XAT 2015 Quant & DI Q5: Mensuration
This aptitude practice question appeared in the quantitative reasoning and data interpretation section of the XAT 2015. Question 5 of a total of 33 questions that appeared in this section in XAT 2015. This one is a medium level difficulty question and tests the concept of computing and comparing flat surface areas of cylinders and cones.
Question
2. Flat surface area of cones2 concepts
A solid metal cylinder of 10 cm 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 the above
Correct Answer Choice D. The percentage change in the flat surface area is 50%.
Explanatory Answer
-
Using information about volumes
Let radius of cylinder, cone 1, cone 2 = R, r1, r2 respectively.
Let their volumes be V, v1, v2 respectively.
Cylinder is melted and recast into cone 1 & cone 2
So V = v1 + v2.
Ratio of volumes of the two cones v1 : v2 is 3 : 4.
If v1 is 3x, v2 will be 4xHence, volume of cylinder V= 7x
-
Compute ratio of the flat surface area
The volumes of the cylinder, cone 1 and cone 2 are πR2h, \\frac{1}{3}\\) πr12h and \\frac{1}{3}\\) πr22h
We know the ratio of the volumes V : v1, v2 is 7 : 3 : 4
So, πR2h : \\frac{1}{3}\\) πr12h : \\frac{1}{3}\\) πr22h is 7 : 3 : 4Cancelling π and h, which are common to all terms, we get R2 : \\frac{1}{3}\\) r12 : \\frac{1}{3}\\) r22 = 7 : 3 : 4
Or R2 : r12 : r22 = 7 : 9 : 12.So, if R2 is 7k, r12 will be 9k and r22 will be 12k.
Flat surface area of cylinder (sum of the areas of the two circles at the top and bottom of the cylinder) = 2 * π * R2
Flat surface area of cone 1 & 2 are: π * r12 & π * r22 respectively (areas of the circle at the bottom of each of the cones).Ratio of the flat surface area of cylinder to that of the two cones is 2 * π * R2 : (π * r12 + π * r22)
Cancelling π on both sides of the ratio we get 2R2 : (r12 + r22)
Or 2(7k) : (9k + 12k)
Or 14k : 21k -
Compute percentage change
Change in flat surface area = 21k – 14k = 7k
% change in flat surface area = \\frac{7k}{14k}\\)*100 = 50%.The correct answer is Choice D.
Video Explanation
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XAT TANCET Practice Questions - Listed Topic wise
- Number Theory
- Permutation Combination
- Probability
- Inequalities
- Geometry
- Coordinate Geometry
- Mensuration
- Trigonometry
- Data Sufficiency
- Percentages
- Profit Loss
- Ratio Proportion
- Mixtures & Alligation
- Speed Distance & Time
- Pipes, Cisterns & Work, Time
- Simple & Compound Interest
- Races
- Averages & Statistics
- Progressions : AP, GP & HP
- Set Theory
- Clocks Calendars
- Linear & Quadratic Equations
- Functions
- English Grammar
- General Awareness