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| Question of the Day (QOTD) - Nov 26, 2008 http://iim-cat-forum.2iim.com/iim-cat-math-quant-questions/question-of-the-day-qotd-nov-26-2008-t96.html |
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| Author: | 2iim [ Fri Mar 06, 2009 3:39 pm ] |
| Post subject: | Remainders and divisors |
I think a similar question actually appeared in one of the CAT papers in early 2000. Correct Answer is Choice (4). 37 is the divisor. |
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| Author: | ram [ Tue Jul 28, 2009 7:03 am ] |
| Post subject: | Re: Question of the Day (QOTD) - Nov 26, 2008 |
Can someone post a detailed explanation to the above question? |
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| Author: | 2iim [ Tue Jul 28, 2009 3:30 pm ] |
| Post subject: | Re: Question of the Day (QOTD) - Nov 26, 2008 |
Let the number be 'x' and the divisor be 'd'. So, when x is divided by d, the remainder is 24. Let the quotient of this division be 'p'. We can therefore, write x = pd + 24. When twice the number i.e., 2x is divided by 'd' the remainder is 11. 2x = 2(pd + 24) = 2pd + 48. So, when 2x is divided by 'd' or when 2pd + 48 is divided by 'd' we know the remainder is 11. 2pd is divisible by 'd' and hence there will not be any remainder. So, the remainder of '11' was obtained when 48 was divided by 'd'. So, subtracting 11 from 48 will make the resultant value divisible by the divisor. i.e., 48 - 11 is divisible by 'd'. i.e., 37 is divisible by 'd'. As 37 is a prime number, the only value that 'd' can take is 37. |
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