Table of Contents
Exercise 10A
Page No 152:
Question 1:
Find each of the following ratios in the simplest form:
(i) 24 to 56
(ii) 84 paise to Rs 3
(iii) 4 kg to 750 g
(iv) 1.8 kg to 6 kg
(v) 48 minutes to 1 hour
(vi) 2.4 km to 900 m
ANSWER:
(i) 24:56 = 24 = 24 ÷ 8 = 3
56 56 ÷ 8 7
As the H.C.F. of 3 and 7 is 1, the simplest form of 24:56 is 3:7.
(ii) 84 paise to Rs 3 = Rs 0.84 to R. 3 = 0.84 = 0.84 ÷ 3 = 0.28 = 28 = 28 ÷ 4 = 7
3 3 ÷ 3 1 100 100 ÷ 4 25
As the H.C.F. of 7 and 25 is 1, the simplest form of 0.84:3 is 7:25.
(iii) 4 kg:750 g = 4000 g:750 g = 4000 ÷ 250 = 16
750 ÷ 250 3
As the H.C.F. of 16 and 3 is 1, the simplest form of 4000:750 is 16:3.
(iv) 1.8 kg:6 kg = 1.8 = 18 = 18 ÷ 6 = 3
6 60 60 ÷ 6 10
As the H.C.F. of 3 and 10 is 1, the simplest form of 1.8:6 is 3:1.
(v) 48 minutes to 1 hour = 48 minutes to 60 minutes = 48:60 = 48 ÷ 12 = 4
60 ÷ 12 5
As the H.C.F. of 4 and 5 is 1, the simplest form of 48:60 is 4:5.
(vi) 2.4 km to 900 m = 2400m:900m = 2400 = 24 = 24 ÷ 3 = 8
900 9 9 ÷ 3 3
As the H.C.F. of 8 and 3 is 1, the simplest form of 2400:900 is 8:3.
Page No 152:
Question 2:
Express each of the following ratios in the simplest form:
(i) 36 : 90
(ii) 324 : 144
(iii) 85 : 561
(iv) 480 : 384
(v) 186 : 403
(vi) 777 : 1147
ANSWER:
(i) 36:90 = 36 = 36 ÷ 18 = 2 (As the H.C.F. of 36 and 90 is 18.)
90 90 ÷ 18 5
Since the H.C.F. of 2 and 5 is 1, the simplest form of 36:90 is 2:5.
(ii) 324:144 = 324 = 324 ÷ 36 = 9 (As the H.C.F. of 324 and 144 is 36.)
144 144 ÷ 36 4
Since the H.C.F. of 9 and 4 is 1, the simplest form of 324:144 is 9:4.
(iii) 85:561 = 85 = 85 ÷ 17 = 5 (As the H.C.F. of 85 and 561 is 17.)
561 561 ÷ 17 33
Since the H.C.F. of 5 and 33 is 1, the simplest form of 85:561 is 5:33.
(iv) 480:384 = 480 = 480 ÷ 96 = 5 (As the H.C.F. of 480 and 384 is 96.)
384 384 ÷ 96 4
Since the H.C.F. of 5 and 4 is 1, the simplest form of 480:384 is 5:4.
(v) 186:403 = 186 = 186 ÷ 31 = 6 (As the H.C.F. of 186 and 403 is 31.)
403 403 ÷ 31 13
Since the H.C.F. of 6 and 13 is 1, the simplest form of 186:403 is 6:13.
(vi) 777:1147 = 777 ÷ 37 = 21 (As the H.C.F. of 777 and 1147 is 37.)
1147 ÷ 37 31
Since the H.C.F. of 21 and 31 is 1, the simplest form of 777:1147 is 21:31.
Page No 152:
Question 3:
Write each of the following ratios in the simplest form:
(i) Rs 6.30 : Rs 16.80
(ii) 3 weeks : 30 days
(iii) 3 m 5 cm : 35 cm
(iv) 48 min : 2 hours 40 min
(v) 1 L 35 mL : 270 mL
(vi) 4 kg : 2 kg 500 g
ANSWER:
(i) Rs 6.30:Rs 16.80
6.30 = 63 = 63 ÷ 21 = 3 (H.C.F. of 63 and 168 is 21.)
16.80 168 168 ÷ 21 8
Ratio = 3 : 8
(ii)3 weeks:30 days = 21days:30 days (1 week = 7 days)
21 = 21 ÷ 3 = 7 (H.C.F. of 21 and 30 is 3.)
30 30 ÷ 3 10
Ratio = 7 : 10
(iii) 3 m 5 cm:35 cm = 305 cm:35 cm (1 m = 100 cm)
305 = 305 ÷ 5 = 61 (H.C.F. of 305 and 35 is 5.)
35 35 ÷ 5 7
Ratio = 61:7
(iv) 48 min:2 hours 40 min = 48 min:160 min (1 hour = 60 mins)
48 = 48 ÷ 16 = 3 (H.C.F. of 48 and 160 is 16.)
160 160 ÷ 16 10
Ratio = 3:10
(v) 1 L 35 mL:270 mL = 1035 mL:270 mL (1 L = 1000 mL)
1035 = 1035 ÷ 45 = 23 (H.C.F. of 1035 and 270 is 45.)
270 270 ÷ 45 6
Ratio = 23:6
(vi) 4 kg:2 kg 500 g = 4000 g:2500 g (1 kg= 1000 g)
4000 = 40 = 40 ÷ 5 = 8 (H.C.F. of 40 and 25 is 5.)
2500 25 25 ÷ 5 5
Ratio = 8:5
Page No 152:
Question 4:
Mr Sahai and his wife are both school teachers and earn Rs 16800 and Rs 10500 per month respectively. Find the ratio of
(i) Mr Sahai’s income to his wife’s income;
(ii) Mrs Sahai’s income to her husband’s income;
(iii) Mr Sahai’s income to the total income of the two.
ANSWER:
Mr Sahai’s earning = Rs 16800
Mrs Sahai’s earning = Rs 10500
(i) Ratio = 16800:10500 = 168:105 = 168 ÷ 21 = 8 (H.C.F. of 168 and 105 is 21.)
105 ÷ 21 5
Mr Sahai’s income:Mrs Sahai’s income = 8:5
(ii)Ratio = 10500:16800 = 105:168 = 105 ÷ 21 = 5 (H.C.F. of 168 and 105 is 21.)
168 ÷ 21 8
Mrs Sahai’s income:Mr Sahai’s income = 5:8
(iii) Total income = 16800 + 10500 = Rs 27300
Ratio = 16800:27300 = 168:273 = 168 = 168 ÷ 21 = 8 (H.C.F. of 168 and 273 is 21.)
273 273 ÷ 21 13
Mrs Sahai’s income:Total income = 8:13
Page No 152:
Question 5:
Rohit earns Rs 15300 and saves Rs 1224 per month. Find the ratio of
(i) his income and savings;
(ii) his income and expenditure;
(iii) his expenditure and savings.
