE (AA) b FH = 25 cm 4 --3 GX -------YX and ∠K = ∠T = 90° ∴ ΔKGX ||| ΔTYX (RHS) e ∠GMK = ∠LHK (given) ∠K is common.4 cm 4 a ∠T = 90 − ∠P (in ΔPWT) ∴ ∠TWN = 180 − [90 + (90 − ∠P)] (angle sum of ΔTWN) = ∠P In ΔPWN and ΔWTN.3˙ c 31.= 0.35 m d T h F Exercise 5-07 3 --2 1 a RHS. ∠XAC = ∠YAC (proved in part a) ∴ ΔXAC ≡ ΔYAC (SAS) c ∴ XC = YC (matching sides of congruent triangles) Also. b AA 4 --5 g SAS c SAS.416 28. 4 --5 i RHS.5 = 624 4 --3 AC -------EC 8 --6 = = BC --------DC and ∠DCE = ∠BCA (vertically opposite) ∴ ΔABC ||| ΔEDC (SAS) c ∠P is common.5 d i ∠A is common. 8 -----15 f AA h SSS. ∠XCA + ∠YCA = 180° (angles on a straight line) and since ∠XCA = ∠YCA (matching angles of congruent triangles) ∠XCA = ∠YCA = 90° ∴ XY ⊥ AB 3 a ∠P is common.3 8 a T b T e F f T c T g T 7 21. ∠AXM = ∠AYN = 90° (given) ∴ ΔAXM ||| ΔAYN (AA) ii AX = 12 e i ∠T is common. 3 --2 3 --5 2 a ∠X is common. ∠FGK = ∠MHK (corresponding angles) ∴ ΔGKF ||| ΔHKM (AA) 7 Exercise 5-06 1 a e 2 a c e ∴ = ii m = 6 6--.2/ 5.16_NCM10extSB_ans Page 624 Thursday.3 Exercise 5-08 1 ∠ABC = 180° − (∠CAB + ∠ACB) (angle sum of ΔABC) ∠CBD = 180° − ∠ABC (angle on a straight line) ∴ ∠CBD = 180° − [180° − (∠CAB + ∠ACB)] = 180° − 180° + (∠CAB + ∠ACB) ∴ ∠CBD = ∠CAB + ∠ACB 2 a XA = YA (equal radii of circle. WT = 3 13 5 a ∠FEG = ∠EHF N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.416 18 12 ˙ ---------. --5. ∠PQR = ∠PMN (corresponding angles. centre A) XB = YB (equal radii of circle. 2006 11:01 AM BQ = DS (given CQ = AS) ∴ ΔPBQ ≡ ΔRDS (SAS) ∴ PQ = RS ii PQRS is a parallelogram because pairs of opposite sides are equal d AC and DB are the diagonals.8 = ∠F is common. ∠XLM = ∠XYW (corresponding angles) ∴ ΔXLM ||| ΔXYW (AA) b AC -------EC = 10 ------7.. Also B 8 2 and D.8 6 6.26˙ b 7. ∠WZY = ∠ZXY = 90° (given) ∴ ΔWYZ ||| ΔZYX (AA) 3 a i ∠K is common. Skillbank 5B b 7 -----25 c 3 -----10 d 7 -----50 e 3 -----50 f 17 -----20 g 8 -----25 h 49 --------100 i 14 -----25 j 9 -----10 k 18 -----25 l 13 -----20 m 1 --5 n 6 -----25 o 53 --------100 4 a 19 -----25 b 1 -----10 c 4 --5 d 9 -----20 e 22 -----25 f 14 -----25 g 3 --4 h 31 --------100 i 17 -----25 j 1 -----20 k 3 --5 l 27 -----50 m 3 -----50 n 49 --------100 o 41 -----50 Yes Yes 16. e SAS. 5 15.= 0. 1 --1. d SSS.5 5. May 18. DO = BO (equal radii small circle) AO = CO (equal radii large circle) ∴ ABCD is a parallelogram because the diagonals bisect each other.2 10.5 e 30 f 22. f Since WD = WE = GY = YF.or 6. QR || MN) ΔPQR ||| ΔPMN (AA) b PR -------PN = PQ --------PM (ratio of matching sides) Since Q is the midpoint of PM. ∠WPN = ∠P = ∠TWN (proved) ∠PNW = ∠WNT = 90° (WN⊥PT) ∴ ΔPWN ||| ΔWTN (AA) b WN = 6. ∠BDC = ∠BCA = 90° (given) ΔBDC ||| ΔBCA (AA) BC -------AB 5 6 7 8 = DB -------BC (matching sides in the same ratio) ∴ BC2 = AB × DB c AC2 + BC2 = AB × AD + AB × DB = AB × (AD + DB) = AB × AB = AB2 In ΔDEG: DE2 = EG2 + DG2 DG2 = DE2 − EG2 In ΔDFG: DG2 = DF2 − GF2 DE2 − EG2 = DF2 − GF2 ∴ DE2 + GF2 = DF2 + EG2 a no b yes c yes a 53 cm b 12 2 m c 5 3 cm d 96 m a In ΔAXB: AB2 = AX2 + BX2 (1) ΔAXD: AD2 = AX2 + DX2 (2) 1 --2 . = -----.2 ------.= 0. Prove ΔCYP ||| ΔCAT (SAS) ∴ YP || AT ∴ PB || AT Similarly. Draw CP to T. Opposite angles of a parallelogram are equal. C Y P X A W B T Medians AX and BY meet at P.2 d 5. BY = DX ∴ BXDY is a parallelogram (opposite sides are equal). so that CP = PT. In ΔABC and ΔEDC: ∠B = ∠D = 90° ∠ACB = ∠ECD (vertically opposite angles) ∴ ΔABC ||| ΔEDC (AA) b m = 16 2--. 9 a 12 6--- b 6 7 c 4. 2 X and Y are midpoints of BC and AY. ∴ ΔDEF ≡ ΔEDG (SSS) ii ∴ ∠DEG = ∠EDF (matching angles of congruent triangles) ∴ ΔDEY is isosceles (two equal angles) iii DY = EY (equal sides of isosceles ΔDEY) and DF = EG (given) By subtraction: YF = YG ∴ ΔFGY is isosceles. α = β ABCD is a parallelogram (opposite angles equal) Also.JL)2 + ( 1--. May 18.4 8 AC 8 -------. 3 4 5 6 ∴ ∠ADB = ∠BAD = 30° (angle sum of isosceles ΔABD) ∴ ∠ADC = ∠ADB + ∠BDC = 30° + 60° = 90° ∴ AD ⊥ CD a Teacher to check b 72 a SAS b AAS c SAS a No. ∠D is a right angle. RHS c No.= 0. sides are not opposite the same angle.6˙ 3 Power plus 1 Teacher to check proofs. b Yes. the angle in the second triangle is not included. iv Let ∠DYE = ∠GYF = x (vertically opposite angles) ∠EDY = ∠MTR = 1 --2 ∠MTR = 1 --2 x × ∠PTR ∴ ∠PTR = 2 × ∠MTR b ∠DBA = 120° (exterior angle of equilateral ΔBCD) (180° − x°) (angle sum of and ∠GFY = Chapter review 1 a ∠BCE = 110° (corresponding angles. 8 D α α φ φ A θ C θ β β B a 2α + 2θ + 2β + 2φ = 360° (angle sum of quadrilateral) α + θ + β + φ = 180° (1) b In ΔABD. β + α + 2φ = 180° φ = 180 − (α + β + φ) (2) From (1): θ = 180 − (α + β + θ) ∴θ=φ c Similarly.4 AB 20 = and } (matching pairs of sides in same ratio) Also ∠DAC = ∠CAB (included angles equal) ∴ ΔACD ||| ΔABC (SAS) 13 a ∠CDA = ∠BAD = 90° (angles in a square) ∠WAB = 90° and ∠YDC = 90° (angles on a straight line) Let ∠AWX = ∠AWB = x° ∠WYX = ∠DYC = (90 − x°) (angle sum of ΔWYX) ∠YCD = x° (angle sum of ΔCYD) In ΔWBA and ΔCYD: ∠WAB = ∠YDC (proved above) ∠AWB = ∠YCD (proved) ∴ ΔWBA ||| ΔCYD (AA) b WA --------CD = AB -------DY (matching pairs of sides in similar triangles) WA × DY = AB × CD But AB = CD = AD (equal sides of a square) ∴ WA × DY = AD × AD and AD2 = WA × DY 14 a M L X K J JK 2 = JX 2 + KX 2 = ( 1--.6 7 12 In ΔACD and ΔABC: AD -------AC 3. BD || CE) ∴ c = 110 (corresponding angles. ∠BCY) DX = BY ∴ ΔDAX = ΔBCY (AAS) AX = CY (matching sides of congruent triangles) AB = CD (opposite sides of parallelogram) AB − AX = CD − CY ∴ BX = DY Also. ΔABC is isosceles (θ = φ) AB = BC d ∴ ABCD is a rhombus (a parallelogram with two adjacent sides equal).KM)2 2 2 JK 2 = 1 --4 JL2 + ∴ + KM 2 JL2 1 --4 KM 2 = 4JK 2 ANSWERS 625 . BC = AD (opposite sides of a parallelogram are equal) BC = DX (given) ∴ AD = DX ∠DAX = ∠BCY (opposite angles in parallelogram) Since ΔDAX. 2006 11:01 AM ΔCXD: CD2 = CX2 + DX2 (3) ΔBXC: BC2 = BX2 + CX2 (4) b Adding (1) and (3): AB2 + CD2 = AX2 + BX2 + CX2 + DX2 = (AX2 + DX2) + (BX2 + CX2) = AD2 + BC2 (using (2) and (4)) ∴ AB2 + CD2 = BC2 + AD2 9 a Since 52 = 32 + 42.5 11 11 3--- 10 10.or 16. BC || DE) b 4w = 114 + w (exterior angles of a triangle) ∴ w = 38 c 2y + 36 = 180 (angle sum of an isosceles triangle) y = 72 2 a Let ∠PTR = ∠PRT = x (equal angles of an isosceles ΔPRT) ∠TRM = 180 − x (angles on a straight line) ∠MTR = ∠RMT = [180 − (180 − x)] ÷ 2 (angle sum of isosceles ΔMRT) 1 --2 isosceles ΔDEY) 7 a b c d 1 --2 (180° − x°) (angle sum of isosceles ΔFGY) ∠EDY = ∠GFY ∴ DE || GF (alternate angles proved equal).16_NCM10extSB_ans Page 625 Thursday. prove PA || BT ∴ APBT is a parallelogram ∴ W is the midpoint of AB (the diagonals of a parallelogram bisect each other). a i KM = PN (sides of a square) MX = NX (sides of an equilateral triangle) ∠KMX = ∠PNX = 30° (angle of a square and angle of an equilateral triangle) ∴ ΔKMX ≡ ΔPNX (SAS) ii XK = XP (matching sides of congruent triangles) ∴ ΔKPX is isosceles (two equal sides) b i DF = EG (given) EF = DG (given) DE is common. ΔBCY are isosceles: ∠DXA = ∠BYC (equal to ∠DAX. 78 Exercise 6-03 1 a c e g i 2 a d g j 3 a b c d e f g h 4 a d $7919. 11 a $2599 b $3576 c $677 d 10.40 $5597.55 v $41.31 ii $55 256.43 $21 832.30 e $63.1% iv 47.99 $42 $0.50 ii $1402.9% iii 53.005 b $1500 $297 500 d $1487.0628 c f i l o c f i l o 0.07 b 0.77 ii $2278. 2006 11:01 AM b Prove ΔNMA = ΔPQB (SAS) Hence show AC = BC (ΔABC is isosceles) Prove NC = PC ∴ ΔNPC is isosceles.045 h 0.99% p.85 0.27.25 2 $17 977. g 11 years 156 days h 2 --1.45 e i $64 ii $576 iii $69.83 i $926 ii $46.07 0.17 ii 52.76 $4448. $66 Exercise 6-05 1 $1589.91 $19 473.73 c $90.75 $70 000 $80 c $53.5 b $4060.3% b 7.59 $30 387.45 b e b e h $9586.04 $4032.9% 9 a $2080 b $8320 c $13 200 d $4880 e 14.10 $1.79 ii $3818.50.90 is a credit (or refund) i $896.50 .a.085 0.88 2 a $270 b $1284 e $173. $101.96 $33.29 0.08 f $837 b $1610 c $538.68 f $1745.09 $480.14 $849.73 $13 433.70 i $11 099.88 b $282.1225 $119.05 iv $1991.4% p.40 ii $3879.18 c $8448.60 $104.50 c $295.84 e $6846. 12 $140.9 0.13 ii $21.75 $156 000 $11 200 $7400 $618 $1.13 $53 366.7 0.88 b 8–9 years c 23.83 $1314.2% e i $403.20 f $73.12 iv $645.33 $68 135.40 $2.50 $52 751.06 iv $729.87 0.30 i $320.06 120 260 d 4 3425.124 0.264 0.40 i $1649 ii $82.9% p. Chapter 6 Start up 1 a d 2 a d 3 a d g j 4 a d 5 a d 6 a d 7 a 8 a 9 a b 10 a 11 a d $100 b $3.50 f $2987.31 i $20 528.a.55.2 0.60 i $322.04 e $4311.648 0.03 i $708 ii $35.3% d 0.29 $963.08 ii $63.a. j 9.80 8 a $1100 b $1440 c $340 d 30.262 0.19 f $120 g $1.50 iii $589.58 i 16.77 i $41 787.99% p.15 0.1825 0.99 f $846.92 $307.years 2 i 1.25 $0.08 i $2823.90 is a purchase charge.89 $47.13 i $729.30 Exercise 6-04 1 a d 2 a d g $2148. $11 366. May 18.70 $640.6% $4576 $20 664 $183 040 $713. $49.50 b $1232.04 iii b $29 100 c 19.10 $217 461.a.45 0.90 $2039.98 0.99 i $563.72iii ii $983.31 b $275.68 7 a i $10 540 ii $8959 iii $6472.06 $777.5% p.75 iii $54.04 ii $4756.a.125% p.31 i $40.a.00 i $25 464.15 626 c $89.29 b $223.52 $12 750.12 v $12.30 c $2946.4 0.50 $3000 $0.20 iii $139.13.50 $294 987.33 ii $123.105 f 0.47 d $1592.80 ii $1199.04 $88 $60 $140 45 b e h k $800 $24 000 48 $22 500 c f i l 900 $800 $550 000 $75.3 1 $731 2 $409 3 $78 4 $125 5 $162.28 $100 902.13 e $35.59 f $19 233.52 b $13.72 $183.19 $118.54 0. July 8.75 $106 600 $131.61 b $12 298.03 v $105.032 $60 b $189 c $58. $9832.20 $2954.31 0.75 ii $456.21 $7155.6 0.2% 8 Yes. since the depreciated value is 0.03 iv $3796.45 Exercise 6-07 1 2 3 4 a a a a b c d e 5 a d $22 080 $9120 $33 280 i $209. ii 13.70 ii $3337. $3488.6% d i $470. $948.96 c $8.50 e $564.07 c i $247.04 0.10 c 357.79 $935.12 $8922.3% 4 a i 90% ii 72.75 0.75 i $861.401 0.98 0.80 c $4795.79 i $32 756.20 e $2957. 2004 The first $34. $2000 + $1800 + $1580 = $5380 $6000 d $620 i $14 700 ii $14 376 $2376 c $2400 $24 0.21 e $4.2% 26% 6. $17 251.76 b i $119.02 0.04 i $7956.7% 10 a $990 b $190 c 23.60 ii $16.75 c $1425 e $100 000 f $2.2% c i $16 120 ii 58.39 f $199.15 f $3856.20 $3540 b $9580 c b $3120 c b $8280 c ii $12 576 iii ii $38 664 iii ii $263 040 iii ii $6963.75% p.0% b i $4416.78 Exercise 6-01 1 a d 2 a d 3 a d g 4 a d g 5 a d 6 a d f $120 $8 $78.35 j $185 196.44 39 weeks c 137 days 4.38 $81.03 ii 46.45 i $250 ii $12. the second $34.89 ii $2199.85 $322.05 0.43.28 $18.52 f 8.79 ii $446.44 0.65 h $5418.61 0.76 $2910.20 $1027.40 $1616.8% b 6–7 years 5 a i $7120 ii $5696 iii $2916.56 c $1337.2 c 600 i 24 ii 60 iii i 52 ii 156 iii 10 b 20 c 2 596.725 0.10 $179.05 $429 c $25.83 ii 60.04 $18 619 $2229.70 6 a i $299.33 i $3846.18 $42.25 ii $675.04 $68 400 Exercise 6-08 Skillbank 6B 2 a d g j Exercise 6-06 1 a b c 2 a b d 3 a c e g 4 a c d e Teacher to check.26 b 2388.31 v $55.76.81 Exercise 6-09 1 a c d e f 2 a b c d e f $10 000 b $6485.15 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.50 $0.49 d i $1281 ii $2989 iii $807.41 7 a $217.16_NCM10extSB_ans Page 626 Thursday.271 0.77% 6 a $11 984.63 f 0. c $6552.085 e 0.21 ii 45.15 f $4.a.27 $13 488.1% c 1152 b 83.2 0.3005 Exercise 6-02 1 a $84 d $1373.50 e $1939. Skillbank 6A 2 a d g j m 4 a d g j m 0.97 c f c f $3411.0675 i 0.91 e $215.47 b $10 666.91.75 ii $47.43 f $10 324.35 b 32.a.64 3 a i $8215.11 c 0.80 d $1528.75% p.80 b $2934.76.01 b $3940 i $39.2/ 5.007 b e h k n b e h k n 0. $2973.48 (or 48%) of the original value.81 v $28.80 h $25 625.60 ii $44.11 iv $1338. ii 16.40 i $38.73 2 years 5 years i $3150 b $1600 e $3645 b $3750 b e h b e h b e b e c $7500 f $60 c $117 000 $1575 c $22 800 $9690 f $9943.55 $1901.13 i $255 256.88 d 3 a b c d e f g h i j 4 a d $294.20 i $1096 i $580. 33 40 $10.7 c −9 5 −1 1--2 5 b 10 ----- 30 a m = 12. 3 d 1 7 b k = 3.85 years 5 a $2249.50 b $2116.60 b $903.) ii 50 iii e i ( 1--. y = 2 b x = 1.71 b $12 838. SAS c No 32 a $1400 b $264 c $1. y = 4 b x = 2.71 4 $63 367. −2.9 or −1.70 42 $68 43 a $373. 2 1--- .16_NCM10extSB_ans Page 627 Thursday. ∠PTW = ∠PRQ (corresponding angles. y = 1 b x=2+ 10 .375 2 TW --------RQ = PT -------PR (matching sides of similar triangles) But T is the midpoint of PR.85 h $22.. $415. y = 6 19 a m = 35 b d = 25 c w = −30 20 a w = 64 (angle sum of an isosceles triangle) b m = 42 (opposite angles of a parallelogram.25 $8.5 b m = ±2. 