Xavier Aptitude Test 2015 Paper

Question : 56 of 113

Marks: +1, -0

A solid metal cylinder of 10cm height and 14 cm diameter is melted and re-cast into two cones in the proportion of 3:4 (volume), keeping the height 10 cm. What would be the percentage change in the flat surface area before and after?

9%
16%
25%
50%
None of these

Validate

Solution:

(d) Volume of the cylinder

=π r^{2} h

Flat surface area

=2 \π r^{2}

Volumes of two cones are

\ (\ {1}/{3}\ ) πr_{1}^{2} h

and

\ (\ {1}/{3}\ ) \πr_{2}^{2} h

(as height of the cones are same as cylinder)
Combined flat surface areaof two cones =

\π r_{1}^{2}+\π r_{2}^{2}

Given,

\ (\ {1}/{3}\ ) \π{r}_{1}^{2} \ {h}+ (\ {1}/{3}\ ) π{r}_{2}^{2} \ {h}=π{r}^{2} {h}

⇒π{r}_{1}^{2}+π{r}_{2}^{2}=3 π{r}^{2}

Therefore, increase in flat surface area

= (π{r}_{1}^{2}+π{r}_{2}^{2} )-2 π{r}^{2}

=3 \π{r}^{2}-2 π{r}^{2}=π{r}^{2}

Hence, percentage increase in surface area = 50%

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