ANSWER:
Rohit’s income = Rs 15300
Rohit’s savings = Rs 1224
(i) Income:Savings = 15300:1224 = 15300 ÷ 612 = 25 (H.C.F. of 15300 and 1224 is 612.)
1224 ÷ 612 2
Income:Savings = 25:2
(ii) Monthly expenditure = Rs (15300 −- 1224) = Rs 14076
Income:Expenditure = 15300:14076 = 15300 ÷ 612 = 25 (H.C.F. of 15300 and 14076 is 612.)
14076 ÷ 612 23
Income:Expenditure = 25:23
(iii) Expenditure : Savings = 14076:1224 = 14076 ÷ 612 = 23 (H.C.F. of 14076 and 1224 is 612.)
1224 ÷ 612 2
Expenditure:Savings = 23:2
Page No 152:
Question 6:
The ratio of the number of male and female workers in a textile mill is 5 : 3. If there are 115 male workers, what is the numkber of female workers in the mill?
ANSWER:
Number of male:Number of female = 5:3
Let the number be x.
Number of male = 5x
Number of female = 3x
Number of male workers = 115
Now, 5x = 115
⇒ x = 115 = 23
5
Number of female workers in the mill = 3x = 3 × 23 = 69
Page No 152:
Question 7:
The bosys and the girls in a school are in the ratio 9 : 5. If the total strength of the school is 448, find the number of girls.
ANSWER:
Boys:Girls = 9:5
Let the number of boys = 9x
Let the number of girls = 5x
Total strength of the school = 448
According to given condition, we have:
9x + 5x = 448
⇒ 14x = 448
⇒ x = 448 = 32
14
Number of boys = 9x = 9 × 32 = 288
Number of girls = 5x = 5 × 32 = 160
Page No 152:
Question 8:
Divide Rs 1575 between Kamal and Madhu in the ratio 7 : 2.
ANSWER:
Kamal:Madhu = 7:2
Sum of the ratio terms = 7 + 2 = 9
Kamal’s share = 7 × 1575 = 11025 = Rs 1225
9 9
Madhu’s share = 2 × 1575 = 3150 = Rs 350
9 9
Page No 152:
Question 9:
Divide Rs 3450 among A, B and C in the ratio 3 : 5 : 7.
ANSWER:
A:B:C = 3:5:7
Sum of the ratio terms = 3 + 5 +7 = 15
A’s share = 3 × 3450 = 10350 = Rs 690
15 15
B’s share = 5 × 3450 = 17250 = Rs 1150
15 15
C’s share = 7 × 3450 = 24150 = Rs 1610
15 15
Page No 152:
Question 10:
Two numbers are in the ratio 11 : 12 and their sum is 460. Find the numbers.
ANSWER:
Two number are in the ratio 11:12.
Let the numbers be 11x and 12x.
Given: 11x + 12x = 460
⇒ 23x = 460
⇒ x = 460 = 20
23
First number = 11x = 11 × 20 = 220
Second number = 12x = 12 × 20 = 240
Hence, the numbers are 220 and 240.
Page No 152:
Question 11:
A 35-cm line segment is divided into two parts in the ratio 4 : 3. Find the length of each part.
ANSWER:
Ratio of the two parts of line segment = 4:3
Sum of the ratio terms = 4 + 3 = 7
First part = 4 × 35 cm = 4 × 5 cm = 20 cm
7
Second part = 3 × 35 cm = 3 × 5 cm = 15 cm
7
Page No 152:
Question 12:
A factory produces electric bulbs. If 1 out of every 10 bulbs is defective and the factory produces 630 bulbs per day, find the number of defective bulbs produced each day.
ANSWER:
Number of bulbs produced each day = 630
Out of 10 bulbs, 1 is defective.
Number of defective bulbs = 630 = 63
10
∴∴ Number of defective bulbs produced each day = 63
Page No 152:
Question 13:
Find the ratio of the price of a pencil to that of a ball pen if pencils cost Rs 96 per score and ball pens cost Rs 50.40 per dozen.
ANSWER:
Price of pencil = Rs 96 per score
Price of ball pen = Rs 50.40 per dozen
Price per unit of pencil = 96 = 4.8
20
Price per unit of ball pen = 50.40 = 4.2
12
Ratio = 4.8 = 48 = 48 ÷ 6 = 8
4.2 42 42 ÷ 6 7
Price of a pencil:Price of a ball pen = 8:7
Page No 152:
Question 14:
The ratio of the length of a field to its width is 5 : 3. Find its length if the width is 42 metres.
ANSWER:
Length:Width = 5:3
Let the length and the width of the field be 5x m and 3x m, respectively.
Width = 42 m
3x = 42
x = 42 = 14
3
∴∴ Length = 5x = 5 × 14 = 70 metres
Page No 152:
Question 15:
The ratio of income to savings of a family is 11 : 2. Find the expenditure if the savings is Rs 1520.
ANSWER:
Income:Savings = 11:2
Let the income and the saving be Rs 11x and Rs 2x, respectively.
Saving = Rs 1520
2x = 1520
x = 1520 = 760
2
∴∴ Income = Rs 11x =Rs (11 × 760) = Rs 8360
Expenditure = Income −- Saving
= Rs (8360 −- 1520 )
= Rs 6840
Page No 152:
Question 16:
The ratio of income to expenditure of a family is 7 : 6. Find the savings if the income is Rs 14000.
ANSWER:
Income:Expenditure = 7:6
Let the income and the expenditure be Rs 7x and Rs 6x, respectively.
Income = Rs 14000
7x = 14000
x = 14000 = 2000
7
Expenditure = Rs 6x = Rs 6 × 2000 = Rs 12000
∴∴ Saving = Income −- Expenditure
= Rs (14000 −- 12000)
= Rs 2000
Page No 152:
Question 17:
The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 11.7 kg find the weight of zinc in it.
ANSWER:
Let the weight of zinc be x kg.
Ratio of zinc and copper = 7:9
Weight of copper in the alloy = 11.7 kg
7 = x
9 11.7
⇒ x = 11.7 × 7 = 81.9 = 9.1
9 9
Weight of zinc = 9.1 kg
Page No 152:
Question 18:
A bus covers 128 km in 2 hours and a train covers 240 km in 3 hours. Find the ratio of their speeds.
ANSWER:
A bus covers 128 km in 2 hours.
Speed of the bus = Distance = 128 km = 64 km/ hr
Time 2 hr
A train covers 240 km in 3 hours.
