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

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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