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