Speed of the train = Distance = 240 = 80 km /hr
Time 3
Ratio of their speeds = 64:80 = 64 = 64 ÷ 16 = 4
80 80 ÷ 16 5
∴∴ Ratio of the speeds of the bus and the train = 4:5
Page No 153:
Question 19:
From each of the given pairs, find which ratio is larger:
(i) (3 : 4) or (9 : 16)
(ii) (5 : 12) or (17 : 30)
(iii) (3 : 7) or (4 : 9)
(iv) (1 : 2) or (13 : 27)
ANSWER:
(i) (3:4) or (9:16)
Making the denominator equal:
3 × 4 = 12 and 12 > 9
4 × 4 16 16 16
∴∴ (3:4) > (9:16)
(ii) (5:12) or (17:30)
Making the denominator equal:
5 × 5 = 25 and 17 × 2 = 34
12 × 5 60 30 × 2 60
⇒ 25 < 34
60 60
∴∴ (5:12) < (17:30)
(iii) (3:7) or (4:9)
Making the denominator equal:
3 × 9 = 27 and 4 × 7 = 28
7 × 9 63 9 × 7 63
⇒ 27 < 28
63 63
(3:7) < (4:9)
(iv) (1:2) or (13:27)
Making the denominator equal:
1× 27 = 27 and 13 × 2 = 26
2 × 27 54 27 × 2 54
⇒ 27 > 26
54 54
(1:2) > (13:27)
Page No 153:
Question 20:
Fill in the place holders:
(i) 2440= 5=12 2440= 5=12
(ii) 3663=4 = 213663=4 = 21
(iii) 57= 28=35 57= 28=35
ANSWER:
(i) 24 = 24 ÷ 8 = 3 = 3 × 4 = 12
40 40 ÷ 8 5 5 × 4 20
(ii) 36 = 36 ÷ 9 = 4 = 4 × 3 = 12
63 63 ÷ 9 7 7 × 3 21
(iii) 5 = 5 × 4 = 20 = 5 × 7 = 35
7 7 × 4 28 7 × 7 49
Page No 155:
Exercise 10B
Question 1:
Determine if the following numbers are in proportion:
(i) 4, 6, 8, 12
(ii) 7, 42, 13, 78
(iii) 33, 121, 9, 96
(iv) 22, 33, 42, 63
(v) 32, 48, 70, 210
(vi) 150, 200, 250, 300
ANSWER:
(i) 4, 6, 8, 12
4 = 4 ÷ 2 = 2 ; 8 = 8 ÷ 4 = 2
6 6 ÷ 2 3 12 12 ÷ 4 3
Hence, 4:9::8:12 are in proportion.
(ii) 7, 42, 13, 78
7 = 7 ÷ 7 = 1 ; 13 = 13 ÷ 13 = 1
42 42 ÷ 7 6 78 78 ÷ 13 6
Hence, 7:42::13:78 are in proportion.
(iii) 33, 121, 9, 96
33 = 33 ÷ 11 = 3 ; 9 = 9 ÷ 3 = 3
121 121 ÷ 11 11 96 96 ÷ 3 32
Hence, 33:121::9:96 are not in proportion.
(iv) 22, 33, 42, 63
2233=22÷1133÷11=23 and 4263=42÷2163÷21=232233=22÷1133÷11=23 and 4263=42÷2163÷21=23
Hence, 22:33 :: 42 : 63 are not in proportion.
(v) 32, 48, 70, 210
32 = 32 ÷ 6 = 7 ; 70 = 70 ÷ 70 = 1
48 48 ÷ 6 8 210 210 ÷ 70 3
Hence, 32:48::70:210 are not in proportion.
(vi) 150, 200, 250, 300
150 = 150 ÷ 50 = 3; 250 = 250 ÷ 50 = 5
200 200 ÷ 50 4 300 300 ÷ 50 6
Hence, 150:200::250:300 are not in proportion.
Page No 155:
Question 2:
Verify the following:
(i) 60 : 105 : : 84 : 147
(ii) 91 : 104 : : 119 : 136
(iii) 108 : 72 : : 129 : 86
(iv) 39 : 65 : : 141 : 235
ANSWER:
(i) 60:105::84:147
60 = 60 ÷ 15 = 4 (H.C.F. of 60 and 105 is 15.)
105 105 ÷ 15 7
84 = 84 ÷ 21 = 4 (H.C.F. of 84 and 147 is 21.)
147 147 ÷ 21 7
Hence, 60:105::84:147 are in proportion.
(ii) 91:104::119:136
91 = 91 ÷ 13 = 7 (H.C.F. of 91 and 104 is 13.)
104 104 ÷ 13 8
119 = 119 ÷ 17 = 7 (H.C.F. of 11 and 136 is 17.)
136 136 ÷ 17 8
Hence, 91:104::119:136 are in proportion.
(iii) 108:72::129:86
108 = 108 ÷ 36 = 3 (H.C.F. of 108 and 72 is 36.)
72 72 ÷ 36 2
129 = 129 ÷ 43 = 3 (H.C.F. of 129 and 86 is 43.)
86 86 ÷ 43 2
Hence, 108:72::129:86 are in proportion.
(iv) 39:65::141:235
39 = 39 ÷ 13 = 3 (H.C.F. of 39 and 65 is 13.)
65 65 ÷ 13 5
141 = 141 ÷ 47 = 3 (H.C.F. of 141 and 235 is 47.)
235 235 ÷ 47 5
Hence, 39:65::141:235 are in proportion.
Page No 155:
Question 3:
Find the value of x in each of the following proportions:
(i) 55 : 11 : : x : 6
(ii) 27 : x : : 63 : 84
(iii) 51 : 85 : : 57 : x
(iv) x : 92 : : 87 : 116
ANSWER:
(i) 55:11::x:6
Product of extremes = Product of means
55 × 6 = 11 × x
⇒ 11x = 330
⇒ x = 330 = 30
11
(ii) 27:x::63:84
Product of extremes = Product of means
27 × 84 = x × 63
⇒ 63x = 2268
⇒ x = 2268 = 36
63
(iii) 51:85::57:x
Product of extremes = Product of means
51 × x = 85 × 57
⇒ 51x = 4845
⇒ x = 4845 = 95
51
(iv) x:92::87:116
Product of extremes = Product of means
x × 116 = 92 × 87
⇒ 116x = 8004
⇒ x = 8004 = 69
116
Page No 155:
Question 4:
Write (T) for true and (F) for false in case of each of the following:
(i) 51 : 68 : : 85 : 102
(ii) 36 : 45 : : 80 : 100
(iii) 30 bags : 18 bags : : Rs 450 : Rs 270
(iv) 81 kg : 45 kg : : 18 men : 10 men
(v) 45 km : 60 km : : 12 h : 15 h
(vi) 32 kg : Rs 36 : : 8 kg : Rs 9
ANSWER:
(i) 51:68::85:102
Product of means = 68 × 85 = 5780
Product of extremes = 51 × 102 = 5202
Product of means ≠ Product of extremes
Hence, (F).
