Rs Aggarwal 2020 2021 Solutions for Class 7 Maths Chapter 8 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among Class 7 students for Maths Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 7 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Express each of the following ratios in simplest form:

(i) 24 : 40
(ii) 13.5 : 15
(iii) $6\frac{2}{3}:7\frac{1}{2}$
(iv) $\frac{1}{6}:\frac{1}{9}$
(v) $4:5:\frac{9}{2}$
(vi) 2.5 : 6.5 : 8

(i) HCF of 24 and 40 is 8.
∴ 24 : 40 = = 3 : 5

Hence, 24 : 40 in its simplest form is 3 : 5.

(ii) HCF of 13.5 and 15 is 1.5.

Hence, 13.5 : 15 in its simplest form is 9 : 10.

(iii)
The HCF of 40 and 45 is 5.

∴ 40 : 45 = = 8 : 9

Hence, in its simplest form is 8 : 9

(iv) 9 : 6
The HCF of 9 and 6 is 3.

9 : 6 = = 3 : 2
Hence, $\frac{1}{6}:\frac{1}{9}$ in its simplest form is 3 : 2.

(v) LCM of the denominators is 2.

4 : 5 : $\frac{9}{2}$ = 8 : 10 : 9
The HCF of these 3 numbers is 1.

∴ 8 : 10 : 9 is the simplest form
.

(vi) 2.5 : 6.5 : 8 = 25 : 65 : 80

The HCF of 25, 65 and 80 is 5.
25 : 65 : 80 = = 5 : 13 : 16

#### Question 2:

Express each of the following ratios in simplest form:

(i) 75 paise : 3 rupees
(ii) 1 m 5 cm : 63 cm
(iii) 1 hour 5 minutes : 45 minutes
(iv) 8 months : 1 year
(v) (2 kg 250 g) : (3 kg)
(vi) 1 km : 750 m

(i) Converting both the quantities into the same unit, we have:
75 paise : (3 $×$ 100) paise = 75 : 300

=     (∵ HCF of 75 and 300 = 75)
= 1 paise : 4 paise

(ii)  Converting both the quantities into the same unit, we have:
105 cm : 63 cm =    (∵ HCF of 105 and 63 = 21)
= 5 cm : 3 cm

(iii) Converting both the quantities into the same unit
65 min : 45 min =   (∵ HCF of 65 and 45 = 5)
= 13 min : 9 min

(iv) Converting both the quantities into the same unit, we get:
8 months : 12 months = $\frac{8}{12}=\frac{8÷4}{12÷4}=\frac{2}{3}$  (∵ HCF of 8 and 12 = 4)
= 2 months : 3 months

(v) Converting both the quantities into the same unit, we get:

2250g : 3000 g =     (∵ HCF of 2250 and 3000 = 750)

= 3 g : 4 g

(vi)  Converting both the quantities into the same unit, we get:
1000 m : 750 m =      (∵ HCF of 1000 and 750 = 250)
= 4 m : 3 m

#### Question 3:

If A : B = 7 : 5 and B : C = 9 : 14, find A : C.

Therefore, we have:

∴ A : C = 9 : 10

#### Question 4:

If A : B = 5 : 8 and B : C = 16 : 25, find A : C.

∴ A : C = 2 : 5

#### Question 5:

If A : B = 3 : 5 and B : C = 10 : 13, find A : B : C.

A : B = 3 : 5

B : C = 10 : 13 =

Now, A : B : C = 3 : 5 : $\frac{13}{2}$

∴ A : B : C = 6 : 10 : 13

#### Question 6:

If A : B = 5 : 6 and B : C = 4 : 7, find A : B : C.

We have the following:

A : B = 5 : 6
B : C = 4 : 7  =

A : B : C =  5 : 6 : $\frac{21}{2}$ =  10 : 12 : 21

#### Question 7:

Divide Rs 360 between Kunal and Mohit in the ratio 7 : 8.

