Scott bought X pens for $\(0.30\)$ each \& Y pencils for $\(0.40\)$ each, so he spent a total of

$\((\mathrm{X} \times 0.30+\mathrm{Y} \times 0.40)=$(0.30 \mathrm{X}+0.40 \mathrm{Y})\)$

Now, the mean of the prices of all the pens \& the pencils Scott bought would be

$\(\frac{Total amount Scott spent }{Number of Pens & Pencils bought}\)\(=\frac{(0.30 \mathrm{X}+0.40 \mathrm{Y}}{\mathrm{X}+\mathrm{Y})}\)$

As $\(\mathrm{Y}>\mathrm{X}\)$, the mean price must be more close to $\(0.40\)$ than to $\(0.30\)$ i.e. the mean price must be greater than $\(0.35\)$ but as the exact values of


$\(\mathrm{X} \& \mathrm{Y}\)$ are not known, a value which would come greater than $\(0.35\)$ (= column A quantity) can't be uniquely compared with 0.36 (= column B quantity)

For example if we take $\(\mathrm{X}=2 \& \mathrm{Y}=3\)$, we get the mean price as $\(\frac{(0.30X+0.40Y)}{X+Y }=\frac{ 0.30 \times 2+0.40 \times 3}{2+3}=\frac{1.80}{5}=0.36\)$ which is equal to column B quantity.

But if we consider $\(\mathrm{X}=2 \& \mathrm{Y}=6\)$, we get the mean price as $\(\frac{(0.30 X+0.40 Y)}{X+Y}=\frac{0.30 \times 2+0.40 \times 6}{2+6}=\frac{3}{8}=0.375\)$ which is greater than column B quantity.

Hence the answer is (D).

Bump for further discussions