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Bunuel
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(X1+x2+.....+x13)/13 = 3

(X1+x2+.....+x13)=39
(X1+x2+.....+x13 -x1-x2 +7)=39-x1-x2 +7
= 46-x1-x2

After performing the removal of two num and addition of 7 the avg is 4

(x3+x4+......+x13+7)/12 = 4
(x3+x4+......+x13+7)=48
So
48= 46-x1-x2
x1+x2= -2

So E

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Question is poorly worded imho, it would be much clearer to mention that only one seven is added

"If two numbers are removed from the set and replaced with the number 7"

At the first run through it sounded to me as if two 7's were added, which would lead to two possible solutions.
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Bunuel
A set of thirteen numbers has an average (arithmetic mean) equal to 3. If two numbers are removed from the set and replaced with the number 7, the new average of the set is equal to 4. What could be the two numbers which were removed from the set?

A. 1.5 and -0.5
B. -1.5 and 0.5
C. 0 and 1.0
D. -1.0 and 0
E. -1.5 and -0.5

The sum of the 13 numbers is 13 x 3 = 39. When two numbers are removed and replaced with a single 7, we have:

(39 - x - y + 7)/12 = 4

46 - x - y = 48

-2 = x + y

Since -1.5 + (-0.5) = -2, answer E is correct.

Answer: E
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