Sum of the ratio terms  = 7 + 8 = 15

Now, we have the following:

Kunal's share = Rs 360 = Rs 168

Mohit's share = Rs 360 = Rs 192

#### Question 8:

Divide Rs 880 between Rajan and Kamal in the ratio $\frac{1}{5}:\frac{1}{6}.$

Sum of the ratio terms = $\frac{1}{5}+\frac{1}{6}=\frac{11}{30}$

Now, we have the following:
Rajan's share = Rs 880   =  Rs 480
Kamal's share = Rs 880 = Rs 400

#### Question 9:

Divide Rs 5600 between A, B and C in the ratio 1 : 3 : 4.

Sum of the ratio terms is (1 + 3 + 4) = 8

We have the following:

A's share =  Rs 5600

B's share =  Rs 5600 = Rs 2100

C's share = Rs 5600 = Rs 2800

#### Question 10:

What number must be added to each term to the ratio 9 : 16 to make the ratio 2 : 3?

Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3

Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.

#### Question 11:

What number must be subtracted from each term of ratio 17 : 33 so that the ratio becomes 7 : 15?

Suppose that x is the number that must be subtracted.
Then, (17 − x) : (33 − x) = 7 : 15

Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15.

#### Question 12:

Two numbers are in the ratio 7 : 11. If added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.

Suppose that the numbers are 7x and 11x.

Then, (7x + 7) : (11x + 7) = 2 : 3

⇒ 21x + 21 = 22x + 14

⇒ x = 7

Hence, the numbers are (7 $×$ 7 =) 49 and (11 $×$ 7 =) 77.

#### Question 13:

Two numbers are in the ratio 5 : 9. On subtracting 3 from each, the ratio becomes 1 : 2. Find the numbers.

Suppose that the numbers are 5x and 9x.
Then, (5x − 3) : (9x − 3) = 1 : 2

⇒ 10x − 6 = 9x − 3
x = 3

Hence, the numbers are (5 $×$ 3 =) 15 and (9 $×$ 3 =) 27.

#### Question 14:

Two numbers are in the ratio 3 : 4. If their LCM is 180, find the numbers.

Let the numbers be 3x and 4x.
Their LCM is 12x.
Then, 12x = 180
x = 15

∴ The numbers are (3 $×$ 15 =) 45 and (4 $×$ 15 =) 60.

#### Question 15:

The ages of A and B are in the ratio 8 : 3. Six years hence, their ages will be in the ratio 9 : 4. Find their present ages.

Suppose that the present ages of A and B are 8x yrs and 3x yrs.
Then, (8x  + 6) : (3x + 6) = 9 : 4

⇒ 32x + 24 = 27x + 54
⇒ 5x = 30
x = 6

Now, present age of A = 8 $×$ 6 yrs = 48 yrs
Present age of  B = 3 $×$ 6 yrs = 18 yrs

#### Question 16:

The ratio of copper and zinc in an alloy is 9 : 5. If the weight of copper in the alloy is 48.6 grams, find the weight of zinc in the alloy.

Suppose that the weight of zinc is x g.

Then, 48.6 : x = 9 : 5

x = $\frac{48.6×5}{9}=\frac{243}{9}$ = 27

Hence, the weight of zinc in the alloy is 27 g.

#### Question 17:

The ratio of boys and girls in a school is 8 : 3. If the total number of girls be 375, find the number of boys in the school.

Suppose that the number of boys is x.
Then, x : 375 = 8 : 3

⇒ x = $\frac{8×375}{3}=8×125$ = 1000

Hence, the number of girls in the school is 1000.

#### Question 18:

The ratio of monthly income to the savings of a family is 11 : 2. If the savings be Rs 2500, find the income and expenditure.

Suppose that the monthly income of the family is Rs x.

Then, x : 2500 = 11 : 2

x = $\frac{11×2500}{2}=11×1250$
x = Rs 13750

Hence, the income is Rs 13,750.
∴ Expenditure = (monthly income − savings)
=Rs (13750 − 2500)
= Rs 11250

#### Question 19:

A bag contains Rs 750 in the form of rupee, 50 P and 25 P coins in the ratio 5 : 8 : 4. Find the number of coins of each type.

Let the numbers one rupee, fifty paise and twenty-five paise coins be 5x, 8x and 4x, respectively.

