RS Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 10 Ratio, Proportion And Unitary Method are provided here with simple step-by-step explanations. These solutions for Ratio, Proportion And Unitary Method are extremely popular among class 6 students for Maths Ratio, Proportion And Unitary Method Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggrawal 2020 2021 Book of class 6 Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggrawal 2020 2021 Solutions. All RS Aggrawal 2020 2021 Solutions for class 6 Maths are prepared by experts and are 100% accurate.

#### Question 1:

(i) 24:56 =     24  =     24 â€‹÷ 8     =    3
56       56 â€‹÷ 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  =  0.84â€‹ ÷ 3    0.28    28    =  28 â€‹÷  4    7
3            3 â€‹÷ 3             1          100      100 â€‹÷ 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 â€‹÷ 250     16
750 â€‹÷ 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 â€‹÷ 6  =     3
6           60      60 â€‹÷ 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 â€‹÷ 12   4
60 â€‹÷ 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 â€‹÷ 3     8
900          9           9 â€‹÷ 3          3
As the H.C.F. of 8 and 3 is 1, the simplest form of 2400:900 is 8:3.

#### Question 2:

(i) 24:56 =     24  =     24 â€‹÷ 8     =    3
56       56 â€‹÷ 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  =  0.84â€‹ ÷ 3    0.28    28    =  28 â€‹÷  4    7
3            3 â€‹÷ 3             1          100      100 â€‹÷ 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 â€‹÷ 250     16
750 â€‹÷ 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 â€‹÷ 6  =     3
6           60      60 â€‹÷ 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 â€‹÷ 12   4
60 â€‹÷ 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 â€‹÷ 3     8
900          9           9 â€‹÷ 3          3
As the H.C.F. of 8 and 3 is 1, the simplest form of 2400:900 is 8:3.

(i) 36:90 =   36    =    36 â€‹÷ 18    =     2          (As the H.C.F. of 36 and 90 is 18.)
90          90 â€‹÷ 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 â€‹÷ 36    =    9       (As the H.C.F. of  324 and 144 is 36.)
144          144 â€‹÷ 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 â€‹÷ 17   =      5         (As the H.C.F. of 85 and 561 is 17.)
561      561 â€‹â€‹÷ 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 â€‹÷ 96     5       (As the H.C.F. of 480 and 384 is 96.)
384          384 â€‹â€‹÷ 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  â€‹÷ 37       21           (As the H.C.F. of 777 and 1147 is 37.)
1147  â€‹÷  37         31
Since the H.C.F. of 21 and 31 is 1, the simplest form of 777:1147 is 21:31.

#### Question 3:

(i) 36:90 =   36    =    36 â€‹÷ 18    =     2          (As the H.C.F. of 36 and 90 is 18.)
90          90 â€‹÷ 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 â€‹÷ 36    =    9       (As the H.C.F. of  324 and 144 is 36.)
144          144 â€‹÷ 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 â€‹÷ 17   =      5         (As the H.C.F. of 85 and 561 is 17.)
561      561 â€‹â€‹÷ 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 â€‹÷ 96     5       (As the H.C.F. of 480 and 384 is 96.)
384          384 â€‹â€‹÷ 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  â€‹÷ 37       21           (As the H.C.F. of 777 and 1147 is 37.)
1147  â€‹÷  37         31
Since the H.C.F. of 21 and 31 is 1, the simplest form of 777:1147 is 21:31.

(i) Rs 6.30:Rs 16.80
6.30     =    63    =   63 â€‹÷ 21      3        (H.C.F. of 63 and 168 is 21.)
16.80         168        168  â€‹÷ 21        8
Ratio = 3 : 8
(ii)3 weeks:30 days = 21days:30 days          (1 week = 7 days)
21     =   21 â€‹÷ 3    7          (H.C.F. of 21 and 30 is 3.)
30          30 â€‹ â€‹÷ 3       10
Ratio = 7 : 10
(iii) 3 m 5 cm:35 cm = 305 cm:35 cm      (1 m = 100 cm)
305   305  â€‹÷ 5    61        (H.C.F. of 305 and 35 is 5.)
35         35  â€‹÷ 5          7
Ratio = 61:7
(iv) 48 min:2 hours 40 min = 48 min:160 min        (1 hour = 60 mins)
48     48  â€‹÷ 16      3         (H.C.F. of 48 and 160 is 16.)
160        160  â€‹÷ 16       10
Ratio = 3:10
(v) 1 L 35 mL:270 mL = 1035 mL:270 mL         (1 L = 1000 mL)
1035    1035  â€‹÷ 45    =   23        (H.C.F. of 1035 and 270 is 45.)
270           270  â€‹÷ 45          6
Ratio = 23:6
(vi) 4 kg:2 kg 500 g = 4000 g:2500 g        (1 kg= 1000 g)
4000   =   40  40  â€‹÷ 5    8     (H.C.F. of 40 and 25 is 5.)
2500         25      25  â€‹÷ 5        5
Ratio = 8:5

