CAT 2020

Slot 2

QA

Question & Solution

Let the number of pencils bought by Aron be $p$, and the cost of each pencil be $a$.
Let the number of sharpeners bought by Aron be $s$, and the cost of each sharpener be $b$.

The amount spent by Aron is:
$$ pa + sb $$

Aditya bought $2p$ pencils and $(s - 10)$ sharpeners.
Amount spent by Aditya is:
$$ 2pa + (s - 10)b $$

Since the total amount spent in both cases is the same:

$$ pa + sb = 2pa + (s - 10)b $$

Rearranging:

$$ pa = 10b $$

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It is given that the cost of a sharpener is 2 more than the cost of a pencil:
$$ b = a + 2 $$

Substitute this into $pa = 10b$:

$$ pa = 10(a + 2) $$

$$ pa = 10a + 20 $$

$$ a = \frac{20}{p - 10} $$

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To ensure both $p$ and $a$ are integers, we need the smallest integer $p$ such that $(p - 10)$ divides 20.

Possible divisors of 20:
$1, 2, 4, 5, 10, 20$

Thus the smallest value of $p$ is:

$$ p - 10 = 1 \Rightarrow p = 11 $$

Total number of pencils bought:

$$ p + 2p = 11 + 22 = \boxed{33} $$