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Set of Five (from Manhattan GMAT)

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Set of Five (from Manhattan GMAT)

by Sprite_TM » Wed May 20, 2009 6:18 pm
hi,

this is from the manhattan CAT test but i don't quite understand their explanation. could someone please explain the answer to me? Thanks!!


Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?


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

OA
a
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by cramya » Wed May 20, 2009 6:34 pm
x y 55 z 3x+20

Since we are asked for the maximum I started out with 78.


Let say 3x+20-x = 78
x=19

3x+20=77

77+19=96 (we are missing 14 from the mean if these were the numbers in the list(

Can I fit in 2 numbers to the right and left side of the median that would satisfy the mean and median being 55 plus compensate for the 14 and provide an ascending sequence.

YES

19 55 55 69 77

A

I am sure others may have a theoritical solution but I couldn't come up with any when I started the prob.

Regards,
CR
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by dumb.doofus » Thu May 21, 2009 12:41 am
I think I can try to give a more theoretical solution.. here it is..

Let's choose the numbers as x, y, 55, z, 3x+20

So range here is 3x + 20 - x = 2x + 20 ------------- (1)

So if the range has to be maximum, than x has to be maximum..

We also know that the average is 55.. that means

x + y + 55 + z + 3x + 20 = 275

or 4x + y + z = 200 ------------- (2)

Now when would x be maximum, only when y and z are minimum.

So
1. What is the minimum value of z? Well, it has to be 55 coz the median is 55, so z cannot be less than 55
2. What is the minimum value of y? Well it has to be equal to x. It can't be less than x..

So with the above deductions, equation (2) becomes

4x + x + 55 = 200

or 5x = 145

or x = 29

Putting this value in (1)
we get 2*(29) + 20 = 78

So answer should be A i.e. 78

Hope this helps..
Last edited by dumb.doofus on Thu May 21, 2009 8:52 pm, edited 1 time in total.
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by Sprite_TM » Thu May 21, 2009 7:26 am
thanks a lot doofus! your explanation is very clear
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by Svedankae » Thu May 21, 2009 12:12 pm
dumb.doofus wrote:I think I can give a try to give a more theoretical solution.. here it is..

Let's choose the numbers as x, y, 55, z, 3x+20

So range here is 3x + 20 - x = 2x + 20 ------------- (1)

So if the range has to be maximum, than x has to be maximum..

We also know that the average is 55.. that means

x + y + 55 + z + 3x + 20 = 275

or 4x + y + z = 200 ------------- (2)

Now when would x be maximum, only when y and z are minimum.

So
1. What is the minimum value of z? Well, it has to be 55 coz the median is 55, so z cannot be less than 55
2. What is the minimum value of y? Well it has to be equal to x. It can't be less than x..

So with the above deductions, equation (2) becomes

4x + x + 55 = 200

or 5x = 145

or x = 29

Putting this value in (1)
we get 2*(29) + 20 = 78

So answer should be A i.e. 78

Hope this helps..

wow youre insanely gifted. thanks a lot
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