#### Question 4:

(i) Rs 6.30:Rs 16.80
6.30     =    63    =   63 â€‹÷ 21      3        (H.C.F. of 63 and 168 is 21.)
16.80         168        168  â€‹÷ 21        8
Ratio = 3 : 8
(ii)3 weeks:30 days = 21days:30 days          (1 week = 7 days)
21     =   21 â€‹÷ 3    7          (H.C.F. of 21 and 30 is 3.)
30          30 â€‹ â€‹÷ 3       10
Ratio = 7 : 10
(iii) 3 m 5 cm:35 cm = 305 cm:35 cm      (1 m = 100 cm)
305   305  â€‹÷ 5    61        (H.C.F. of 305 and 35 is 5.)
35         35  â€‹÷ 5          7
Ratio = 61:7
(iv) 48 min:2 hours 40 min = 48 min:160 min        (1 hour = 60 mins)
48     48  â€‹÷ 16      3         (H.C.F. of 48 and 160 is 16.)
160        160  â€‹÷ 16       10
Ratio = 3:10
(v) 1 L 35 mL:270 mL = 1035 mL:270 mL         (1 L = 1000 mL)
1035    1035  â€‹÷ 45    =   23        (H.C.F. of 1035 and 270 is 45.)
270           270  â€‹÷ 45          6
Ratio = 23:6
(vi) 4 kg:2 kg 500 g = 4000 g:2500 g        (1 kg= 1000 g)
4000   =   40  40  â€‹÷ 5    8     (H.C.F. of 40 and 25 is 5.)
2500         25      25  â€‹÷ 5        5
Ratio = 8:5

Mr Sahai's earning = Rs 16800
Mrs Sahai's earning = Rs 10500
(i) Ratio = 16800:10500 = 168:105 =  168  â€‹÷ 21  =           (H.C.F. of 168 and 105 is 21.)
105 â€‹ â€‹÷ 21           5
Mr Sahai's income:Mrs Sahai's income = 8:5
(ii)Ratio = 10500:16800 = 105:168 =  105  â€‹÷ 21   =    5        (H.C.F. of 168 and 105 is 21.)
168 â€‹ â€‹÷ 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  â€‹÷ 21   (H.C.F. of 168 and 273 is 21.)
273        273  â€‹÷ 21     13
Mrs Sahai's income:Total income = 8:13

#### Question 5:

Mr Sahai's earning = Rs 16800
Mrs Sahai's earning = Rs 10500
(i) Ratio = 16800:10500 = 168:105 =  168  â€‹÷ 21  =           (H.C.F. of 168 and 105 is 21.)
105 â€‹ â€‹÷ 21           5
Mr Sahai's income:Mrs Sahai's income = 8:5
(ii)Ratio = 10500:16800 = 105:168 =  105  â€‹÷ 21   =    5        (H.C.F. of 168 and 105 is 21.)
168 â€‹ â€‹÷ 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  â€‹÷ 21   (H.C.F. of 168 and 273 is 21.)
273        273  â€‹÷ 21     13
Mrs Sahai's income:Total income = 8:13

Rohit's income = Rs 15300
Rohit's savings = Rs 1224
(i) Income:Savings = 15300:1224 = 15300 â€‹÷ 612   25        (H.C.F. of 15300 and 1224 is 612.)
1224  â€‹÷ 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  â€‹÷ 612        23
Income:Expenditure = 25:23
(iii) Expenditure : Savings = 14076:1224 =  14076  ÷ 612  23       (H.C.F. of 14076 and 1224 is 612.)
1224  â€‹÷ 612         2
Expenditure:Savings = 23:2

#### Question 6:

Rohit's income = Rs 15300
Rohit's savings = Rs 1224
(i) Income:Savings = 15300:1224 = 15300 â€‹÷ 612   25        (H.C.F. of 15300 and 1224 is 612.)
1224  â€‹÷ 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  â€‹÷ 612        23
Income:Expenditure = 25:23
(iii) Expenditure : Savings = 14076:1224 =  14076  ÷ 612  23       (H.C.F. of 14076 and 1224 is 612.)
1224  â€‹÷ 612         2
Expenditure:Savings = 23:2

Number of male:Number of female = 5:3
Let the number be x.
Number of male = 5x
â€‹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

#### Question 7:

Number of male:Number of female = 5:3
Let the number be x.
Number of male = 5x
â€‹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

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 â€‹× 32 = 160

#### Question 8:

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 â€‹× 32 = 160

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

#### Question 9:

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

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

#### Question 10:

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

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.

