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What is the average (arithmetic mean) of A, B, and 4C?

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What is the average (arithmetic mean) of A, B, and 4C?

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What is the average (arithmetic mean) of A, B, and 4C?

(1) A + B = 17

(2) C^2 = 49

The OA is the option E.

Using both statements together is not sufficient to get an answer? Experts, can you show me what is the best way to solve this PS question?

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Hello Gmat_mission.

We are asked for $$\frac{A+B+4C}{3}=?$$

(1) A+B=17.

This implies that $$\frac{A+B+4C}{3}=\frac{17+4C}{3}=?$$ Hence, we cannot determine the average.

$$\left(2\right)\ \ C^2=49.$$

This implies that C=7 or C=-7. Hence we get $$\frac{A+B+4C}{3}=\frac{A+B+4\left(\pm7\right)}{3}=\frac{A+B\pm28}{3}=?$$ Hence, we cannot determine the average.

Using both statements together we get $$\frac{A+B+4C}{3}=\frac{17\pm28}{3}.$$ We have two options $$\frac{A+B+4C}{3}=\frac{17+28}{3}=\frac{45}{3}=15$$ or $$\frac{A+B+4C}{3}=\frac{17-28}{3}=-\frac{11}{3}.$$ Since the options are different, hence this option is NOT sufficient.

Therefore, the correct answer is E.

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$$mean\ of\ A,\ B\ and\ 4C\ is\ =\frac{A+B+4C}{3}$$
$$If\ c^2=49$$
$$then\ c=\sqrt{49}=\left(\frac{+}{-}7\right)$$
$$i.e\ c=+7\ and\ c=-7$$
[A+B=17], when c=+7
$$mean=\ \frac{17+4\left(7\right)}{3}=15$$
when c=-7
$$mean=\ \frac{17+4\left(-7\right)}{3}=\frac{11}{3}$$

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M7MBA wrote:
What is the average (arithmetic mean) of A, B, and 4C?

(1) A + B = 17

(2) C^2 = 49

The OA is the option E.

Using both statements together is not sufficient to get an answer? Experts, can you show me what is the best way to solve this PS question?
You are supposed to calculate the value of (A + B + 4C)/3.

Certainly (1) and (2) alone are not sufficient. While (1) does not have the value of C, (2) does not have the value of A and B.

Even combining the two statements will not help as C^2 = 49 gives two values of C = 4 and -4. Thus, with each value of C, you get different values of (A + B + 4C)/3. In DS, we want a unique value, which is not the case here. Insufficient.

The correct answer: E

Hope this helps!

-Jay
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