(ii) 36:45::80:100
Product of means = 45 × 80 = 3600
Product of extremes = 36 × 100 = 3600
Product of means = Product of extremes
Hence, (T).
(iii) 30 bags:18 bags::Rs 450:Rs 270
or 30:18::450:270
Product of means = 18 × 450 = 8100
Product of extremes = 30 × 270 = 8100
Product of means = Product of extremes
Hence, (T).
(iv) 81 kg:45 kg::18 men:10 men
or 81:45::18:10
Product of means = 45 × 18 = 810
Product of extremes = 81 × 10 = 810
Product of means = Product of extremes
Hence, (T).
(v) 45 km:60 km::12 h:15 h
or,45:60::12:15
Product of means = 60 × 12 = 720
Product of extremes = 45 × 15 = 675
Product of means ≠ Product of extremes
Hence, (F).
(vi) 32 kg:Rs 36::8 kg:Rs 9
Product of means = 36 × 8 = 288
Product of extremes = 32 × 9 = 288
Product of means = Product of extremes
Hence, (T).
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Question 5:
Determine if the following ratios form a proportion:
(i) 25 cm : 1 m and Rs 40 : Rs 160
(ii) 39 litres : 65 litres and 6 bottles : 10 bottles
(iii) 200 mL : 2.5 L and Rs 4 : Rs 50
(iv) 2 kg : 80 kg and 25 g : 625 kg
ANSWER:
(i) 25 cm:1 m and Rs 40:Rs 160 (or) 25 cm:100 cm and Rs 40:Rs 160
25 = 25 ÷ 25 = 1 and 40 = 40 ÷ 40 = 1
100 100 ÷ 25 4 160 160 ÷ 40 4
Hence, they are in proportion.
(ii) 39 litres:65 litres and 6 bottles:10 bottles
39 = 39 ÷ 13 = 3 and 6 = 6 ÷ 2 = 3
65 65 ÷ 13 5 10 10 ÷ 2 5
Hence they are in proportion.
(iii) 200 mL:2.5 L and Rs 4:Rs 50 (or) 200 mL:2500 mL and Rs 4:Rs 50
200 = 2 and 4 = 4 ÷ 2 = 2
2500 25 50 50 ÷ 2 25
Hence, they are in proportion.
(iv) 2 kg:80 kg and 25 g:625 kg (or) 2 kg:80 kg and 25 g:625000 g
2 = 2 ÷ 2 = 1 and 25 = 25 ÷ 25 = 1
80 80 ÷ 2 40 625000 625000 ÷ 25 25000
Hence, they are not in proportion.
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Question 6:
In a proportion, the 1st, 2nd and 4th terms are 51, 68 and 108 respectively. Find the 3rd term.
ANSWER:
Let the 3rd term be x.
Thus, 51:68::x:108
We know:
Product of extremes = Product of means
51 × 108 = 68 × x
⇒ 5508 = 68x
⇒ x = 5508 = 81
68
Hence, the third term is 81.
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Question 7:
The 1st, 3rd and 4th terms of a proportion are 12, 8 and 14 respectively. Find the 2nd term.
ANSWER:
Let the second term be x.
Then. 12:x::8:14
We know:
Product of extremes = Product of means
12 × 14 = 8x
⇒ 168 = 8x
⇒ x = 168 = 21
8
Hence, the second term is 21.
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Question 8:
Show that the following numbers are in continued proportion:
(i) 48, 60, 75
(ii) 36, 90, 225
(iii) 16, 84, 441
ANSWER:
(i) 48:60, 60:75
Product of means = 60 × 60 = 3600
Product of extremes = 48 × 75 = 3600
Product of means = Product of extremes
Hence, 48:60::60:75 are in continued proportion.
(ii) 36:90, 90:225
Product of means = 90 × 90 = 8100
Product of extremes = 36 × 225 = 8100
Product of means = Product of extremes
Hence, 36:90::90:225 are in continued proportion.
(iii) 16:84, 84:441
Product of means = 84 × 84 = 7056
Product of extremes = 16 × 441 = 7056
Product of means = Product of extremes
Hence, 16:84::84:441 are in continued proportion.
Page No 155:
Question 9:
If 9, x, x 49 are in proportion, find the value of x.
ANSWER:
Given: 9:x::x:49
We know:
Product of means = Product of extremes
x × x = 9 × 49
⇒ x2 = 441
⇒ x2 = (21)2
⇒ x = 21
Page No 155:
Question 10:
An electric pole casts a shadow of length 20 m at a time when a tree 6 m high casts a shadow of length 8 m. Find the height of the pole.
ANSWER:
Let the height of the pole = x m
Then, we have:
x:20::6:8
Now, we know:
Product of extremes = Product of means
8x = 20 × 6
x = 120 = 15
8
Hence, the height of the pole is 15 m.
Page No 155:
Question 11:
Find the value of x if 5 : 3 : : x : 6.
ANSWER:
5:3::x:6
We know:
Product of means = Product of extremes
3x = 5 × 6
⇒ x = 30 = 10
3
∴∴ x = 10
Page No 157:
Exercise 10C
Question 1:
If the cost of 14 m of cloth is Rs 1890, find the cost of 6 m of cloth.
ANSWER:
Cost of 14 m of cloth = Rs 1890
Cost of 1 m of cloth = 1890 = Rs 135
14
Cost of 6 m of cloth = 6 × 135 = Rs 810
Page No 157:
Question 2:
If the cost of a dozen soaps is Rs 285.60, what wil be the cost of 15 such soaps?
ANSWER:
Cost of dozen soaps = Rs 285.60
Cost of 1 soap = 285.60
12
Cost of 15 soaps = 15 × 285.60 = 4284 = Rs 357
12 12
Page No 157:
Question 3:
If 9 kg of rice costs Rs 327.60, what will be the cost of 50 kg of rice?
ANSWER:
Cost of 9 kg of rice = Rs 327.60
Cost of 1 kg of rice = 327.60
9
Cost of 50 kg of rice = 50 × 327.60 = 16380 = Rs 1820
9 9
Hence, the cost of 50 kg of rice is Rs 1820.
Page No 157:
Question 4:
If 22.5 m of a uniform iron rod weighs 85.5 kg, what will be the weight of 5 m of the same rod?
ANSWER:
Weight of 22.5 m of uniform iron rod = 85.5 kg
Weight of 1 m of uniform iron rod = 85.5 kg
22.5
Weight of 5 m of uniform iron rod = 5 × 85.5 = 427.5 = 19 kg
22.5 22.5
Thus, the weight of 5 m of iron rod is 19 kg.
Page No 157:
Question 5:
If 15 tins of the same size contain 234 kg of oil, how much oil will there be in 10 such tins?
ANSWER:
Oil contained by 15 tins = 234 kg
Oil contained by 1 tin = 234 kg
15
Oil contained by 10 tins = 10 × 234 = 2340 = 156 kg
15 15
Page No 157:
Question 6:
If 12 L of diesel is consumed by a car in covering a distance of 222 km, how many kilometres will it go in 22 L of diesel?