#### Question 11:

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.

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

#### Question 12:

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

Number of bulbs produced each day = 630
Out of 10 bulbs, 1 is defective.
Number of defective bulbs =  630  = 63
10

$\therefore$ Number of defective bulbs produced each day = 63

#### Question 13:

Number of bulbs produced each day = 630
Out of 10 bulbs, 1 is defective.
Number of defective bulbs =  630  = 63
10

$\therefore$ Number of defective bulbs produced each day = 63

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  â€‹÷ 6      8
4.2        42        42  â€‹÷  6         7
Price of a pencil:Price of a ball pen = 8:7

#### Question 14:

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  â€‹÷ 6      8
4.2        42        42  â€‹÷  6         7
Price of a pencil:Price of a ball pen = 8:7

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
$\therefore$ Length = 5x = 5 × 14 = 70 metres

#### Question 15:

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
$\therefore$ Length = 5x = 5 × 14 = 70 metres

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
$\therefore$ Income = Rs 11x =Rs (11 × 760) = Rs 8360
Expenditure = Income $-$ Saving
= Rs (8360 $-$ 1520 )
= Rs 6840

#### Question 16:

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
$\therefore$ Income = Rs 11x =Rs (11 × 760) = Rs 8360
Expenditure = Income $-$ Saving
= Rs (8360 $-$ 1520 )
= Rs 6840

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
$\therefore$ Saving = Income $-$ Expenditure
= Rs (14000 $-$ 12000)
= Rs 2000

#### Question 17:

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
$\therefore$ Saving = Income $-$ Expenditure
= Rs (14000 $-$ 12000)
= Rs 2000

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

#### Question 18:

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

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
$\therefore$ Ratio of the speeds of the bus and the train = 4:5

#### Question 19:

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
$\therefore$ Ratio of the speeds of the bus and the train = 4:5

(i) (3:4) or (9:16)

Making the denominator equal:

3 × 4   12  and 12    9
4 × 4       16         16       16

$\therefore$ (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
$\therefore$ (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 â€‹× 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 â€‹× 2        54

⇒   27  26
54       54

(1:2) > (13:27)

#### Question 20:

(i) (3:4) or (9:16)

Making the denominator equal:

3 × 4   12  and 12    9
4 × 4       16         16       16

$\therefore$ (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
$\therefore$ (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 â€‹× 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 â€‹× 2        54

⇒   27  26
54       54

(1:2) > (13:27)

(i)   24   =   24 â€‹÷ 8    3   =    3 × 4  12
40         40 â€‹÷ 8      5          5  × 4      20

(ii)    36    36  â€‹÷ 9    =   4 × 3    12
63        63  â€‹÷ 9      7        7 × 3        21

(iii)   5    5 × 4   20   5 × 7    35
7        7 × 4       28        7 × 7         49

#### Question 1:

(i)   24   =   24 â€‹÷ 8    3   =    3 × 4  12
40         40 â€‹÷ 8      5          5  × 4      20

(ii)    36    36  â€‹÷ 9    =   4 × 3    12
63        63  â€‹÷ 9      7        7 × 3        21

(iii)   5    5 × 4   20   5 × 7    35
7        7 × 4       28        7 × 7         49

(i) 4, 6, 8, 12
4  =    4 â€‹÷ 2   =   2 ;     8  =    8  â€‹÷ 4    2
6        6  â€‹÷ 2       3       12       12  â€‹÷ 4       3
Hence, 4:9::8:12 are in proportion.

(ii) 7, 42, 13, 78
7    7  â€‹÷ 7    1 ;     13   =  13  â€‹÷ 13     =    1
42      42  â€‹÷ 7      6        78        78  â€‹÷ 13           6
Hence, 7:42::13:78 are in proportion.

(iii) 33, 121, 9, 96
33    33  â€‹÷ 11    3   ;     9    9  â€‹÷ 3    3
121      121  â€‹÷ 11      11         96       96  â€‹÷ 3      32
Hence, 33:121::9:96 are not in proportion.

(iv) 22, 33, 42, 63

Hence, 22:33 :: 42 : 63 are not in proportion.

