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

Exercise 5D

Q1 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 1:

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If 5x6 is exactly divisible by 3, then the least value of x is
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1:

(b) 1

If a number is exactly divisible by 3, the sum of the digits must also be divisible by 3.
5+x+6=11+x must be divisible by 3.
The smallest value of x is 1.
x=1 
 x+11 = 12 is divisible by 3.


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Q2 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 2:

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If 64y8 is exactly divisible by 3, then the least value of y is
(a) 0
(b) 1
(c) 2
(d) 3

Answer 2:

(a) 0

If a number is divisible by 3, then the sum of the digits is also divisible by 3.
6+4+y+8=18+y
This is divisible by 3 as y is equal to 0.


Q3 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 3:

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If 7x8 is exactly divisible by 9, then the least value of x is
(a) 0
(b) 2
(c) 3
(d) 5

Answer 3:

(c) 3

If a number is exactly divisible by 9, the sum of the digits must also be divisible by 9.
7+x+8=15+x
18 is divisible by 9.
15+x=18  x=3


Q4 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 4:

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If 37y4 is exactly divisible by 9, then the least value of y is
(a) 2
(b) 3
(c) 1
(d) 4

Answer 4:

(d) 4

A number is divisible by 9 if the sum of the digits is divisible by 9.

3+7+y+4=14+y
For this sum to be divisible by 9:
14+y=18  y=4


Q5 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 5:

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If 4xy7 is exactly divisible by 3, then the least value of (x + y) is
(a) 1
(b) 4
(c) 5
(d) 7

Answer 5:

(a) 1
If a number is divisible by 3, the sum of the digits is also divisible by 3.
4+x+y+7=11+(x+y)
For the sum to be divisible by 3:
11+(x+y)=12  (x+y) = 1

Q6 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 6:

Tick (✓) the correct answer
If x7y5 is exactly divisible by 3, then the least value of (x + y) is
(a) 6
(b) 0
(c) 4
(d) 3

Answer 6:

(d) 3

When a number is divisible by 3, the sum of the digits must also be divisible by 3.
x+7+y+5=(x+y)+12
This sum is divisible by 3 if  x+y+12 is 12 or 15.
For x+y+12 = 12:
x+y=0
But x+y cannot be 0 because then x and y both will have to be 0.
Since x is the first digit, it cannot be 0.
∴ x+y+12 = 15
or x+y = 15-12=3


Q7 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 7:

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If x4y5z is exactly divisible by 9, then the least value of (x + y + z) is
(a) 3
(b) 6
(c) 9
(d) 0

Answer 7:

(c) 9

A number is divisible by 9 if the sum of the digits is divisible by 9.

x+4+y+5+z=9+(x+y+z)
The lowest value of (x+y+z) is equal to 0 for the number x4y5z to be divisible by 9.
In this case, all x, y and z will be 0.
But x is the first digit, so it cannot be 0.
∴ x+4+y+5+z = 18
or x+y+z+9 = 18
or x+y+z = 9


Q8 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 8:

Tick (✓) the correct answer
If 1A2B5 is exactly divisible by 9, then the least value of (A + B) is
(a) 0
(b) 1
(c) 2
(d) 10

Answer 8:

(b) 1

For a number to be divisible by 9, the sum of the digits must also be divisible by 9.

1+A+2+B+5=(A+B)+8

The number will be divisible by 9 if (A+B) =1.


Q9 | Ex-5D | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

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Question 9:

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If the 4-digit number x27y is exactly divisible by 9, then the least value of (x + y) is
(a) 0
(b) 3
(c) 6
(d) 9

Answer 9:

(d) 9

If a number is divisible by 9, then the sum of the digits is divisible by 9.

x+2+7+y=(x+y)+9
For this to be divisible by 9, the least value of (x+y) is 0.
But for x+y = 0, x and y both will be zero.
Since x is the first digit, it can never be 0.
∴ x + y + 9 = 18
or  x + y = 9

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