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Quant Problems

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
Expert replies
by sukh » Sun Jul 17, 2011 8:25 am
1.After taking N tests each containing 100 questions john had an average of 70 % correct answers . how much john needs to score on next test to make his average to 72 %
2.15 chess players take part in a tournament if every player plays twice with each of his opponents how many games will be played
3.For every dollar jim earns jaon earns 1.5 $ and for every 2$ jaon earns sam earns 1.6$ . if sam earned $ 300 how much did jim earn
4. How many integers are there between 11 and 55 inclusive such that is divisible by both 2 and 7?
5. What is the largest value of 98 ^ 98 such that is divisible by 7 ^ N ?
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Source: — Quantitative Reasoning |

by Tani » Sun Jul 17, 2011 1:22 pm
1. So far John has earned .70 * 100 * N or 70N points.
With the next exam he will have (N+1) exams and needs to average .72*100 points. Therefore he needs a total of 72(N+1) points.
His score on the last test needs to be the difference between those two figures:
72(N+1) - 70N = 2N + 72

Another way to look at it is that John is down 2 points for every test he has taken so far. (Net down = 2N). To reach a 72 average he has to score 72 on the last test PLUS make up for the points missed so far so: 72 + 2N.

2. Each of 15 players plays the other 14 players twice. That might look like 15*14*2, BUT - you need to divide by 2 since a game with players A and B is the same as a game with players B and A. So (15*14*2)/2 = 210

3. Sam earns $300. That represents $1.60 for every $2 Jason earned. So (Jason/2)*1.6 = Sam. That means .8*Jason = $300. So Jason earned $375. But Jason earned $1.5 for every dollar Jim earned so we have to divide Jason's earnings by 1.5. We find that Jim earned $250.

4. To be divisible by 2 and 7, an integer must be a multiple of 14. Between 11 and 55 we have 14, 28, and 42. Total = 3

5. This is unclear. I assume you are looking for the largest value of N such that 98*98 is divisible by 7^N. If so, you attack it this way. 98 = 7*7*2. Therefore 98*98 = (7*7*2)*(7*7*2) = (7^4)*(2^2). Answer = 4.
Tani Wolff
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by rohangupta83 » Mon Jul 18, 2011 6:04 am
About part 5

I think the question meant 98^98 and not 98*98 :).

I'd say factors of 98 are 7x7x2

therefore, we have (7x7x2)^(98)

or 7^98 x 7^98 x 2^98

or 7^(2*98) x 2^98

or 7^196 x 2^98

so 7^196 would be the largest number and 196 will be the largest value of N which can divide 98^98.

I hope I interpreted the question right.
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by Tani » Mon Jul 18, 2011 10:04 am
If the question is in fact 98^98 you are correct. :-)
Tani Wolff
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