(v) 32, 48, 70, 210
32    32  â€‹÷ 6   7 ;    70    70  â€‹÷ 70   1
48        48  â€‹÷ 6       8     210      210  â€‹÷ 70      3
Hence, 32:48::70:210 are not in proportion.

(vi) 150, 200, 250, 300
150    150  â€‹÷ 50   3;   250  250  â€‹÷ 50  =   5
200         200  â€‹÷ 50        4   300      300  â€‹÷ 50       6
Hence, 150:200::250:300 are not in proportion.

#### Question 2:

(i) 4, 6, 8, 12
4  =    4 â€‹÷ 2   =   2 ;     8  =    8  â€‹÷ 4    2
6        6  â€‹÷ 2       3       12       12  â€‹÷ 4       3
Hence, 4:9::8:12 are in proportion.

(ii) 7, 42, 13, 78
7    7  â€‹÷ 7    1 ;     13   =  13  â€‹÷ 13     =    1
42      42  â€‹÷ 7      6        78        78  â€‹÷ 13           6
Hence, 7:42::13:78 are in proportion.

(iii) 33, 121, 9, 96
33    33  â€‹÷ 11    3   ;     9    9  â€‹÷ 3    3
121      121  â€‹÷ 11      11         96       96  â€‹÷ 3      32
Hence, 33:121::9:96 are not in proportion.

(iv) 22, 33, 42, 63

Hence, 22:33 :: 42 : 63 are not in proportion.

(v) 32, 48, 70, 210
32    32  â€‹÷ 6   7 ;    70    70  â€‹÷ 70   1
48        48  â€‹÷ 6       8     210      210  â€‹÷ 70      3
Hence, 32:48::70:210 are not in proportion.

(vi) 150, 200, 250, 300
150    150  â€‹÷ 50   3;   250  250  â€‹÷ 50  =   5
200         200  â€‹÷ 50        4   300      300  â€‹÷ 50       6
Hence, 150:200::250:300 are not in proportion.

(i) 60:105::84:147
60    60  â€‹÷ 15    4         (H.C.F. of 60 and 105 is 15.)
105       105  â€‹÷ 15       7
84    84  â€‹÷ 21   4         (H.C.F. of 84 and 147 is 21.)
147     147  â€‹÷ 21        7
Hence, 60:105::84:147 are in proportion.
(ii) 91:104::119:136
91   91  â€‹÷ 13    7         (H.C.F. of 91 and 104 is 13.)
104       104  â€‹÷ 13      8
119   119  â€‹÷ 17   7      (H.C.F. of 11 and 136 is 17.)
136        136  â€‹÷ 17       8
Hence, 91:104::119:136 are in proportion.
(iii) 108:72::129:86
108   108  â€‹÷ 36    3        (H.C.F. of 108 and 72 is 36.)
72           72 â€‹ â€‹÷ 36        2
129    â€‹129  â€‹÷  43  3       (H.C.F. of 129 and 86 is 43.)
86          86   â€‹÷ 43        2
Hence, 108:72::129:86 are in proportion.
(iv) 39:65::141:235
39    39  â€‹÷ 13   3        (H.C.F. of 39 and 65 is 13.)
65         65  â€‹÷ 13      5
141    141  â€‹÷  47    3     (H.C.F. of 141 and 235 is 47.)
235         235  â€‹÷ 47        5
Hence, 39:65::141:235 are in proportion.

#### Question 3:

(i) 60:105::84:147
60    60  â€‹÷ 15    4         (H.C.F. of 60 and 105 is 15.)
105       105  â€‹÷ 15       7
84    84  â€‹÷ 21   4         (H.C.F. of 84 and 147 is 21.)
147     147  â€‹÷ 21        7
Hence, 60:105::84:147 are in proportion.
(ii) 91:104::119:136
91   91  â€‹÷ 13    7         (H.C.F. of 91 and 104 is 13.)
104       104  â€‹÷ 13      8
119   119  â€‹÷ 17   7      (H.C.F. of 11 and 136 is 17.)
136        136  â€‹÷ 17       8
Hence, 91:104::119:136 are in proportion.
(iii) 108:72::129:86
108   108  â€‹÷ 36    3        (H.C.F. of 108 and 72 is 36.)
72           72 â€‹ â€‹÷ 36        2
129    â€‹129  â€‹÷  43  3       (H.C.F. of 129 and 86 is 43.)
86          86   â€‹÷ 43        2
Hence, 108:72::129:86 are in proportion.
(iv) 39:65::141:235
39    39  â€‹÷ 13   3        (H.C.F. of 39 and 65 is 13.)
65         65  â€‹÷ 13      5
141    141  â€‹÷  47    3     (H.C.F. of 141 and 235 is 47.)
235         235  â€‹÷ 47        5
Hence, 39:65::141:235 are in proportion.

