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Q.

The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number.

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a

46

b

48

c

24

d

97 

answer is C.

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Detailed Solution

It is given that the sum of a two-digit number and the number obtained by interchanging the digits is 66 and the digits of the number differ by 2.
Let the unit’s digit and the ten’s digit in the two-digit number be y and x.
Thus, the original number will be, 10y + x.
The number after reversing will be 10x + y.
According to question, we have,
⇒10y+x+10x+y=66 ⇒11y+11x=66 ⇒11(x+y)=66 ⇒x+y= 66 11 ⇒x+y=6   ...(i)  
And
x−y=2   ...(ii)  
Adding equation (i) and (ii), we have,
⇒x+y+x−y=6+2 ⇒2x+0=8 ⇒2x=8 ⇒x= 8 2 ⇒x=4  
Now substituting the value of x in equation (ii), we have,
⇒4−y=2 ⇒y=4−2 ⇒y=2  
Hence, the number is:
⇒Required number=10y+x ⇒Required number=10(2)+4 ⇒Required number=24  
Hence, the number is 24.
Therefore, option 3 is correct.
 
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