catxprt said – Tue, 03 Nov 2009 16:14:05 -0000 ( Link )
Hey CAT Aspirants
Here’s a question which might be helpful for CAT preparation and question like this is likely to be in the upcoming exam.
If the integers a and b are chosen at random from integers from integers 1 to 100 with replacement, then the probability that a number of the form is divisible by 5 equals?
1. 1 / 2
2. 1 / 4
3. 1 / 8
4. 1 / 16
5. None of these
Be on lookout for more such important questions in my future discussions. Post your answers and we will discuss the solution in coming days.
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catxprt said – Thu, 05 Nov 2009 04:52:56 -0000 ( Flag Edit Link )
Correct answer to this question is 1 / 8
shrii said – Fri, 06 Nov 2009 06:09:30 -0000 ( Flag Edit Link )
Hi jstplayin,
can u explain how 7a+7b became s2+2b?
kimkimy said – Sun, 08 Nov 2009 18:21:19 -0000 ( Flag Edit Link )
i didnt get it at al….......
shaunakc said – Tue, 10 Nov 2009 12:58:09 -0000 ( Flag Edit Link )
the answer is 1/8. There are just four types of nos – 4k+1, 4k+2, 4k+3, 4k. now if a=4k+1 type of no the remainder is 2. similarly for 4k+2 rem is 4; 4k+3 rem is 3 & for 4k rem is 1. If you take a as 4k+1 type then necessarily b has to be 4k+3 type of no. Similar explanation for 4k+2 & 4k type of nos. Now required probability is 2525/(100100)+ 2525/(100100) = 1/16+1/16 = 1/8