(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 â€‹× 84 = â€‹× 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 â€‹× 87
⇒                      116x = 8004
⇒                           x  =   8004  = 69
116

#### Question 4:

(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 â€‹× 84 = â€‹× 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 â€‹× 87
⇒                      116x = 8004
⇒                           x  =   8004  = 69
116

(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 â€‹× 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 â€‹× 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).

#### Question 5:

(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 â€‹× 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 â€‹× 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).

(i) 25 cm:1 m and Rs 40:Rs 160 (or) 25 cm:100 cm and Rs 40:Rs 160
25  25 â€‹÷ 25  and  40  40 ÷ 40  1
100      100 â€‹â€‹÷ 25     4         160    160 â€‹÷ 40      4
Hence, they are in proportion.

(ii) 39 litres:65 litres and 6 bottles:10 bottles
39   39 â€‹÷ 13    3    and   6    6 â€‹÷ 2    3
65        65 â€‹â€‹÷ 13       5            10      10 â€‹÷ 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 â€‹÷ 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 â€‹÷ 2     1    and   25       25 â€‹÷ 25        1
80      80 â€‹÷ 2       40         625000    625000 â€‹â€‹÷ 25     25000
Hence, they are not in proportion.

#### Question 6:

(i) 25 cm:1 m and Rs 40:Rs 160 (or) 25 cm:100 cm and Rs 40:Rs 160
25  25 â€‹÷ 25  and  40  40 ÷ 40  1
100      100 â€‹â€‹÷ 25     4         160    160 â€‹÷ 40      4
Hence, they are in proportion.

(ii) 39 litres:65 litres and 6 bottles:10 bottles
39   39 â€‹÷ 13    3    and   6    6 â€‹÷ 2    3
65        65 â€‹â€‹÷ 13       5            10      10 â€‹÷ 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 â€‹÷ 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 â€‹÷ 2     1    and   25       25 â€‹÷ 25        1
80      80 â€‹÷ 2       40         625000    625000 â€‹â€‹÷ 25     25000
Hence, they are not in proportion.

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.

#### Question 7:

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.

Let the second term be x.
Then. 12:x::8:14
We know:
Product of extremes = Product of means
12 × 14 = 8x
⇒                       168 = 8x
â€‹            ⇒                           x  168  = 21
8
Hence, the second term is 21.

#### Question 8:

Let the second term be x.
Then. 12:x::8:14
We know:
Product of extremes = Product of means
12 × 14 = 8x
⇒                       168 = 8x
â€‹            ⇒                           x  168  = 21
8
Hence, the second term is 21.

(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.

#### Question 9:

(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.

Given: 9:x::x:49
We know:
Product of means = Product of extremes
x × x = 9 × 49
⇒               x2 = 441
⇒               x2 = (21)2
⇒                x = 21

#### Question 10:

Given: 9:x::x:49
We know:
Product of means = Product of extremes
x × x = 9 × 49
⇒               x2 = 441
⇒               x2 = (21)2
⇒                x = 21

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 = 20â€‹ × 6
x 120  = 15
8
â€‹Hence, the height of the pole is 15 m.

#### Question 11:

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 = 20â€‹ × 6
x 120  = 15
8
â€‹Hence, the height of the pole is 15 m.

5:3::x:6
We know:
Product of means = Product of extremes
3x = 5 â€‹× 6
⇒ x 30  = 10
3
$\therefore$x = 10

#### Question 1:

5:3::x:6
We know:
Product of means = Product of extremes
3x = 5 â€‹× 6
⇒ x 30  = 10
3
$\therefore$x = 10

Cost of 14 m of cloth = Rs 1890
Cost of 1 m of cloth =  1890  = Rs 135
14
Cost of 6 m of cloth = 6â€‹ × 135 = Rs 810

#### Question 2:

Cost of 14 m of cloth = Rs 1890
Cost of 1 m of cloth =  1890  = Rs 135
14
Cost of 6 m of cloth = 6â€‹ × 135 = Rs 810

Cost of dozen soaps = Rs 285.60
Cost of 1 soap =  285.60
12
Cost of 15 soaps = 15â€‹ ×  285.60  4284  = Rs 357
12            12

#### Question 3:

Cost of dozen soaps = Rs 285.60
Cost of 1 soap =  285.60
12
Cost of 15 soaps = 15â€‹ ×  285.60  4284  = Rs 357
12            12

Cost of 9 kg of rice = Rs 327.60
Cost of 1 kg of rice =  327.60
9
Cost of 50 kg of rice = 50â€‹ ×  327.60  16380  = Rs 1820
9              9
Hence, the cost of 50 kg of rice is Rs 1820.

