RS Aggarwal solution class 8 chapter 5 Playing With Numbers Exercise 5A

Exercise 5A

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Q1| Ex-5A | CLASS 8 | RS AGGARWAL | chapter 5 | Playing With Numbers

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Question 1:

The units digit of a two-digit number is 3 and seven times the sum of the digits is the number itself. Find the number.

Answer 1:

Let the tens place digit be x.

The units place digit is 3.

∴ Number = (10x  + 3)              ... (1)
 
Given:
7( x + 3) = (10 x + 3)

7 x + 21 = 10 x + 3

∴ 10 x - 7x = 21 - 3

⇒ 3 x  = 18

or x = 6

Using x = 6 in equation (1):

The number is 63.

Q2| Ex-5A | CLASS 8 | RS AGGARWAL | chapter 5 | Playing With Numbers

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Question 2:

In a two-digit number, the digit at the units place is double the digit in the tens place. The number exceeds the sum of its digits by 18. Find the number.

Answer 2:

Let the tens digit be x.

The digit in the units place is 2x.

Number = 10x + 2x

Given:
(x + 2x) + 18 = (10x + 2x)

∴ 3x + 18 = 12x

12x - 3x = 18

9x =18

x = $\frac{18}{2}$ =9

The digit in the tens place is 9

The digit in the units place is twice the digit in the tens place.
The digit in the units place is 4.

Therefore, the number is 94.

Q3| Ex-5A | CLASS 8 | RS AGGARWAL | chapter 5 | Playing With Numbers

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Question 3:

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

Answer 3:

Let the tens place digit be a and the units place digit be b.
Then, number is (10a + b).

According to the question:

4(a + b) + 3 = (10 a + b) 
4a + 4b + 3 = 10a + b
6a - 3b = 3
3(2a - b) = 3
2a - b =1            ... (1)

Given:
If 18 is added to the number, its digits are reversed.
The reverse of the number is (10b + a).

∴ (10a + b) + 18 = 10b + a
10a - a + b -10b = -18
9a - 9b = -18
9(a - b) = -18
a - b = -2            ... (2)

Subtracting equation (2) from equation (1):

 2a  - b   =  1
  a   - b   = -2
-     +        +     
  a            = 3


Using a = 3 in equation (1):

2(3) - b = 1
6 - b = 1
∴ b = 5

Number = 10a+b = 10 $\times$ 3 + 5 = 35

Q4| Ex-5A | CLASS 8 | RS AGGARWAL | chapter 5 | Playing With Numbers

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Question 4:

The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.

Answer 4:

Let the tens place digit be a and the units place digit be b.
Then, the number is (10a + b).

​Given:
a + b = 15         ... (1)

When the digits are interchanged the number will be (10 b + a).
Given:
10a + b + 9 = 10 b + a
∴ 10a - a + b - 10b = -9
9a - 9b = -9​
a - b = -1            ... (2)

Adding equations (1) and (2):

a + b = 15
a - b  = -1 
2a     = 14
 ∴ a = 7

Using a = 7 in equation (2):
7 - b = -1
∴ b = 8

Original number = 10a+b = 10 $\times$ 7 + 8 = 78

Q5| Ex-5A | CLASS 8 | RS AGGARWAL | chapter 5 | Playing With Numbers

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Question 5:

The difference between a 2-digit number and the number obtained by interchanging its digits is 63. What is the difference between the digits of the number?

Answer 5:

Let the tens place digit be 'x' and the units place digit be 'y'.
∴ Number =  (10x + y)

Number obtained by interchanging the digits = (10y + x)

Given: (10x + y) - (10y + x) = 63

∴ 10x - x + y - 10 y = 63
9x - 9y = 63
9(x - y) = 63
x - y = 7

Therefore, the difference between the digits of the number is 7.

Q6| Ex-5A | CLASS 8 | RS AGGARWAL | chapter 5 | Playing With Numbers

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Question 6:

In a 3-digit number, the tens digit is thrice the units digit and the hundreds digit is four times the units digit. Also, the sum of its digits is 16. Find the number.

Answer 6:

Let the units place digit be x.
Then, the tens place digit will be 3x and the hundreds place digit will be 4x.

Given:
4x + 3x + x = 16
or 8x = 16
or x =2
Units place digit = 2
Tens place digit =  3 $\times$ 2 = 6
Hundreds place digit =  4 $\times$ 2 = 8

Therefore, the number is 862.