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IMO:-C i.e 4

Lets first term be a

The general term for this equation will be = a + 4 (n-1); where n is the nth term.

Here n will range from 1 to 10, so Summation of General term will be = 10a + 180----(1)

The arithmetic mean will be = a+18

The highest term will be a+36 and a +32

Subtracting two highest terms from equation 1, resultant summation will be 8a +112

The arithmetic mean for these 8 Nos will be a +14

Hence difference of two AM will be 4
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Bunuel
Set P consists of 10 positive integers arranged in order of increasing magnitude. The difference between any two successive terms of the set is 4. If the two largest terms of the set are removed, what is the decrease in the average(arithmetic mean) of the set?

A. 0
B. 2
C. 4
D. 6
E. 8

Solution:

If we let x = the smallest member in the set, the three largest members of the set are x + 28, x + 32 and x + 36. Recall that the average of an evenly spaced set is equal to the average of the smallest member and the largest member. Therefore, the (original) average of the set is:

(x + x + 36)/2 = (2x + 36)/2 = x + 18

When x + 32 and x + 36 are removed, the new average of the set is:

(x + x + 28)/2 = (2x + 28)/2 = x + 14

Therefore, the average of the set decreases by (x + 18) - (x + 14) = 4 when the two largest members of the set are removed.

Answer: C
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Since this is a consecutive set, can't we simply say that the total will decrease by 2*4=8 overall? And that since we are removing 2 of the integers the average will change by -8/2=-4. Which brings us to the answer of a decrease in 4....

Is it wrong to go about this as simply as I am thinking you can....?
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Ans C

If let x = the smallest member in the set, the three largest members of the set are x + 28, x + 32 and x + 36.
the (original) average of the set is:

(x + x + 36)/2 = (2x + 36)/2 = x + 18
x + 32 and x + 36 are removed, the new average of the set is:

(x + x + 28)/2 = (2x + 28)/2 = x + 14

the average of the set decreases by (x + 18) - (x + 14) =4

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