#### Question 4:

Cost of 9 kg of rice = Rs 327.60
Cost of 1 kg of rice =  327.60
9
Cost of 50 kg of rice = 50â€‹ ×  327.60  16380  = Rs 1820
9              9
Hence, the cost of 50 kg of rice is Rs 1820.

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 = 5â€‹ ×  85.5  427.5  = 19 kg
22.5       22.5
Thus, the weight of 5 m of iron rod is 19 kg.

#### Question 5:

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 = 5â€‹ ×  85.5  427.5  = 19 kg
22.5       22.5
Thus, the weight of 5 m of iron rod is 19 kg.

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

#### Question 6:

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

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

#### Question 7:

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

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 = 35â€‹ ×  540  18900  = Rs 756
25          25

#### Question 8:

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 = 35â€‹ ×  540  18900  = Rs 756
25          25

Let the weight of copper be x g.
â€‹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.

#### Question 9:

Let the weight of copper be x g.
â€‹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.

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 = 315â€‹ ×   35    11025  = 126
87.50      87.50
Hence, we can buy 126 inland letters for Rs 315.

#### Question 10:

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 = 315â€‹ ×   35    11025  = 126
87.50      87.50
Hence, we can buy 126 inland letters for Rs 315.

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.

#### Question 11:

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.

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

#### Question 12:

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

(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

#### Question 13:

(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

(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 = 7â€‹ × 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

#### Question 14:

(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 = 7â€‹ × 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

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

#### Question 15:

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

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.

#### Question 16:

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.

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

#### Question 17:

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

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

#### Question 18:

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

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.

#### Question 19:

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.

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

#### Question 1:

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

(d) 4 : 5
92:115 =   92 â€‹÷ 23   (As H.C.F. of 92 and 115 is 23.)
115 â€‹÷ 23       5

#### Question 2:

(d) 4 : 5
92:115 =   92 â€‹÷ 23   (As H.C.F. of 92 and 115 is 23.)
115 â€‹÷ 23       5

(a) 95
57:x::51:85
57   51
x        85
x 57 × 85
51
x 4845  = 95
51

#### Question 3:

(a) 95
57:x::51:85
57   51
x        85
x 57 × 85
51
x 4845  = 95
51

(a) 63
25:35::45:x
25  45
35       x
x 35 × 45  1575  = 63
25             25

#### Question 4:

(a) 63
25:35::45:x
25  45
35       x
x 35 × 45  1575  = 63
25             25

(c) 28
4:5::x:35
⇒  x
5      35
x 4 × 35  = 4 × 7 = 28
5

#### Question 5:

(c) 28
4:5::x:35
⇒  x
5      35
x 4 × 35  = 4 × 7 = 28
5

Given:
a, b, c, d are in proportion.
a:b::c:d
a  c
b      d

#### Question 6:

Given:
a, b, c, d are in proportion.
a:b::c:d
a  c
b      d

(b) b2 = ac
Given:
a, b, c are in proportion.
a:b::b:c
Product of means = Product of extremes
⇒â€‹ bâ€‹2 = ac

#### Question 7:

(b) b2 = ac
Given:
a, b, c are in proportion.
a:b::b:c
Product of means = Product of extremes
⇒â€‹ bâ€‹2 = ac

(b) (5 : 8) < (3 : 4)

We can write

Making the denominator equal:
5   and   3 × 2   6
8           4 × 2        8
As 6 > 5,   5    3
8        4

#### Question 8:

(b) (5 : 8) < (3 : 4)

We can write

Making the denominator equal:
5   and   3 × 2   6
8           4 × 2        8
As 6 > 5,   5    3
8        4

(a) Rs 440
A:B = 8:11
Sum of ratio terms = 8 + 11 = 19
B's share =  11  × 760 =  8360  = Rs 440
19                  19

#### Question 9:

(a) Rs 440
A:B = 8:11
Sum of ratio terms = 8 + 11 = 19
B's share =  11  × 760 =  8360  = Rs 440
19                  19

(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.

#### Question 10:

(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.

(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
â€‹⇒ x 90  = 10
9
First side = 1x = 1 â€‹× 10 = 10 cm
Second side = 3x = 3 â€‹× 10 = 30 cm
Third side = 5x = 5 × 10 = 50 cm
The length of the largest side is 50 cm.