ANSWER:
Distance covered by a car in 12 L diesel = 222 km
Distance covered by it in 1 L diesel = 222 km
12
Distance covered by it in 22 L diesel = 22 × 222 = 4884 = 407 km
12 12
Page No 157:
Question 7:
A transport company charges Rs 540 to carry 25 tonnes of weight. What will it charge to carry 35 tonnes?
ANSWER:
Cost of transporting 25 tonnes of weight = Rs 540
Cost of transporting 1 tone of weight = 540
25
Cost of transporting 35 tonnes of weight = 35 × 540 = 18900 = Rs 756
25 25
Page No 158:
Question 8:
4.5 g of an alloy of copper and zinc contains 3.5 g of copper. What weight of copper will there be in 18.9 g of the alloy?
ANSWER:
Let the weight of copper be x g.
Then, 4.5:3.5::18.9:x
Product of extremes = Product of means
4.5 × x = 3.5 × 18.9
⇒ x = 66.15 = 14.7
4.5
So, the weight of copper is 14.7 g.
Page No 158:
Question 9:
35 inland letters cost Rs 87.50. How many such letters can we buy for 315?
ANSWER:
Number of inland letters whose total cost is Rs 87.50 = 35
Number of inland letters of whose cost is Re 1 = 35
87.50
Number of inland letters whose cost is Rs 315 = 315 × 35 = 11025 = 126
87.50 87.50
Hence, we can buy 126 inland letters for Rs 315.
Page No 158:
Question 10:
Cost of 4 dozen bananas is Rs 104. How many bananas can be purchased for Rs 6.50?
ANSWER:
Number of bananas that can be purchased for Rs 104 = 48 (4 dozen)
Number of bananas that can be purchased for Re 1 = 48
104
Number of bananas that can be purchased for Rs 6.50 = 6.50 × 48 = 312 = 3
104 104
Hence, 3 bananas can be purchased for Rs 6.50.
Page No 158:
Question 11:
The cost of 18 chairs is Rs 22770. How many such chairs can be bought for Rs 10120?
ANSWER:
Number of chairs that can be bought for Rs 22770 = 18
Number of chairs that can be bought for Re 1 = 18
22770
Number of chairs that can be bought for Rs 10120 = 10120 × 18 = 182160 = 8
22770 22770
Page No 158:
Question 12:
A car travels 195 km in 3 hours.
(i) How long will it take to travel 520 km?
(ii) How far will it travel in 7 hours with the same speed?
ANSWER:
(i) Time taken by the car to travel 195 km = 3 hours
Time taken by it to travel 1 km = 3 hours
195
Time taken by it to travel 520 km = 520 × 3 = 1560 = 8 hours
195 195
(ii) Distance covered by the car in 3 hours = 195 km
Distance covered by it in 1 hour = 195 = 65 km
3
Distance covered by it in 7 hours = 7 × 65 = 455 km
Page No 158:
Question 13:
A labourer earns Rs 1980 in 12 days.
(i) How much does he earn in 7 days?
(ii) In how many days will he earn Rs 2640?
ANSWER:
(i) Earning of a labourer in 12 days = Rs 1980
Earning of the labourer in 1 day = 1980 = Rs 165
12
Earning of the labourer in 7 days = 7 × 165 = Rs 1155
(ii) Number of days taken by the labourer to earn Rs 1980 = 12 days
Number of days taken by him to earn Re 1 = 12 days
1980
Number of days taken by him to earn Rs 2640 = 2640 × 12 = 31680 = 16 days
1980 1980
Page No 158:
Question 14:
The weight of 65 books is 13 kg.
(i) What is the weight of 80 such books?
(ii) How many such books weigh 6.4 kg?
ANSWER:
Weight of 65 books = 13 kg
(i) Weight of 1 book = 13 kg
65
Weight of 80 books = 80 × 13 = 1040 = 16 kg
65 65
(ii) Number of books weighing 13 kg = 65
Number of books weighing 1 kg = 65 = 5
13
Number of books weighing 6.4 kg = 6.4 × 5 = 32
Page No 158:
Question 15:
If 48 boxes contain 6000 pens, how many such boxes will be needed for 1875 pens?
ANSWER:
Number of boxes containing 6000 pens = 48
Number of boxes containing 1 pen = 48
6000
Number of boxes containing 1875 pens = 1875 × 48 = 90000 = 15
6000 6000
15 boxes are needed for 1875 pens.
Page No 158:
Question 16:
24 workers can build a wall in 15 days. How many days will 9 workers take to build a similar wall?
ANSWER:
Number of days taken by 24 workers to build a wall = 15 days
Number of days taken by 1 worker to build the wall = 15 × 24 = 360 days (less worker means more days)
Number of days taken by 9 workers to build the wall = 360 = 40 days
9
Page No 158:
Question 17:
40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?
ANSWER:
Number of men required to complete the work in 26 days = 40
Number of men required to complete the work in 1 day = 40 × 26 = 1040 men (less men more days)
Number of men required to complete the work in 16 days = 1040 = 65
16
Page No 158:
Question 18:
In an army capm, there were provisions for 550 men for 28 days. But, 700 men attended the camp. How long did the provisions last?
ANSWER:
Number of days the provisions will last for 550 men = 28 days
Number of days the provisions will last for 1 man = 28 × 550 = 15400 days (less men means more days)
Number of days the provisions will last for 700 men = 15400 = 22 days
700
The provision will last for 22 days.
Page No 158:
Question 19:
A given quantity of rice is sufficient for 60 persons for 3 days. How many days would the rice last for 18 persons?
ANSWER:
Number of days for which the given quantity of rice is sufficient for 60 persons = 3 days
Number of days for which it is sufficient for 1 person = 3 × 60 = 180 days (less men means more days )
Number of days for which it is sufficient for 18 persons = 180 = 10 days
18
Page No 158:
Exercise 10D
Question 1:
The ratio 92 : 115 in its simplest for is
(a) 23 : 25
(b) 18 : 23
(c) 3 : 5
(d) 4 : 5
ANSWER:
(d) 4 : 5
92:115 = 92 ÷ 23 = 4 (As H.C.F. of 92 and 115 is 23.)
