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A group of 20 friends went out for a lunch. Five of them spent $21 eac : Quantitative Comparison Questions

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A group of 20 friends went out for a lunch. Five of them spent $21 eac [#permalink]
Five of them spent \(5 \times 21 = 105\)

Let the \(\text{ arithmetic mean = a } \), then the rest of them spent \(15 \times (a-3)\).

Their average spend (Quantity A) is now given by \(\frac{105 + 15 \times (a-3)}{20}\)

Now I will take the value in Quantity B as the benchmark and assume it to be the average value or arithmetic mean and substitute its value for \(a\) in the above formula and see the result. Why am I doing it? The reason is, it will simplify the calculations and enable me to estimate the greater of the two quantities quickly, without doing any algebraic manipulation.

Their average spend (Quantity A) is now given by \(\frac{105 + 15 \times (12-3)}{20}\)

\(=> \frac{105+180-45}{20}\)

\(=> \frac{240}{20}\)

\(=> 12\)

Now we got the very same value under Quantity B

Therefore,

BOTH quantities are EQUAL

The answer is C.
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A group of 20 friends went out for a lunch. Five of them spent $21 eac [#permalink]
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