2) ii iii (−1. −2 1--. angles on a straight line) c x = 48 (angles at centre is 90° since the shape is a rhombus.) ii 34 iii − 3--- d i (2 1--.6 87.90 i $711. y = 10 − 2 10 2 11 r = ± π R2 – A -------------------π 12 a 1.57 c $81.50 Mixed Revision 2 b 1 --3- 1 a 8 e −1 2--3 f 39 . y = −3 18 a x = 1.. n = 1 0 16 x = ±2. 4 1. 89 6 $35 a 0. − 6 1--2 ) ANSWERS 627 5 1 --7 5 9 .4 b −0. 6) ii − --23 x 0 1 2 3 y −3 −2 −1 0 x −2 −1 0 1 y −4 −1 2 5 x −1 0 1 2 y 3 1 −1 −3 1 b 6 7 4 − 36 h − --1.58 $10 b $2. −2 − 10 . 6 1 ± 21 -------------------5 c ± 17 ----------------------4 −5 3± 3 ----------------3 9 a x = 4. −4) 2 units (6.70 b $878. x = ±3 b ii x = 4...04 b $36.6 25 165° 26 a ∠LPM = ∠NMP.61% 10 $440 11 a i $320 ii $88. 2) e f i 2 a 1 --3 (6.80 b i $176 ii $48. 1 8 a e 2 23 a 12 6--- d 2 --1- 20 mL a x = ±4. y = 5 c x = −1.= --PR 2 TW --------RQ 1 --2 = ∴ TW = 1 --2 RQ b c d 3 a N 4 a −2 x −2 0 2 4 y −4 −1 2 5 c P c −1 d P d 5 b X b 8 e N f X e 64 f 64 5 a − 2--. TW || RQ) ∴ ΔPWT ||| ΔPQR (AA) b 15 a x 1 −2 3 Chapter 7 Start up 1 a i (3.625% p.94 9 65.4.5 units.06 e $0. − 6) iv b i 6 units ii iii 6 units c 4.33 c $495. angle sum of right-angled triangle) e a = 108 (equal angles of an isosceles triangle.= − 1--- b − 2--.71.66 b 4 5--.a.73 b $9274. DF⊥AB) ∠COE = ∠DOF (vertically opposite angles) ∴ ΔCEO ≡ ΔDFO (AAS) 29 a ∠P is common.) 2 2 1 --2 b i (8. 1 1--. and x = 2 − 10 −2 + 2 . y = 2 17 a ii x = 3. 1) (−5. y = 1 c x = − 2--. −3) 8 4 8 c 8. y = −3 3 −1 0 4 5 24 10.67 years c No (when compounded annually it doesn’t double till the end of the 18th year) Chapter 6 review 1 $1440 2 a $45 b $208.2% 15 a $8. angle sum of a triangle) d h = 23 (the shape is a rectangle. 2006 11:01 AM g h 3 a d g i $3738 ii $186.54 33 a $7941.25 b x6 −1 0 1 2 3 4 5 6 7 c x 3 1--4 Power plus 1 4 years.12 41 a $8.≈ 4. −7) ii 29 iii − 2--- ii 85 iii − --2- ii 2 iii 1 1 a i (3 1--.6 13 a 4w2 − 5w − 1071 = 0 b 17 m × 63 m 14 a x = 1.04 b $391.59 f $2955. ) 2 2 2 2 2 f i (2 --12 g i (−1 1--2 .63 d $2582. bisect 27 a PT = RQ (given) RT = PQ (given) PR is common.43 f $10. g 7. CD || AB) ∠DWC = ∠BWA (vertically opposite angles) ∴ ΔCDW ||| ΔABW (AA) 22 31 a No b Yes.49 5 $5839. −2 -------41 d 1 --3 ± 5 -------------------2 −1 b f −1.5 2 2 4 --5 i −2.84 12 23. 61 days 2 $4444.2 c a = ±6.29 7 a $2497..a.d= c y = 1 ----41 3 4 5 7 3 --4 .19 6 a $215.04 8 $16 638. 1) . y = 16 and x = −1.71.44 3 a $5415.50 ii $35.25 4 Exercise 7-01 ii 26 iii 1 --5 ii 41 iii 5 --4 c i (−2 1--.13 e $107.= − 1--- 6 y=4 8 a x=7 7 x = −2 b y = −3 c (7. 36 D 37 B 38 45 months 39 $232.23% 14 a $15 120 b $5120 c 10. angles on a straight line) f y = 52 (angle sum of an isosceles triangle) 21 160° 22 a ∠CDW = ∠ABW (alternate angles.5% p.67 years b 17. ∠PLN = ∠MNL b Opposite sides of a rectangle are equal c AAS d Matching sides of congruent triangles LPT and MTN e Rectangle. May 18. 88.19 d $67. y = 10 + 2 10 . ΔPRT ≡ ΔRPQ (SSS) b ∠PRT = ∠RPQ (matching angles of congruent triangles) TR || QP (alternate angles proved equal in part b) c PQRT is a parallelogam because it has one pair of opposite sides equal and parallel.8% 13 a $2600 b $3200 c $400 d 9.50 c $465.. isosceles d (−2. PT 1 -------.78 6 3 970 000 7 a 17.86 c $21 216 d $6080 e $2060 34 a i $20 400 ii $17 340 iii $14 739 b 61.56 d $7. $2838. 28 CO = DO (equal radii of circle) ∠CEO = ∠DFO (CE⊥AB.87 b $2063. 3 b no solutions c 1.4% 35 a $15 900 b $3900 c 6.85 $56.59 3 0. 3 1--. h = −2 2 a m = 2. mAD = − 1--- 34 . because − 5 --6 k 21 6 a mPQ = mRT = 50 . because opposite sides are equal and diagonals are equal e 68 units2 2 a c e g x −2 0 34 3 --7 y = 4 + 2x y = 4 − 2x b mLM = − --5. because all sides are equal and the diagonals are also equal . WV = scalene b − 6--- 5 3 c 7 --5 2x + 2y = 5 2. isosceles f DX = 10. mLP = 3 3 --5 WX = 32 . c 5 6 17 . 2) 2 a AB = 20 . XP = 80 .5 = 1 2--5 7 e N f P N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. mJK = 7 --3 . AL = 13.5 0 Exercise 7-03 1 a − 1--- b 10 -----3 = 3 1--- d − --2- e 20 -----7 = 2 --6- c N d L 628 −6 y−x=6 y = 2x + 10 n = 8m − 6 z=r+9 w = 9t − 9 1 a No b No c Yes d Yes e Yes f Yes g No h No i Yes j Yes 2 b (2.5 0 2 x . −3) ii 20 iii −2 3 a − 6--- i ii 272 iii −4 4 a − 1--- b 4 --5 -----c − 20 17 d − 1--- e − --b- f n --3 10 g − ----- h m i (1. XY = 72 .. BC = 1. mKL = 3 --7 . WY = 104 i 104 ii 104 They are the same. scalene 5 . AC = b XW = 41 . 2006 11:01 AM h i (4. because all sides are equal and the diagonals are not equal y = 3x + 4 b = 5a − 2 p = 10k − 3 h = 4 − 2d 0 58 units. b N x 2 3x − y = 6 0 Exercise 7-02 3 0 y 6 c Skillbank 7A b d f h y i x 0 x 3 5x + 3y − 15 = 0 2 7 --3 c HK = 200 . isosceles 3 a AB = 41 . PR = 116 . 2) lies on each of the three lines. because opposite sides are equal 4 a LM = 6 − x = 2y 3 c A parallelogram 73 . A right-angled triangle (using Pythagoras’ theorem) 6 a The length of each side is 40 units. 5 a Yes c PT = d MN = 5.5 units c mAB = 4 --5 b Yes 3 4 --5 8 --3 LP = 136 . QR = 1 a . mNP = − --5. DP = 10. mCD = y f b mPT = mQR = − 8--- 2 -----11 d No. JL = 32 d A rhombus. NP = 0 3 d A parallelogram. MT = 5. 2) 3 They intersect at one point because (1..16_NCM10extSB_ans Page 628 Thursday. mBC = − 1---. XV = c PQ = 26 .3 l x 2 y x x − 4y − 2 = 0 −0. mHL = 2x = 4y − 8 −4 j 0 −6 d y x 0 −4 10 x 2x − 5y − 20 = 0 y 4 x k 3x − 2y = 12 y −6 4 4x + 2y − 8 = 0 e y 0 2. isosceles e AF = 13. b DF = EG = 80 c A square. BC = 40 . AD = 40 b 25. QR = scalene a 52 . 3 --5 . MN = × 73 2 -----11 4 Exercise 7-04 y 3 4 3 0 136 . FL = 208 .2/ 5. CD = 41 . May 18. mMN = b mHJ = y g ≠ −1 x 6 e A parallelogram . 7 a The length of each side is 2 a L 5 x 2 y b 5 a b c d 5 y h c LN = 170 = MP d A rectangle. NT = 20 . m = − 4--. b = 3 b y = − --x.+ 1 3 a m= 4 b y = −5x − 2 y ii 4 1 j m = --3. b = 2 --1- y = 2x + 5 8 .. b = − 6 y = −2x + 1. D e D 6 a B. b = −5 c m = − --1..b=6 f m = 2. b = 4. b = 8 −4 −2 0 4 2 6 x 8 f i m = 6. C b A c B. b = 4 1 --2 y ii d i m = − 1--. b = − 4 y ii b m = −3. b = 3 .+ 3 2 −2 −2 x −6 x --2 3 xc y = ----−2 6 4x -----3 2x -----3 +2 4 2 f y= d m = −1.. m = 3--.. b = 5 4 −8 y 0 x 4 2 −6 −3 −8 2 x −4 Exercise 7-06 6 0 −4 1 3 −4 −2 −2 −2 −2 4 2x . m = −2.− 5 2 8 +2 d y= 2 h i m = − --3. b = g m = 2. y = −x − 5 4 a y= 0 −4 7 . b = −7 i m= 5 --4 d m= g i m = 2. May 18. b = 2. b = 1 ii y 8 6 d y= 6 4 −4 −2 3x -----2 + 2. b = −5. y = x --2 −6 −2 c y= 3 --4 +4 2 2 −4 −2 −2 − 4-----x3 0 4 2 6 −6 i i m = − --5.. m = 4..16_NCM10extSB_ans Page 629 Thursday. b = 1 8 2 +2 y ii 6 2 −1 0 2 −4 −2 2 −2 1 a i m = 1. b = 1 1 2 4 6 8 x g y = − 4-----x. m = −2.b=2 5 --4 4 x --3 e y= 5x -----2 g y= 3x -----10 +8 +7 2 --3 .. y = − ----1 --2 2 d y= j m = 4x + 3 6 4 −4 −2 −2 h y = − 4x + 2 i y = − 5-----x. 2006 11:01 AM Exercise 7-05 1 a m = 6.+5 b m = − --2. b = −5 f y = 2x − 6. b = − --1- 2 2 a y = 3x + 7 c y= 6 − 1--2 h m = − 4. m = 3--. m = −1. b = − 6 2 0 x+y−2=0 b 3x − y + 2 = 0 5x − y + 8 = 0 d x − 2y + 3 = 0 x − 5y + 1 = 0 f x − 2y − 6 = 0 8x − y + 2 = 0 h 6x − y − 3 = 0 x + y − 10 = 0 j 2x − y + 4 = 0 8x − 2y − 1 = 0 l x − 2y − 6 = 0 3x − 5y + 10 = 0 n x + 2y + 10 = 0 15x − 3y − 2 = 0 p 3x − 24y − 8 = 0 4x − 3y − 2 = 0 r 4x − 3y + 15 = 0 3x − 2y − 12 = 0 t 2x + 5y + 3 = 0 y = −2x − 5.. D d C.+ 4. y = 2 −4 7 c m= 2 x+4 -----------5 2x .... b = 9 2 e m= − 2--3 . D 7 a y = 5x + 6 b y = − 4x − 1 − --x2 0 c i m = −2. b = 2 2 e y = 4x − 5.−2 f y = − ----- 3 y ii 6 d y = −x − 4 e y = − --x. b = − 7--. b = 4 3 h y= 3x -----2 3 − 7--. b = 5. m = 2.(or −3 1--. b = − 4 ii 4 4 6 2 x 0 −6 −4 −2 2 −2 −5 −4 4 −6 −4 y y=x−4 4 e i m = − --2.) 2 ANSWERS 2 2 629 2 . b = 7 3 ii 2 −4 1 −6 2 4 6 8 x 8 x Exercise 7-07 −2 6 1 1 a c e g i k m o q s 2 a b c 3 4 2 −8 b i m = 2. b = − 4 y ii 3 +4 x 2 4 4 −8 5 a C b B c B d C. b = −5 y = −x − 6. 0) 3 5 a CE = ii (3.− 2. 2). B(3 1--. 2 1---⎞⎠ . e Opposite sides are equal and parallel.. May 18.20 e $3. m = 0 For DG and EF. −1) e No. 0) 11 -----3 × 3 − ----11 = −1 x e D e C 1 --3 . mCD = − 4--.16_NCM10extSB_ans Page 630 Thursday. --1-⎞ ⎝ 2 2⎠ 3 mPR = − ----- N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. m = − 1--.84 f $4. − 4) x − 5y + 20 = 0 x + 2y + 6 = 0 (0.25 units2 Area ΔRQB = 14. opposite sides are parallel and equal 7 a AC = BD = 104 units b Midpoint of AC = (1. M and N are collinear points.50 c d $20. mJM = − --5- c 1 a c e 2 a c e 3 a c 4 a 5 a c 2 a 19 b $7. m = 1--. 2 1---⎞⎠ 2 d Parallelogram.72 e 160.95 k $0. 2006 11:01 AM 7 a c e 8 a i y = − --x. mBE = − 4--- 5 $490 $255 8. --1..6 i j 9. diagonals bisect each other at right angles b 2 3 ∴ XY || CB c XY = 1 --2 117 .60 c 2..6 l 42. 7). DF = 130 units and 11 -----. 10 b= m = 2. −1) d (−1. adjacent sides are equal.25 units2 Area ΔRQB − Area ΔAOB = 2 units2 Exercise 7-08 1 a 2x − y + 1 = 0 c 4x − y − 20 = 0 e 2x − 3y − 4 = 0 g 3x + y − 4 = 0 i 3x − 4y + 10 = 0 2 i and ii y b d f h j x+y+2=0 x − 2y − 10 = 0 x + 5y + 38 = 0 4x + y + 1 = 0 2x + y + 10 = 0 x−y−5=0 a x + 3y + 3 = 0 0 x P d 87. CB = 117 ∴ CB || 2. 3 2 ⎛ --1-.. 1⎞ --2 2⎠ mKM × mLN = 1 × −1 = −1 9 Teacher to check. 14 a Midpoint of TU = A(4. midpoint of BD = (1. −1) Midpoint of UV = B(0.XY 1 --3 16 a i mLM = ii mLN = 1 --3 iii mMN = b L. CD = BE = 3 k: x + 2y − 7 = 0. m = 2 Since 2 40 units. 3 --1-) 2 3 d Midpoint of DF = (0. BD = 65 units ∴ ABCD is a parallelogram. 2 l y = 2x − m= 1 -----.. −1) 6 c Midpoint of BD = ⎛⎝ 1 1---. −2) b 3 a b d 4 a c e g i d Square. −3) 3x − y − 46 = 0 7x + 15y + 66 = 0 iii (15 --1-.) Y(−1 --1. 12 Trapezium 37 units TW = SX = 2 37 units XS ⊥ ST because mXS = − 1--. mST = 6 6 ∴ STWX is a rectangle because opposite sides are equal and angles are right angles.76 4 a 10 b 124 c d $1.9 Exercise 7-10 c 5 b d b b d 61 units. b JK = LM = 2 ii (6. 4) iv (0. −1) Midpoint of ST = D(−1. 7 units2 2 a 5x − 2y − 25 = 0 b 5x + 7y − 25 = 0 d t = −10 --25 3 a DE = EF = FG = DG = 5 units b For DE and GF.. 2) iii 7x − 3y − 1 = 0 c i m = −2 ii (−2..23 g 4c i $0.5) iii 3x − 4y + 6 = 0 b i m= − --37 ii (1.4 h $0.48 j $6. 0) v -----123 17 21 v 26 units2 ii (2.76 i j $1.. --1-⎞ ⎝ 2 2⎠ c mCE = = 29 units ∴ JKLM is a parallelogram because opposite sides are parallel.4 f g 79.3 0 11 ∴ CE ⊥ DF because 2 b mXY = − 2--. 0) The diagonals bisect each other because their midpoints are the same. 1) 7 6 7 2 2 Midpoint of CE = ⎛⎝ 1 1---. diagonals are equal and bisect each other but not at right angles ⎛ --1-.76 l m 31.01 o p $2431. 1) i x − 2y + 4 = 0 ii (− 4.76 f g $0. 0) Midpoint of EG = (0. 3) Midpoint of SV = C(−5. mDE = 5--. m = − --1− 1--2 3 13 ST = WX = v 35 units2 iv (−9 --3-.35 h 6.6 Exercise 7-09 6 a i m = − 4--3 b x + 2y − 11 = 0 d 4x − 2y − 25 = 0 b x + 3y − 4 = 0 d 2x + y − 8 = 0 3x − 4y + 22 = 0 x + 2y − 16 = 0 3x + y − 11 = 0 A(10.7c (10c) k $152.. 0) 2 − 9--2 b Area ΔAOB = 12. diagonals are equal and bisect each other at right angles 2 Skillbank 7B 4x + y − 10 = 0 x − 5y − 13 = 0 x − 4y + 10 = 0 b x + 2y + 4 = 0 x − 4y − 2 = 0 d x−y+1=0 (6. −3) iii 3x − y + 3 = 0 e i m = − 1--2 iii 2x − y − 7 = 0 f i m = − 1--8 iii 8x − y − 15 = 0 630 1 a i ii iii b i ii iii c i ii 5x + 2y − 18 = 0 3x − 4y − 16 = 0 (0. mKL = − --5..54 n $1..5 40 $270 $8.40 (3. 6) 65 units b mBC = 5--.26 42. mLM = − --1.. −12.8c (9c) 64 d $1. l: 3x − y + 7 = 0 4x + y − 20 = 0 c 5x − 7y − 30 = 0 2x − 3y + 18 = 0 e 3x + 5y − 30 = 0 x−y−4=0 b 4x − 5y + 18 = 0 5x − 6y + 23 = 0 d 8x + 3y − 10 = 0 3x + 2y − 6 = 0 f 5x − 3y − 1 = 0 6x + 11y + 38 = 0 h 13x − 10y + 24 = 0 x+y−3=0 f 4x − 3y − 11 = 0 2x − y + 4 = 0 3x + y − 13 = 0 5x + 4y + 37 = 0 2x + y + 2 = 0 2x − 3y − 19 = 0 2x + 9y − 8 = 0 x+y−9=0 2x + y − 11 = 0 x − 3y + 3 = 0 4x + 5y − 40 = 0 5x − 4y − 50 = 0 6 a BC = DE = 10 a mJK = − --1. m = − 4--3 × 2 = −1 it is true that DF ⊥ EG 4 a 6 5 units b 6 5 units c (−1. 15 a X(3.. −5) b Gradient of AB = −1 = gradient of CD Gradient of AD = --4. 0) B(0.45 k $768 l 6 a 100 b $0. 9) iv (0. 10 + 9 --. 