#### Question 11:

(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
â€‹⇒ x 90  = 10
9
First side = 1x = 1 â€‹× 10 = 10 cm
Second side = 3x = 3 â€‹× 10 = 30 cm
Third side = 5x = 5 × 10 = 50 cm
The length of the largest side is 50 cm.

(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

#### Question 12:

(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

(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

#### Question 13:

(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

(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

#### Question 14:

(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

(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

#### Question 15:

(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

(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

#### Question 16:

(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

(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

#### Question 17:

(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

(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
â€‹      ⇒ x 180  = 30
6
$\therefore$  (3x )$°$ = (3 â€‹× 30)$°$ = 90o
â€‹     (1x)$°$ = (1â€‹ × 30)$°$ = 30o
â€‹     (2x)$°$ = (2 × 30)$°$ = 60o
The measure of the largest angle is 90oâ€‹.

#### Question 18:

(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
â€‹      ⇒ x 180  = 30
6
$\therefore$  (3x )$°$ = (3 â€‹× 30)$°$ = 90o
â€‹     (1x)$°$ = (1â€‹ × 30)$°$ = 30o
â€‹     (2x)$°$ = (2 × 30)$°$ = 60o
The measure of the largest angle is 90oâ€‹.

(b) 45 m
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

#### Question 19:

(b) 45 m
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

(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

#### Question 20:

(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

(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

#### Question 21:

(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

(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 = 12â€‹ ×  20  240  = Rs 30
8         8
Cost of 25 packets of 12 pencils each = 25 × 30 = Rs 750

#### Question 22:

(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 = 12â€‹ ×  20  240  = Rs 30
8         8
Cost of 25 packets of 12 pencils each = 25 × 30 = Rs 750

(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 = 120â€‹ ×  215   = 25800 = Rs 344
75          75

#### Question 23:

(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 = 120â€‹ ×  215   = 25800 = Rs 344
75          75

(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

#### Question 24:

(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

(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

#### Question 1:

(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

(a) 90 cm:1.05 m (or) 90 cm:105 cm             (1 m = 100 cm)
90   90 â€‹÷ 15   6         (H.C.F. of 90 and 105 is 15.)
105      105 â€‹÷ 15      7
$\therefore$ 6:7

(b) 35 minutes to an hour (or) 35 minutes:60 minutes      (1 hour = 60 minutes)
35  35 â€‹÷ 5  7       (H.C.F. of 35 and 60 is 5.)
60       60 â€‹÷ 5     12
$\therefore$  7:12

(c) 150 mL to 2 L (or) 150 L:2000 L                (1 L= 1000 mL)
150   150 â€‹÷ 50   3    (HCF of 150 and 2000 is 50)
2000      2000â€‹ â€‹÷50       40
$\therefore$ 3:40

(d) 2 dozens to a score (or) 24:20       (1 dozen = 12 and 1 score = 20)
24   24 â€‹÷ 4     (H.C.F. of 24 and 20 is 4)
20        20â€‹ â€‹÷ 4      5
$\therefore$ 6:5

#### Question 2:

(a) 90 cm:1.05 m (or) 90 cm:105 cm             (1 m = 100 cm)
90   90 â€‹÷ 15   6         (H.C.F. of 90 and 105 is 15.)
105      105 â€‹÷ 15      7
$\therefore$ 6:7

(b) 35 minutes to an hour (or) 35 minutes:60 minutes      (1 hour = 60 minutes)
35  35 â€‹÷ 5  7       (H.C.F. of 35 and 60 is 5.)
60       60 â€‹÷ 5     12
$\therefore$  7:12

(c) 150 mL to 2 L (or) 150 L:2000 L                (1 L= 1000 mL)
150   150 â€‹÷ 50   3    (HCF of 150 and 2000 is 50)
2000      2000â€‹ â€‹÷50       40
$\therefore$ 3:40

(d) 2 dozens to a score (or) 24:20       (1 dozen = 12 and 1 score = 20)
24   24 â€‹÷ 4     (H.C.F. of 24 and 20 is 4)
20        20â€‹ â€‹÷ 4      5
$\therefore$ 6:5

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 = 7â€‹ × 1.4 = 9.8 kg

#### Question 3:

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 = 7â€‹ × 1.4 = 9.8 kg

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 =  â€‹× Rs 1400 = Rs  4200  = Rs 420
10                             10
Share of C =  â€‹× Rs 1400 = Rs  7000  = Rs 700
10                            10

#### Question 4:

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 =  â€‹× Rs 1400 = Rs  4200  = Rs 420
10                             10
Share of C =  â€‹× Rs 1400 = Rs  7000  = Rs 700
10                            10

We can write:

By making their denominators same: (Taking the L.C.M. of 6 and 4, which is 24.)
Consider, 5:6
5 â€‹× 4  20
6 â€‹× 4      24

And,  3 â€‹× 6  18
4 â€‹× 6       24
As 20 > 18
Clearly, (5:6) > (3:4)

#### Question 5:

We can write:

By making their denominators same: (Taking the L.C.M. of 6 and 4, which is 24.)
Consider, 5:6
5 â€‹× 4  20
6 â€‹× 4      24

And,  3 â€‹× 6  18
4 â€‹× 6       24
As 20 > 18
Clearly, (5:6) > (3:4)

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

#### Question 6:

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

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.

