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The average (arithmetic mean) weight of two men is x, which is twice

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The average (arithmetic mean) weight of two men is x, which is twice  [#permalink]

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New post 16 Oct 2019, 23:31
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

68% (01:26) correct 32% (01:22) wrong based on 25 sessions

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Re: The average (arithmetic mean) weight of two men is x, which is twice  [#permalink]

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New post 16 Oct 2019, 23:50
Let the five men be A, B, C, D & E

Average of A & B = x
Average of B, C and D = x/2

Sum of weights of A and B = 2x
Sum of weights of B, C and D = 3x/2

Sum of weights of A, B, C, D and E = 2x+3x/2=7x/2

Average weight of A, B, C, D and E = 7x/2/5 = 7x/10

Answer is (D)
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Re: The average (arithmetic mean) weight of two men is x, which is twice  [#permalink]

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New post 16 Oct 2019, 23:55
Bunuel wrote:
The average (arithmetic mean) weight of two men is x, which is twice the average weight of three different men. Which of the following expresses the average weight of all five men in terms of x?

A. x/2
B. 3x/5
C. 7x/2
D. 7x/10
E. 8x/5


Total weight of 5 men = 2x + 3(x/2) = 3.5x

Average weight of 5 men = 3.5x/5 = 7x/10

IMO D
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Re: The average (arithmetic mean) weight of two men is x, which is twice  [#permalink]

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New post 16 Oct 2019, 23:57
Bunuel wrote:
The average (arithmetic mean) weight of two men is x, which is twice the average weight of three different men. Which of the following expresses the average weight of all five men in terms of x?

A. x/2
B. 3x/5
C. 7x/2
D. 7x/10
E. 8x/5


Let the weights of men be a, b, c, d and e respectively

Average weight of 2 men = x
--> (a + b)/2 = x
--> a + b = 2x ....... (1)

Given x is twice the average weight of three different men
--> x = 2(c + d + e)/3
--> c + d + e = 3x/2 ....... (2)

Average weight of 5 men = (a + b + c + d + e)/5 = (2x + 1.5x)/5 = 3.5x/5 = 7x/10

IMO Option D
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Re: The average (arithmetic mean) weight of two men is x, which is twice   [#permalink] 16 Oct 2019, 23:57
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