2) c mAC = −5. ∴ AC ⊥ BD d The diagonals are equal and bisect each other at right angles. b = −2 2 2 j y= x --3 + 4. since mPR × mQS ≠ −1 f Rectangle. 11 Teacher to check. 8 Midpoint of KM = Midpoint of LN = ⎛⎝ 2 1---. Exercise 7-11 1 a C 2 a B 3 a b A b A c B c C d C d D y 130 units. 1) iii x − 2y + 4 = 0 d i m = − 1--3 ii (−2.2/ 5. b = 4 k y= x -----10 3 7 ------.. mBD = --1. mCB = − 2--- Diagonal EG..93 h 89.5 e $1. b = 7 -----10 iii 2x + y − 7 = 0 iv A(0. KL = JM 2 c Diagonal DF.= gradient of BC 5 AC = 9 units.. −1) 0 2 x 0 y c x 0 −4 y i 0 x−y=6 x 3 −2 0 x 0 y 4 a y d y g 4 y = 2x y h x 0 −4 0 2 x = 2y − 4 x y b y y=7−x 7 0 y i x 0 −2 x 0 −4 e x 6 −6 12 = x − 2y x 7 0 x 12 −6 f y y c 4 x −1 0 1 y=1−x 0 1 5 a i y −x and y x --2 ii y −x and y x --2 x b i y = − 1--. q: x = − 4 13 – 2 x ------------------7 and x − 4 and x 3 and y0 iv 14 units2 2 x 2 c i y = 3 − 3x ii y = x + 3 iii A: y > x + 3 and y 3 − 3x B: y < x + 3 and y 3 − 3x C: y < x + 3 and y 3 − 3x D: y > x + 3 and y 3 − 3x x Exercise 7-12 1 a y x −3 −2 −1 0 1 2 3 y 9 4 1 0 1 4 9 2 0 2 x ANSWERS 631 .16_NCM10extSB_ans Page 631 Thursday.− 1 and y < 2 2 g y d y y = 4x + 2 5 d i y= (1.x − 1 ii y − --x. 6) 1 0 iii y 0 e y=2−x 13 – 2 x ------------------7 ii p: x = 3. May 18. 2006 11:01 AM y b y h y = 3x − 4 f y 1 x (−1. 4 is the y-coordinate 3 a k=5 b k = −2 x 4 2 2 --3 2 --3 208 units b i m=3 c i m= 1 D(−3.16_NCM10extSB_ans Page 632 Thursday. −4 4 a x−2 b C(9. 2) 2 a T(−1. 5) 4 d mAC = 5 a i m = −2 8 6 68 units c m=4 b Z(5. 2x + y + 3 = 0. 4). x − y − 1 = 0.(15 2 6 Chapter 7 review 2 10 4 2 = 5 × 20 = 100 units2 14 12 0 −8 x = −6 y y x_ 2 2 x b y=6− 4 x 5 a b 8 0 −4 −2 2 y 8 8 0 7 y b y = x2 16 x .3 3 y 6 a b 632 1 --2 ii m = − 1--- e i m= − 3-----x2 ii m = − 1--3 5 B(2. May 18. 6) −4 −2 3 a 4 2 x 6 3 c mTZ = 2 2 --3 4 -----11 e Length of AC = −2 −1 0 1 2 y −8 −1 −4 −2 −2 0 1 8 −4 0 4 x Power plus 2 a y= 2 −4 −2 −2 0 4 a − 3--2 b 3x + 2y + 2 = 0 or y = ii m = 3 d i m=2 ii m = − --1- − 1--4 3 --2 ii m = 4 f i m= −1 −6 Point of intersection is −8 x −3 −2 −1 0 1 2 3 y 1 --8 1 --4 1 --2 1 2 4 8 2 ii m = − 2--3 4 ⎛ − --2-. 10) f Length of TZ = 52 units ∴ AC || TZ and AC = 2. − --5-⎞ ⎝ 3 3⎠ x−y=4 0 x 4 −4 y x + 2y = 16 8 0 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.2/ 5. 2006 11:01 AM y b 10 6 6 4 4 2 2 −4 −2 6 4 2 a x −3 −2 −1 0 1 2 3 4 y 15 8 3 0 −1 0 3 8 15 5 x 4 2 −8 −6 −4 −2 −2 −3 −2 −1 0 1 2 3 −4 y 8 4 2 1 1 --2 1 --4 1 --8 −6 8 6 16 4 Area = 0 −4 −2 8 6 a 6 −3 −2 −1 y 4 -----11 2 --3 1 1--3- 1 a (4. −1) 6 x + 5y + 9 = 0.TZ 3 a No b Yes c No d Yes e Yes f No 4 a Neither b Parallel c Perpendicular d Parallel e Neither f Parallel y b x=4 = 2 52 units 2 x y = x2_ − 3 2 --3 2 1 1--3- 2 x b 1 4 0 8 + 5) 0 y b 2 x 4 2 x 4 10 -----. .2 m ANSWERS c f i c c 633 54. b = 5 y = 5 − 3x y 4 3 12 a y = 2x -----5 b y= 3x -----5 c y 2 1 −4 −3 −2 −1 −1 0 1 2 3 x −2 −3 −4 b m = --1.+ 5 8 a y = 2x − 4 −6 −4 −2 2 9 a m = −3.1 cm g 48.25 units2 21 a 2 −3 −2 −1 −1 5 11 a 3x + y − 9 = 0 c 2x + 7y − 3 = 0 e x + 4y − 10 = 0 5 x 0 3 8 7 a m = −3.3 mm 55°7′ c 34°53′ 6 1 −2 + 9x -+6 = –-------2 1 y = − 25_x + 5 + 2.60 m .b= 0 3 x 6 --5 . 3 b = −4 e m = − 5--.5 −2 13 a 3x − y − 10 = 0 b 2x − 3y + 26 = 0 c x + 5y + 33 = 0 d x − 4y − 10 = 0 14 a 3x − 5y − 20 = 0 b x + y + 3 = 0 c 2x − 5y − 16 = 0 d 8x − 5y + 22 = 0 15 a 4x − y − 19 = 0 b 3x − 2y − 18 = 0 c x − 2y − 10 = 0 d 5x − 2y − 26 = 0 16 8x + 3y − 95 = 0 17 5x − 4y + 4 = 0 18 a x + y − 6 = 0 b Q(0. −3.2 cm d 18.2 m 29°46′ ii ii b b 5. b = 10 b = −2 4 f m = 2.m= − 9--2 .5 70° iii 28° 164. m = 6 --5 −1.5 cm 43° 45°1′ 8 3. 3 b y = − --x. b = 2 6 b m = 6.b=6 34 units.7 cm b 14..0 iii 70. b = 3 2 y= y 4 x_ 2 +3 1 3 2 −3 0 2 4 6 8 b 2 --5 0 x y x+y=3 3 . b = −5 2 d m= 1 --2 2 --..m= 3 --5 . May 18. mPL = − 5--.6 m b 11.5 cm c 270.4 cm 17. PN = PL = 34 units.16_NCM10extSB_ans Page 633 Thursday. b = −3 y y 3 y b 2 3x − 2y = 6 8 6 0 x 4 3x + 4y − 12 = 0 y d 0 2 −6 −4 −2 −2 5x − 2y = 40 0 32 4 −4 0 6 8 x y c 2 3 −6 x 8 −8 10 m = − 2--. 20 A only y y 2x − 3y = 7 c 3. 6) c 5x − 8y − 30 = 0 d P(0.5 cm e 5.. 2006 11:01 AM c c m = 3--.b=2 . b = 5 x 2 4 Exercise 8-01 0 1 2 3 x 0 x 1 a 64.5 cm 17. 3 --5 y = 2x + 3 x b 5x − 4y + 4 = 0 d 3x − 2y − 36 = 0 f x − 6y − 12 = 0 19 PN = LM = mPN = y .8 cm h 59 cm 2 a 39° b 56° 3 a 52°57′ b 64°37′ 4 a 73° b 5..6 i 65° 37. b = 5 5 −20 22 a y c m= 1 --.75) e 29.7 m 5 10° 6 65 m 7 60° 9 a 11... ∴ PN ⊥ PL 3 ∴ LMNP is a square because all sides are equal and PN ⊥ PL.5 x 0 −2 13_ Chapter 8 Start up 1 a b 2 a 3 a i 17. 91 c c = 12. Exercise 8-03 1 a P b P c N d N e 2 a 10° b 70° c 50° d 83° e 3 a − 0.4 c Results are the same. tan 5 sin -----------cos tan = 5 -----5 13 -----.114 km or 114 m b 38°48′ c 61.89 b 0.9° c 125.6 m2 e 93. sin E = 9 -----41 e cos Y = 5 ------3 . 13 5 -----12 cos so = sin -----------cos 5+1 ----------------.9 mm2 320.3° e 5 a 149°7′ b d 135°33′ e 6 a k = 6.2/ 5.26 g 0.0237 e 38. 12 12 -----13 = cos = 3 --3 5 --.6 f 69.2° i j −tan 39°50′ k sin 4.8 m b 112.3 d 11.8065 h 90.9 mm2 d 132.6 3 ------2 P f N 65° f 12° d − 0. sin β = b 1 4 a sin so 12 -----13 g − ------3- 6 a 123° b 143° c 110° d 130° e 173° f 135° g 100° h 155° i 114° j 147° k 105° l 118° 7 a 34°51′.05 i − 0.4 mm 81°54′ or 98°6′ b77°24′ or 102°36′ 49°37′ 6.2 m b 16.4 b 15.1 c 44.0° e 60.58 e = 30.8 -----------------sin 56° (since sin 90° = 1) which is the same result when using the sine ratio.3 m 7 a 46° or 134° c 55° or 125° e 51° 8 a 75° or 21° b 9 79 m 11 a 113° 12 a 110° 13 235° or 205° 14 d 124..33 e d = 19. 145°9′ b 48°3′.8 cm c 136. 173°48′ e 27°2′.2 m 2 a 32° b 142° (or 38°) c 29° d 55° e 37° f 125° 3 105 m 4 a 32°+23° = 55° (exterior angle of a triangle) b 108. 5 Teacher to check.51 b d d = 4.10 e 3 a 27° b 4 a 44.1 m e 3.1 m2 97.7 cm2 8 a 112° b 37 cm2 c 740 cm3 Exercise 8-07 1 a 10.4066 g 5.0° c d 23.4 cm2 f 0.0° f 5 20.11 h − 0.5° l 5 a 1 --2 b −1 634 c 1 ------2 d − 1--2 h − 3 i 0 1 j − -----2 1 a 18. 2 a 331 b 157 c 1587 d 255 e 421 f 203 g 413 h 734 i 6723 j 15 744 k 276 l 72.45° e 51°43′ f 72°22′ b 0.5° c 4 a −cos 38° d −tan 22° e −cos 27..3 3 a 46 cm2 b 20 cm2 c 294 cm2 d 321 cm2 4 a 225 m b 2770 m2 5 a 130° b 82 m c 766. 159°26′ d 6°12′..3° Chapter 8 Review 1 2 3 4 a a a a 10.4967 b 416.5° b d 67.2° 82.6957 d 5.6 cm 95.36 f 4.8 -----------------sin 56° becomes: d= 12.5 m2 f 152.17 e − 0.64 m 11 11 m 12 79° 13 480 m 14 48 m 15 a 052° b 206° c 300° d 125° e 238° f 338° 16 a 68 km b 015° 17 a 58 km b 301° 18 2757 km 19 367 km 20 a 14.43 b −tan 59. sin X = = 4--. 154°37′ g 136°39′ h 136°28′ i 61°3′.6° c 32.30 b c = 54. 13 5–1 ---------------2 3 = 5 = sin -----------cos = cos 2 Exercise 8-04 5 ------3 c 1 = 3--.19° d sin 84°43′ 23°19′ b 64°28′ 25°32′ or 154°28′ d 117°2′ 114°27′ f 27°2′ or 152°58′ 0.0° 18.6° −tan 67.16_NCM10extSB_ans Page 634 Thursday.0 m 21 389 m f b 16° c 87.8° 8 47 km Exercise 8-06 1 a c e 2 a d 413. May 18. 5+1 = 5–1 ----------------5+1 12 -----.4 b 2 a a = 20. 2 3 = tan 60 -----91 b tan Y = 1.19 f 0.8° 6 64°40′ 7 99° 35.44 f 3 a 70° b 33° c 109° 4 a 111..07 m2 6 852.8° f 31.5 e 0.1999 Exercise 8-05 1 a 5.1 c 2 a a = 8. 126°52′ f 25°23′.6 b 13.4 m c 11.9 m 64°59′ 12° 27° b 4.1 cm2 326.4 m2 b 463.4 m2 b 463.6 cm2 b 418. When using the sine rule: d -----------------sin 90° = 12.4 m b 14. 11 Teacher to check.1 mm c 17.75 c cos β = 5 -----13 .55 37° c 54° 46.7° b 55.5 km 7 486 km Power plus 1 a x = 12 b y = 16 3–1 b 2 a c g = 10 1 --.63° 5 a cos α = 60 -----61 b tan A = 6 a 30° b 45° c 60° 7 a 24 b 48 3 c 8 a c 9 a c e 10 a 11 a c 12 a 13 a 14 a 48 ------2 39 -----80 = 24 2 −cos 17°13′ b sin 25.52 c d b = 16.43 (to two decimal places) 4 a 177.95 c −1..2 m2 227. Exercise 8-02 1 a 43° d 34.60 sin 79°25′ −cos 59°35′ −cos 45° sin 75° e 0 The following are the exact answers. 4 4 --5 = = 5–1 ----------------. 115°51′ 8 a 53°8′ b 126°52′ c 16°42′ d 163°18′ e 53°8′.8° 2 a 0. 152°58′ f 64°9′.7° or 159.1 c 105.8° b 108. 131°57′ c 20°34′. d and e Teacher to check.7 m2 7 a 1256.0 cm c p = 8.72 d 131° 121. sin φ = 11 ---------.6 10 Teacher to check.2 m b 39° d 44° or 136° 41° c 84° 10 106°31′ b 1042 cm b 131.( 2 3 – 1) 3 a 45° b 60° c 30° d 150° e 120° f 135° 4 a 0 b 1 c. 6 7.1 cm d 13. cos α = d sin F = 40 -----41 .9 m f 18.0 b = 11. Check how close your estimates were..752 c 16.5 b 23. 2006 11:01 AM 10 33. sin Y = 2 --3 f cos φ = 5 ------4 .6 m Skillbank 8 5 a tan A = d cos X = 3 --4 = tan c sin 5+1 ----------------2 3 11 ---------4 = tan 5 -----. 118°57′ j 69°18′ k 41°49′.2 cm2 d 245.1 mm c 7.41 f = 40.85 f f = 3.8 e = 88. 4 3 a 0 12 -----13 .3˙ c tan X = d tan P = 9 -----40 e tan Q = 40 ---------3 6 a sin X = 11 -----61 b cos X = 7 -----25 c sin X = 2 --3 2 --3 1 ------3 7 2 8 a 45° b 30° c 30° 9 a 8 b 12 c 3 2 10 35 3 m 11 Teacher to check.50 m c 89 m 5 a 15. cos b sin so .8° 129° c 142°8′ 129°29′ f 162°13′ b w = 29.9734 i 8. 138°11′ l 143°8′ 9 a 30.1° 165 cm2 b 286 m2 c 30 m2 .2 m2 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5..6 m2 c 246..3° f g sin 85° h −tan 9.1 m b 53.9 cm2 c 173.7 m 15 b 595 m 21. 7 a 67°1′ or 112°59′ b 30° or 150° c 20.. 2 3 tan = 5–1 ----------------. cos F = 9 -----41 .5 cm b 48°59′ c 57°12′ b 0. 6 Teacher to check. 8 symmetrical no outliers ii no clustering symmetrical no outliers ii clustered near 7 negatively skewed one outlier at 25 clustered near 18. 6 a No. 7 The opposition team in this game was missing some of its better players. 16 not symmetrical. ANSWERS 635 9 . 5 a Yes. mode = 5 d range = 5. 22.5 iv 15 b i 6 ii 3.5 iv 6. 12. mean = 5. mean = 17. median = 29.5. 4 and 7. median = 5. no change. Richard who frequently scores 0. 1 are not far from the main body of data which begins at 5.5 iii 27. or most students did not study. Exercise 9-03 1 a b c d Richard 167. c Skewed towards the higher marks d The students practised spelling the words before the test.5 iii 17.4. discrete categorical quantitative.4 iii 15 iv 15 ii 9.1 iii 3 iv 2 c i 6 ii 6. continuous quantitative. median = 17. 14. 90. median and mode have increased by 5 a i 26 ii 20. However. because results of 3. 6 symmetrical no outliers ii clustered near 12 positively skewed one outlier (24) clustered near 5.42 b i 7 4 a ii 8 Stem Leaf 5 5 Frequency 1 5 6 0 2 2 4 4 7 8 8 8 4 7 2 2 3 3 4 5 7 7 8 0 1 9 1 3 4 9 5 7 6 3 7 1 b 5 a b c d b Frequency histogram and frequency polygon 6 7 8 9 10 10 8 6 4 11 2 0 1 c 4 5 a 30 6 a i 5 b i 9 c i 9 2 3 4 5 6 Length of queue 7 d 11 b 2 c 27% ii 5. such as 20.4. 2. 9 symmetrical (if outlier is ignored) no outliers ii clustered near 51 symmetrical no outliers clustered near 20. median = 13.3 iii 70 iv 68 i 21 ii 74 iii 74. a 342 b 38 Exercise 9-02 1 a i ii iii b i ii iii c i ii iii d i ii iii e i iii f i iii g i ii iii h i iii i i iii j i ii iii k i iii l i ii iii 2 a no outliers some clustering near 5 and 11 symmetrical outlier at 24 clustered near 15 to 18 positively skewed no outliers clustered near 13. continuous quantitative.8 Richard 7.5 iv 26 a 120 b 154 c 34 85% 11 or 42 a 12. 40s and 50s c Skewed towards lower marks d The test was difficult.3 iii 5 iv 5 ii 16. James 41. mode = 28. James 46. 26 symmetrical 67 68 69 70 71 72 73 74 75 b no outliers c i clustered near 69–70 and 72–74 ii not symmetrical. 2006 11:01 AM 3 a Start up 1 a b c d e f g h i j 2 a 3 a b 4 a quantitative. discrete quantitative. 29 c range = 6.5 iv 7 Exercise 9-01 1 a i 5 ii 16. possibly through injury. 81. continuous categorical quantitative. continuous quantitative.5 iv 81 Range. 11. The team is consistent in the number of goals it scores.5 2 a range = 5.6 iii 7.2 iii 8. 78. not skewed 3 a iii 8 Length of queue 2 Frequency x Frequency histogram 10 Frequency Chapter 9 8 6 4 2 0 1 2 3 4 5 6 7 Number of children 8 b no outliers c i clustered around 2–3 ii positively skewed 4 a 26 b Clustering in 30s. 10. not skewed no outliers clustered near 3. mean = 12. May 18.6 ii 7. mean = 28. or the words were very easy to spell. b Clustering occurs around 8–9. 15 b 17. 17 c Any three scores who have a sum of 62.125 iii 15. mode = 13 i 36 ii 70. 