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

Exercise 5C

Page-79

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

OPEN IN YOUTUBE

Question 1:

Replace A, B, C by suitable numerals.
$$\begin{gathered} \begin{array}{ccc}& 5& A\\ +& 8& 7\end{array} \\ \overline{\underline{\begin{array}{ccc}C& B& 3\end{array}}} \end{gathered}$$

Answer 1:

 $$\begin{gathered} A=6 \\ \therefore A+7=6+7=13 \end{gathered}$$

1 is carried over.
$(1+5+8)=14$

1 is carried over.
∴ $B=4$
and $C=1$

∴ $A=6, B=4 and C=1$

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

OPEN IN YOUTUBE

Question 2:

Replace A, B, C by suitable numerals.
$$\begin{gathered} \begin{array}{ccccc}& 4& C& B& 6\\ +& 3& 6& 9& A\end{array} \\ \overline{\underline{\overline{\underline{\begin{array}{ccccc} & 8& 1& 7& 3 \end{array}}}}} \end{gathered}$$

Answer 2:

 $A=7, A+6=7+6=13$       (1 is carried over)

$(1+B+9)=17, or B =7$       (1 is carried over)

$A=7, B=7 and C=4$          (1 is carried over)

∴ $A=7, B=7 and C=4$

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

OPEN IN YOUTUBE

Question 3:

Replace A, B, by suitable numerals.
$$\begin{gathered} \begin{array}{cc}& A\\ +& A\\ +& A\end{array} \\ \overline{\underline{\begin{array}{cc}B& A \end{array}}} \end{gathered}$$

Answer 3:

$A+A+A=A$     (with 1 being carried over)
This is satisfied if $A$ is equal to $5$.

When $A=5$:

$A+A+A=15$           (1 is carried over)

Or $B=1$
∴ $A=5 and B=1$

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

OPEN IN YOUTUBE

Question 4:

Replace A, B by suitable numerals.
$$\begin{gathered} \begin{array}{ccc}& 6& A\\ -& A& B\end{array} \\ \overline{\underline{\begin{array}{ccc} & 3& 7\end{array} }} \end{gathered}$$

Answer 4:

First look at the left column, which is:
 $6-A=3$

This implies that the maximum value of A can be 3. 

$A\le 3$                 ... (1)

The next column has the following:
$A-B=7$

To reconcile this with equation (1), borrowing is involved.

We know:
$12-5=7$

∴ $A=2 and B=5$

Page-80

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

OPEN IN YOUTUBE

Question 5:

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

Answer 5:

$5-A=9$ 
This implies that 1 is borrowed.
We know:
$15-6=9$
∴ $A=6$

$B-5=8$ 
This implies that 1 is borrowed.
$13-5=8$
But 1 has also been lent
∴$B=4$

$C-2=2$ 
This implies that 1 has been lent.
∴ $C=5$

∴ $A=6, B=4 and C=5$

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

OPEN IN YOUTUBE

Question 6:

Replace A, B, C by suitable numerals.
$$\begin{gathered} \begin{array}{ccc}& A& B\\ & \times & 3\end{array} \\ \overline{\underline{\begin{array}{ccc}C& A& B\end{array}}} \end{gathered}$$

Answer 6:

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

OPEN IN YOUTUBE

Question 7:

Replace A, B, C by suitable numerals.
$$\begin{gathered} \begin{array}{ccc} & A& B\\ \times & B& A\end{array} \\ \overline{\underline{\begin{array}{ccc} (B+1)& C& B\end{array}}} \end{gathered}$$

Answer 7:

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



In the question:
First digit = B+1
Thus, 1 will be carried from 1+B² and becomes (B+1) (B² -9) B.
∴ C = B² -1
Now, all B, B+1 and B² -9  are one digit number.

This condition is satisfied for B=3 or B=4.