Total value of these coins = ()

However, the total value is Rs 750.
∴ 750 = 10x
x = 75

Hence, number of one rupee coins = 5 $×$ 75 = 375
Number of fifty paise coins = 8 $×$ 75 = 600
Number of twenty-five paise coins = 4 $×$ 75 = 300

#### Question 20:

If (4x + 5) : (3x + 11) = 13 : 17, find the value of x.

(4x + 5) : (3x + 11) = 13 : 17

#### Question 21:

If x : y = 3 : 4, find (3x + 4y) : (5x + 6y).

Now, we have (3x + 4y) : (5x + 6y)

= 25 : 39

#### Question 22:

If x : y = 6 : 11, find (8x − 3y) : (3x + 2y).

Now, we have:

∴ (8x − 3y) : (3x + 2y) = 3 : 8

#### Question 23:

Two numbers are in the ratio 5 : 7. If the sum of the numbers is 720, find the numbers.

Suppose that the numbers are 5x and 7x.
The sum of the numbers is 720.
i.e., 5x + 7x = 720
⇒ 12x = 720
x = 60

Hence, the numbers are (5 $×$ 60 =) 300 and (7 $×$ 60 =) 420.

#### Question 24:

Which ratio is greater?

(i) (5 : 6) or (7 : 9)
(ii) (2 : 3) or (4 : 7)
(iii) (1 : 2) or (4 : 7)
(iv) (3 : 5) or (8 : 13)

(i) The LCM of 6 and 9 is 18.

∴ (7 : 9) $<$ (5 : 6)

(ii)  The LCM of 3 and 7 is 21.

∴ (4 : 7) $<$ (2 : 3)

(iii)  The LCM of 2 and 7 is 14.

Clearly, $\frac{7}{14}<\frac{8}{14}$

∴ (1 : 2) $<$ (4 : 7)

(iv) The LCM of 5 and 13 is 65.

∴ (3 : 5) $<$ (8 : 13)

#### Question 25:

Arrange the following ratios in ascending order:

(i) (5 : 6), (8 : 9), (11 : 18)
(ii) (11 : 14), (17 : 21), (5 : 7) and (2 : 3)

(i) We have

The LCM of 6, 9 and 18 is 18. Therefore, we have:

Hence, (11 : 18) $<$ (5 : 6) $<$ (8 : 9)

(ii)

The LCM of 14, 21, 7 and 3 is 42.

#### Question 1:

Show that 30, 40, 45, 60 are in proportion.

We have:

Product of the extremes = 30 60 = 1800
Product of the means = 40 45 = 1800
Product of extremes = Product of means

Hence, 30 : 40 :: 45 : 60

#### Question 2:

Show that 36, 49, 6, 7 are not in proportion.

We have:
Product of the extremes = 36 $×$ 7 = 252
Product of the means = 49 $×$ 6 = 294
Product of the extremes $\ne$ Product of the means

Hence, 36, 49, 6 and 7 are not in proportion.

#### Question 3:

If 2 : 9 :: x : 27, find the value of x.

Product of the extremes = 2 27 = 54
Product of the means  = 9 x = 9x

Since 2 : 9 :: x : 27, we have:
Product of the extremes = Product of the means
⇒ 54 = 9x
⇒ x = 6

#### Question 4:

If 8 : x :: 16 : 35, find the value of x.

Product of the extremes = 8 35 = 280
Product of the means = 16 x = 16x

Since 8 : x :: 16 : 35, we have:
Product of the extremes = Product of the means
⇒ 280 = 16x
x = 17.5

#### Question 5:

If x : 35 :: 48 : 60, find the value of x.

Product of the extremes = x $×$ 60 = 60x
Product of the means = 35 $×$ 48 = 1680

Since x : 35 :: 48 : 60, we have:
Product of the extremes = Product of the means
⇒ 60x= 1680
x = 28

#### Question 6:

Find the fourth proportional to the numbers:

(i) 8, 36, 6
(ii) 5, 7, 30
(iii) 2.8, 14, 3.5

(i) Let the fourth proportional be x.
Then, 8 : 36 :: 6 : x

8                                    [Product of extremes = Product of means]
⇒ 8x = 216
x = 27
Hence, the fourth proportional is 27.

