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

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Source: — Data Sufficiency |
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by Vincen » Thu Jun 14, 2018 2:13 am

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VJesus12 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.

Why is the correct answer the option E? Could anyone give me some help here? Please.
Hello Vjesus12.

We have to find $$\frac{A+B+4C}{3}.$$

First Statement
(1) A + B = 17
If A+B=17 then we get $$\frac{A+B+4C}{3}=\frac{17+4C}{3}$$ but we can't determine the value. Therefore, this statement is not sufficient.

Second Statement
(2) C^2 = 49
If C^2=49 then C=7 or C=-7. Then we get $$\frac{A+B+4C}{3}=\frac{A+B\pm4\left(7\right)}{3}=\frac{A+B\pm28}{3}$$ but we can't determine the value. Therefore, this statement is not sufficient.

Both Statements Together
(1) A + B = 17
(2) C^2 = 49
Now we know that A+B=17 and C=7 or C=-7. Then we get $$\frac{A+B+4C}{3}=\frac{17\pm28}{3}$$ $$\Rightarrow\ \ \frac{17+28}{3}=\frac{45}{3}=15\ \ \ \ \ or\ \ \ \ \ \ \ \ \frac{17-28}{3}=-\frac{11}{3}.$$ Since we get two different answers, then we conclude again that using both statements together is not sufficient.

Finally, we conclude that the correct answer is the option E.

I hope it helps.
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