115 ÷ 23 5
Page No 158:
Question 2:
If 57 : x : : 51 : 85, then the value of x is
(a) 95
(b) 76
(c) 114
(d) none of these
ANSWER:
(a) 95
57:x::51:85
57 = 51
x 85
⇒ x = 57 × 85
51
⇒ x = 4845 = 95
51
Page No 158:
Question 3:
If 25 : 35 : : 45 : x, then the value of x is
(a) 63
(b) 72
(c) 54
(d) none of these
ANSWER:
(a) 63
25:35::45:x
25 = 45
35 x
⇒ x = 35 × 45 = 1575 = 63
25 25
Page No 158:
Question 4:
If 4 : 5 : : x : 35, then the value of x is
(a) 42
(b) 32
(c) 28
(d) none of these
ANSWER:
(c) 28
4:5::x:35
⇒ 4 = x
5 35
⇒ x = 4 × 35 = 4 × 7 = 28
5
Page No 158:
Question 5:
If a, b, c, d are in proportion, then
(a) ac = bd
(b) ad = bc
(c) ab = cd
(d) none of these
ANSWER:
(b) ad = bc
Given:
a, b, c, d are in proportion.
a:b::c:d
a = c
b d
⇒ ad = bc
Page No 158:
Question 6:
If a, b, c are in proportion, then
(a) a2 = bc
(b) b2 = ac
(c) c2 = ab
(d) none of these
ANSWER:
(b) b2 = ac
Given:
a, b, c are in proportion.
a:b::b:c
Product of means = Product of extremes
⇒ b2 = ac
Page No 158:
Question 7:
Choose the correct statement:
(a) (5 : 8) > (3 : 4)
(b) (5 : 8) < (3 : 4)
(c) two ratios cannot be compared
ANSWER:
(b) (5 : 8) < (3 : 4)
We can write
(5:8) = 58 and (3:4) = 34(5:8) = 58 and (3:4) = 34
Making the denominator equal:
5 and 3 × 2 = 6
8 4 × 2 8
As 6 > 5, 5 < 3
8 4
Page No 159:
Question 8:
If Rs 760 is divided between A and B in the ratio 8 : 11, then B’s share is
(a) Rs 440
(b) Rs 320
(c) Rs 430
(d) Rs 330
ANSWER:
(a) Rs 440
A:B = 8:11
Sum of ratio terms = 8 + 11 = 19
B’s share = 11 × 760 = 8360 = Rs 440
19 19
Page No 159:
Question 9:
Two numbers are in the ratio 5 : 7 and the sum of these numbers is 252. The larger of these numbers is
(a) 85
(b) 119
(c) 105
(d) 147
ANSWER:
(d) 147
Ratio = 5:7
Let x be any number such that we have:
5x + 7x = 252
⇒ 12x = 252
⇒ x = 252 = 21
12
Now, 5x = 5 × 21= 105
7x = 7 × 21 = 147
The largest number is 147.
Page No 159:
Question 10:
The sides of a triangle are in the ratio 1 : 3 : 5 and its perimeter is 90 cm. The length of its largest side is
(a) 40 cm
(b) 50 cm
(c) 36 cm
(d) 54 cm
ANSWER:
(b) 50 cm
The sides of the triangle are in the ratio 1:3:5.
Let x be any number such that the sides are 1x cm, 3x cm and 5x cm.
1x + 3x + 5x = 90
⇒ 9x = 90
⇒ x = 90 = 10
9
First side = 1x = 1 × 10 = 10 cm
Second side = 3x = 3 × 10 = 30 cm
Third side = 5x = 5 × 10 = 50 cm
The length of the largest side is 50 cm.
Page No 159:
Question 11:
The ratio of boys and girls in a school is 12 : 5. If the number of girls is 840, the total strength of the school is
(a) 1190
(b) 2380
(c) 2856
(d) 2142
ANSWER:
(c) 2856
Ratio of boys and girls = 12:5
Let x be any number such that the number of boys and girls are 12x and 5x, respectively.
Number of girls = 840
5x = 840
⇒ x = 840 = 168
5
Number of boys = 12x = 12 × 168 = 2016
Number of girls = 840
Total strength of the school = 2016 + 840 = 2856
Page No 159:
Question 12:
If the cost of 12 pens is Rs 138, then the cost of 14 such pens is
(a) Rs 164
(b) Rs 161
(c) Rs 118.30
(d) Rs 123.50
ANSWER:
(b) Rs 161
Cost of 12 pens = Rs 138
Cost of 1 pen = Rs 138
12
Cost of 14 pens = Rs 138 × 14 = Rs 1932 = Rs 161
12 12
Page No 159:
Question 13:
If 24 workers can build a wall in 15 days, how many days will 8 workers take to build a similar wall?
(a) 42 days
(b) 45 days
(c) 48 days
(d) none of these
ANSWER:
(b) 45 days
Time taken by 24 workers to build a wall = 15 days
Time taken by 1 worker to build a wall = 24 × 15 = 360 days (clearly less workers will take more time to build a wall)
Time taken by 8 workers to build a wall = 360 = 45 days
8
Page No 159:
Question 14:
If 40 men can finish a piece of work in 26 days, how many men will be required to finish it in 20 days?
(a) 52
(b) 31
(c) 13
(d) 65
ANSWER:
(a) 52
Number of men required to finish the work in 26 days = 40
Number of men required to finish it in 1 day = 40 × 26 = 1040 men (More men means less days)
Number of men required to finish it in 20 days = 1040 = 52
20
Page No 159:
Question 15:
In covering 111 km, a car consumes 6 L of petrol. How many kilometres will it go in 10 L of petrol?
(a) 172 km
(b) 185 km
(c) 205 km
(d) 266.4 km
ANSWER:
(b) 185 km
Distance covered in 6 L of petrol = 111 km
Distance covered in 1 L of petrol = 111 km
6
Distance covered in 10 L of petrol = 111 × 10 = 1110 = 185 km
6 6
Page No 159:
Question 16:
In a fort, 550 men had provisions for 28 days. How many days will it last for 700 men?
(a) 22 days
(b) 3571135711 days
(c) 34 days
(d) none of these
ANSWER:
(a) 22 days
Number of days for which 550 men had provisions = 28 days
Number of days for which 1 man had provisions = 28 × 550 = 15400 days (more men means less days)
Number of days for which 700 men had provisions = 15400 = 22 days
700
Page No 159:
Question 17:
The angles of a triangle are in the ratio 3 : 1 : 2. The measure of the largest angle is
(a) 30°
(b) 60°
(c) 90°
(d) 120°
ANSWER:
(c) 90°
Ratio of the angles of a triangle is 3:1: 2
Let x be any number such that the three angles are (3x)°°, (1x)°° and (2x)°°.
We know, the sum of the angles of a triangle is 180°°.
3x + 1x + 2x = 180
⇒ 6x = 180
⇒ x = 180 = 30
6
∴∴ (3x )°° = (3 × 30)°° = 90o
(1x)°° = (1 × 30)°° = 30o
(2x)°° = (2 × 30)°° = 60o
The measure of the largest angle is 90o.
Page No 159:
Question 18:
Length and breadth of a rectangular field are in the ratio 5 : 4. If the width of the field is 36 m, what is its length?