20.16_NCM10extSB_ans Page 635 Thursday. 13. 19 positively skewed two outliers ii clustered near 4. discrete categorical 34 b 8 c 4 d 44% i 26% ii 7% iii 5% i 312 or 313 ii 187 or 188 iii 75 f xf 4 4 16 5 3 15 6 6 36 7 10 70 8 16 128 9 6 54 10 3 30 11 2 22 50 371 x = 7.5 iv 76 69. 2 a i $95 250 ii $91 500 b The median because the value of the mean has been affected by $170 000 which is much higher than the other salaries. 75.5 iv 16 b i 26 ii 30.75 iii 6.325.5 James because he is the more consistent batsman. 9. mean. 81 i 21 ii 79 iii 79. b Clustered around 2 c Symmetrical (especially without the outlier). has had two very good scores. James 45 Richard 46. mode = 17 b range = 52. 8 d i 1.5 iv 7. 2006 11:01 AM Exercise 9-04 30 25 20 15 10 5 0 13 14 15 16 17 18 Ages e Median = 15 Number of calls f cf 0 5 5 1 6 11 70 2 6 17 60 3 6 23 4 9 32 5 6 38 6 1 39 7 1 40 3 a Cumulative frequency histogram and polygram Cumulative frequency 1 a Cumulative frequency histogram and polygon d Cumulative frequency 3 a i 1. the average number of accidents is 1. 4 a Toyota b The data is categorical.5 4 5 . median 73 b Amy because she is more consistent. apart from the score of 23. Amy’s mean of 64. mean 66.5 per workplace although many of the workplaces had no accidents.2/ 5.3 1 2 3 Score Median = 2. May 18.5 ii 0. median 69 Amy: range 62. c The range. 5 a Jade: range 39. mean 64.6 has been affected by the mark of 23. mean and median can only be found for numerical data. which could be due to illness. 6 a i 64 ii 63 b i 70–79 (the 70s) ii 60–69 (the 60s) Σf = 40 b 50 40 30 20 10 0 Cumulative frequency histogram and polygon 17 18 19 20 21 22 Score Median = 20 40 35 Cumulative frequency histogram and polygram b 25 30 20 15 10 5 0 0 1 2 3 4 5 6 Number of calls/minute 7 Ages f cf 13 1 1 14 6 7 15 10 17 16 7 24 17 5 29 18 1 30 b 17 c 80% 636 25 20 15 10 5 0 0 Median = 3 2 a Cumulative frequency Frequency 30 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.6.16_NCM10extSB_ans Page 636 Thursday.5 iii 0 b A combination of the mean and mode. 07 12 22 0.05–0.16_NCM10extSB_ans Page 637 Thursday.02 10 10 0.53 0.22 6 46 1.1 Σf = 45 Heights of students (cm) Class centre.2 0.5 3 a Class interval x 0.15–0.17 9 40 1.08 0.04 0.05 b Mean = 0.00–0.09 37 42 47 52 57 62 67 72 77 82 87 Marks in maths exam ANSWERS 637 .32 0.08 f cf Σf = 50 fx Σfx = 6.4 c Class centre Frequency fx cf 35–39 37 5 185 5 40–44 42 4 168 9 45–49 47 6 282 15 50–54 52 12 624 27 55–59 57 8 456 35 60–64 62 3 186 38 65–69 67 5 335 43 70–74 72 6 432 49 75–79 77 0 0 49 80–84 82 0 0 49 85–89 87 1 50 87 50 2755 Frequency histogram and polygon Frequency b 12 10 8 6 4 2 0 2 fx cf 396 3 324 40 5815 ii 140–144 Cumulative frequency histogram and polygon Cumulative frequency Class interval 3 Σf = 40 Exercise 9-05 1 a f 40 30 20 10 0 132 137 142 147 152 157 162 Heights (class centres) Median ≈ 144. 2006 11:01 AM f cf 4 4 4 5 2 6 6 4 10 7 8 18 8 10 28 9 9 37 10 4 41 11 3 44 12 1 45 c 50–54 Cumulative frequency histogram and polygon d Cumulative frequency 4 a Number of seedlings 50 40 30 20 10 0 37 42 47 52 57 62 67 72 77 82 87 Marks (class centres) Median ≈ 54 e 55.14 0. x 130–134 132 50 135–139 137 7 959 10 40 140–144 142 10 1420 20 30 145–149 147 8 1176 28 20 150–154 152 6 912 34 10 155–159 157 4 628 38 160–164 162 b 2 a Cumulative frequency Cumulative frequency histogram and polygon 0 4 5 6 7 8 9 10 11 12 Number of seedlings/punnet c Median = 8 b i 145.25–0.05–0.27 4 50 1. May 18.84 0.19 0.12 9 31 1.24 0.29 0.121 c Modal class = 0.20–0.09 0.10–0. 9 km/h 5 a 40 74.5 10 1145 50 Σf = 50 4985 b 100–109 638 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. Interquartile range for X is 5 and for Y is 3.5 104. . 4 a Cumulative frequency histogram and polygon c 50 Cumulative frequency Cumulative frequency d Class interval Class centre 70–79 74.60 d $470 h $105 l $237. since the interquartile range is 3 compared with 5 for X.07 0.5 40 110–119 114.5 9 − 6 = 3 b 50 − 48 = 2 c 6 − 3 = 3 24 − 22 = 2 Range for X is 10.5. Also the median speed is lower.22 0.5 84.12 f The mean and median are the most useful measures of location. Grouping of 70–79.17 0.5 80–89 90–99 f fx cf 2 149 2 84.50 Exercise 9-06 117 50 b 105–109 Cumulative frequency histogram and polygon Cumulative frequency 30 Skillbank 9 Σf = 50 4995 50 40 30 1 a b c d 2 a 3 a d 4 a b c d 20 5 a 10 Q1 = 4.7 e Both methods of grouping give similar results. 2 a d g j 4 a e i $408 $408 $517 364 330 $175 68 b 132 c e $672 f h $1170 i k $262. Technician Y’s spread of calls is less. Q3 = 16 Q1 = 18.50 l b $240 c 1600 f 370 g $51 j $388 k $108 200 81 525 $954. and for B is 71 km/h. Q2 = 8.3 b c d e A 5 9887433320 6 0012355789 9965544332211000 7 00223355566 200 8 02345558 9 0 Median speed for A is 73 km/h.2/ 5. Range for A is 30 km/h and for B is 24 km/h. x 75–79 77 80–84 82 85–89 90–94 f fx cf 2 154 2 4 328 6 87 5 435 11 92 3 276 14 95–99 97 7 679 21 102 7 714 28 105–109 107 12 1284 40 110–114 112 9 1008 49 115–119 117 1 20 10 94.5.5 19 1985. Q3 = 22. and for Y is 10.5 Speeds (class centres) Median ≈ 102 km/h d x ≈ 99. The mean and median are approximately equal. Q2 = 21.5 114. Q3 = 13 Q1 = 55.5 10 945 21 100–109 104. B 8 0 77 82 87 92 97 102 107 112 117 Speed (km/h) (class centres) Median ≈ 102 km/h d x ≈ 99. 2006 11:01 AM 40 30 20 10 0 0.16_NCM10extSB_ans Page 638 Thursday. Q2 = 9. Interquartile range for A is 15 km/h and for B is 8 km/h. Q2 = 71.12 0. Suburb B is much safer for driving because the speeds are more clustered (interquartile range of B about half that of A). Q3 = 78 9 b 3 c 15 d 2 e 3 f 22.5 9 760. … condenses data more than grouping of 75–79.27 Blood alcohol levels (class centres) Class interval Class centre.5 11 94. both are much larger than the mode. … and so some information about actual speeds of vehicles may be lost. Median for X is 5.5 100–104 c 50 0 e Median ≈ 0. and for Y is 9. May 18.5 Q1 = 6. while the range is significantly lower.02 0. This means the results are higher and not as spread. c The interquartile range. σ ≈ 2.6.3. σ ≈ 13. d Day 2.4 Exercise 9-09 35 25 1 a Men: x ≈ 71. b Mean mark for science is 63. c The interquartile range for Day 1 is 14.6.5 c 8 d 2.5.5 Test 2: x ≈ 67.0 for Team A. and for Class 2 is 4. range and standard deviation for English are all lower than those for science. Class 1 is more consistent. Female 45 ANSWERS 639 . σ ≈ 14.4.7 b x ≈ 23. σ ≈ 1. May 18.41 iii 1. so the class has benefitted from the remedial work. 3 The first set. σ ≈ 1. c Bushrangers are more consistent.33 4 ii 4 4 ii 4 4 ii 4 4 a x ≈ 7. and for Day 2 is 44. Exercise 9-10 1 a i The range for science is 47.3 b The mean for Test 2 is higher than for Test 1 and the standard deviation for Test 2 is lower than the Test 1. σ ≈ 2. since their mean is much larger than the girls’ mean. compared with 27. σ ≈ 0.0 Team B: x ≈ 120.5 e 4 4 a Team A: x ≈ 122. σ ≈ 2.0.29.8. Their interquartile range is lower and the range of the Bushrangers is 152 compared to 190 for the Ghosts.09 iii 1. σ ≈ 13. and for Class 2 is 5. σ ≈ 23. σ ≈ 1.0. 5 a Vatha: x ≈ 13. σ ≈ 9.9. σ ≈ 2. 20 2 a i x ≈ 150.16_NCM10extSB_ans Page 639 Thursday. σ ≈ 27. Exercise 9-07 1 a 7 ii x ≈ 165.0 iii 1. because the interquartile range and the range are significantly larger than for Day 1.4 c x ≈ 49. Female 46 b Range: Male 52.83 50–59 4 38 60–69 2 40 Cumulative frequency histogram and polygon b 40 c x ≈ 22. and for English is 44.4.5. 2006 11:01 AM Day 1 6 a Day 2 0 8 98765554221 1 234556667889 9876422110 2 24788 885100 3 1235677 760 4 0238 5 2 3 a e 4 a b c d e 5 C 7 a 21 hours b 16 hours c 24 hours d 8 hours i 75% ii 25% i 9 ii 10 The median mark for Class 1 is 6. the mean of women’s pulse rates is much higher. and for Day 2 is 20. 6 a Test 1: x ≈ 58.63 b x ≈ 46. 5 6 7 8 9 Number of hours 10 2 a Team 1 Team 2 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 Number of goals b Team 1 is more consistent because the interquartile range (the length of the box) is smaller than that of team 2.5. σ ≈ 1. 0–9 1 10–19 8 9 20–29 5 14 Exercise 9-08 30–39 9 23 1 a x ≈ 6.2 iii x ≈ 158.5 Ana: x ≈ 14.07. and for English is 14.8. and for the Ghosts is 73.73. so student marks for English are better than those for science. 2 a Male 53. σ ≈ 7. σ ≈ 13.9. ii The interquartile range for science is 22. 7 a Class interval Frequency Cumulative frequency 1 40 60 80 100 120 140 160 180 200 220 240 260 b The interquartile range for the Bushrangers is 60. σ ≈ 0.90 e 3 a b c x i i i d x ≈ 12. iii The standard deviation for science is 14.19 ≈ 4. Also the mean is higher for English. and for English is 10. and for English is 65. σ ≈ 11.7. σ ≈ 0.12 b x ≈ 7. i 75% ii less than 50% 6 C Bushrangers b The range for Day 1 is 36.0 b Yes. The interquartile range for Class 1 is 3.9.71.3. The mean of the class is the average of the girls’ and boys’ means.4.7.8 Women: x ≈ 77.0 b The boys are much taller than the girls.7 b Vatha is more consistent as the standard deviation for her times is significantly lower than the standard deviation for Ana’s times. σ ≈ 6.6 40–49 11 34 2 a x ≈ 13.5 30 10 Q1 5 9 –1 10 20 0– 9 0 Median Q3 15 –2 30 9 –3 40 9 –4 50 9 –5 60 9 –6 9 Cumulative frequency Ghosts Number of SMS communications c Median ≈ 37 Interquartile range ≈ 45 – 21 = 24 b 5.1. since the standard deviation is less than that of the second set. which may be due to the stresses involved in shopping (and looking after children at the same time).1.9.5.6 b Team B is slightly more consistent as its standard deviation is 23. 63 3 a 12 11 13 15 14 13 10 14 15 b x ≈ 13. d Neither.25. 14 d 40.5 standard deviations above the mean while maths is only 4 standard deviations above the mean.27 Class interval Class centre x f cf 85–89 87 3 3 90–94 92 2 5 95–99 97 7 12 100–104 102 7 19 105–109 107 15 34 110–114 112 13 47 115–119 117 3 50 50 . 0 9 1--3 x Cumulative frequency histogram and polygon Cumulative frequency Interquartile range: Male 22. 13.6. 3 a A 75. 94. Also females as a group are more consistent in their spending because the range.3 Cumulative frequency Centre A 640 10 c 7 1 a Range = 10. 4 a i Team 1 51. 107.08.94 i 2.16_NCM10extSB_ans Page 640 Thursday. interquartile range and standard deviation are all less than Group B’s.5. c Neither. Female 10.2/ 5. 103. 12. 2 x ≈ 8.5. median = 23. σ ≈ 1. 108 and 112 2 Teacher to check. b Modal class 105–109 Cumulative frequency histogram and polygon c Centre B 9754 0 578889 866521 1 2233456778 65521 2 01245 9885100 3 023 10 4 b Centre A: x = 23. Y 12 c 12. median = 17. 1.48 c Males spend more than females at the restaurant. Y 73. since it is 4.5 Standard deviation: Male 13. 2006 11:01 AM 4 a b Power plus 1 a Maths.06 b A 38. σ ≈ 3.9 b Team 2 appears to be more consistent since its range and standard deviation are lower than Team 1’s. 9. However Team 1’s range and standard deviation have been affected by the score of 73. 11. since they are both 3 standard deviations above the mean. 4 a 14 12 16 20 18 16 10 18 20 c 5 a b 6 a b cf 5 5 6 8 13 7 15 28 8 13 41 9 8 49 10 1 50 50 40 30 20 10 6 7 8 Score 9 5 a b x ≈ 16.5. Female 16. b Maths. May 18. 9 c 6.39 c Group A has less spread since the range. σ is unchanged. σ ≈ 1.5 7 a X 81.5. since it is 1 standard deviation above the mean while 65 for English is less than 1 standard deviation above the mean.5 b X 9.63 Because it is a measure of the ‘distance’ of a score from the mean. N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.67 ii 2. B 74.64 B 42.5 .5. Team 2 36 ii Team 1 22. The mean of the waiting times of Centre B is much lower than that of Centre A. 50 40 30 20 10 0 87 92 97 102 107 112 117 Speeds (km/h) d 107 km/h e x = 105 6 a 6 b 2. since they are both 2 standard deviations below the mean. The waiting times for Centre B also have less spread than the waiting times for Centre A. 17.7. since it is less than 1 standard deviation below the mean while 48 for English is more than 1 standard deviation below the mean. mode = 18 3 5 5 Chapter 9 Review b 4. Team 2 23 iii Team 1 14. 