ML AGGARWAL CLASS 8 CHAPTER 5 Playing with Numbers Exercise 5.2

Exercise 5.2


Question 1 

Sol :

4  A 
3  5
--------
B 2
--------

A+5= we get a number whose unit's

digit in 2, A+5 should not exceed 14

$\begin{aligned} \therefore A+5 &=12 \\ A &=12-5 \\ A &=7 \end{aligned}$

4  7
3   5
-------
8   2
-------

B = 8

∴ A = 7 , B = 8 

Question 2

5     A 
7     9
---------
CB  3
---------

A+9 Should Not exceed 18 . and for A+9= a number

whose unit's place is 3 this:s possible when A=44







∴ A = 4 , C = 1 , B = 3


Question 3

Sol :
4  2  A 
2  A  5 
---------
A  0  2 
---------

A+5 sum should not exceed 14 and it is a

number whose Unit's digit is 2 This is possible

when A=7 








A = 7 

satisfies given condition 

 A = 7 


Question 4

Sol :
     A  A
+   A  A
-----------
   BA  8 

case(i)

 $\begin{aligned} A+A &=8 \\ 2 A &=8 \\ A &=4 \end{aligned}$

check 

        4  4 

        4  4 
-------------
         8  8 

but given the sum 

 as there digit number 

case (ii)

A + A 18 

2A = 18 

A = 9 

check :   9  9 

              9  9 
         ---------------
               198
         ----------------

      A = 9 ,  B = 1 

      A = 9 ,   B = 1 



Question 5

Sol :

  1  8  A 
+B  A  7
-----------
   C B   2 
-----------

A + 7 sum should not exceed 16 and it is a 
 
number whose unit, s digit is 2 

 ∴ A + 7 = 12 

A =  12 - 7 
 
A =  5

$1+8+A \Rightarrow 1+8+5=13$

1+B=C

C=1+4

C=5

 ∴ A = 5 , B = 4 , C = 5 


Question 6

Sol :
 A  2  1  B 
+1  C  A  B 
--------------
   B 4   9   6 
 
B +  B  sum should not exceed 18 and its units digits is 6 

case (i) 

B + B = 6 

2B = 6 

B = 3 

 Then 1 + A = 9 

A = 8 

A + 1 = B 

8 + 1 = B 

B = 9 but we got B = 3 earlier 

so B ≠ 3


case (ii) 

B+B=16

2 B=16

$B=\frac{16}{2}$

B=8

7   2  1   8 
1   2  7   8 
--------------
 8  4   9   6 

A + 1 = B 

A   + 1 = 8 

A  = 8 - 1 

 A = 7      A = 7 

 ஃ A = 7  B = 8  C = 2 



Question 7

Sol :

B  3  4  5
C  9  B  A 
--------------
8   B  A  2 
--------------

 Sum 5 + A = 12 

A = 12 - 5 

A  = 7

1 + 4  + B = A 

1 +  4  + B = 7 

B = 7 - 5

B = 2 

$\begin{aligned} B+C &=8 \\ 2+C &=8 \\ C &=8-2 \end{aligned}$

c=6 

$A=7, \quad B=2, C=6$



Question 8

Sol :

$\begin{array}{r}A B \\(-) B 6 \\\hline 4 7 \\\hline\end{array}$

$\begin{aligned} B-6 &=7 \\ B &=7+6 \\ B &=13 \end{aligned}$

means  B = 3 , because B should be single digit 

$\begin{aligned} A-B-1 &=4 \\ A-3-1 &=4 \\ A-4 &=4 \\ A &=8 \\ \therefore \quad A=8, \quad B &=3 \end{aligned}$


Question 9

Sol :

$\begin{array}{r}2 A \\\times 3 A \\\hline B 7 A\end{array}$

A x A = A means A should be 1 or 5 or 0 

case (i) A = 0 

$\begin{array}{r}20 \\30 \\\hline 600 \\\hline\end{array}$ 
 ≠ B7A 

∴ This is not given number so A  ≠ 0

case (ii) A = 1 

$\begin{array}{l}21 \\31 \\\hline 21 \\\hline 23 \\\hline 251 \\\hline\end{array}$
 ≠ B71

∴ B ≠ 1
 
case (iii)

$\begin{array}{l}25 \\35 \\\hline 125 \\75 \\\hline 875\end{array}$

∴ B = 8 

$\therefore \quad A=5,  B=8

Question 10

Sol :

$\begin{array}{r}A B \\A B \\\hline 6 A B\end{array}$

B should either 0 , 1 or  5 

case (i) B = 0 

$\begin{array}{r}A 0 \\B 0 \\\hline 00 \\A B O \\\hline A B B \quad 0 \end{array}$
≠ 6AB

∴ A ≠ 0

case (ii) B = 1


 






case (iii)

B = 5 

A 5 
A 5 
------
X25   =  6AB   (∵ may be any number)

A = 2 (∵from rules by square number)


Question 11

Sol :

$\begin{array}{l}A A \\4 A \\\hline 9 A 4\end{array}$

A = 2

A = 2 is correct 


Question 12

Sol :











Question 13

Sol :











Question 14

Sol :


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