For B< 3, B² -9 will be negative.
For B>3, B² -9 will become a two digit number.
For B=3 , C = 3² - 9 = 9-9 = 0
For B = 4, C = 4² -9 = 16-9 = 7
 

Required answer:

A=1, B=3, C = 0

or

A=1, B=4, C = 7

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

OPEN IN YOUTUBE

Question 8:

Replace A, B, C by suitable numerals.
$$\begin{gathered} \begin{array}{cccccc}6)& 5& A& B& (9& C\\ -& 5& 4& & & \end{array} \\ \overline{\begin{array}{cccc}& & 3& B \\ & -& 3& 6 \end{array}} \\ \overline{\underline{\begin{array}{ccc}& \times & \end{array}}} \end{gathered}$$

Answer 8:

$(A-4) = 3\Rightarrow A=7$
​Also, $6\times 6=36 \Rightarrow C=6$
 $36-36=0 \Rightarrow B= 6$

∴ $$\begin{gathered} A=7 \\ B=C=6 \end{gathered}$$

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

OPEN IN YOUTUBE

Question 9:

Find two numbers whose product is a 1-digit number and the sum is a 2-digit number.

Answer 9:

$1 and 9$ are two numbers, whose product is a single digit number.
∴ $1\times 9=9$

Sum of the numbers is a two digit number. 
∴ $1+9=10$

Q10 | Ex-5C | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

OPEN IN YOUTUBE

Question 10:

Find three whole numbers whose product and sum are equal.

Answer 10:

The three whole numbers are 1, 2 and 3.

$1+2+3=6=1\times 2\times 3$

Q11 | Ex-5C | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

OPEN IN YOUTUBE

Question 11:

complete the magic square given below, so that the sum of the numbers in each row or in each column or along each diagonal is 15.

6	1
	5

Answer 11:

Taking the diagonal that starts with 6:
$6+5+x=15 \Rightarrow x=4$
 

6	1
	5
		4

Now, taking the first row:
$6+1+x=15 \Rightarrow x=8$
 

6	1	8
	5
		4

Taking the last column:
$8+x+4=15 \Rightarrow x=3$
 

6	1	8
	5	3
		4

Taking the second column:
$1+5+x=15 \Rightarrow x=9$
 

6	1	8
	5	3
	9	4

Taking the second row:
$x+5+3=15 \Rightarrow x=7$
 

6	1	8
7	5	3
	9	4

Taking the diagonal that begins with 8:
$8+5+x=15 \Rightarrow x=2$
 

6	1	8
7	5	3
2	9	4

Q12 | Ex-5C | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

OPEN IN YOUTUBE

Question 12:

Fill in the numbers from 1 to 6 without repetition, so that each side of the triangle adds up to 12.

Answer 12:

6+2+4 = 12
4+3+5 = 12
6+1+5 = 12

Q13 | Ex-5C | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

OPEN IN YOUTUBE

Question 13:

Fibonacci numbers Take 10 numbers as shown below:
a, b, (a + b), (a + 2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21a + 34b). Sum of all these numbers = 11(5a + 8b) = 11 × 7th number.
Taking a = 8, b = 13; write 10 Fibonacci numbers and verify that sum of all these numbers = 11 × 7th number.

Answer 13:

Given:
 $a=8 and b=13$
The numbers in the Fibonnaci sequence are arranged in the following manner:
1st, 2nd, (1st+2nd), (2nd+3th), (3th+4th), (4th+5th), (5th+6th), (6th+7th), (7th+8th), (8th+9th), (9th+10th)

The numbers are $8, 13, 21, 34, 55, 89, 144, 233, 377 \mathrm{and} 610$.
Sum of the numbers = ​$8+13+21+34+55+89+144+233+377+610$
                               $=1584$
$11\times 7th number = 11\times 144=1584$

Q14 | Ex-5C | Class 8 | RS AGGARWAL | Playing With Numbers | Chapter 5 | myhelper

OPEN IN YOUTUBE

Question 14:

Complete the magic square:

	14		0
8		6	11
4		.	7
	2	1	12

Answer 14:

The magic square is completed assuming that the sum of the row, columns and diagonals is 30. This is because the sum of all the number of the last column is 30.
 

3	14	13	0
8	5	6	11
4	9	10	7
15	2	1	12