## RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8A
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8C
- RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion CCE Test Paper

**Question 1.**

**Solution:**

We know that a, b, c, d are in proportion if ad = bc

Now 30, 40, 45, 60 are in proportion

if 30 x 60 = 40 x 45

if 1800= 1800

which is true

30, 40, 45, 60 are in proportion.

**Question 2.**

**Solution:**

We know that if a, b, c, d are in proportion if ad = bc

Now 36, 49, 6, 7 are in proportion

if 36 x 7 = 49 x 6

if 252 = 294

But 252 ≠ 294

36, 49, 6, 7 are not in proportion

**Question 3.**

**Solution:**

2 : 9 : : x : 27

9 x x = 2 x 27

x = \(\frac { 2 x 27 }{ 9 }\) = 2 x 3 = 6

**Question 4.**

**Solution:**

8 : x : : 16 : 35

x x 16 = 8 x 35

x = \(\frac { 8 x 35 }{ 6 }\) = \(\frac { 35 }{ 2 }\) = 17.5

**Question 5.**

**Solution:**

x : 35 : : 48 : 60

x x 60 = 35 x 48

x = 7 x 4 = 28

**Question 6.**

**Solution:**

Let x be the fourth proportional, then

x = \(\frac { 35 }{ 2 }\) = 17.5

Fourth proportional = 17.5

**Question 7.**

**Solution:**

36, 54, x are in continued proportion

36 : 54 : : 54 : x

⇒ 36 x x = 54 x 54

**Question 8.**

**Solution:**

27, 36, x are in continued proportion

27 : 36 :: 36 : x

27 x x = 36 x 36

**Question 9.**

**Solution:**

Let x be the third proportional, then

(i) 8 : 12 : : 12 : x

**Question 10.**

**Solution:**

Third proportional = 28 then

7 : x :: x : 28

⇒ 7 x 28 = x x x

⇒ x^{2} = 28 x 7 = 196

⇒ x = √196 = 14

x = 14

**Question 11.**

**Solution:**

Let x be the mean proportional, then

(i) 6 : x :: x : 24

⇒ x^{2} = 6 x 24 = 144

x = √144 = 12

Mean proportional = 12

(ii) 3 : x : : x : 27

⇒ x^{2} = 3 x 27 = 81

x = √81 = 9

Mean proportional = 9

(iii) 0.4 : x :: x : 0.9

⇒ x^{2} = 0.4 x 0.9

x = √o.36 = 0.6

Mean proportional = 0.6

**Question 12.**

**Solution:**

Let x be added to each of the given numbers then

5 + x, 9 + x, 7 + x, 12 + x are in proportion

\(\frac { 5 + x }{ 9 + x }\) = \(\frac { 7 + x }{ 12 + x }\)

By cross multiplication :

(5 + x) (12 + x) = (7 + x) (9 + x)

⇒ 60 + 5x + 12x + x^{2} = 63 + 7x + 9x + x^{2}

⇒ 60 + 17x + x^{2} = 63 + 16x + x^{2}

⇒ 17x + x^{2} – 16x – x^{2} = 63 – 60

⇒ x = 3

Required number = 3

**Question 13.**

**Solution:**

Let x be subtracted from each of the given number, then

10 – x, 12 – x, 19 – x and 24 – x are in proportion

\(\frac { 10 – x }{ 12 – x }\) = \(\frac { 19 – x }{ 24 – x }\)

By cross multiplication :

(10 – x) (24 – x) = (19 – x) (12 – x)

⇒ 240 – 10x – 24x + x^{2} = 228 – 19x – 12x + x^{2}

⇒ 240 – 34x + x^{2} = 228 – 31x + x^{2}

⇒ -34x + x^{2} + 31x – x^{2} = 228 – 240

⇒ -3x = -2

⇒ 3x = 12

⇒ x = 4

Required number = 4

**Question 14.**

**Solution:**

Scale of map = 1 : 5000000

Distance between two town on the map = 4 cm

**Question 15.**

**Solution:**

Height of a tree = 6 cm

and its shadow at same time = 8 m

Shadow of a pole = 20 m

Let height of pole = x m

6 : 8 = x : 20

⇒ x= \(\frac { 6 x 20 }{ 8 }\) = 15 m

Height of pole = 15 m

Hope given RS Aggarwal Solutions Class 7 Chapter 8 Ratio and Proportion Ex 8B are helpful to complete your math homework.

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