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Steve took 5 history exams during a semester. Each score was

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Steve took 5 history exams during a semester. Each score was [#permalink] New post 08 Nov 2004, 06:43
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Steve took 5 history exams during a semester. Each score was one of the integers 1 through 100 inclusive. What was Steve's median score for these 5 exams?

1- Steve's scores on the first 4 exams were 76, 87, 73 and 96, respectively, and Steve's average (arithmetic mean) score on the last 3 exams was 89
2- Steve's score on the first 3 exams were 76, 87, and 73, respectively, and Steve's average (arithmetic mean) on all 5 exams was 86
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 [#permalink] New post 08 Nov 2004, 07:10
I think the answer is D

Statement 1: Gives you the first four scores and the average scores for the last three exams. So, using the average the last score can be determined. Once we have all the five scores, we can tell the median is 87.

Statement 2: Gives the first three scores and the average for all the five scores. So we have two unknowns the last two exam scores.

Using the average, we can see that 76+87+73+x+y = 86*5
=> x+y = 194.

But the scores have a range of 1 to 100 inclusive.

Assuming x =100, y = 94, so you can tell he did not get any lower than 94 in both the exams.

ie, the median is 87.
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 [#permalink] New post 08 Nov 2004, 08:39
well...what if x is 100 and y is 1....median will change.....!!!!
I guess it should be A
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 [#permalink] New post 08 Nov 2004, 09:56
Key info is # of scores is an odd number.
Statement 1 is sufficient for reasons mentioned above.
As far as statement 2 is concerned all you need to know is whether the remaining unknowns (x & y) fall within or outside the [73,87] interval.
As x+y = 194 you know that min value for x or y is 94>87. Hence median is 87.

Either statement is sufficient.
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 [#permalink] New post 08 Nov 2004, 19:47
OA is D. Well demonstrated :)
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Re: DS mean-median [#permalink] New post 28 Aug 2011, 11:18
I have a small doubt regarding Statement 1.
5 scores are 73,76,87,96,x
Avg of last three=89
87+96+x/3=89
x+183=267
x=84
5 scores are 73,76,84,87,96. Median is 84.

According to statement 2, median is 87.

Now, in DS two statements cannot have conflicting answers. So how come we are getting different answers?

btw, this is from Kaplan 800.

Thanks.
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Re: DS mean-median [#permalink] New post 28 Aug 2011, 11:53
jamifahad wrote:
I have a small doubt regarding Statement 1.
5 scores are 73,76,87,96,x
Avg of last three=89
87+96+x/3=89
x+183=267
x=84
5 scores are 73,76,84,87,96. Median is 84.

According to statement 2, median is 87.

Now, in DS two statements cannot have conflicting answers. So how come we are getting different answers?

btw, this is from Kaplan 800.

Thanks.


You've reordered the set in the increasing order.

1 says: Steve's scores on the first 4 exams were 76, 87, 73 and 96, respectively. Means, the order must not be disturbed.
1st test:76
2nd test:87
3rd test:73
4th test:96
5th test:x

Average of last three:
(73+96+x)/3=89
169+x=267
x=98

Now, everything will fit in.
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Re: DS mean-median [#permalink] New post 28 Aug 2011, 11:58
Paul wrote:
Steve's average (arithmetic mean) score on the last 3 exams was 89


Oh! How did i miss that? :roll:

Thanks fluke.
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Re: DS mean-median [#permalink] New post 28 Aug 2011, 12:11
1. Sufficient

we were given,first four test scores and average of last three . this is enough to find out the last scores.

=> enough to find out the median.

2. Sufficient

(76+87+73+x+y)/5 = 86

=> x+y = 194

when one of these two is maximum allowed value 100 , other number would be 94.

=> median is 87.

Answer is D.
Re: DS mean-median   [#permalink] 28 Aug 2011, 12:11
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