Rs Aggarwal 2019 Solutions for Class 7 Math 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 Math Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 7 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 7 Math 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

#### Answer:

(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

#### Answer:

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

=     (âˆµ 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 =    (âˆµ HCF of 105 and 63 = 21)
= 5 cm : 3 cm

(iii) Converting both the quantities into the same unit
65 min : 45 min =   (âˆµ 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 = $\frac{8}{12}=\frac{8÷4}{12÷4}=\frac{2}{3}$  (âˆµ HCF of 8 and 12 = 4)
= 2 months : 3 months

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

2250g : 3000 g =     (âˆµ 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 =      (âˆµ 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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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 $\frac{1}{5}:\frac{1}{6}.$

#### Answer:

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.

#### Answer:

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?

#### Answer:

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?

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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

#### Question 21:

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

#### Answer:

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

= 25 : 39

#### Question 22:

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

#### Answer:

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.

#### Answer:

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)

#### Answer:

(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)

#### Answer:

(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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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

#### Answer:

(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.

#### Answer:

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.

#### Answer:

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

#### Answer:

(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.

#### Answer:

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

#### Answer:

(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?

#### Answer:

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?

#### Answer:

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?

#### Answer:

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.

#### Answer:

(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 (âœ“) 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

#### Answer:

The correct option is (d).

Hence, a : c = 2 : 3

#### Question 2:

Mark (âœ“) 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 (âœ“) 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

#### Answer:

The correct option is (d).

Hence, A : C = 15 : 8

#### Question 4:

Mark (âœ“) 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

#### Answer:

The correct option is (b).

Hence, A : B = 4 : 3

#### Question 5:

Mark (âœ“) 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 (âœ“) 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

#### Answer:

(b)  30 : 42 : 77

#### Question 7:

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

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

#### Answer:

(a) 3 : 4 : 5

= 3 : 4 : 5

#### Question 9:

Mark (âœ“) 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 (âœ“) 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

#### Answer:

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).

#### Question 11:

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

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

(c)  9

#### Question 13:

Mark (âœ“) 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

#### Answer:

(b) 7

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

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

#### Question 14:

Mark (âœ“) 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

#### Answer:

(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 (âœ“) 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

#### Answer:

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

#### Question 16:

Mark (âœ“) 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 (âœ“) 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

#### Answer:

(d) 416

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

#### Question 18:

Mark (âœ“) 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

#### Answer:

(a) (2 :3)

LCM of 3 and 7 = $7×3=$21

#### Question 19:

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

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

#### Answer:

(c) 16

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

#### Question 20:

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

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

#### Answer:

(b) 12

Suppose that the mean proportional is x.

Then, 9 : x :: x : 16

#### Question 21:

Mark (âœ“) 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

#### Answer:

(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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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.

#### Answer:

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?

#### Answer:

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.

#### Answer:

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?

#### Answer:

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 (âœ“) 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 (âœ“) 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 (âœ“) 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 (âœ“) 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

#### Answer:

(a) 3

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

#### Question 14:

Mark (âœ“) 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

#### Answer:

(b) 360

Sum of the ratio terms = 4 + 3 = 7

∴ B's share = = Rs 360

#### Question 15:

Mark (âœ“) 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

#### Answer:

(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 (âœ“) 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

#### Answer:

(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 = ...... .

#### Answer:

(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

#### Answer:

(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

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