(ii) Let the fourth proportional be x.
Then, 5 : 7 :: 30 : x
[Product of extremes = Product of means]
⇒ 8x = 216
⇒ 5x = 210
x = 42

Hence, the fourth proportional is 42.

(iii) Let the fourth proportional be x.
Then, 2.8                                 [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 2.8x = 49
x = 17.5
Hence, the fourth proportional is 17.5.

#### Question 7:

If 36, 54, x are in continued proportion, find the value of x.

36, 54 and x are in continued proportion.
Then, 36 : 54 :: 54 : x
[Product of extremes = Product of means]
⇒ 36x = 2916
x = 81

#### Question 8:

If 27, 36, x are in continued proportion, find the value of x.

27, 36 and x are in continued proportion.
Then, 27 : 36 :: 36 : x
[Product of extremes = Product of means]
⇒ 27x = 1296
x = 48

Hence, the value of x is 48.

#### Question 9:

Find the third proportional to:

(i) 8 and 12
(ii) 12 and 18
(iii) 4.5 and 6

(i)  Suppose that x is the third proportional to 8 and 12.
Then, 8 :12 :: 12 : x
⇒ 8                                             (Product of extremes = Product of means )
⇒ 8x = 144
x = 18

Hence, the required third proportional is 18.

(ii) Suppose that x is the third proportional to 12 and 18.
Then, 12 : 18 :: 18 : x
(Product of extremes = Product of means )
⇒ 12x = 324
x = 27

Hence, the third proportional is 27.

(iii) Suppose that x is the third proportional to 4.5 and 6.
Then, 4.5 : 6:: 6 : x
(Product of extremes = Product of means )
⇒ 4.5x = 36
x = 8

Hence, the third proportional is 8.

#### Question 10:

If the third proportional to 7 and x is 28, find the value of x.

The third proportional to 7 and x is 28.
Then, 7 : x :: x : 28
⇒ 7 $×$ 28 = ${x}^{2}$           (Product of extremes = Product of means)
x = 14

#### Question 11:

Find the mean proportional between:

(i) 6 and 24
(ii) 3 and 27
(iii) 0.4 and 0.9

(i)  Suppose that x is the mean proportional.

Then, 6 : x :: x : 24

(Product of extremes = Product of means)

x = 12

Hence, the mean proportional to 6 and 24 is 12.

(ii)  Suppose that x is the mean proportional.

Then, 3 : x :: x : 27
(Product of extremes =Product of means)
x = 9

Hence, the mean proportional to 3 and 27 is 9.

(iii)  Suppose that x is the mean proportional.

Then, 0.4 : x :: x : 0.9

(Product of extremes =Product of means)
⇒x = 0.6

Hence, the mean proportional to 0.4 and 0.9 is 0.6.

#### Question 12:

What number must be added to each of the numbers 5, 9, 7, 12 to get the numbers which are in proportion?

Suppose that the number is x.

Then, (5 + x) : ( 9 + x) :: (7 + x) : (12 + x)

Hence, 3 must be added to each of the numbers: 5, 9, 7 and 12, to get the numbers which are in proportion.

#### Question 13:

What number must be subtracted from each of the numbers 10, 12, 19, 24 to get the numbers which are in proportion?

Suppose that x is the number that is to be subtracted.

Then, (10 − x) : (12 − x) :: (19 − x) : (24 − x)

.
Hence, 4 must be subtracted from each of the numbers: 10, 12, 19 and 24, to get the numbers which are in proportion.

#### Question 14:

The scale of a map is 1 : 5000000. What is the actual distance between two towns, if they are 4 cm apart on the map?

Distance represented by 1 cm on the map = 5000000 cm = 50 km

Distance represented by 3 cm on the map = 50 $×$ 4 km = 200 km

∴ The actual distance is 200 km.

#### Question 15:

At a certain time a tree 6 m high casts a shadow of length 8 metres. At the same time a pole casts a shadow of length 20 metres. Find the height of the pole.

(Height of tree) : (height of its shadow) = (height of the pole) : (height of its shadow)

Suppose that the height of pole is x cm.

