Latest TCS Aptitude Question SOLUTION: A 2 digit number is greater than 4 then 8 times the sum of its digits. If 7times the 10's place the digit subtracted from the given number the new number to be obtained is that by

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(#M40008383) TCS question Two Digit Problem Keep an EYE

A 2 digit number is greater than 4 then 8 times the sum of its digits. If 7times the 10's place the digit subtracted from the given number the new number to be obtained is that by interchanging the digits of the original number.

Asked In TCS K Suryakanth (12 years ago)

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Rakesh Kumar Kohli  (12 years ago)

Ans. Original number is 92

Solution:
Let digit at 10's place is x and at unit's place is y in original number,
Then original number = 10x + y
The number by interchanging the digits of the original number = 10y + x
As per given conditions
10x + y = 8(x + y)+ 4
=> 2x = 7y + 4 (i)
and 10x + y - 7x = 10y + x
=> 2x = 9y (ii)
Solving equation (i)and(ii)
We get y = 2 and x = 9
So original number is 92 and number obtained by interchanging the digits of the original number is 29.

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Vadivelan M  (12 years ago)

92 is the number.

Explanation:
Sum of digits = 9+2 = 11
8 times sum of digits = 8*11 = 88
2 digit number greater than 4 than 8 times sum of digits = 88 + 4 = 92

7 times 10's place = 7*9 = 63
above subtracted from given number = 92 - 63 = 29
new number is 29 which is got interchanging the digits of original number.

Is this solution Helpfull? Yes (7) | No (4)  

Naunihal Maurya  (12 years ago)

Let digit at 10's place is X, and digit at unit place is Y. Then number is 10X+Y.
Now according to first condition
8*(X+Y)+4=10X+Y that is
2X-7Y=4 ...(1)
and according to second condition
10X+Y-7X= 10Y-X that is
2X-9Y=0 ...(2)
After solving equation (1) and (2), we get X=9, Y=2, then original
number is (10X+Y)=92.

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