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Arithmetic mean

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Arithmetic mean

by pareekbharat86 » Fri Nov 22, 2013 12:49 am
If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be

a. 6x
b. 4x
c. 5x/3
d. 3x/2
e. 3x/5

OA is B.
Thanks,
Bharat.
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by vipulgoyal » Fri Nov 22, 2013 1:36 am
6x - not possible, max 5x - no
4x - could be
5x/3 - if all are equal, but we need sum of greatest three - no
3x/2 too small - no
3x/5 - too small - no
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by Mathsbuddy » Fri Nov 22, 2013 1:52 am
If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be

Total = A + B + C + D + E = 5x where A to D are in increasing order of maximum size

Let C + D + E = Px

then A + B + Px = 5x

So Px = 5x - (A+B)

So Px <= 5x


Also, since (A,B,C,D,E) = (x,x,x,x,x) is the limiting case where (A+B)<=(C+D+E)
then 3x <= Px

So

3x <= Px <= 5x

The only value on the list that complies is 4x.
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by pareekbharat86 » Fri Nov 22, 2013 1:53 am
Read a good reply to the exact same question online.

1. Sum of the remaining 2 nos. cannot be greater than the sum of the 3 greatest nos.(obviously)&
2. Since all the nos. are +ve, no no. can be > 5x

Based on (1), C,D,E are ruled out.
Based on (2), A is ruled out.

That leaves us with B.
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Bharat.
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by Uva@90 » Fri Nov 22, 2013 1:58 am
pareekbharat86 wrote:If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be

a. 6x
b. 4x
c. 5x/3
d. 3x/2
e. 3x/5

OA is B.
Hi Bharat,
From question we can derive that
a+b+c+d+e = 5x (Let a,b,c,d,e are temperatures in increasing order)

c,d,e are greatest Temperatures.
So, c+d+e must be greater than 3x and less than 5x

So only option matching is B

Answer is B

Hope it helps you.

Regards,
Uva.
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by theCodeToGMAT » Fri Nov 22, 2013 2:08 am
_ + _ + _ + _ + _ = 5x

Just for assumption, assume two lowest be 0

So, sum of top3 = 5x

Therefore, the sum of top will be slight lower than 5x

Only {B}
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by GMATGuruNY » Fri Nov 22, 2013 3:23 am
pareekbharat86 wrote:If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be

a. 6x
b. 4x
c. 5x/3
d. 3x/2
e. 3x/5

OA is B.
Let the 5 temperatures be 1, 2, 3, 4, 5.
Since the values are consecutive, the average = the median.
Thus, x=3.
Here, the sum of the 3 greatest temperatures = 3+4+5 = 12.
The case above illustrates that the sum of the 3 greatest integers can be equal to answer choice B:
4x = 4*3 = 12.

The correct answer is B.
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by Brent@GMATPrepNow » Fri Nov 22, 2013 8:15 am
pareekbharat86 wrote:If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be

a. 6x
b. 4x
c. 5x/3
d. 3x/2
e. 3x/5
Let's let the 5 temperatures = J, K, L, M, and N, and let's say that J < K < L < M < N

If the mean of the 5 numbers is x, then (J+K+L+M+N)/5 = x
Multiply both sides of equation by 5, we get J+K+L+M+N = 5x
We want to find a possible sum of the 3 greatest numbers (i.e., L+M+N)

L, M and N represent 3 of the 5 numbers. Since they are the 3 greatest values, we know that their sum must be greater than or equal to 3/5 of the sum of J+K+L+M+N
So, L+M+N > 3/5(J+K+L+M+N)

Since J+K+L+M+N = 5x we can say L+M+N > 3/5(5x)
In other words, L+M+N > 3x
Also, since J+K+L+M+N = 5x, we can conclude that L+M+N < 5x

This tells us that 3x < L+M+N < 5x
When we check the answer choices, only B works.

Cheers,
Brent
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