## RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A.

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A
- RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B
- RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C
- RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

**Question 1.**

**Solution:**

(i) 24 to 56

= \(\\ \frac { 24 }{ 56 } \)

= \(\frac { 24\div 8 }{ 56\div 8 } \)

**Question 2.**

**Solution:**

(i) 36 : 90

= \(\\ \frac { 36 }{ 90 } \)

= \(\frac { 36\div 18 }{ 90\div 18 } \)

**Question 3.**

**Solution:**

(i) The given ratio = Rs. 6.30 : Rs. 16.80

= \(\\ \frac { Rs.6.30 }{ Rs.16.80 } \)

= \(\\ \frac { 630 }{ 1680 } \)

**Question 4.**

**Solution:**

Earning of Sahai = Rs. 16800

and of his wife = Rs. 10500

Then total income = Rs. 16800 + 10500

= Rs. 27300

**Question 5.**

**Solution:**

Rohit monthly earnings = Rs. 15300

and his savings = Rs. 1224

So, his expenditure = Rs. 15300 – 1224

= Rs. 14076

Now,

**Question 6.**

**Solution:**

Let the number of male and female workers in the mill be 5x and 3x respectively. Then,

5x = 115

=> \(\\ \frac { 5x }{ 5 } \) = \(\\ \frac { 115 }{ 5 } \)

(Dividing both sides by 5)

=> x = 23

Number of female workers in the mill

= 3x

= 3 x 23 = 69.

**Question 7.**

**Solution:**

Let the number of boys and girls in the school be 9x and 5x respectively.

According to the question,

9x + 5x = 44

=> 14x = 448

=> \(\\ \frac { 14x }{ 14 } \) = \(\\ \frac { 448 }{ 14 } \)

(Dividing both sides by 14)

=> x = 32.

Number of girls =5x

= 5 x 32

= 160

**Question 8.**

**Solution:**

Total amount = Rs. 1575

Ratio in Kamal and Madhu’s share = 7 : 2

Sum of ratios = 7 + 2 = 9

**Question 9.**

**Solution:**

Total amount = Rs. 3450

Ratio in A, B and C shares = 3 : 5 : 7

Sum of share = 3 + 5 + 7 = 15

**Question 10.**

**Solution:**

Let the numbers be 11x and 12x.

Then. 11x + 12x = 460

=> 23x = 460

**Question 11.**

**Solution:**

Length of line segment = 35 cm

Ratio = 4 : 3

Sum of ratio = 4 + 3 = 7

**Question 12.**

**Solution:**

Total bulbs produced per day = 630

Out of every 10 bulbs, defective bulb = 1

Out of every 10 bulbs, lighting bulbs = 10 – 1 = 9

**Question 13.**

**Solution:**

Price of 20 pencils = Rs. 96

(1 score = 20 pencils)

Price of 1 pencil = Rs. (96 ÷ 20)

= Rs. 4.80

Price of 12 ball pens = Rs. 50.40

(1 dozen = 12)

Price of 1 ball pen = Rs. (50.40 ÷ 12)

= Rs. 4.20.

Ratio of the price of a pencil to that of a ball pen = Rs. 4.80 : Rs. 4.20

= 480 paise : 420 paise

= 480 : 420

= 48 : 42

= 8 : 7.

Required ratio = 8 : 7.

**Question 14.**

**Solution:**

It is given that the ratio of the length of a field to its width is 5 : 3.

If the width of the field is 3 metres then length = 5 metres.

If the width of the field is 1 metres than length = \(\\ \frac { 5 }{ 3 } \) metres.

If the width of the field is 42 metres then length

= \(\\ \frac { 5 }{ 3 } \) x 42 metres

= 5 x 14 metres

= 70 metres.

**Question 15.**

**Solution:**

Ratio in income and savings of a family = 11 : 2

But Total savings = Rs. 1520

Let income = x

11 : 2 = x : 1520

=> x = \(\\ \frac { 11\times 1520 }{ 2 } \) = 11 x 760

= Rs 8360

Expenditure = total income – savings

= Rs 8360 – 1520

= Rs 6840

**Question 16.**

**Solution:**

Ratio in income and expenditure = 7 : 6

Total income = Rs. 14000

Let expenditure = x, then

7 : 6 :: 14000 : x

=>x = \(\\ \frac { 6\times 14000 }{ 7 } \) = Rs. 12000

Savings = Total income – Expenditure

= Rs. 14000 – 12000

= Rs. 2000

**Question 17.**

**Solution:**

It is given that the ratio of zinc and copper in an alloy is 7 : 9.

If the weight of zinc in the alloy is 7 kg then the weight of copper in the alloy is 9 kg.

**Question 18.**

**Solution:**

A bus covers in 2 hours = 128 km

128 It will cover in 1 hour = \(\\ \frac { 128 }{ 2 } \) = 64 km

A train cover in 3 hours = 240 km

It will cover in 1 hour = \(\\ \frac { 240 }{ 3 } \)

= 80 km

Ratio in their speeds = 64: 80

= 4 : 5

{Dividing by 16, the LCM of 64, 80}

**Question 19.**

**Solution:**

(i) (3 : 4) or (9 : 16)

LCM of 4, 16 = 16

**Question 20.**

**Solution:**

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A are helpful to complete your math homework.

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