(a) 40 m
(b) 45 m
(c) 54 m
(d) 50 m
ANSWER:
(b) 45 m
Length:Breadth = 5:4
Let x be any number such that the length and the breadth are 5x and 4x, respectively.
Now , 4x = 36
x = 36 = 9
4
Length = 5x = 5 × 9 = 45 m
Page No 159:
Question 19:
If a bus covers 195 km in 3 hours and a train covers 300 km in 4 hours, then the ratio of their speeds is
(a) 13 : 15
(b) 15 : 13
(c) 13 : 12
(d) 12 : 13
ANSWER:
(a) 13 : 15
Speed = Distance
Time
Speed of the bus = 195 km = 65 km/hr
3 hr
Speed of the train = 300 km = 75 km/hr
4 hr
Ratio = 65 = 65 ÷ 5 = 13 = 13:15
75 75 ÷ 5 15
Page No 159:
Question 20:
If the cost of 5 bars of soap is Rs 82.50, then the cost of one dozen such bars is
(a) Rs 208
(b) Rs 192
(c) Rs 198
(d) Rs 204
ANSWER:
(c) Rs 198
Cost of 5 bars of soap = Rs 82.50
Cost of 1 bar of soap = 82.50 = Rs 16.5
5
Cost of 12 (1 dozen) bars of soap = 16.5 × 12 = Rs 198
Page No 159:
Question 21:
If the cost of 30 packets of 8 pencils each is Rs 600, what is the cost of 25 packets of 12 pencils each?
(a) Rs 725
(b) Rs 750
(c) Rs 480
(d) Rs 720
ANSWER:
(b) Rs 750
Cost of 30 packets of 8 pencils each = Rs 600
Cost of 1 packet of 8 pencils = 600 = Rs 20
30
Cost of 1 pencil = Rs 20
8
Cost of 1 packet of 12 pencils = 12 × 20 = 240 = Rs 30
8 8
Cost of 25 packets of 12 pencils each = 25 × 30 = Rs 750
Page No 159:
Question 22:
A rail journey of 75 km costs Rs 215. How much will a journey of 120 km cost?
(a) Rs 344
(b) Rs 324
(c) Rs 268.75
(d) none of these
ANSWER:
(a) Rs 344
Cost of rail journey of 75 km = Rs 215
Cost of rail journey of 1 km = Rs 215
75
Cost of rail journey of 120 km = 120 × 215 = 25800 = Rs 344
75 75
Page No 159:
Question 23:
The 1st, 2nd and 4th terms of a proportion are 12, 21 and 14 respectively. Its third term is
(a) 16
(b) 18
(c) 21
(d) 8
ANSWER:
(d) 8
Let the third term be x.
Then, we have:
12:21::x:14
We know:
Product of means = Product of extremes
21x = 12 × 14
⇒ 21x = 168
⇒ x = 168 = 8
21
The third term is 8
Page No 159:
Question 24:
10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?
(a) 9 h 36 min
(b) 15 h
(c) 6 h 40 min
(d) 13 h 20 min
ANSWER:
(b) 15 h
Time taken by 10 boys to dig a pitch = 12 hours
Time taken by 1 boy to dig a pitch = 12 × 10 = 120 hours (less boys means more time)
Time taken by 8 boys to dig a pitch = 120 = 15 hours
8
Page No 161:
Exercise 10E
Question 1:
Find the ratio of:
(a) 90 cm to 1.05 m
(b) 35 minutes to an hour
(c) 150 mL to 2 L
(d) 2 dozens to a score
ANSWER:
(a) 90 cm:1.05 m (or) 90 cm:105 cm (1 m = 100 cm)
90 = 90 ÷ 15 = 6 (H.C.F. of 90 and 105 is 15.)
105 105 ÷ 15 7
∴∴ 6:7
(b) 35 minutes to an hour (or) 35 minutes:60 minutes (1 hour = 60 minutes)
35 = 35 ÷ 5 = 7 (H.C.F. of 35 and 60 is 5.)
60 60 ÷ 5 12
∴∴ 7:12
(c) 150 mL to 2 L (or) 150 L:2000 L (1 L= 1000 mL)
150 = 150 ÷ 50 = 3 (HCF of 150 and 2000 is 50)
2000 2000 ÷50 40
∴∴ 3:40
(d) 2 dozens to a score (or) 24:20 (1 dozen = 12 and 1 score = 20)
24 = 24 ÷ 4 = 6 (H.C.F. of 24 and 20 is 4)
20 20 ÷ 4 5
∴∴ 6:5
Page No 161:
Question 2:
The ratio of zinc and copper in an alloy is 7 : 9. If the weight of copper in the alloy is 12.6 kg, find the weight of zinc in it.
ANSWER:
Ratio of zinc and copper in an alloy is = 7:9
Let the weight of zinc and copper in it be (7x) and (9x), respectively.
Now, the weight of a copper = 12.6 kg (given)
∴ 9x = 12.6
⇒ x = 12.6 = 1.4
9
∴ Weight of zinc = 7x = 7 × 1.4 = 9.8 kg
Page No 161:
Question 3:
Divide Rs 1400 among A. B and C in the ratio 2 : 3 : 5.
ANSWER:
Given:
A:B:C = 2:3:5
Sum of ratio = 2 + 3 + 5 = 10
Total money = Rs 1400
Then, share of A = 2 × Rs 1400 = Rs 2800 = Rs 280
10 10
Share of B = 3 × Rs 1400 = Rs 4200 = Rs 420
10 10
Share of C = 5 × Rs 1400 = Rs 7000 = Rs 700
10 10
Page No 161:
Question 4:
Prove that (5 : 6) > (3 : 4).
ANSWER:
We can write:
(5:6) = 56 and (3:4) = 34(5:6) = 56 and (3:4) = 34
By making their denominators same: (Taking the L.C.M. of 6 and 4, which is 24.)
Consider, 5:6
5 × 4 = 20
6 × 4 24
And, 3 × 6 = 18
4 × 6 24
As 20 > 18
Clearly, (5:6) > (3:4)
Page No 161:
Question 5:
40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?
ANSWER:
Number of men needed to finish a piece of work in 26 days = 40
Number of men needed to finish it in 1 day = 26 × 40 = 1040 (less days means more men)
Number of men needed to finish it in 16 days = 1040 = 65
16
Page No 161:
Question 6:
In an army camp, there were provisions for 425 men for 30 days. How long did the provisions last for 375 men?
ANSWER:
Number of days for which provisions last for 425 men = 30 days
Number of days for which provisions last for 1 men = 30 × 425 = 12750 days. (less men means more days)
Number of days for which provisions last for 375 men = 12750 = 34 days
375
Hence, provisions will last for 34 days for 375 men.
Page No 161:
Question 7:
Find the value of x when 36 : x : : x : 16.
ANSWER:
Given:
36:x::x:16
We know:
Product of means = Product of extremes
x × x = 36 × 16
⇒ x2 = 576
⇒ x2 = 242
⇒ x = 24
Page No 161:
Question 8:
Show that 48, 60, 75 are in continued proportion.
