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by pradeepfriend2000 » Wed Apr 17, 2013 11:42 pm
Hi,

I encountered below probelm in manhattan tests. Can anybody give me easy explanation?
-----
List R contains five numbers that have an average value of 55. If the median of the numbers in the list is equal to the mean, and the largest number is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the list?


a) 78
b) 77 1/5
c) 66 1/7
d) 55 1/7
e) 52

Correct answer is: A
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by Anju@Gurome » Thu Apr 18, 2013 12:35 am
pradeepfriend2000 wrote:List R contains five numbers that have an average value of 55. If the median of the numbers in the list is equal to the mean, and the largest number is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the list?
Say the numbers are a, b, c, d, and e; where a ≤ b ≤ c ≤ d ≤ e
Now, median = c = 55
and, e = (3a + 20)

Therefore, range = (max - min) = (e - a) = (2a + 20)
Therefore to maximize the range we have to maximize a.

Also, (a + b + c + d + e) = 5*55 = 275
--> (a + b + 55 + d + 3a + 20) = 275
--> (4a + b + d) = 200
--> 4a = (200 - b - d)

Now, a will be maximum when both b and d will be minimum.
Minimum value of b is a and that of d is c, i.e. 55.

Hence, when a is minimum,
  • 4a = (200 - a - 55) --> 5a = 145 --> a = 29
Hence maximum value of the range is (2*29 + 20) = 78

The correct answer is A.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
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by pradeepfriend2000 » Thu Apr 18, 2013 1:41 am
Thank you very much. Now I realised where I went wrong .
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