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.
$x=1$
$\Rightarrow x+11 = 12$ 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.

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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.
∴ $15+x=18 \Rightarrow x=3$

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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:
$14+y=18 \Rightarrow y=4$

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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:
$11+(x+y)=12 \Rightarrow (x+y) = 1$

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Question 6:

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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

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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 $(x+y+z) \mathrm{is} \mathrm{equal} \mathrm{to} 0$ 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

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Question 8:

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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 $(A+B) =1$.

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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 $(x+y) \mathrm{is} 0$.
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