Rs Aggarwal 2017 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 2017 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 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math 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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