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a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e -
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Bunuel
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a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e - [#permalink] New post  15 Feb 2021, 00:15
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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 = 34
(2) c = a + 10

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chetan2u
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Re: a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e - [#permalink] New post  15 Feb 2021, 10:08
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Bunuel wrote:
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 = 34
(2) c = a + 10

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The median will be c as per the given condition, so we have to find if A>c?

(1) e + c = 34
e-c=4
Two equations so we can find e and c, but nothing about a and b

(2) c = a + 10
Now e-c=4
For taking the max value of mean, let us take all unknown values b and d as maximum possible. => b=c and d=e
So mean = \(\frac{a+b+c+d+e}{5}=\frac{c-10+c+c+c+4+c+4}{5}=c-\frac{2}{5}\)<c.
Thus even the maximum possible value of mean, A, is less than c, that is M.
So A<M always
Sufficient

B
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TarunKumar1234
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Re: a, b, c, d, and e are five numbers such that a ≤ b ≤ c ≤ d ≤ e and e - [#permalink] New post  15 Feb 2021, 10:46
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Kudos
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?

We need to check, average (arithmetic mean) > median (M), if a, b, c, d, and e are equidistant or not.

Stat1: e + c = 34
So, e= 19 and c= 23, but no idea about a and b, they can be equidistant or not. Not sufficient

Stat2: c = a + 10 or, c-a =10, given, e - c = 4. So we can saw, a, b, c, d, and e are not equidistant. So, definitely, A <M. Sufficient.

So, I think B. :)
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