ANSWER:
Consider 48:60::60:75
Product of means = 60 × 60 = 3600
Product of extremes = 48 × 75 = 3600
So product of means = Product of extremes
Hence, 48, 60, 75 are in continued proportion.
Page No 161:
Question 9:
Two numbers are in the ratio 3 : 5 and their sum is 96. The larger number is
(a) 36
(b) 42
(c) 60
(d) 70
ANSWER:
(c) 60
Ratio = 3:5
Let x be any number such that we have:
3x + 5x = 96
⇒ 8x = 96
⇒ x = 96 = 12
8
The numbers are:
3x = 3 × 12 = 36
5x = 5 × 12 = 60
The largest number = 60
Page No 161:
Question 10:
A car travels 288 km is 4 hours and a train travels 540 km in 6 hours. The ratio of their speeds is
(a) 5 : 4
(b) 4 : 5
(c) 5 : 6
(d) 3 : 5
ANSWER:
(b) 4 : 5
Speed of the car = Distance = 288 km = 72 km/hr
Time 4 hr
Speed of the train = Distance = 540 km = 90 km/hr
Time 6 hr
Ratio of their speeds = 72:90
where, 72 = 72 ÷ 18 = 4 (H.C.F. of 72 and 90 is 18.)
90 90 ÷ 18 5
Page No 161:
Question 11:
The first three terms of a proportion are 12, 21 and 8 respectively. The 4th term is
(a) 18
(b) 16
(c) 14
(d) 20
ANSWER:
(c) 14
Let the 4th term be x, such that we have:
12:21::8:x
Now, we know:
Product of extremes = Product of means
12x = 21 × 8
x = 168 = 14
12
Page No 161:
Question 12:
The ratio 92 : 115 in simplest form is
(a) 23 : 25
(b) 18 : 23
(c) 3 : 5
(d) 4 : 5
ANSWER:
(d) 4 : 5
92:115
92 = 92 ÷ 23 = 4 (H.C.F. of 92 and 115 is 23)
115 115 ÷ 23 5
Page No 161:
Question 13:
If 57 : x : : 51 : 85, then the value of x is
(a) 95
(b) 76
(c) 114
(d) none of these
ANSWER:
(a) 95
Given :
57:x::51:85
We know:
Product of means = Product of extremes
51x = 57 × 85
x = 4845 = 95
51
Page No 161:
Question 14:
If 4 : 5 : : x : 45, then the value of x is
(a) 54
(b) 60
(c) 36
(d) 30
ANSWER:
(c) 36
Given:
4:5::x:45
We know:
Product of mean = Product of extremes
5x = 4 × 45
x = 180 = 36
5
Page No 161:
Question 15:
If a, b, c are in proportion, then
(a) a2 = bc
(b) b2 = ac
(c) c2 = ab
(d) none of these
ANSWER:
(b) b2 = ac
Given:
a, b, c are in proportion, such that we have:
a:b::b:c
Now, we know:
Product of means = Product of extremes
b × b = a × c
b2 = ac
Page No 161:
Question 16:
10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?
(a) 9 hrs 36 min
(b) 15 hrs
(c) 6 hrs 40 min
(d) 13 hrs 10 min
ANSWER:
(b) 15 hrs
Time taken by 10 boys to dig a pitch = 12 hours
Time taken by 1 boy to dig a pitch = 12 × 10 = 120 hours (Less boys would take more hours.)
Time taken by 8 boys to dig a pitch = 120 = 15 hours
8
Page No 161:
Question 17:
In covering 148 km, a car consumes 8 litres of petrol. How many kilometres will it go in 10 litres of petrol?
(a) 172 km
(b) 185 km
(c) 205 km
(d) 266.4 km
ANSWER:
(b) 185 km
Distance covered by a car in 8 litres of petrol = 148 km
Distance covered by it in 1 litre of petrol = 148 km
8
Distance covered by it in 10 litres of petrol = 10 × 148 = 1480 = 185 km
8 8
Page No 161:
Question 18:
Fill in the blanks.
(i) 1421= 3=6 1421= 3=6
(ii) 90 cm : 1.5 m = …… .
(iii) If 36 : 81 : : x : 63, then x = …… .
(iv) If 25, 35, x are in proportion, then x = …… .
(v) If 9, x, x, 49 are in proportion, then x = …… .
ANSWER:
(i)
Let 1421 = x3Thus, we have: 21x = 14 × 3 ⇒ x = 14 × 321 = 2∴ 1421 = 23Again, let 23=6yThus, we have: 2y = 6 × 3 ⇒ y = 6 × 32 = 9∴ 23 = 69∴ 1421 = 23 = 69Let 1421 = x3Thus, we have: 21x = 14 × 3 ⇒ x = 14 × 321 = 2∴ 1421 = 23Again, let 23=6yThus, we have: 2y = 6 × 3 ⇒ y = 6 × 32 = 9∴ 23 = 69∴ 1421 = 23 = 69
(ii) 90 cm:1.5 m (or) 90 cm:150 cm (1 m = 100 cm)
90 = 9 = 9 ÷ 3 = 3 (H.C.F. of 9 and 15 is 3.)
150 15 15 ÷ 3 5
(iii) If 36:81::x:63
Product of means = Product of extremes
81x = 36 × 63
x = 2268
81
x = 28
(iv) Given:
25, 35, x are in proportion.
25:35::35:x
Now, we know:
Product of extremes = Product of means
25 × x = 35 × 35
25x = 1225
x = 1225 = 49
25
(v) Given:
9, x, x, 49 are in proportion.
9:x::x:49
Now, we know:
Product of means = Product of extremes
x × x = 9 × 49
x2 = 441
x2 = 212
x = 21
Page No 162:
Question 19:
Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) 30, 40, 45, 60 are in proportion.
(ii) 6 : 8 and 9 : 12 are equivalent ratios of 3 : 4.
(iii) a dozen : a score = 5 : 3.
(iv) 60 p : Rs 3 = 1 : 5.
ANSWER:
(i) 30, 40, 45, 60
30 = 3 , 45 = 45 ÷ 15 = 3 They are in proportion.
40 4 60 60 ÷ 15 4
Hence, true.
(ii) 6 = 6 ÷ 2 = 3 , 9 = 9 ÷ 3 = 3 Hence, they are equivalent to 3:4.
8 8 ÷ 2 4 2 12 ÷ 3 4
Hence, true.
(iii) 1 dozen:1 score = 12:20
12 = 12 ÷ 4 = 3
20 20 ÷ 4 5
Hence, false.
(iv) 60p:Rs 3 = 60p:300p (1 Re = 100 p)
60 = 6 = 6 ÷ 6 = 1
300 30 30 ÷ 6 5
Hence, true.
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