RS Aggarwal solution class 8 chapter 5 Playing With Numbers Test Paper 5

Test Paper 5

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Q1 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 1:

Find all possible values of x for which the 4-digit number 320x is divisible by 3. Also, find the numbers.

Answer 1:

If a number is divisible by 3, then the sum of the digits is also divisible by 3.

$3+2+0+x=5+x$ must be divisible by 3.
This is possible in the following cases:
$$\begin{gathered} i) x=1 \\ \therefore 5+x=6 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 3201. \end{gathered}$$

$$\begin{gathered} ii) x=4 \\ \therefore 5+x=9 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 3204. \end{gathered}$$

$$\begin{gathered} iii) x=7 \\ \therefore 5+x=12 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 3207. \end{gathered}$$

Q2 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 2:

Find all possible values of y for which the 4-digit number 64y3 is divisible by 9. Also, find the numbers.

Answer 2:

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

$6+4+y+3=13+y$

For this to be divisible by 9:
$y=5$

The number will be 6453.

Q3 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 3:

The sum of the digits of a 2-digit number is 6. The number obtained by interchanging its digits is 18 more than the original number. Find the original number.

Answer 3:

Let the two numbers of the two-digit number be 'a' and 'b'.

$a+b=6$              ... (1)

The number can be written as $(10a+b)$.
After interchanging the digits, the number becomes $(10b+a)$.

$$\begin{gathered} (10a+b)+18=(10b+a) \\ 9a-9b=-18 \end{gathered}$$
$a-b=-2$            ... (2)

Adding equations (1) and (2):
$2a=4\Rightarrow a =2$

Using $a=2$ in equation (1):
$b=6-a=6-2=4$

Therefore, the original number is 24.

Q4 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 4:

Which of the following numbers are divisible by 9?
(i) 524618
(ii) 7345845
(iii) 8987148

Answer 4:

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

Number	Sum of the digits	Divisible by 9?
524618	26	No
7345845	36	Yes
8987148	45	Yes

 

Q5 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 5:

Replace A, B, C by suitable numerals:
$$\begin{gathered} \begin{array}{cccc}& 5& 7& A\\ -& C& B& 8\end{array} \\ \overline{\underline{\begin{array}{cccc} & 2& 9 & 3\end{array}}} \end{gathered}$$

Answer 5:

$A-8=3$
​This implies that 1 is borrowed.
$$\begin{gathered} 11-8=3 \\ \Rightarrow A = 1 \end{gathered}$$

Then, $7-B=9$
1 is borrowed from 7.
∴ $$\begin{gathered} 16-B=9 \\ \Rightarrow B = 7 \end{gathered}$$

Further, $5-C=2$
But 1 has been borrowed from 5.
∴ 4 - C = 2
$\Rightarrow C=2$

∴ $A=1, B=7 and C=2$

Q6 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 6:

Replace A, B, C by suitable numerals:

$$\begin{gathered} \begin{array}{cccccc}7)& 6& A& B& (8& C\\ -& 5& 6& & & \end{array} \\ \overline{\underline{\begin{array}{ccc}& 6& B \\ -& 6& 3 \end{array}}} \\ \times \end{gathered}$$

Answer 6:

Here, $$\begin{gathered} A-6 = 6 \\ \Rightarrow A =2 (\mathrm{with} 1 \mathrm{being} \mathrm{borrowed}) \end{gathered}$$
$B=3$
Since $7\times 9=63$, $C=9$
∴ $A=2, B=3 and C=9$

Q7 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 7:

Find the values of A, B, C when
$$\begin{gathered} \begin{array}{ccc}& A& B\\ \times & B& A\end{array} \\ \overline{\begin{array}{ccc} B& C& B\end{array}} \end{gathered}$$

Answer 7:

$A\times B=B\Rightarrow A=1$

$$\begin{gathered} \begin{array}{ccc}& 1& B\\ \times & B& 1\end{array} \\ \begin{array}{ccc}& 1& B\\ B& B^2& \times \end{array} \\ \begin{array}{ccc}B& (1+B^2)& B\end{array} \end{gathered}$$

Now, $B\ne A=1$ and $(1+B^2)$ is a single digit number.
∴ $B=2$
$C=(1+B^2) = (1+4) =5$
∴ $A=1, B=2 and C=5$

Q8 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 8:

Mark (✓) against the correct answer
If 7x8 is exactly divisible by 3, then the least value of x is
(a) 3
(b) 0
(c) 6
(d) 9  

Answer 8:

(b) 0
If a number is exactly divisible by 3, the sum of its digits is also divisible by 3.

$7+x+8=15+x$

$15+x$ can be divisible by 3 even if x is equal to 0.

Q9 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 9:

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If 6x5 is exactly divisible by 9, then the least value of x is
(a) 1
(b) 4
(c) 7
(d) 0

Answer 9:

(c) 7

When a number is divisible by 9, the sum of the digits is also divisible by 9.

$6+x+5=11+x$

To be divisible by 9:
$11+x=18\Rightarrow x=7$

Q10 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 10:

Mark (✓) against the correct answer
If x48y is exactly divisible by 9, then the least value of (x + y) is
(a) 4
(b) 0
(c) 6
(d) 7

Answer 10:

(c) 6

When a number is divisible by 9, the sum of its digits is also divisible by 9.
$x+4+8+y=12+(x+y)$

For $12+(x+y)$ to be divisible by 9:
$12+(x+y)=18 \Rightarrow (x+y)=6$

Q11 Test Paper 5 Class 8 RS AGGARWAL chapter 5 Playing With Numbers

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Question 11:

If 486*7 is divisible by 9, then the least value of * is
(a) 0
(b) 1
(c) 3
(d) 2

Answer 11:

(d) 2

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

$4+8+6+*+7=25+*$

Now, $25+*=27 (\mathrm{if} *=2$ and 27 is divisible by 9)