May 16, 2010

Quantitative Ability Question of the Day

Question:-

The number 3190623X1Y6, where X and Y are odd digits, is divisible by 88. It is also divisible by which of the following?

OPTIONS
1)16
2)256
3)792
4)None of these

Solution

Since the given number is divisible by 88, it has to be divisible by both 8 and 11.
For the number to be divisible by 8, its last three digits must be a multiple of 8.
This will happen if the last three digits of the given number are 136 or 176.
So Y = 3 or 7
For the number to be divisible by 11, the difference of the sum of the digits in even and odd places should be either 0 or a multiple of 11.
Sum of the digits in odd places = 3 + 9 + 6 + 3 + 1 + 6 = 28.
Sum of digits in even places = 1 + 0 + 2 + X + Y = 3 + X + Y.
The difference in the sum of digits in even and odd places= 25 – X – Y.
If (25 – X – Y) = 0, X + Y = 25. But this is not possible because both X and Y are single digit numbers.
If (25 – X – Y) = 11, X + Y = 14. In this case if Y = 3, X = 11. So it is not possible. If Y = 7, then X = 7. This is possible.
If (25 – X – Y) = 22, X + Y = 3. Only possible case is when Y = 3 and X = 0. But since X and Y are both odd digits, it is not possible.
Thus the only possible solution is X = Y = 7. So the number is 31906237176.
Since the last four digits of the number are not divisible by 16, the number is not divisible by 16 and hence not by 256.
The number is divisible by 9 because the sum of its digit = 45. Since 88 × 9 = 792 and we already know that the number is divisible by 88, the number must be divisible by 792 as well.
Hence, option 3.


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