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Suppose the height of a pyramid with a square base is decreased by p% and length of the side of its square base is increased by p% (where p>0). If the volume remains the same, then

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50 < p < 55

55 < p < 60

60 < p < 65

65 < p < 70

answer is C.

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Detailed Solution

height is decreased by p%, then the new height given by h will become,

h = y - \frac{p}{100} y

sides of the base are increased by p%, so, the new length of base of the pyramid given by l becomes,

l = x + \frac{p}{100} x

b is the new breadth of the base of the pyramid which is equal to

b = x + \frac{p}{100} x

Volume of pyramid

V = \frac{1}{3} x^{2} y

New volume of pyramid =

\frac{1}{3} lbh = \frac{1}{3} \times {(x + \frac{p}{100} x)}^{2} (y - \frac{p}{100} y) = \frac{1}{3} x^{2} y {(1 + \frac{p}{100})}^{2} (1 - \frac{p}{100})

Since volume becomes same

\Rightarrow \frac{1}{3} x^{2} y = \frac{1}{3} x^{2} y {(1 + \frac{p}{100})}^{2} (1 - \frac{p}{100})

\Rightarrow {(1 + \frac{p}{100})}^{2} (1 - \frac{p}{100}) = 1

\Rightarrow p \approx 61.80

Hence, Option 3 is correct, 60 < p < 65.

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