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by artstudent » Thu Jul 07, 2011 9:26 pm
a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?

(1) e+c=3
(2) c=a+10
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Source: — Data Sufficiency |
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by artstudent » Thu Jul 07, 2011 9:42 pm
my approach:

I look at deviation. For A= M the deviation must zero out

I set the middle term (C) = 0; Since e-a=14 and e-c=4. e is 4 higher than c and a is less than c. Thus:
-10,b,0,d,+4. The left side has the min deficit of -10 if b=c. The the highest b can be is +4. So there's no way to make it balance. Thus average has to be less than median.

Please give me feedback on my approach. Is this correct?
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by vikram4689 » Thu Jul 07, 2011 10:34 pm
IMO B

1. Tried with various no. and got both A>M and A<M....so not suff.

2. c=a+10 , e=c+10=a+14, max value of d=a+13, max value of b=a+9
a+b+c+d+e = a + a+9 + a+10 + a+13 + a+14 = 5a+46

A=(5a+46)/5 = a+9.2 ...which less than M=a+10......hence suff.
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by Frankenstein » Thu Jul 07, 2011 10:40 pm
vikram4689 wrote:IMO B

1. Tried with various no. and got both A>M and A<M....so not suff.

2. c=a+10 , e=c+10=a+14, max value of d=a+13, max value of b=a+9
a+b+c+d+e = a + a+9 + a+10 + a+13 + a+14 = 5a+46

A=(5a+46)/5 = a+9.2 ...which less than M=a+10......hence suff.
Hi,
I will add a minor correction although it will not change the answer
max. value of b = a+10 and max. value of d = a+14
So, max.A = (5a+48)/5 = a+9.6 < M =a+10
Cheers!

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by artstudent » Thu Jul 07, 2011 10:56 pm
hi frankenstein,

can you look at my method. i dont have a strong background. i just did what seems logical. please let me know if my solution is correct.
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by newgmattest » Thu Jul 07, 2011 10:56 pm
Hi GMAT Experts,

Please help.

Thanks.
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by Frankenstein » Thu Jul 07, 2011 11:03 pm
artstudent wrote:hi frankenstein,

can you look at my method. i dont have a strong background. i just did what seems logical. please let me know if my solution is correct.
Hi,
Your method is fine except for the typo. It should be d instead of b.
my approach:

I look at deviation. For A= M the deviation must zero out

I set the middle term (C) = 0; Since e-a=14 and e-c=4. e is 4 higher than c and a is less than c. Thus:
-10,b,0,d,+4. The left side has the min deficit of -10 if b=c. The the highest b can be is +4. So there's no way to make it balance. Thus average has to be less than median.

Please give me feedback on my approach. Is this correct?
Cheers!

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by amit2k9 » Fri Jul 08, 2011 2:04 am
taking 3 different numbers a,b,c. median = b and mean = m
m>b when c-b > b-a
and vice versa.

using the same principle here..
b fits in well.
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