13. e English. range = 28 c Yes. f 0 x is doubled and so is σ.84.5. Team 2 11.63 c Mean increases by 5. interquartile range and standard deviation are all significantly lower than those of the males. mean = 17.63. 9. range = 37 Centre B: x = 16. median = 15.45. May 18. 2006 11:01 AM c c i m = − 4--. b = −2 ii c y = − --1.65. σ ≈ 8. 1 --2 1 --2 ) ) 15 a y = − 1--. y 4 8 a Girls: x ≈ 55. The mean of reaction times of men is much lower than that of women.8 b The mean mark for boys was just higher than the mean for girls. 12 y = −4x + 3 ii y 13 a y = 4 −4 x − 3 b y = − 3--.6 Women 9 a 4 Men 2 6 8 9 9 9 9 9 8 7 7 6 5 5 4 2 3 0 1 1 2 2 3 7 8 8 5 3 2 0 0 4 3 4 7 8 9 −4 7 4 1 1 5 3 5 −6 0 d −6 −4 −2−2 6 y = − 1--. 29 units. σ ≈ 19.98 e Yes. BC = 32 x −4 4 2 9 3 × (−1) − 3 + 6 = 0 2 4 x 10 AB = AC = 11 All sides have the same length of BD both have length b i m = −3. b = 2 7 a y = − --3.. m = − 1--.16_NCM10extSB_ans Page 641 Thursday.94 1 a Neither c Parallel 3 --2 b y = 3x − 2 4 25 30 35 40 45 50 55 60 1 --2 4 2 5 3x − y − 5 = 0 Men 3 a i m= a 2 b Women 2 c 6 Boys: x ≈ 55.. .2.8.x + 3 --2 2 x 6 b . ii 34 iii − --5- ii 125 iii − 1--- 2 2 0 −2 −2 1 --2 2 4 2 x c i (−1. as it shows the difference in spread between men and women and the positions of the medians b y = 2x + 1 0 −4 −2 −2 2 4 2x + y = 3 2 0 −4 −2 −2 x −4 2 4 x −4 y c 4 b Parallel d Perpendicular x + 2y − 4 = 0 2 2 --3 0 −4 −2 −2 y 2 4 40 .x + 5 c Box plot.x + 6 y 8 a d Women: x ≈ 41. σ ≈ 6.x + 1 c y = 5 4 14 a i (−1.x + 1 2 16 17 18 19 20 C a 2x − y − 3 = 0 6x + 8y + 7 = 0 a x+y−1=0 a 2x + 3y − 6 = 0 b y = −x − 4 9 --4 3 2 c y= 2 --5 x−2 d y = 2x + 2 b 3x + 4y − 6 = 0 b 4x − y − 13 = 0 b 3x − 2y − 5 = 0 ANSWERS 641 . b = 1 y 4 2 Mixed revision 3 0 −2 −2 2 4 Men: x ≈ 37. but the reaction times for men have more spread than the women’s reaction times. σ ≈ 23.) ii 97 iii b i (3 --1. b = 2 ii y 3 Group X 4 Group Y 2 0 −2 −2 62 64 66 68 70 72 74 76 78 80 82 84 86 88 Pulse rates 4 2 x −4 d Group Y had greater spread since its range and interquartile range were both larger than those for group X. and diagonals AC and 58 units. − 1--.. but the spread of marks for boys was much greater. May 18.5 m c 5.3. ii Interquartile range for Runner A is 0. 40 a i No ii Clustered around 3 and 4 iii Nearly symmetrical b i No ii Clustered around 15–17 iii Negatively skewed c i Yes. 128° a 116° b 43° c 41° 019° a i 8 ii 6.7 iii 7 iv 7 b i 5 ii 5. The median best represents the data because the mean is affected by the score of 20 and the mode is the lowest score. Runner A is no better.67) 21 a 3x − 2y + 4 = 0 b 5x − 2y − 17 = 0 22 a 39.2.5 24. 125 ii Clustered in 70–80s iii Skewed d i No ii No iii Symmetrical 41 a i Range for Runner A is 3. median = 6.6 c Runner B is more consistent because the range and standard deviation are much lower than for Runner A and the interquartile ranges are nearly the same.4° b 3 ------2 e 1 ------2 f 1 b − ------3e 27 13° 2 1 c − ------ 3 ------2 1 f − ------ 2 2 12 28 3 8 36 4 7 43 5 4 47 6 2 49 1 50 7 Σf = 50 3 g 1 h 0 i 0 a 20.3 iii 22 iv 23 37 a 8 − 5 = 3 b 6−4=2 c 19 − 17 = 2 d 128 − 102.29.7 30 20 10 0 1 2 3 4 5 6 Number of DVDs 44 x = 7.0 iii 5 iv 4 c i 6 ii 18. 50 50–59 d 1 ------2 26 50 m 1 ------3 16 40–49 28 a −1 25 C c 6 10 30–39 24 72 m 1 ------2 6 1 20–29 d 3 ------2 cf 0 Cumulative frequency 23 a f .5.55. 45 a Marks in test x 10–19 14.5 29 30 31 32 33 34 35 36 Cumulative frequency histogram and polygon b and c b x ≈ 50.2/ 5.4° b 118° c 95. Runner A’s results have been affected by the outlier of 15. ------⎞ ⎝ 13 13⎠ c 35 4 ----52 d −2 1--- 43 a 2 Number of DVDs square units (≈4.4 m b 32.5 = 1.6 40 0 39 a A: x ≈ 396.7 iii 119 iv 134 e i 4 ii 22.5 e 23 − 21. σ ≈ 13. iii The standard deviation for Runner A is 1. d Although Runner B is more consistent.2 (which may be an error in timing or recording).3 436 Σfx = 1965 4 642 7 Median = 2 Frequency 42 a x ≈ 55.5 40–49 44.5 20 42 890 50–59 54.55.8 b Brand A is the longer-lasting brand because the mean for Brand A is nearly 30 hours above that of Brand B. σ ≈ 1. σ ≈ 34.3. mode = 2.5 m a 41 m2 b 340 mm2 a 43°54′ b 95°12′ 52°.03 and for B is 0.5 10 11 245 34.5 = 25.6° or 159.7. B 12. σ ≈ 27.3° b 51. 2006 11:01 AM c 8 ⎛ 27 ------.1 iii 18 iv 18 d i 39 ii 117.2.4° a 5.5 20–29 30–39 f cf fx 1 1 14.5 8 50 Σf = 50 Frequency histogram and polygon b 20 16 12 8 10–19 0 Marks in test c 40–49 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.5 and for B is 0.2 B: x ≈ 369.5 38 a 10 b i 6 ii 11. Range for Runner B is 0.16_NCM10extSB_ans Page 642 Thursday.8.9° c 44.7. b A 12.5 11 22 379. 2. Sunil $900 Lisa $160. or it is a draw c Player A wins.3 e 42 46 4 2 -----13 = 7 a Drawing a blue marble (or any colour other than green or red) b 1 8 a 0 b 1 30 20 8 -----52 3 --7 + ii 4 --7 =1 12 a i b 1 a even chance b likely or unlikely (depending on today’s weather) c even chance d very unlikely e unlikely f very unlikely g even chance h likely i certain j impossible 2 a 1. 4 of hearts. 1. 4.4 Skillbank 10 Mark $100. 3. I c i 1 -----10 or 0. 3. 1. 2. 6.155 v 0. 1. 5 of hearts.486 12 a 0.7675 Obtaining an odd number Obtaining a tail Selecting a diamond. 5 or 6 b Team A wins. John $1000 Carol $550.7 14 a 40% b 60 15 a Missing the bullseye 2 a b c d e f g h i j k l 10 -----31 1 a Each outcome to appear 5 times 1 --5 ii 0. 5.845 b 0. g i Dice 1 Dice 2 1 2 3 4 5 6 1 0 1 2 3 4 5 2 1 0 1 2 3 4 3 2 1 0 1 2 3 4 3 2 1 0 1 2 5 4 3 2 1 0 1 6 5 4 3 2 1 0 ii Difference Frequency Probability 0 6 6 -----36 = 1 --6 1 10 10 -----36 = 5 -----18 2 8 8 -----36 = 2 --9 3 6 6 -----36 = 1 --6 4 4 4 -----36 = 1 --9 5 2 2 -----36 = 1 -----18 Total 36 Exercise 10-02 1 a i 41 --------100 ii 21 --------100 38 --------100 iii or b 1 c i 102 or 103 ii 52 or 53 2 a 10 b to e Teacher to check. 3. 6.515 1--4 c to f Teacher to check. Queen of diamonds. 3 4 --6 or 2 --3 ii 2 --6 or 1 --3 5 a {Jack of hearts. 5. 3.505 vi 0. iv A 67 B 222 v Teacher to check. 2. 3. 4 a Teacher to check. 6 b i 1 --6 ii C 1111 1 --2 b i 3 -----40 ii 11 -----40 iii 7 -----40 iv 6 -----40 = 3 -----20 b i v 5 -----40 vi 8 -----40 = 1 --5 c Yes d 3. 3. or it lands tails e HH. 2006 11:01 AM d d Cumulative frequency Cumulative frequency histogram and polygon 50 40 9 a i 3 --7 iv 7 --7 b 10 10 a b c d e f 50–59 40–49 30–39 20–29 10–19 0 Marks in test f x ≈ 39. May 18. 5. 3. 6. club or spade Selecting a vowel Missing the bullseye Selecting a red disc 11 a {1. Louis $440 Yvette $800. 2.01 b Exercise 10-03 c Obtaining a number other than 1 Start up 26 -----52 b 7 -----31 11 a 1000 b i 0. Jack of spades} b {4 of clubs. 8 c 1 7 a 2 -----15 = 7 -----10 = b P(R) = P(G) = P(Y) = P(B) = 13 0.1 ANSWERS 1 --8 or iii 0. Andre $3200 Arden $2100. 4. 3. 5. 4. 6. 2. 5} Chapter 10 6 a iii 0 =1 d {2. 4. 3. 1. 5.2325 iii 0. 5 of spades. HT. 5 of clubs. 5.5 = 1 --8 19 --------100 d 97 --------100 b 15 --------100 = 3 -----20 iii 5 --6 e 4 --8 4 a 1. 4. Jack of diamonds. 3. 1. 6. since the coins are not identical in size. 5. 4. 3 a 1. 6. Mai $1200 ii 0. 6.3 ii 7 -----10 or 0. 3. 3. TT f 1. 2. 2 a to f Teacher to check. 4 of spades. 5. 6. 4 ii 1. 5 of diamonds} c {Jack of hearts. 3. 2. 4. 5. 4.669 c 0. Bree $560 William $500. 4. 2. 5 iii 2. 1. 2.125 iv 1 ii 0. 7. TH. 4. 6.375 vi 0 b 3 -----10 or 0. 5 a i 55 --------100 11 -----20 = ii 45 --------100 9 -----20 = b 1 6 a 40 19 -----50 1 iii Teacher to check. 4. Ivan $2800 Tan $2000. 1. 3 a 5 b 1 --5 c i 1 --5 ii 0 iii 1 --5 d 2 --5 4 a i 2. 3. 4 of diamonds. King of diamonds} = 1 --2 b 12 -----52 = 3 -----13 c 4 -----52 = 1 -----13 99 --------100 1 --------100 9 a 3 -----20 b 2 -----20 d 11 -----20 e 3 -----20 10 a 11 -----60 b 21 -----60 d 42 -----60 c 12 -----31 = 1 -----10 c 1 -----20 = 7 -----20 c 8 -----60 4 --5 1 --5 f + 4 --5 =1 iii 1% = 0. Pete $1800 Sharanya $900. Player B wins d It lands heads. 2. 4. 1. 3. Asam $2100 Cindy $3000. 6 f i 0. 4. 5. 3. 2.7 643 1 --2 . 1.99 or 99% b 0. 5. E. Team B wins. 6 3 a A b B c B d B 4 a C b C c A d D 5 a C b D Exercise 10-01 1 a 26 b Yes 2 a 5 b No. 3. 1. 4.16_NCM10extSB_ans Page 643 Thursday. 4. Adriana $1500 Ed $2700. 4.5 v 0. 5. Carmen $600 Nancy $600. 3. Jack of diamonds. 6. 5} e b i 4 --7 8 a c 57 --------100 1 --8 ii 5 a A. 4. b 1 c and d Teacher to check. 1. 5. King of hearts. 3 a Teacher to check.17 iv 0. 2. 2. 1. 6. b 1 c and d Teacher to check. Jack of clubs. 4. 5. 2. 2.2525 d 0. 6. 4. 5. 2. 3. 5 iv 1. 2. Jenni $50 Simon $1200. Queen of hearts. 1. 6.62 or 62 --------100 or b 0. 3 2. 3. 1 6. 4 1. 4. 5 4. 5. 1 1.3 6. 2. 1 5. 6 3 3.16_NCM10extSB_ans Page 644 Thursday. 6 5 5. 2 5. 2 2. 4 6. 5 5. 3 3. 6 b 11 a 7 BB BB BR BB BB BR RB RB RR 1st die d = 1 -----10 C J M L S CJ CM CL CS F FJ FM FL FS M MJ MM ML MS H HJ HM HL HS . 2 1. 6 18 a a 5 5. 4 4.07 or 7 --------100 or 7% e 1 --3 d 1 --3 g 0 h 2 --5 c 1 --6 d 5 --6 c 7 -----11 d 3 -----11 c 1 --5 d 1 -----10 c 31 -----50 1 --2 b B R B R B B B B R or 62% 2 --7 b i 3 1 --2 ii 4 --7 b 2 --3 4 --7 iii iv c 0 3 --7 v d 3 --7 4 5 --6 19 C 20 a 1 --4 b 50 6 21 27 or 28 Exercise 10-04 1 1 --2 2 3 a Independent c Dependent e Dependent 1--6 7 b Dependent d Independent 4 a i 5 --8 ii 4 --7 b i 3 --8 ii 5 --7 c i 5 --8 ii 3 --7 d i 3 --8 ii 2 --7 2 3 4 5 6 1 1. 2 1. 3 5.3 c 2 -----20 1 --6 = 5 -----12 a 2 -----12 = 1 --6 b 2 -----12 = 1 --6 c 4 -----12 = 1 --3 d 4 -----12 = 1 --3 ii 9 -----25 ii 6 -----20 = 3 -----10 10 a i 4 -----25 b i 2 -----20 = 1 -----10 Exercise 10-06 1 People 4 -----25 ii without replacement a = K spades Q clubs Q spades K clubs Q spades Q clubs K spades K clubs Q spades K spades K clubs Q clubs 9 Car 4 -----25 18 2 Q spades HH HT TH TT 1 = ----- 1 Q clubs i with replacement 6 6. 4 4. 6 K clubs 5 Independent. May 18. 2 6. 2 6. 1 2. 3 6. 4 5. 1 6. 3 3. 3 1. 2 3. 2 4. 6 644 6 -----36 K spades b No H T H T H 2 2. 6 c 11 b i 8 34 36 37 43 46 47 63 64 67 73 74 76 2nd toss 1 1. 1 3. because the outcome for the coin will not affect the outcome for the die. 6 b 36 33 34 36 37 43 44 46 47 63 64 66 67 73 74 76 77 6 6. 4 2. 5 4. 5 2. 5 2. 4 5. 6 2 2. since the number of balls from which the draw is made is decreasing. 3 2. 3 1. 4 3. 6 6 = 1 --6 ii 11 -----36 iii 16 -----36 = 4 --9 iv 2 -----36 a a 4 -----25 b 4 -----25 c 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 b i 3 -----36 = 1 -----12 ii 6 -----36 iii 3 -----36 = 1 -----12 iv 35 -----36 v 10 -----36 = 5 -----18 vi 7 -----36 vii 1 -----36 viii 15 -----36 d 16 -----25 e 2 -----20 = 1 -----10 b 4 -----20 = 1 --5 d 12 -----20 = 3 --5 e 4 -----20 = 1 --5 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. 5 1. Blue die 2nd die 1 BB BR BB BR RB RB 3 4 6 7 3 4 6 7 3 4 6 7 3 4 6 7 4 a 16 D 17 a 1. 1 1. 5 3. 4 1. 5 3. 5 1. 4 3. 5 6. 1 2. 6 Dependent. 2 4. 2006 11:01 AM 6 a 3 -----10 7 -----10 + 2 --6 = =1 B B R B B R B B R 3 a 1 --3 b 4 --6 or B 2 --3 7 6 red marbles 1 --4 b 1 --2 c 1 -----13 d 2 -----13 e 1 -----13 f 3 -----13 g 1 -----52 h 3 -----13 9 a 5 -----11 8 a 10 b 4 -----11 B 9 -----11 c R 3 --7 1 --2 3 -----10 f 1 --5 12 a 1 --3 b 1 --3 13 a 1 -----11 b 4 -----11 14 a 1 --2 b 1 --2 15 a 0.2/ 5. 2 3. 1 5. 3 4. 6 4 4. 5 5. 3 4. 3 5. 6 4 4. 1 4. 5 6. 2 2. 1 4. 4 2. 1 3. 4 6. 2 5. 6 4 6 7 3 6 7 3 4 7 3 4 6 b 3 4 6 Exercise 10-05 1 BA BR BC BM 2 AB AR AC AM RB RA RC RM MB MA MR MC 7 1st toss 5 a a Red die 1 2 3 4 5 6 T 3 3. 3 is twice as likely.139 iii 2 --5 d 3 --4 2 -----13 c 12 -----13 d 1 --4 e + 3 --4 =1 iii 0.09 b i 0. 8. 1 is 0. 5 3. 7.54 c 0. 3 is 0.41. 5 4. 4 2. 5 2.465 9 a 640 b 260 c i 0. 2006 11:01 AM Coin 2 1 2 3 4 5 6 1 2 3 4 5 6 H T 1st coin 3 a 7 a Teacher to check. 5 iv ≈ 0. green 11 a i H a 1 --9 1 --4 b 2. green 1. blue. 10 v c 4 a 70% 5 a 0 is 0. Die 2nd coin H H T H T T 1 --8 b i ii 3 --8 1 --8 iii iv 1 2 a 4 a 3rd coin Sample space b p3 b (1 − p)p2 = p2 − p3 d 3p2(1 − p) = 3p2 − 3p3 H T H T H T H T HHH HHT HTH HTT THH THT TTH TTT v 1 --8 vi 3 4 b 6 1. 4 1 --6 48 --------115 b b Yes c i 1. 6 3 3. 3 4. 4. 