Then, 6 : 8 = x : 20

x =
∴ Height of the pole = 15 cm

#### Question 1:

Mark (✓) against the correct answer
If a : b = 3 : 4 and b : c = 8 : 9, then a : c = ?

(a) 1 : 2
(b) 3 : 2
(c) 1 : 3
(d) 2 : 3

The correct option is (d).

Hence, a : c = 2 : 3

#### Question 2:

Mark (✓) against the correct answer
If A : B = 2 : 3 and B : C = 4 : 5, then C : A = ?

(a) 15 : 8
(b) 6 : 5
(c) 8 : 5
(d) 8 : 15

(a) 15 : 8

#### Question 3:

Mark (✓) against the correct answer
If 2A = 3B and 4B = 5C, then A : C = ?

(a) 4 : 3
(b) 8 : 15
(c) 3 : 4
(d) 15 : 8

The correct option is (d).

Hence, A : C = 15 : 8

#### Question 4:

Mark (✓) against the correct answer
If 15% of A = 20% of B, then A : B = ?

(a) 3 : 4
(b) 4 : 3
(c) 17 : 16
(d) 16 : 17

The correct option is (b).

Hence, A : B = 4 : 3

#### Question 5:

Mark (✓) against the correct answer
If then A : B : C = ?

(a) 1 : 3 : 6
(b) 2 : 3 : 6
(c) 3 : 2 : 6
(d) 3 : 1 : 2

(a)  1 : 3 : 6

#### Question 6:

Mark (✓) against the correct answer
If A : B = 5 : 7 and B : C = 6 : 11, then A : B : C = ?

(a) 30 : 42 : 55
(b) 30 : 42 : 77
(c) 35 : 49 : 66
(d) none of these

(b)  30 : 42 : 77

#### Question 7:

Mark (✓) against the correct answer
If 2A = 3B = 4C, then A : B : C = ?

(a) 2 : 3 : 4
(b) 4 : 3 : 2
(c) 6 : 4 : 3
(d) 3 : 4 : 6

(c)  6 : 4 : 3

#### Question 8:

Mark (✓) against the correct answer
then A : B : C = ?

(a) 3 : 4 : 5
(b) 4 : 3 : 5
(c) 5 : 4 : 3
(d) 20 : 15 : 12

(a) 3 : 4 : 5

= 3 : 4 : 5

#### Question 9:

Mark (✓) against the correct answer
$\mathrm{If}\frac{1}{\mathrm{x}}:\frac{1}{\mathrm{y}}:\frac{1}{\mathrm{z}}=2:3:5,$ then, x : y : z = ?

(a) 2 : 3 : 5
(b) 15 : 10 : 6
(c) 5 : 3 : 2
(d) 6 : 10 : 15

(b)  15 : 10 : 6

#### Question 10:

Mark (✓) against the correct answer
If x : y = 3 : 4, then (7x + 3y) : (7x − 3y) = ?

(a) 4 : 3
(b) 5 : 2
(c) 11 : 3
(d) 37 : 39

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).

#### Question 11:

Mark (✓) against the correct answer
If (3a + 5b) : (3a − 5b) = 5 : 1, then a : b = ?

(a) 2 : 1
(b) 3 : 2
(c) 5 : 2
(d) 5 : 3

(c) 5 : 2

a : b = 5 : 2

#### Question 12:

Mark (✓) against the correct answer
If 7 : x :: 35 : 45, then x = ?

(a) 11
(b) 15
(c) 9
(d) 5

(c)  9

#### Question 13:

Mark (✓) against the correct answer
What number has to be added to each term of 3 : 5 to make the ratio 5 : 6?

(a) 6
(b) 7
(c) 12
(d) 11

(b) 7

Suppose that x is the number that is to be added.

Then, (3 + x) : (5 + x) = 5 : 6

#### Question 14:

Mark (✓) against the correct answer
Two numbers are in the ratio 3 : 5. If each number is increased by 10, the ratio becomes 5 : 7. The sum of the numbers is

(a) 8
(b) 16
(c) 35
(d) 40

(d) 40

Suppose that the numbers are x and y.

