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Bunuel
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Easier way:

the average of the components added were:

[3-5-1][/3] = -1

Adding this to the original average of 13, we come up with 12 as the final average. Answer D
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Solution



Given:
Arithmetic mean of a, b, and 7 is 13.

To find:
The average of a + 3, b – 5, and 6

Approach and Working:
Average of a + 3, b – 5, and 6 = \(\frac{(a+3+b-5+6)}{3}\) = \(\frac{(a+b+4)}{3}\) ------------(1)
We need the value of a+b to find the average of a+3, b-5, and 6.
Arithmetic mean of a, b, and 7 is 13.
\(\frac{(a+b+7)}{3}\) = 13
a + b + 7 = 39
a + b =32

Hence,\(\frac{(a+b+4)}{3}\)=\(\frac{(32 + 4)}{3}\)= 12
Thus, the correct answer is option D.

Answer: D
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Bunuel
If the average (arithmetic mean) of a, b, and 7 is 13, what is the average of a + 3, b – 5, and 6?

A. 7
B. 9
C. 10
D. 12
E. 16

We are given that (a + b + 7)/3 = 13 → a + b + 7 = 39 → a + b = 32. We need to determine the value of (a + 3 + b - 5 + 6)/3, or (a + b + 4)/3.

Since a + b = 32, then we have (a + b + 4)/3 = (32 + 4)/3 = 36/3 = 12.

Answer: D
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