6 1 --4 vi 1 --8 vii 23 -----24 1st coin 5 viii 2nd coin b c 7 a b 1 --4 c 5 --6 8 a 2 --5 b 3 --5 c 3 --5 9 a 1 -----52 10 d i 10 -----64 H H T T H T T 6 a 50 b 1 --4 c b 300 3 --8 d 1 --4 c 750 e 1 -----16 4th coin Sample space H T H T H T H T H T H T H T H T HHHH HHH T HH T H HH T T H T HH HTHT HT TH HT T T T HHH T HH T THTH THT T T T HH T THT T T TH TTTT f 15 -----16 ≈ 0.417 3 a Red. 3 iii ≈ 0. 1 3. 1 ii 1 --8 iv 1 --6 3 a c (1 − p)3 1 1 -----24 3 --8 1 Teacher to check. blue. 6 4 4. 1 4. 4 3.2 or 1 --5 B B G B B G G B B G G B G G a 1 --2 b i 3 --8 ii B G B G B G B G B G B G B G B G BBBB B B BG B BGB B B GG BGB B BGBG B GG B B GGG GB B B GB BG GBGB G B GG GG B B GG B G GGG B GGGG 15 -----16 ANSWERS 645 f 3 --4 . 9. 2 1 --6 d 0. 2 3. 1 --2 5 22 --------115 1 a 5 1. 3. 6 2 2. 6. 2 is 0. 4 4.383 ii 0. May 18. Tetrahedral die c i iii Power plus 1 --2 2 1 --8 ii 8 a 0.333 ii 0. 1 2.41 1.173 4 a Normal die 1 --8 b i H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6 12 a 1 --4 5 -----32 = 5 -----32 = b 13 a 0.1 or 1 --6 1 -----10 ii 20 -----56 = 5 -----14 ii 10 -----56 = 5 -----28 ii 6 -----56 = 3 -----28 ii 10 -----56 = 5 -----28 c 1 -----12 b 0.277 b 0.883 1 -----26 e 1 -----13 1 --9 23 -----24 T 1 -----16 b 9 -----64 T 16 --------115 5 --9 iv 7 -----12 c i H 1 --9 6 a 1 --6 T iii 5 --6 10 -----64 H 2 --9 ii 1 --3 b i H 1 --5 ii 2 --3 25 -----64 3rd coin 1 --5 c i c Yellow.45 Chapter 10 review 1. 2 2.2 or 14 1--5 d 1 --6 c 0. yellow. 2 4. 3 2. 3 3.117.16_NCM10extSB_ans Page 645 Thursday.383. 5. as there are two 3s.191 25 --------216 2 a 9 b No. b During the 3rd hour (from t = 2 to t = 3). ii Rate of change of speed (acceleration) iii −3 c i Independent variable is quantity. No. then cycles back towards the starting point at a speed of 30 km/h for 1 h.16_NCM10extSB_ans Page 646 Thursday. ii Teacher to check. Kate 133 m/min. 0 km/h 3 a i 12. Dependent variable is cost. because the slope of the graph changes. Nooreen 9 min 2 a No. b = 2. (Other answers possible.5 km/h ii 25 km/h b i 12. travels at a speed of 20 km/h for 1 h. Colleen 114 m/min Kate. At D. i 10 km/h ii From C to D. 12:25pm 10:50pm 0610 hours 9:10am 1100 hours 12:20am 6:05pm 12:10pm 1245 hours 10:50pm 1545 hours 1:35pm Time d iii 1:10am 10:55pm 0010 hours 3:15am 2305 hours 11:35am 6:40am 2:50am 0355 hours 12:15pm 0400 hours 7:20am 1 --3 d i Independent variable is time.3 3 Teacher to check. May 18.2/ 5. stops for 1 h. b The person’s speed is fast initially. Dependent variable is height of tide. ii cm/min iii −3 1--9 Time 5 a i C b d f h j l b d f h j l 1 a i Independent variable is time. y = 1 --2 0 2 4 6 8 10 12 4 a x−1 Distance d Exercise 11-01 1 a i 160 m/min ii 100 m/min b i 160 ii 100 c The steepness of the graph. Dependent variable is speed. since the gradients of the intervals are all different. and continues for another hour at a speed of 10 km/h. 2 a H b D c A–B d F e E f C 3 120 100 Skillbank 11A 80 60 40 20 3 a m = 2. Dependent variable is volume of petrol. in 9 minutes c 1. then slows down (the graph becomes less steep) before increasing speed again (the graph becomes steeper). c Jade stopped to talk to a friend. ii Teacher to check. ii Cost per person iii 1 b i Independent variable is time. y = 2x + 2 b m = −2.5 ii −25 c i Yes ii No. Dependent variable is water level. b i Independent variable is time. opposite in sign d Car A is moving away from the starting point (positive gradient) while car B is moving towards the starting point (negative gradient). c i Independent variable is distance. 4 a The cyclist leaves the starting point. iii The slope of this graph is between the other two (Kiet). d i Independent variable is age. 4 Time b c d 5 a b d f 1 --2 646 Distance b Time Distance c h. slowing down gradually to a stop. Dependent variable is height. Dependent variable is profit. ii Is the least steep (the smallest gradient) and must be the slowest (Cameron). then slows down and stops (the graph is horizontal).min. iii A . ii Profit/item Distance cyclist stops for 2 a c e g i k 4 a c e g i k Exercise 11-03 d Zaid 11 --1.5 min 690 m e 110 m The graph shows the distance they move down the slope and this increases as more time passes. the b i Is the steepest (has the greatest gradient) and must be the fastest (Jade). Dependent variable is temperature. The cyclist is moving back towards the starting point. 6 No.) d This person speeds up slightly and maintains speed for a while. 2006 11:01 AM Chapter 11 Exercise 11-02 Start up 1 a i −2 b i −1 iv −12 iv 29 ii 13 ii 95 iii 23 iii −3 ii −2 iii d i −2 ii −10 iii 5 iv −10 e i 1 ii iii 81 iv c i 4 2 a b c 1 --3 8 --5 iv −8 1 --9 n −2 −1 0 1 4 c 1 --4 1 --2 1 2 16 B 1 4 10 32 80 L 40 10 4 1 --14- 1 --2 x −3 −1 0 1 2 y −27 −1 0 1 8 x −3 0 1 4 9 y 9 0 1 16 81 1 a The person starts the journey fast (the graph is steep). 2 a i Independent variable is number of persons. because it does not account for variable speeds when leaving and arriving at home. 7 a C b D c E d F e B f A ii B N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. b = −1. c The person starts the journey at a high speed and then gradually slows down to a stop. ii Teacher to check. ii Teacher to check. b = 2. y = −2x + 2 c m= 1 --2 . May 18.16_NCM10extSB_ans Page 647 Thursday. 2006 11:01 AM j Height Distance b Time t c Time k Distance b H Distance from shop 4 a H Time Time t d Sound level l Distance c H Time t e Exercise 11-04 Litres d H Time t f 1 a f 2 a f 3 a 4 a 5 a 6 a b 7 a c iii h v c A d vi i vii d B g f H Distance Time Speed b Time t h Same as g 5 a Time Water level Distance i e ix e A c C d B e A c A d F e C F d D e A f B iii E iv F iii F iv C B d C e A f D Speed Distance t viii b i iv g ii C b B B g A B b A B b H C b E c i C ii A i A ii D E b F c 8 a Litres e H Time Time Time ANSWERS 647 . 2006 11:01 AM d e f y y 1_ 2 Speed Speed c 0 Time x 1_ 3 0 1 −2 2_ 5 Time e f 1 a y = 3x2 d y = −2x2 Speed Speed Exercise 11-06 2 a i Time b y = 3x2 e y = −3x2 c y = 0.16_NCM10extSB_ans Page 648 Thursday.5x2 f y = −10x2 y y = 4x2 ii (0. 0) y = x2 3 1 −8 −12_ 0 x 0 x y = − 12_ x2 648 ii (0. 0) iii y = −18 y = x2 10 Speed 0 x y = −2x2 Distance from home c i y y = x2 ii (0. May 18. 0) iii y = 54 x −6 c x N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. 0) iii y = 36 Time c Speed y = x2 b Speed Speed 9 a Time 0 Time b i Time x y ii (0.3 iii y = −4 --12 x .2/ 5. 0) iii y = 3 Skillbank 11B 2 a d g j 8 h 30 min 8 h 25 min 5 h 10 min 7 h 25 min b 5 h 40 min e 11 h 25 min h 5 h 45 min c 3 h 25 min f 1 h 40 min i 7 h 55 min y = 13_ x2 0 Exercise 11-05 1 a b y y d i 12 0 3 y = 6x2 y y = x2 0 4 x 0 y d y e i 0 2 x x y ii (0. 2) 0 x y = 14_ x2 0 x Axis of symmetry is the y-axis (or x = 0) for all parabolas. 6) 0 V = (0. −6) −6 x x y = −3x2 y 10 i h i y ii (0.16_NCM10extSB_ans Page 649 Thursday. 10) 0 c −10 0 x d 4 a b c d e d y x V = (0. −3) x k i y V = (0. 0) iii y = −27 h 6 0 x y V = (0. −10) 0 −10 x V = (0. 2 a vi b ix c i d xi e x f iii g ii h xii i viii j v k vii l iv Exercise 11-07 1 a 3 a y = −x2 y b 10 V = (0. 10) y i ii (0. 0) iii y = 27 y = 3x2 j 0 y= 0 0 x x2 x −3 V = (0. −10) c y = −x2 + 1 --2 y = x2 + 3 e y = −x2 + 9 f y = x2 − 8 i narrower ii moved down i wider ii moved up i wider ii moved up i wider ii moved down i the same ii moved down f i narrower x y 0 y 10 V = (0. 6) 0 0 0 x −6 x V = (0. 10) b y = x2 ii moved down iii iii iii iii iii −1 3 4 −4 −3 iii − 1--2 5 a ii b vii c iv d x e v f ix g i h xi i viii j xii k vi l iii 6 a ii y = x2 b ii y = x2 − 1 c ii y = −x2 d ii y = −x2 + 3 7 b 80 m c 43.2 m d Approximately 4. 0) iii y = 18 y = 2x2 y = x2 6 V = (0.1 seconds. −5) 4 2 V = (0. May 18. −6) x y g i y f y y = x2 0 y g ii (0. 0) 0 iii y = 2 1--- y = x2 l y y x −5 V = (0. 8 a to c Teacher to check. 2006 11:01 AM y f i e ii (0. d i parabola ii 100 m2 iii 10 m × 10 m ANSWERS 649 . 2 y 2 0 iii x = 1 3--4 g y 5 −1 0 650 iv c i None ii 4 iii x = 5 x y − 3--4 4 8 (− 34_..55) 0 0 −40 b i 0.. −49) 0. −1) (3. 4 0 x=2 (2. 6) d (3. −2 14_) i i 1 1--. 18_) ii −6 ) −6 112_ 2 x .2/ 5.16_NCM10extSB_ans Page 650 Thursday.. 14) −20 x 4 Approx. 25) (4.) 4 a i ii iii iv v x 4 4 4 Approx.7.) 3 a (−2.2 ii 21 y b (3.) −4 −2 0 2 x 0 4 v Up 3 x 1_ (12 . 8) y 0 4 x y iii x = 1 1--2 iv ( 1 1--. −13) 4 6 c (− 2 1--.7 5 y x=3 (3. 14) Down 5 21 iv (− 1 --1. 10 −40 x=3 (3.7.) −1 − 2_ 3 0 e (1. −3) x 1 f i ii iii iv v f x=3 h i x = − 1--6 2 2 2_ 2 1 1--2 ii ii ii ii ii ii ii 4 0 c x = 2 1--- b x=0 h i ii iii iv v 10 x (3. 30) 2 v Down 0 x Exercise 11-09 c 1 a x=3 y d x = −1 −3 0 −5 x −15 y d 10 2 a b c d e f g i i i i i i i e x= x=3 x=3 x=3 x=4 x=4 x = −4 x=4 e ii (− --1. −80) (4.. 30) 2 −2 x e i Approx. 8) Down x (2. −13) 3 Up 0 j i x = − --1- 2 y (3. 1) (4. 1.6 y 3 x=4 (4. 3 --1. −9) (−4. 1 --1. 2006 11:01 AM Exercise 11-08 d i ii iii iv v y h (Diagrams not to scale. −9) −2 f (4.) x (134_ . − 10 1--. 2 78_ ) v Up N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. . 1. −4. 3 ii 0 8 y None 4 x = −0.) 2 −4. −49) Up −4 4 4 (−0.7 (−0. 2 7--. 7. 1.4. −0. 0) (3. 1) f ( 1 --1.. 80) ii (− 1--..3 0 ( 1 3--4 .7.. 0.. − 1 1--.2 1--. 1 --8 v Down 4 iv (− --3. May 18.2.) 1 a y 0 −5 0 1 1 3_ x 0 y i iii x = − 1 1--- 0 0 1_ 2 −2 x x 2 y (−112_ . 6.55) Up y x y g i ii iii iv v b (1. 4) x 0 ii y = −13. 4) y Exercise 11-11 1 a 0 x (1. 2006 11:01 AM Exercise 11-10 f i y y b 1 a y = 3x3 is narrower than y = x3 b y = 3x3 is narrower than y = 2x3 c y = 0.5 0 y x h y (−1. 5) y 0 0 x x 0 x y e ii y = −54 c i y ii y = 81 i x i y 0 (2.5x3 is narrower than y = x 1 3 --.x3 5 1 3 -x f y = −10x3 is narrower than y = − ----10 2 a i 0 x y c ii y = 54 g i y (4.75 narrower wider wider wider same f i narrower 2 a i f iv ii y = 162 e i b v g viii x ii ii ii ii ii 0 moved down 1 moved up 3 moved up 4 moved down 4 moved down 3 ii moved down c vii h ix d iii i ii 1 --2 y g e vi (3. −2) − 1--.x 4 0 d y = −2x3 is narrower than y = −x3 e y= −3x3 is narrower than y = (1. −4) 0 x x 0 y f ii y = 9 d i (−4. 5) y 0 0 x 0 ii y = 108 b i x x y d ii y = −81 y h i (1. 1) x 0 ANSWERS 651 .16_NCM10extSB_ans Page 651 Thursday. May 18. 4) y 0 x 3 a b c d e ii i i i i i y = 6. 3 or 5. 0). 2) x x y b a−x 1 y y e −1 2 y = 2x 1 1 x c y x 1– 2 1– 2 y − 1–2 y x − 1–2 2 y = 3x −1 1 0 −3 −2 −1 −1 1 x y = −3x −2 x (−1. r = 6 i Centre (0. May 18. 0). d y = −a x 4 a increasing b decreasing c increasing d decreasing 1 8 x − 8 x y e 5 y h (1. 2 a 1 a b c d e f N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. r = 16 g Centre (0. 3) c x2 + y2 = 1 y = 3x 4 y d −4 4 x −4 1 b i y= 0 3− x 1 2 ii y = 3 a −1 (−1. −5) 8 y g − 8 (−1. r = 3 Centre (0. r = 5 2 a x 2 + y 2 = 144 b x 2 + y 2 = 25 y y 4 Centre (0. 2006 11:01 AM Exercise 11-12 Exercise 11-13 y b 4 3 y = 5x y = 3x y 1 a 2 (1. r = b 1 c For y = ax where a = 2. r = 7 Centre (0. r = 3 Centre (0. r = 20 Centre (0.16_NCM10extSB_ans Page 652 Thursday. as a increases the graph increases more rapidly as x becomes larger. 2) y y d c y = −2x and y = −3x are reflections of y = 2x and y = 3x in the x-axis. 4) 0 −3 −2 −1 −1 y = −2x −2 b d x2 + y2 = 7 y a 2 −2 −1 1 --9 3 (Diagrams not to scale. 0). 0). 2) −1 1 x −5 5 x −5 x (−1.3 .2/ 5. 0). −4) y = 2x 1 0 −2 −1 x −1 1 2 c x y = 3−x (−1. r = 10 Centre (0. 0).) x 3 1 --2 h Centre (0. 0). 0). 0). −10) 652 x 1 f 5 a 1 −1 (−1. −3). 25) 5 a i Power plus 1 1 i 0 −4 −2 0 −2 (−2. 3) −11 11 x 0 y = 2x x y = 1–2 × 2x 5 −11 1 Exercise 11-14 1 a f k 2 a f k 3 a H L P vii i v b g l b g l P H E xii iv ix c h m c h L L Q x ii d i n d i E P L xi vi e j o e j 0. −2). For y = 5x − 2 asymptote is y = −2. For y = 5x asymptote is y = 0. 11) 3 2 y=0 1 y 0 −47 x 4 x 0 2 −4 −2 −2 4 x 2 −1 y = −2 ii For y = + 1 asymptote is y = 1. 2 a Centre (1. −1) −11–2 −12 c 12 x −12 (2. the x-axis. 3) 4 x ii 0 0 2 −4 y y=1 y 4 2 y = 5x + 1 y = 5x y x y = 5x − 2 e 5 x 3 −45 y d 0 b Centre (−4. r = 6 x 2 y 4 x −4 −2 0 −2 5x (5. r = 2 −4 iii y 6 4 2 ANSWERS 653 . 5) 0 x x 40 y h x −3 4 0 −5 0 b 2 3 4 x x 2 b y y y i 0 12 x (1.16_NCM10extSB_ans Page 653 Thursday. May 18. 2006 11:01 AM y f f 3 i y y = 5 × 2x y 11 (2. 3) y y 0 (2.5 y g L L C viii iii 0 53 ii The asymptote for each graph is y = 0. 