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Q.

The sum of digits of two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of digits of the number. Find the number.

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a

24

b

18

c

42 

d

83

answer is B.

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Detailed Solution

It is given that in a two-digit number, the sum of two-digit number is 9. Nine times the number is twice the number obtained by reversing the order of digits of the number.
Let, the ten’s place digit be x  , and the one’s place digit be y  .
Then, the number is 10x+y  .
According to the first condition, the sum of two-digit number is 9.
So,
x+y=9  
According to the second condition, nine times the number is twice the number obtained by reversing the order of digits of the number.
⇒9(10x+y)=2(10y+x) ⇒90x+9y=20y+2x ⇒88x=11y  
Multiplying the equation x+y=9   by 11 on both the sides.
So, we get,
11x+11y=99  
Adding the equation 88x−11y=0  , and 11x+11y=99  .
So,
⇒88x+11x-11y+11y=0+99
⇒99x=99
⇒x=1
Substituting x=1 in the equation x+y=9  .
So, we get,
⇒x+y=9 ⇒1+y=9 ⇒y=8  
Substituting x=1 and y=8 in the number 10x+y  .
Number =10x+y  
=10 1 +8 =10+8 =18  
The two-digit number is 18.
Hence, the correct option is 2.
 
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