Then, x : y = 3 : 5 and (x + 10) : (y + 10) = 5 : 7

Hence, sum of numbers = 15 + 25 = 40

#### Question 15:

Mark (✓) against the correct answer
What least number is to be subtracted from each term of the ratio 15 : 19 to make the ratio 3 : 4?

(a) 3
(b) 5
(c) 6
(d) 9

(a)  3
Suppose that x is the number that is to be subtracted.
Then, (15 − x) : (19 − x) = 3 : 4

#### Question 16:

Mark (✓) against the correct answer
If Rs 420 is divided between A and B in the ratio 3 : 4, then A's share is

(a) Rs 180
(b) Rs 240
(c) Rs 270
(d) Rs 210

(a)  Rs 180

A's share =

#### Question 17:

Mark (✓) against the correct answer
The boys and girls in a school are in the ratio 8 : 5. If the number of girls is 160, what is the total strength of the school?

(a) 250
(b) 260
(c) 356
(d) 416

(d) 416

Let x be the number of boys.
Then, 8 : 5 = x : 160

#### Question 18:

Mark (✓) against the correct answer
Which one is greater out of (2 : 3) and (4 : 7)?

(a) 2 : 3
(b) 4 : 7
(c) both are equal

(a) (2 :3)

LCM of 3 and 7 = $7×3=$21

#### Question 19:

Mark (✓) against the correct answer
The third proportional to 9 and 12 is

(a) 10.5
(b) 8
(c) 16
(d) 21

(c) 16

Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x

#### Question 20:

Mark (✓) against the correct answer
The mean proportional between 9 and 16 is

(a) 12.5
(b) 12
(c) 5
(d) none of these

(b) 12

Suppose that the mean proportional is x.

Then, 9 : x :: x : 16

#### Question 21:

Mark (✓) against the correct answer
The ages of A and B are in the ratio 3 : 8. Six years hence, their ages will be in the ratio 4 : 9. The present age of A is

(a) 18 years
(b) 15 years
(c) 12 years
(d) 21 years

(a)  18 years

Suppose that the present ages of A and B are 3x yrs and 8x yrs, respectively.
After six years, the age of A will be (3x+6) yrs and that of B will be (8x+6) yrs.
Then, (3x +6) : (8x + 6) = 4 : 9

#### Question 1:

Compare 4 : 5 and 7 : 9.

The given fractions are .
LCM of 5 and 9 = 5 $×$ 9 = 45

#### Question 2:

Divide Rs 1100 among A, B and C in the ratio 2 : 3 : 5.

The sum of ratio terms is 10.

Then, we have:

A's share = Rs

#### Question 3:

Show that the numbers 25, 36, 5, 6 are not in proportion.

Product of the extremes = 25 $×$ 6 = 150
Product of the means = 36 $×$ 5 = 180

The product of the extremes is not equal to that of the means.

Hence, 25, 36, 5 and 6 are not in proportion.

#### Question 4:

If x,18,108 are in continued proportion, find the value of x.

x : 18 :: 18 : 108

#### Question 5:

Two numbers are in the ratio 5 : 7. If the sum of these numbers is 84, find the numbers.

Suppose that the numbers are 5x and 7x.
Then, 5x + 7x = 84
⇒ 12x = 84
x = 7

Hence, the numbers are (5 $×$ 7 =) 35 and (7 $×$ 7 =) 49.

#### Question 6:

The ages of A and B are in the ratio 4 : 3. Eight years ago, their ages were in the ratio 10 : 7. Find their present ages.

Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively.
Eight years ago, age of A = (4x − 8) yrs
Eight years ago, age of B = (3x − 8) yrs
Then, (4x − 8) : (3x − 8) = 10 : 7

#### Question 7:

If a car covers 54 km in an hour, how much distance will it cover in 40 minutes?

Distance covered in 60 min = 54 km
Distance covered in 1 min =

∴ Distance covered in 40 min =

#### Question 8:

Find the third proportional to 8 and 12.

Suppose that the third proportional to 8 and 12 is x.
Then, 8 : 12 :: 12 : x

⇒ 8x = 144  (Product of extremes = Product of means)
x = 18

Hence, the third proportional is 18 .

#### Question 9:

If 40 men can finish a piece of work in 60 days, in how many days will 75 men finish the same work?