4 a (5. y = −2 b i x = 1. y = 0 iii x = −3. Anita 16 km/h c 1 h 22. his graph is steeper b Ben 20 km/h. 0) Radius 2 0 y 9 a 4 x (1. y = 0 y 8 a y ii x = −1. May 18.5 min 2 0 4 0 x −2 Semi-circle Centre (0.16_NCM10extSB_ans Page 654 Thursday. 0) Radius 5 c y c 2 1–2 4 x −8 y − 2 0 y ii x 8 −2 1–2 x 2 Semi-circle Centre (0.2/ 5. 0) Radius 4 b 2 x Distance −4 4 y Time 1 3 a i Time ii Distance b Teacher to check. 4 a ii b iii c i y −5 0 5 0 −1 x 1 x y 5 a i −5 0 Semi-circle Centre (0.3 x 1 2 y = − __ 10 x . y = 1 iii x = 0. y = 3 6 a ii x = 0. 2006 11:01 AM b i x = 0. y = 0 4 b 2 Chapter 11 review 1 a Ben. 1) y 2 7 a i 0 −2 1–2 2 0 y iii x 4 0 −2 2 x y b −2 x 0 −8 2 y 2 ii 6 0 2 x y −2 0 −2 y c a y = x2 3 0 y 2 iii −3 0 x −3 −2 0 x −2 654 c 2 x −2 −4 y = 4x2 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. 0). −4) x-intercepts: −2. 3) 0 x b Centre (0. (5. −1) y −5 2 1–2 x-intercepts: 0. −1) and axis of symmetry x = 0. radius 10 c 15 a e i y (1. Both are 4 concave up. 2 --12 y-intercept: 0 x 12 a 14 a Centre (0. −40) 10 a y = 2x3 y (−2. y = 3 − 2x2 is narrower than y = 3 − 1---x2. 4) 4 2 2 y b b i x = 1 1--- x 0 y = x3 x −1 y b y c Both have the same vertex (0.. 2006 11:01 AM 7 a y ii (1. −38) c x −1 2 −20 y d y 0 increasing x −1 (−1.x2 − 1. 3) and axis of symmetry x = 0. −4) 0 −1 10 x x (2. radius 5 Centre (0. 10 y-intercept: −20 c decreasing (2.) y = 1–4 x2 − 1 c i x = −4 ii (−4. 23) 1 b x y = x3 y 8 a increasing y 0 x (2. −7) 9 a i x=1 y = 4x2 − 1 ii (1 1--. May 18. 2 a 4 --6 = 2 --3 b ANSWERS 3 --6 = 1 --2 c 655 3 --6 = 1 --2 . radius 7 ix b vii c vi i f iv g viii v j xi k x d ii h iii l xii Mixed Revision 4 1 Teacher to check. 1 y-intercept: 1 −2 x x b (1. 4) x y = 3 − 1–2 x2 y = 3 − 2x2 Both have the same vertex (0. 14 3--. 1 y = − 1–2 x3 0 11 a y b y (−1. 0). 8) x 0 x b 3 y 13 a 0 x (1. 0). 4) 0 −1 −1 1 x 0 1 c 0 y decreasing y y x-intercepts: −1.16_NCM10extSB_ans Page 655 Thursday. y = 4x2 − 1 is narrower than y = 1--. Both are 2 concave down. 3 ii (0.8 7 a 1 -----26 b 3 -----13 c 12 -----13 1 -----36 b 4 --9 c 5 --6 d 5 -----36 e 35 -----36 1 --4 g 1 --2 h 1 --4 i 11 -----36 j 5 --6 28 --------121 b 56 --------121 c 49 --------121 d 105 --------121 8 B 9 a f 10 a 11 a i b ii 5 --8 25 -----64 c i 12 3 --8 27 --------512 ii 135 --------512 iii 125 --------512 iv 387 --------512 1 --2 13 a −4 −3 −2 −1−1 0 1 2 3 4 x −2 −3 −4 1 --8 b 3 --8 c 18 a Naim. 4) y d i −1 1 1 0 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5.511 b The theoretical probability is the expected probability.5 0.535 0.5 0.58 0. Naim’s graph is steeper than Jenna’s.51 0. 1) x . c 6 d 6 km/h 19 a C 20 a i 7 --8 y ii (0. b Naim is stationary.6 0.5 as the number of trials is increased. 5 a 1 -----12 b 1 --6 −1 x 0 1 2 3 4 5 −2 −3 16 C y 4 3 2 1 17 6 0. whereas the probabilities in the table are based on an experiment.545 0. 3) 14 B y 3 2 1 15 a 3 0 0 −3 −2 1 2 3 4 5 x b i −2 −3 −4 −5 0 x y c i ii (0. The experimental probability will approach 0. 2) 2 0 1 2 3 x x −2 −3 −4 −5 656 ii (0. 2006 11:01 AM 3 a 4 --5 b i 2 --3 ii 1 --3 y 5 4 3 2 1 c 4 a Number 10 20 30 of trials 50 80 100 200 400 1000 Number of heads 29 41 51 109 214 511 6 P(H) 10 15 0. May 18.2/ 5.16_NCM10extSB_ans Page 656 Thursday.5125 0. −2) y 3 2 1 −3 −2 x 0 −2 y b (1. 23 x-intercepts are y −2. d Independent variable is temperature.98 y 5 a (114_ . 7. σ = 12. 0).19 iv 0. 6) 0 1 x 0 y f i b x y −5 0 (1.47 v 0. 0).. −5) 0 b i x= 2 0 x (2.2 m3 5747. May 18.or y = 3 3 −4 y b 1 _ ( 16_ .6 m c 4.72 14 a y ≥ 1. dependent variable is height. dependent variable is cost. b Independent variable is air leaking out. 3) 4 4 (2.15 ii 0. 3 18_) 0 b m = 1. − 1--.unit 3 1 a k = 6 1--2 b y = 10 c m = 7 1--- d k = 3 3--- e x = ±3 f y = ±5 g x = −11.5 cm3 b x = 3. p = −1 3 2 C 3 a 4 4 a $5618 1 13_ x 2x + 3y + 5 = 0 x + 2y − 9 = 0 $4742. 29 a vii b vi c v d xii e iii f x g ii h iv i xi j i k ix l viii 30 a Centre (0. −5 −4 15 a 19. increasing before decreasing and finally stopping without reaching home. −2 7 x −14 24 a i x = − 3--- ii (− 3--. ⎝4 1 --18-⎞⎠ ii (0.5 c $1597. 25 700 mm2 b 145 125 mm3.6.1. 4) y c y (1. r = 1--.2 11 a 9 2 b 7 2 c 7 3 – 15 2 3 5 ---------5 12 a i 3+ 2 ----------------2 ii b i −172 −2 2– 5 ---------------3 iii ii 1 + 9 3 iii 98 + 24 10 13 a 270 b i 0. − 21 1 ---⎞ ⎝ 3 3⎠ iii −5 y ii 1 1--- 0 ii − 1--. -412 ) b 2.8 cm 16 a In ΔABD and ΔACD: AB = AC (given) AD is common ∠ADB = ∠ADC = 90° (AD ⊥ BC) ∴ ΔABD ≡ ΔACD (RHS) b i ∴ BD = CD (matching sides of congruent triangles) ∴ AD bisects BC ii ∴ ∠BAD = ∠CAD (matching angles of congruent triangles) ∴ AD bisects ∠BAC 17 a 396 mm2 b 476 mm2 c 792 mm2 d 3750 mm2 e 785 mm2 f 6792 mm2 18 a x = 2 iii 9 b i (4. 10) 5 --4 ⎛ 1 --1-. 2 h y = −1 2--. 0). c = − 1 -----c a = 1 18 7 25 a x ii (0. −2) x 19 a 1 --4 -7 ± 41 ---------------------2 b b x= 5 -----32 c -1 ± 34 ---------------------3 5 -----16 d 169 --------256 20 a 81°45′ b 37°45′ c 142°49′ 21 a 360 498 mm3.47 1 2 3 b x < 1 1--. σ = 10.40 69% 78.or d = 1 --1- −1 2x − y + 7 = 0 3x + 7y − 2 = 0 i $851.3. 2 −3 −2 −1 0 1 2 3 −3 −2 −1 0 1 c x < −3. σ = 1. −1 0 iii 0. 17 604 mm2 22 a i 13 . −49) x b c 27. −6) 23 28 a Independent variable is distance.20 $1200 7. ⎝ 2 0⎞⎠ ⎛2 1 ---. m = 2 --3 ANSWERS ii 40 . m = − 1--3 657 . 2006 11:01 AM y e i 27 a ii (0.. y = 2 11 -----. The journey begins again towards home.88 1 --2 −3 0 11 x −33 1 2 2_ 26 a i ⎛1 1 ---.0 cm3 x = 2. y-intercept is −14. r = 1 unit c Centre (0. c Independent variable is number of people.1 m b 10.6 General revision 5 0 b d ii c b 10 a x = 63.) 2 3 --4 ii c i x=0 d i x= 2 4 ⎛3 ---.. dependent variable is amount of fuel. dependent variable is diameter.16_NCM10extSB_ans Page 657 Thursday. r = 6 units b Centre (0. 5 3 23 c x = 50.3 i d = − --2.5 b $935.. −7) 21 y = x2 + 3 is the graph of y = x2 moved up 3 units (along the y-axis) 22 Speed on the journey away from home increases then decreases to a stop. − 1 --1-⎞ ⎝ 4 8⎠ ii y 6 a c 7 a b 8 a c 9 a x −1 (−1. and Pythagoras’ theorem) c c = 134 (a tangent is perpendicular to the radius. m = 2 --3 b i 5. We show that A lies on the O circle. 1) 2 2 c A parallelogram. OD and let A′ be a point on the B D circumference. then ∠BCD = 180° − x (given) ∴ Reflex ∠BOD = 2 × (180° − x) = 360° − 2x ∴ ∠BOD = 2x ∴ ∠A′ = x (angle at centre and circumference of a circle) ∴ ∠A = ∠A′ (both equal x) ∴ A lies on the circle ∴ A.5 m (line from the centre is the perpendicular bisector of the chord and the chords are equal because they are the same distance from centre) b DE = 12 m (sides opposite equal angles) c ∠UVO = 58° (angle sum of isosceles Δ. chords of equal length subtend equal angles) d PQ = 30 mm (Pythagoras: the line from the centre is the perpendicular bisector of the chord) e OM = 21 cm (Pythagoras: the line from the centre is the perpendicular bisector of the chord) f OD = 18 2 (Pythagoras: the line from the centre is the perpendicular bisector of the chord) 7 a i 52 cm ii 96 cm b i 58 cm ii 8 cm c 18. 7 In ΔPBC: sin P = where R is the radius of the circumcircle of ΔABC Exercise 12-03 1–3 Teacher to check. construction diameters from A and B. angle sum of a quadrilateral) d g = 67 (alternate segment theorem) 5 a 15 b 5 c 9 d 7 e 20 f 4 6 a x = 7 cm b i XP = 10 cm ii AB = 24 cm Exercise 12-04 1 ∴ a ------------sin A 2 3 4 5 a ------2R a ------2R = 2R Similarly. B. (2 1--. C If ∠A = x. May 18. 20 cm c i AB = 30 cm ii area = 840 cm2 d i 52 cm ii area = 1920 cm2 Exercise 12-02 1–3 4 a b c d e f g h i j k l Teacher to check.16_NCM10extSB_ans Page 658 Thursday. and by subtraction) f x = 62 (angle in semi-circle theorem) y = 118 (opposite angles of cyclic quadrilateral theorem) z = 31 (base angles of isosceles Δ) 6 a iii WXYZ is a cyclic quadrilateral because ∠W + ∠Y = 180° and ∠X = ∠Z = 180° i. twice 232 ⎪ ⎬ the angle at the circumference 40 ⎪ ⎪ standing on the same arc) 74 ⎪ ⎪ 63 ⎪ ⎪ 104 ⎭ 90 (angle in semi-circle theorem) 48 ⎫ (angles at the circumference of ⎪ 36 ⎬ a circle standing on the same ⎪ 30 ⎭ arc are equal) 658 m 23 ⎫ (angle in semi-circle.3 (angles at the circumference standing on the same arc are equal) ∠RYP = ∠QYS (vertically opposite angles) ∴ ΔPYR ||| ΔSYQ ∴ but ∠P = ∠A (angles at the circumference standing on the same arc are equal) ∴ sin A = ∠R = ∠Q ∠P = ∠S 6 PY -------YS = RY -------QY ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ (ratio of sides opposite equal angles) ∴ PY × YQ = RY × YS ∠ADC = ∠BEC (opposite angles of a parallelogram equal) ∠ADC = ∠CBE (exterior angle of cyclic quadrilateral) ∴ ΔCBE is isosceles as ∠BEC = ∠CBE (base angles equal) ∠TZY = ∠X (alternate segment theorem) ∠X = ∠Y (base angles of isosceles ΔZXY (XZ = YZ) ) ∴ ∠TZY = ∠Y now XY || ST as alternate angles ∠TZY and ∠Y are equal Construction: Draw a perpendicular from O to meet DG at P. Since the perpendicular from the centre to a chord bisects the chord: DP = GP and EP = FP ∴ DE = DP − ED = GP − FP = FG ∠THJ = ∠HIJ (alternate segment theorem) ∠THJ = ∠HPI (equal alternate angles) ∴ ∠HIJ = ∠HPI In ΔHIP and ΔHJI ∠HIJ = ∠HPI (proved above) ∠IHJ = ∠IHP (common angles) ∴ ΔHIP ||| ΔHJI (equiangular) ∴ ∠HIP = ∠HJI (third pair of equal angles in similar triangles) In ΔUVX and ΔUWX.. 45 ⎫ ⎪ 112 ⎪ ⎪ 120 ⎪ ⎪ (the angle at the centre. straight line) UV = UW (given) . 1) 40 .2/ 5. C. 4 a a = 56 (the angle between the radius and the tangent is a right angle) b b = 21 (radius is perpendicular to a tangent. 6 a UC = 4. Join OB. y 23 0 −11–3 24 a 32 b 4 2 x c 1 --9 d 1 -----32 Chapter 12 Start up 1 a d 2 a d SSS AAS SSS SSS b e b e SAS SAS AA SAS or AA c f c f RHS AAS RHS AA Exercise 12-01 1–5 Teacher to check.. opposite angles are supplementary b Draw the A′ circumcircle of A ΔBCD. ⎪ n 9 ⎬ ⎪ angle sum triangle) o 45 ⎭ p 63 (opposite angles of a cyclic quadrilateral are supplementary) q 75 ⎫ (exterior angle of cyclic ⎬ r 88 ⎭ quadrilateral theorem) 5 a x = 75 (angle at centre theorem) y = 33 (angles at circumference theorem) z = 72 (angle sum triangle) b x = 108 (angle at centre theorem) y = 126 (opposite angles of a cyclic quadrilateral theorem) z = 252 (angle sum at a point or angle at centre) c x = 70 (straight line) y = 110 (exterior angle cyclic quadrilateral theorem) z = 70 (straight line) d x = 96 (angle at centre theorem) y = 42 (base angles of isosceles Δ) z = 264 (angle sum at a point) e x = 140 (base angles of isosceles Δ) y = 70 (angle at centre theorem) z = 35 (angle sum of isosceles Δ. ∴ b ------------sin B = 2R and c ------------sin C = 2R a ------------sin A = b ------------sin B c ------------sin C = 2R = N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. m = − 1--- iv 3 ii 9.4 km 8 a 77 cm b 34 cm.e. 2006 11:01 AM iii 13 . ∠UXV = 90° = ∠UXW (angle in a semicircle. (2 1--. D are concyclic. −4) (1. −6) b x = − 5--. −5) y e 10 Let ∠PTX = a ∴ ∠R = a (alternate segment theorem) now ∠QTY = a (vertically opposite angles) ∴ ∠S = a (alternate segment theorem) ∴ ∠R = ∠S = a ∴ PR || SQ because alternate angles R and S are equal.or x = 2 2 i x = −3 or x = y x −1 Chapter 13 1 a 13 b 3 c 12 2 a (x − 4)(x + 4) b x(x − 4)(x + 4) c 3(x − 3)(x + 3) d 3x(x − 3)(x + 3) e (x − 5)(x + 3) f (x + 8)(x − 3) g (x − 2)(2x + 5) h x(x − 10)(x + 7) i (x − 5)(x + 5)(x − 2)(x + 2) y x 0 −1 Start up e y = − –21 x 3 AB 3 --2 g y (2. 2) x 0 x 0 y d d y = x3 y 3 x 0 x 0 PQ -------AB 1 --2 = PQ -------AB 1 --2 x 0 x 0 y=x+2 (ratio of sides opposite equal angles in similar Δs) XA (radius half diameter) = ∴ PQ = −2 y 1 a 2 PQ -------AB 1 --. May 18. −3) y b y b y = x2 x 0 1 −1 x 0 y c c y= y −x 2 (1. 2006 11:01 AM UX is common ∴ ΔUVX ≡ ΔUWX (RHS) ∴ VX = VW (matching sides in congruent triangles) ∴ circle bisects base of triangle 7 Let ∠QRP = x (alternate segment theorem) ∴ ∠SRP = x (PR bisects ∠QRS) ∴ ∠PSQ = x = ∠QRP (angles at the circumference standing on the same arc) ∠PQS = x = ∠SRP (angles at the circumference standing on the same arc) ∴ ∠PQS = ∠PSQ = x ∴ ΔSPQ is isosceles because base angles PQS and PSQ are equal 8 Now AP 2 = YP × PX (intersecting tangent and secant theorem) and BP 2 = YP × PX (intersecting tangent and secant theorem) ∴ AP 2 = ΒP 2 ∴ AP = ΒP 9 Join APX and BQX In ΔXPQ and ΔXAB ∠X is common ∠XPQ = ∠XAB (corresponding angles) ∠XQP = ∠XBA (corresponding angles) ∴ ΔXPQ ||| ΔXAB (AAA) ∴ XP -------XA = but XP = ∴ 1 --2 −2 (−1.XA 2 ----------XA ∴ Exercise 13-01 y 3 a (2. −3) y f x2 1 + y2 1 0 =1 f 2 x 0 4 a x = − 5--- (2. 