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.

#### Page No 124:

(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

#### Page No 124:

(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

#### Page No 124:

Therefore, we have:

∴ A : C = 9 : 10

∴ A : C = 2 : 5

#### Page No 124:

A : B = 3 : 5

B : C = 10 : 13 =

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

∴ A : B : C = 6 : 10 : 13

#### Page No 124:

We have the following:

A : B = 5 : 6
B : C = 4 : 7  =

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

#### Page No 124:

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

#### Page No 125:

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

#### Page No 125:

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

#### Page No 125:

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.

#### Page No 125:

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.

#### Page No 125:

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.

#### Page No 125:

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.

#### Page No 125:

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.

#### Page No 125:

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

#### Page No 125:

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.

#### Page No 125:

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.

#### Page No 125:

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

#### Page No 125:

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

#### Page No 125:

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

#### Page No 125:

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

= 25 : 39

#### Page No 125:

Now, we have:

∴ (8x − 3y) : (3x + 2y) = 3 : 8

#### Page No 125:

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.

#### Page No 125:

(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)

#### Page No 125:

(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.

#### Page No 128:

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

#### Page No 128:

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.

#### Page No 128:

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

#### Page No 128:

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

#### Page No 128:

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

#### Page No 128:

(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.

#### Page No 128:

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

#### Page No 128:

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.

#### Page No 128:

(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.

#### Page No 128:

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

#### Page No 128:

(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.

#### Page No 128:

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.

#### Page No 128:

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.

#### Page No 128:

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.

#### Page No 128:

(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

#### Page No 128:

The correct option is (d).

Hence, a : c = 2 : 3

(a) 15 : 8

#### Page No 128:

The correct option is (d).

Hence, A : C = 15 : 8

#### Page No 128:

The correct option is (b).

Hence, A : B = 4 : 3

(a)  1 : 3 : 6

#### Page No 129:

(b)  30 : 42 : 77

(c)  6 : 4 : 3

#### Page No 129:

(a) 3 : 4 : 5

= 3 : 4 : 5

(b)  15 : 10 : 6

#### Page No 129:

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).

(c) 5 : 2

a : b = 5 : 2

(c)  9

#### Page No 129:

(b) 7

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

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

#### Page No 129:

(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

#### Page No 129:

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

(a)  Rs 180

A's share =

#### Page No 129:

(d) 416

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

#### Page No 129:

(a) (2 :3)

LCM of 3 and 7 = $7×3=$21

#### Page No 129:

(c) 16

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

#### Page No 129:

(b) 12

Suppose that the mean proportional is x.

Then, 9 : x :: x : 16

#### Page No 129:

(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

#### Page No 131:

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

#### Page No 131:

The sum of ratio terms is 10.

Then, we have:

A's share = Rs

#### Page No 131:

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.

#### Page No 131:

x : 18 :: 18 : 108

#### Page No 131:

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.

#### Page No 131:

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

#### Page No 131:

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

∴ Distance covered in 40 min =

#### Page No 131:

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 .

#### Page No 131:

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.

(d)  6 : 4 : 3

(a)  2 : 3 : 4

(c) 11 : 3

#### Page No 131:

(a) 3

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

#### Page No 131:

(b) 360

Sum of the ratio terms = 4 + 3 = 7

∴ B's share = = Rs 360

#### Page No 131:

(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.

#### Page No 131:

(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

#### Page No 131:

(i) 15 : 8

∴ C : A=15 : 8

(ii) 5 : 4

(iii) 1 : 3 : 6

(iv)  30 : 42 : 77

#### Page No 131:

(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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