40 men can finish the work in 60 days.
1 man can finish the work in 60 $×$ 40 days.     [Less men, more days]
75 men can finish the work in

Hence, 75 men will finish the same work in 32 days.

#### Question 10:

Mark (✓) against the correct answer
If 2A = 3B = 4C then A : B : C = ?

(a) 2 : 3 : 4
(b) 3 : 4 : 6
(c) 4 : 3 : 2
(d) 6 : 4 : 3

(d)  6 : 4 : 3

#### Question 11:

Mark (✓) against the correct answer
$\mathrm{If}\frac{A}{2}=\frac{B}{3}=\frac{C}{4}$ then A : B : C = ?
(a) 2 : 3 : 4
(b) 4 : 3 : 2
(c) 3 : 2 : 4
(d) none of these

(a)  2 : 3 : 4

#### Question 12:

Mark (✓) against the correct answer
If (x : y) = 3 : 4, then (7x + 3y) : (7x − 3y) = ?

(a) 7 : 3
(b) 5 : 2
(c) 11 : 3
(d) 14 : 9

(c) 11 : 3

#### Question 13:

Mark (✓) against the correct answer
What least number must be subtracted from each term of the ratio 15 : 19 to make the ratio 3 : 4?

(a) 3
(b) 5
(c) 6
(d) 9

(a) 3

Suppose that the number to be subtracted is x.
Then, (15 − x) : (19 − x) = 3 : 4

#### Question 14:

Mark (✓) against the correct answer
If Rs 840 is divided between A and B in the ratio 4 : 3, then B's share is

(a) Rs 480
(b) Rs 360
(c) Rs 320
(d) Rs 540

(b) 360

Sum of the ratio terms = 4 + 3 = 7

∴ B's share = = Rs 360

#### Question 15:

Mark (✓) against the correct answer
The ages of A and B are in the ratio 5 : 2. After 5 years, their ages will be in the ratio 15 : 7. The present age of A is

(a) 48 years
(b) 36 years
(c) 40 years
(d) 35 years

(c) 40 years

Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively.
After 5 years, the ages of A and B will be (5x+5) yrs and (2x+5) yrs, respectively.

Then, (5x + 5) : (2x + 5) = 15 : 7

Cross multiplying, we get:

35x + 35 = 30x + 75
⇒ 5x = 40
x = 8

Hence, the present age of A is 5 $×$ 8 = 40 yrs.

#### Question 16:

Mark (✓) against the correct answer
The boys and girls in a school are in the ratio 9 : 5. If the number of girls is 320, then the total strengh of the school is

(a) 840
(b) 896
(c) 920
(d) 576

(b)  896

Suppose that the number of boys in the school is x.
Then, x : 320 = 9 : 5
⇒ 5x = 2880
x = 576

Hence, total strength of the school = 576 + 320 = 896

#### Question 17:

Fill in the blanks.

(i) If A : B = 2 : 3 and B : C = 4 : 5, then C : A = ...... .
(ii) If 16% of A = 20% of B, then A : B = ...... .
(iii) If then A : B : C = ...... .
(iv) If A : B = 5 : 7 and B : C = 6 : 11, then A : B : C = ...... .

(i) 15 : 8

∴ C : A=15 : 8

(ii) 5 : 4

(iii) 1 : 3 : 6

(iv)  30 : 42 : 77

#### Question 18:

Write 'T' for true and 'F' for false

(i) Mean proportional between 0.4 and 0.9 is 6.
(ii) The third proportional to 9 and 12 is 10.5.
(iii) If 8 : x :: 48 : 18, then x = 3.
(iv) If (3a + 5b) : (3a − 5b) = 5 : 1, then a : b = 5 : 2

(i) F
Suppose that the mean proportional is x.
Then, 0.4 : x :: x : 0.9

(ii)  F
Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x
⇒ 9x = 144                      (Product of extremes = Product of means)
x = 16

(iii)  T
8 : x :: 48 : 18
⇒ 144 = 48x               (Product of extremes = Product of means)
x = 3

(iv) T

⇒ 12a = 30b

a : b = 5 : 2

View NCERT Solutions for all chapters of Class 7