11) 4 0 −5 ANSWERS x 659 . x 0 (2.or x = 2 2 2 3 --5 c x = 0 or x = 10 d x = 0 or x = e x = −1 or x = −5 f x = −10 or x = 12 g x = −2 or x = − 3--- h x = − 5--.16_NCM10extSB_ans Page 659 Thursday. −2. 2006 11:01 AM y h y o 4 0 (1. 5) ii 6 y 6 iii −1 0 x 0 −4 ii −3 y iii e i −3. 4. 5 660 1 x d i −6. 1. 3. 3 x ii 0 y iii x 0 d i −1. 3 y iii (2. 3) 0 2 x 3 (2. 2. May 18. 8) 2 5 x 0 x 0 i y iii ii −12 y iii 4 j (−3. 1 y 3 x 0 (−1. 0. 6 iii (−2. 1) (1. −1. 1 ii 4 . 2 ii 4 y 4 iii 0 1 c i −1. −1) x −3 −3 −30 2 a i −3. 7) −3 −2 −1 x 0 b i −1. 1. −2.16_NCM10extSB_ans Page 660 Thursday. 3 −20 (3.2/ 5. 0. 2 2 3 x −2 3 a i −2. 3 2 x −2 ii 3 0 −1 0 1 2 3 b i −1. 0. 1. 8) 3 x ii 0 ii 0 y iii y 4 −6 x −1 0 1 2 g i −3. 2 y ii −12 h i −2. 5 y ii −20 −1 y iii 0 x 0 −1 1 2 3 4 5 6 x c i −1. 1 y −3 −2 −1 0 1 2 f i 0. 1.3 3 ii −30 4 5 6 1 x 0 x e i − 4. 1) n 1 y iii 3 m 3 x ii 0 y −1 y 1 0 −12 x 0 −2 l 1 −12 y iii k 0 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. −3) (−6. 2006 11:01 AM c x-intercepts are −2 and 3 y-intercept is 18 y iii 4 x f i −2. 0. 12) (1. 0) i (4. 0) y 1 0 Exercise 13-03 3 a b x-intercept is −1 y-intercept is 1 x 1 y f y iii y = (x − 1)4 1 −2 −2 y y 0 iii y = x4 18 −1 −4 e x x x y = −2(x + 2)4 y = −2x4 ANSWERS 661 . −1) (−5. 2 x 1 13 a b c d e f ii − 4 y Exercise 13-02 −6 −1 1 −4 x 1 a v f i b viii g vi c ix h iv y 2 a d iii i ii y = (x − 2)2 4 h i − 4. 4) (3. 4) ii 5 n i (0. −3) ii ii ii ii ii ii ii ii ii ii ii 5 13 5 1 15 2 1 4 11 1 8 ii 2 --12 m i (3. −1) ii 2 a b c d e f g h 1 --3 x2 + y2 = 16 (x − 1)2 + (y + 2)2 = 9 x2 + (y + 11)2 = 4 (x + 3)2 + (y − 2)2 = 100 (x + 1)2 + (y + 1)2 = 1 (x − 7)2 + y2 = 81 (x + 6)2 + (y − 2)2 = 5 (x + 1)2 + y2 = 8 y 2 1 0 −2 0 y = −x5 (1.16_NCM10extSB_ans Page 661 Thursday. May 18. 1. 4) (5. −1) −1 −2 x Move up 4 units Move right 5 units Move left 3 units Move up 4 units Move right 3 units Move left 2 units 1 a b c d e f g h i j k l i i i i i i i i i i i (−3. 4 ii 48 y = x2 y iii −4 2 0 48 4x 1 y= 4 a x-intercepts are −2. 1) (9. −1. 1 and 1 --12 y-intercept is 6 x y y = 3x2 + 1 b 3x2 1 x 0 y c 6 −2 1 2 x y 2 y = −x3 y = −x3 + 2 x 0 e vii d y (−1. 0) (−2. 3 1 0 3 x ii 0 1 --2 d x-intercepts are . −2 y-intercept is 4 y = −x5 − 2 y 4 3 x 0 −2 g i −6. 1) (2. −12) (−2. −8) (0. 2 --3 2 l x=3 . R = 17 (2x − 1)(3x + 2) (2x − 1)(x2 + x + 1) (2x − 1)(4x + 7) (2x − 1)(3x2 + 2x + 1) (2x − 1)(x3 − 3x2 − 4x + 2) (2x − 1)(x3 − x + 3) (2x − 1)(3x2 + 1) (2x − 1)(4 − 3x − x5) Exercise 13-08 47 -----64 5 b −181 c −7 f 1709 g 54 b 2 c 14 174 g 0 h 6 −1 −16 d −2 i −1 1 a B. x = 1 5 a x3 + 4x2 − 5x − 20 y 1 a b 2 3–8 c x = −5. −2. −1. 10). 2). r = 7 d (−10. (2. not monic c Yes. −3). 3) (1. r = 5 c (−2.) + 8 1--2 e x3 + 6x2 + 5x − 4 = (x − 3)(x2 + 9x + 32) + 92 f 4x3 + 2x2 + x = (x + 4)(4x2 − 14x + 57) − 228 g 2x3 − x2 + 5x + 3 = (x + 6)(2x2 − 13x + 83) − 495 h 3x3 − x2 + 11 = (x + 2)(3x2 − 7x + 14) − 17 i x5 − x4 + 8x3 + 2x2 − x − 1 = (x + 1)(x4 − 2x3 + 10x2 − 8x + 7) − 8 j x4 − x2 − 10 = (x + 3)(x3 − 3x2 + 8x − 24) + 62 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. −4) (3. x = 1 −4 Exercise 13-07 −2 x 1 a x2 + 7x + 4 = (x + 2)(x + 5) − 6 b x2 − 6x + 2 = (x − 3)(x − 3) − 7 c 4x2 + 3x + 10 = (x − 1)(4x + 7) + 17 d 8x2 + 9x + 11 = (2x + 1)(4x + 2 1--. r = 6 e (2. B. r = 2 3 Exercise 13-04 1 a c e g i k (0. (1. 3 2 c A b x = −4 d x = ±5 e x = ±4. R = 3 b (x + 7).1 Exercise 13-10 b 24 – 21 2 4 a x2 − 7x + 6 c x = 6. 0). r = 1 f (6. −1). 1. −1) i 3 j i 4 3 a 17 d 7 2–5 y (3. R = 42 (4x + 6). c x = −2. May 18. R = 14 (3x3 + 14x2 − 2x + 21).. 3 2 5 --. − 4). 4) (−3. −2) 4 a (−3. monic 1 a b c d e f g h i j 2 a c e 3 a c e g h (3x − 1). x = 0. −3). 5 j x = − --1. −3. −1.. 1 --. 3 a x(x + 2)(x + 4) b x(x − 2)(x + 1) c (x − 1)(x + 1)(x + 2) d (x − 2)(2x − 1)(x + 4) e (x − 1)(x − 2)(x − 3) f (x − 2)(x + 8)(x − 5) g (x − 6)(x + 1)(3x − 1) h (x − 2)(3x + 1)(2x − 1) i x2(2x − 1)(x − 2) j (x − 3)(2x + 1)(2x + 3) 9x3 + 8x2 + 6x − 2 3x4 + 2x3 − 3x2 −2x − 4x2 − 3x − 2 −x4 − 3x3 − 5x2 + 4 x6 + 2x5 + 11x4 + 10x3 + 25x2 6x4 + x3 − 2x2 + 2x + 11 7x6 + x5 − 3x4 − x3 + 2x2 7x4 − 16x3 − 48x2 + 6x + 4 6x3 − 8x2 + 4x + 6 − 4x4 − 8x3 + 5x2 + x − 2 x2 + 11x − 1 b −x2 − 3x − 5 x2 + 3x + 5 d 2x2 + 26x − 5 4x3 + 25x2 − 13x − 6 −x2 + 7x + 3 b x2 + 4x − 15 −3x2 + 11x + 6 d 3x2 − 15x + 3 3x2 − 7x − 15 f 2x2 − 15x + 9 − 4x3 + 9x2 + 24x − 54 8x3 − 62x2 + 99x 4 a x = − 4. −1) b d f h j l (1.2/ 5. not monic h No Yes. 3 f x = −7. −1). − 3 --1. 2 g x = −3. Exercise 13-05 1 a Yes. 5). (−6. 3). 4) x 0 y (1. 3) (1.16_NCM10extSB_ans Page 662 Thursday. not monic f Yes. 1) (−8.. (1. 3). r = 4 h (4. 1) (12. 4 i x = − 4. (−1. 1). (3. 3) 0 x 2 Yes. −3) (− 4. 1) (3. B 2 Teacher to check. 3). r = 9 b (4. 0) −3 0 f 5 x 4 −2 y x 0 2 −2 (−2.3 d 179 h −1 e −12 j 115 Exercise 13-09 Exercise 13-06 5 e 4 1 --3 b −3 4 a 1 d ii − --5- 2 a c d 3 a b c d e f g h y b 2 −2 −1 1 −2 x . 9) (2. monic l Yes. not monic i 5 ii 9 iii 1 i 5 ii −6 iii 3 i 2 ii 11 iii −10 i 1 ii −6 iii 0 i 5 ii 7 iii 3 i 0 ii 9 iii 9 i 6 ii 1 iii −11 h i 1 y c i x 0 2 (−1. 3 k x= 1 --4 . −12) 2 a ( 2 --2. 0) and (1. (−5. 8) (2. C d A. monic e No 662 iii 22 iii 0 ii 2 2 b −1 iii 8 c −7 ii 1 a e 2 a f e − 5--8 c 0 d 13 – 5 3 e 14 f g 56 h −7 b No d Yes. 2006 11:01 AM b g i k 2 a b c d e f g y 2 3 (2. r = 5 g (−3. 9). 5).) 5 5 b The algebraic method is more accurate. 6). 0. −3). C e A. 1) (3.. 2 h x = −2. (3. 0) (−1. monic j No Yes. (0. C b B. 16_NCM10extSB_ans Page 663 Thursday. May 18. 2006 11:01 AM y c 3 x 2 2 −2 x 0 3 4x 0 −4 2 x −24 y d y b −3 y e 1 x 0 −1 1_ 2 −5 −4 3 x −3 y e −2 y f 0 x −20 y c 0 1 2 x −5 −8 −2 1 −12 6 x y f 4 x y g −1 2 −64 3 x −1 0 4 x −64 −2 y 2x 0 −32 −2 −1 0 x −48 y 4 i −4 y f −36 2 a 4x y h 2 y −1 x 0 −4 −3 −2 2x −1 0 e y h 1 x −2 0 1 3 x 4 4 y −2 0 y d 0 g x y 3 a −1 1 2 4 −64 y c 0 −2 0 −18 y d x 0 1_ 2 y j −2 8 −4 y b 1 2 x ANSWERS 663 . 16_NCM10extSB_ans Page 664 Thursday. y −2 −1 0 1 2 y = P(−x) 2 x y = P(x) x 0 Chapter 14 Start-up y 1 a y iii y y = P(x) −1 0 −2 y y = P(−x) 1 y = P(x) −1 0 2 x 2 0 −1 0 −2 x −3 x y = P(x − 3) y b y = −P(x) iv y y = P(x) 0 1 −1 0 x 2 x y = P(x) y = 2P(x) d i x −3 y c y y = P(x) 9 0 −9 664 x 3 y = 2P(x) 4 −2 x 3 y = P(x) y = P(x) ii y 2 x 0 ii 1 y = −P(x) y 3 x −12 y 2 −2 y = 2P(x) 1 y = P(x) 0 −9 x −4 c i b i 1 y = P(x) −1 0 −9 y y = P(x) 2 iv iii y = P(−x) y iii y = P(x) −3 y iv ii y = P(x) y = P(−x) x 1 −5 y = P(x) − 3 −2 y = −P(x) y ii −1 0 −2 2 0 y y = P(x) 1 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. 3 Teacher to check. 2006 11:01 AM Exercise 13-11 1 a i y = P(x) iii y −2 x −1 0 −2 −2 x 0 iv −1 0 y = P(x) − 3 y = P(x) − 3 y = 2P(x) y 0 −9 x 1 y = P(x) −18 2 Teacher to check.2/ 5. May 18.3 3 y = −P(x) −3 x 0 x . 1) x 0 1 a i 6 b i −2 c i 12 1 --7 d i b y x 1 0 7 --2 h i −2 2 a i 16 ii (1.31 i All x Exercise 14-01 1 a e 2 a e i y 3 a (1.16_NCM10extSB_ans Page 665 Thursday. 2) 1 iii iii −1 iii −2 2 x ii 0 iv x = iii 5 --2 y d 1 --2 (1. 2) c 3 i x≠0 2 3 --0 1 has no value. d x = − 1--- 6 a i −4 b −8 ii y > 0 i All x 3 --2 b Teacher to check.y + 1--- b x= c x= ± y–3 6 a 2.47 b 3.5 − 1 has no value.58 2 1 x 0 −1 2 3 --y 1 0 Yes No Yes No Yes b f b f j No Yes Yes Yes No c g c g k x –1 d 3. 2006 11:01 AM y 2 a Exercise 14-03 y b (2.83 c 1. 0 7 a 8 b 11 c 40 d 3k 4 − 2k2 + 3 8 a −26 b −7 c 1 + y3 d −2y3 (1. 8) (−1. 3) 1 a b c d e f g h i j k l 1 x 0 x 0 y c y b (1. May 18. c x = 1 ± 2 x 0 3 ii 6 iii 1 ii −14 iii 6x + 16 iv x = 2 ii − --1- y c iii 1 1 --2 ii y ≤ 1 i All x 3 a i 11 ii 15 iii 10 b 9 − 2x c −2 d x = 16 e x = 3 4 a i 11 ii 4 iii 20 5 a 0 1 --3 ii 1 iv t = −2 or t = y 1_ 2 4 a iii 1 f i 3 c i 5 −2 ii −1 ii b i 4 c iii 0 iii 1 iii 0 e i 27 g i 0 −1 ii −2 ii 2 ii 0 1– 2 i All x ANSWERS x ii All y 665 . 4) 0 −1 x i i i i i i i i i i i i −3 x 3 −4 x 0 x0 x −1 x −1 No restrictions No restrictions No restrictions No restrictions No restrictions No restrictions No restrictions ii ii ii ii ii ii ii ii ii ii ii ii 0y3 0y4 y5 y0 y −3 y = −3 y>0 y=3 y > −2 −1 y 1 y −4 0<y4 y 2 a (1. e y x 0 b x = 7. 2) 1 x ii y ≠ 0 y e ii 4 c x = 1 or x = 3 d Teacher to check. 4) x 0 y c 5 a x = --1. −4) ii All y y b No No No No Yes d h d h l Yes Yes No Yes Yes 1 −1 0 x 1 Exercise 14-02 (1. 2006 11:01 AM y f y 2 iv (2. c.2/ 5. h 2 a i f −1(x) = 3x x y vi 0 −1 Exercise 14-05 ii 0 x y v 4x y c 1 -10 c i ii iii iv 0 x x 0 ii All y Exercise 14-04 y b y 0 −2 x 0 4 a x y = 2x − 5 x x . −1) x y y = 3x 0 y = _x3 ii y = 2 − _x y 4 x 0 c i −3 0 f −1(x) = 3 x y 1 y b i 0 2 ii x 0 x 1 y 3 y= x 2 0 d i f −1(x) = x ii −4 y iii −3 −1 0 −1 N E W C E N T U R Y M A T H S 1 0 : S T AGE S 5. −1) 0 (−2. 2) y = f(x) is moved up 1 unit y = f(x) is moved down 3 units y = f(x) is moved 2 units to the right y = f(x) is moved 1 unit to the left 5 a b y 3 0 −3 x 0 x y y 2 a c x 0 y b i y y ii 0 −2 x 0 x y b i (1. 1) 666 x ii y iii x y = 8 − 4x y 3 a x b i y = 8 − 4x 3 x 1 b.3 1x x+5 -----------2 y 0 0 y = x3 x+5 y = __ 2 (−2.16_NCM10extSB_ans Page 666 Thursday. 8) −1 i All x y 1 a (1. May 18. 3) 0 2 x −3 0 x 1 y iii x 0 iv 0 −1 y ii y 3 0 (1. 75 x = 1.121 j x = 1. y 2 b 3 3 g 3 6 l 3 log5 25 = 2 c log10 10 000 = 4 x e 4 a h Exercise 14-06 2x + 3 ---------------2 (logm a + logm x) f −(logm x + logm y) or −logm x − logmy g 2 logm x − logm y c x2 y=1− 1 --2 g x= 5 --4 b x= 5 --3 c x= e x= 7 --2 -----f x = − 13 6 h x = −2 4 a x=8 b x = 1000 1 --3 1 -----25 d x= e x= g x= 1 -----------1000 h x = 16 2 i x= 1 ---------10 j x= 1 --8 k x = 128 l x= 1 -----25 c x= 1 --2 5 a x=2 d g j 6 a 7 a c b x= 81 -----16 c x= 1 -----64 1 --5 f x = 32 x = 0.6021 0.519 d x = −0.943 x = 0.975 l k = −2.3979 f 2. 0 c 2 h 2 = −4 g log8 4 = 3 a Teacher to check. b The graph of f(x) = 2 − x is itself symmetrical about the line y = x.5 i d = 2.915 h x = 23.5 26 1 a 7 d f −1(x) = − x + 3 −3 1 --2 = −2 1 log6 -----6 1 --2 1 --.7345 iv 0.30105 e −0.682 l x = 19.80105 a+b b a+b+c 2b + c d c − (a + b) 2c + a + b f −a b+c -----------2 i (a + c) − 2b y − 1--.1192 b.1 e x = 256 f x=2 x = 3. May 18.43 ≈ 23 months A ≈ 106 g b t = 20 days t ≈ 58 days d Teacher to check.555 h x = −0.180 11.3979 h 0.9138 v 0.661 x = 7.069 3 a x=2 d x = − 1--- 2 a 1 g x i − --1. h −(a + c) j 5 --4 (b − c) ANSWERS 667 .428 f x = −0.107 x = −1.5 f a = 3.6021 c 3.(logm x − logm y) or 2 i 3 j 2 k −1 3 a logx 30 b logx 5 d logx 2 e logx 40 y −3 5 a f logm x − 3 logm y Exercise 14-08 d 8 2 = 23. x 0 y y 1 --4 3 --2 1 --- x+3 −3 1 --------125 3 e 5 j 8 h log10 0.687 k x = 1.011 y = 0. Exercise 14-09 1 a d g 2 a c e g i k k = 10 b m=8 c d = 10 x = 2.2042 0.3010 ii 2.7973 iii 3. c. x 0 f log3 b 10 = 6 k 2 = 83 y = (x + 3) x 0 d log25 5 = l i 10 = 100 2 y = x2 − 3.227 x = 2.16_NCM10extSB_ans Page 667 Thursday.846 x = 6.2042 b 2.5 k = 1. d Teacher to check.01 = −2 53 c 27 = x d 4 i 6 b log4 64 = 3 2 --3 2 = i log4 b No.5 e y = −4.logm x 2 1–5 Teacher to check. 6 a i 1.5 h n = −1.425 b x = 2. 2006 11:01 AM e i f −1(x) = y − 2x y = 2_____ 3 y ii 6 a logm x + logm y + logm z b logm a +logm b − logm c c logm a − (logm b + logm c) or logm a − logm b − logm c d logm a + logm (x + y) b f −1(x) = y = − 2 – x 2 – 2x ---------------3 2 0 0 x y = − (2 − x) e x f i y= 3x __ 2 1 a f k 2 a y ii +3 _____ y = 2x 2 0 −3 _____ y = 2x 2 1 log2 ----16 k log9 27 = 3 a 125 = e 64 = −3 c f −1(x) = g 0 = 5−3 h 2 x 1 -----81 1 --2 = − 1--2 101 = 3− 4 1 --- 2 = 86 3 --- j 5 5 = 52 l 1 --------100 1 --2 = 100 −1 b 3 c 2 d −1 f −2 g −4 h −4 b 2 c 1 d 3 e 3 f 2 g −1 h y = − (x + 3) j 4 a d g 5 a c e g 2 logx 1--4 h logx 1--5 1 --2 l −4 c logx 8 f logx 10 i logx 12 logx 2 1.3979 vi −0.04 i x = 3.89 ≈ 12 years b 22.logm 2 Exercise 14-07 −3 0 1 --9 j log16 4 = 1 --- e y = x2 − 3.