Question

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 same, then
A. 50 < p < 55
B. 55 < p < 60
C. 60 < p < 65
D. 65 < p < 70

Answer 1

Hint: In this question, we are given that the height of the pyramid with square base is decreased and length of sides are increased and we have to find an approximate value of percentage with which it increased and decreased. For this, we will use formula as: $\text{New height}=h\pm \dfrac{p}{100}\times h$ where h is old height, p is the increased or decreased percentage and $\pm $ is determined according as height is increased (+) or decreased (-). We will solve quadratic equation formed using discriminant given by $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for equation $a{{x}^{2}}+bx+x=0$. We will also use formula for volume of pyramid given as $\text{Volume}=\dfrac{1}{3}\times l\times b\times h$ where l, b are sides of base of pyramid and h is height of pyramid.

Complete step by step answer:
Here, we are given that, height of the pyramid is decreased by p% and lengths of the square base are increased by p%. Let us assume that, earlier length of base was x and the height of the pyramid was y. Since, the base of the pyramid is square, so the breadth of the base will also be x.
Now, height is decreased by p%, then the new height given by h will become $\Rightarrow h=y-\dfrac{p}{100}y$.
Also, the sides of the base are increased by p%, so, the new length of base of the pyramid given by l becomes $\Rightarrow l=x+\dfrac{p}{100}\times x$.
Similarly, b is the new breadth of the base of the pyramid which is equal to $\Rightarrow b=x+\dfrac{p}{100}\times x$.
As we know, volume of pyramid is given by $\Rightarrow \text{V}=\dfrac{1}{3}\times x\times x\times y=\dfrac{1}{3}{{x}^{2}}y$ where, x is length or breadth of base and y is height of pyramid.
New volume of pyramid becomes,
\[\begin{align}
  & \text{V}=\dfrac{1}{3}\times l\times b\times h \\
 & \Rightarrow V=\dfrac{1}{3}\times \left( x+\dfrac{p}{100}x \right)\left( x+\dfrac{p}{100}x \right)\left( x+\dfrac{p}{100} \right)y \\
 & \Rightarrow V=\dfrac{1}{3}{{x}^{2}}y{{\left( 1+\dfrac{p}{100} \right)}^{2}}\left( 1-\dfrac{p}{100} \right) \\
\end{align}\]
Since volume becomes same, therefore,
\[\Rightarrow \dfrac{1}{3}{{x}^{2}}y=\dfrac{1}{3}{{x}^{2}}y{{\left( 1+\dfrac{p}{100} \right)}^{2}}\left( 1-\dfrac{p}{100} \right)\]
Cancelling $\dfrac{1}{3}{{x}^{2}}y$ from both sides, we get:
\[\Rightarrow 1={{\left( 1+\dfrac{p}{100} \right)}^{2}}\left( 1-\dfrac{p}{100} \right)\]
Taking LCM in brackets, we get:
\[\begin{align}
  & \Rightarrow 1={{\left( \dfrac{100+p}{100} \right)}^{2}}\left( \dfrac{100-p}{100} \right) \\
 & \Rightarrow 1000000=\left( 10000+{{p}^{2}}+200p \right)\left( 100-p \right) \\
 & \Rightarrow 1000000=1000000-10000p+100{{p}^{2}}+20000p-{{p}^{3}}-200{{p}^{3}} \\
\end{align}\]
Simplifying we get:
\[\Rightarrow 0=-{{p}^{3}}-100{{p}^{2}}+10000p\]
Which can be reduced to get,
\[\Rightarrow {{p}^{2}}+100p-{{\left( 100 \right)}^{2}}=0\]
Using quadratic formula we get:
\[\begin{align}
  & \Rightarrow p=\dfrac{-100\pm \sqrt{{{\left( 100 \right)}^{2}}+4{{\left( 100 \right)}^{2}}}}{2} \\
 & \Rightarrow p=\dfrac{-100\pm \sqrt{10000+40000}}{2} \\
 & \Rightarrow p=\dfrac{-100\pm \sqrt{50000}}{2} \\
 & \Rightarrow p=-50\pm \sqrt{12500} \\
 & \Rightarrow p=-50\pm 111.803 \\
\end{align}\]
Since, it is given that p>0, so neglecting negative terms, we get $\Rightarrow p=-50\pm 111.803\Rightarrow p\simeq 61.80$ and hence lies between 60 and 65.

So, the correct answer is “Option C”.

Note: Students should take care of (+) and (-) signs while taking increased or decreased heights. While solving quadratic equations, make sure of signs and reject negative signs at last because p is given to be greater than 0 and hence cannot be negative. Students should know the basic formula for calculating the volume of the pyramid.