#### Question 7:

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.

Given:
36:x::x:16
We know:
Product of means = Product of extremes
× x = 36 × 16
x2 = 576
x2 = 242
x = 24

#### Question 8:

Given:
36:x::x:16
We know:
Product of means = Product of extremes
× x = 36 × 16
x2 = 576
x2 = 242
x = 24

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.

#### Question 9:

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.

(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 â€‹× 12 = 36
5x = 5 â€‹× 12 = 60
The largest number = 60

#### Question 10:

(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 â€‹× 12 = 36
5x = 5 â€‹× 12 = 60
The largest number = 60

(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 â€‹÷ 18      (H.C.F. of 72 and 90 is 18.)
90      90 â€‹÷ 18      5

#### Question 11:

(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 â€‹÷ 18      (H.C.F. of 72 and 90 is 18.)
90      90 â€‹÷ 18      5

(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

#### Question 12:

(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

(d) 4 : 5
92:115
92  92 â€‹÷ 23          (H.C.F. of 92 and 115 is 23)
115     115 â€‹÷ 23      5

#### Question 13:

(d) 4 : 5
92:115
92  92 â€‹÷ 23          (H.C.F. of 92 and 115 is 23)
115     115 â€‹÷ 23      5

(a) 95
Given :
57:x::51:85
We know:
Product of means = Product of extremes
51x = 57 × 85
x =   4845  = 95
51

#### Question 14:

(a) 95
Given :
57:x::51:85
We know:
Product of means = Product of extremes
51x = 57 × 85
x =   4845  = 95
51

(c) 36
Given:
4:5::x:45
We know:
Product of mean = Product of extremes
5x = 4 â€‹× 45
x =    180   = 36
5

#### Question 15:

(c) 36
Given:
4:5::x:45
We know:
Product of mean = Product of extremes
5x = 4 â€‹× 45
x =    180   = 36
5

(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 â€‹× b = a â€‹× c
b2 = ac

#### Question 16:

(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 â€‹× b = a â€‹× c
b2 = ac

(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

#### Question 17:

(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

(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

#### Question 18:

(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

(i)

(ii) 90 cm:1.5 m (or) 90 cm:150 cm          (1 m = 100 cm)
90  9 ÷ 3              (H.C.F. of 9 and 15 is 3.)
150     15    15 â€‹â€‹÷ 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 â€‹× 35
25x1225
x =  1225  = 49
25

(v) Given:
9, xx, 49 are in proportion.
9:x::x:49
Now, we know:
Product of means = Product of extremes
x â€‹× = 9 â€‹× 49
x2 = 441
x2 = 212
x = 21

#### Question 19:

(i)

(ii) 90 cm:1.5 m (or) 90 cm:150 cm          (1 m = 100 cm)
90  9 ÷ 3              (H.C.F. of 9 and 15 is 3.)
150     15    15 â€‹â€‹÷ 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 â€‹× 35
25x1225
x =  1225  = 49
25

(v) Given:
9, xx, 49 are in proportion.
9:x::x:49
Now, we know:
Product of means = Product of extremes
x â€‹× = 9 â€‹× 49
x2 = 441
x2 = 212
x = 21

(i) 30, 40, 45, 60
30  =    3 ,    45  =   45 â€‹÷ 15  =         They are in proportion.
40        4      60        60 â€‹÷ 15         4
Hence, true.

(ii)  6  6 â€‹÷ 2  3 9 â€‹÷ 3  3     Hence, they are equivalent to 3:4.
8       8 â€‹÷ 2      4    2      12 â€‹÷ 3     4
Hence, true.
(iii) 1 dozen:1 score = 12:20
12  12 â€‹÷ 4
20      20 â€‹÷ 4       5
Hence, false.
(iv) 60p:Rs 3 = 60p:300p                        (1 Re = 100 p)
60  6 â€‹÷ 6
300     30    30 â€‹÷ 6      5

Hence, true.

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