Question

An ice-cream company makes a popular brand of ice-cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company decides to reduce the volume of the block by 20% by keeping the bar thickness constant and decreasing the length and the breadth by the same percentage amount. Which of the following conditions is satisfied by the new bar length?
[a] 5.5 < l < 6
[b] 5 < l < 5.5
[c] 4.5 < l < 5
[d] 4 < l < 4.5

Answer 1

Hint: Assume that the fractional decrease in length and breadth be x(Note that this is possible since both length and breadth are decreased by same percent). Hence find the new length and breadth of the ice cream block. Use the fact that the volume of a cuboid of length l, breadth b and height h is given by V = lbh and hence find the volume of the original ice cream block and the new ice cream block. Use the fact that the new volume is 20% less that the original and hence form an equation in x. Solve for x and hence find the new length of the ice cream. Verify which of the options are satisfied by the new length.

Complete step-by-step answer:
Let the new length be x times of the original length and the new breadth be x times of the original breadth.
Hence, we have $l=6x,b=5x,h=2$
Note that this is possible since the length and the breadth are decreased by equal percentage
Now, we know that the volume of the cuboid of length l , breadth b and height h is given by V = lbh.
Hence, we have
Original volume of the ice-cream block is given by ${{V}_{o}}=6\times 5\times 2=60c{{m}^{3}}$
And the new volume of the ice-cream block is given by
Now, given that the new volume is 20% less than the original volume
Hence, we have
${{V}_{n}}={{V}_{o}}-\dfrac{20}{100}\times {{V}_{o}}=0.8{{V}_{o}}$
Hence, we have
$\dfrac{{{V}_{n}}}{{{V}_{o}}}=0.8$
Substituting the values of ${{V}_{n}}$ and ${{V}_{o}}$, we get
$\begin{align}
  & \dfrac{60{{x}^{2}}}{60}=0.8 \\
 & \Rightarrow {{x}^{2}}=0.8 \\
 & \Rightarrow x=\pm \sqrt{0.8} \\
\end{align}$
Since length is positive, we have x>0
Hence, we have
$x=\sqrt{0.8}=0.894$
Hence, we have
$l=6x=5.367$
Hence option [b] is correct.

Note:Alternatively, we can solve the question directly as follows without finding the individual volumes.
We have
$V=lbh$
So, if h is constant, we have
\[\begin{align}
  & V\propto lb \\
 & \Rightarrow \dfrac{{{V}_{o}}}{{{V}_{n}}}=\dfrac{{{l}_{o}}{{b}_{o}}}{{{l}_{n}}{{b}_{n}}} \\
\end{align}\]
Now since
 $\begin{align}
  & {{l}_{n}}={{l}_{o}}\times x \\
 & \Rightarrow \dfrac{{{l}_{o}}}{{{l}_{n}}}=\dfrac{1}{x} \\
\end{align}$
Similarly, we have
$\dfrac{{{b}_{o}}}{{{b}_{n}}}=\dfrac{1}{x}$
Hence, we have
$\begin{align}
  & \dfrac{{{V}_{o}}}{{{V}_{n}}}=\dfrac{1}{x}\times \dfrac{1}{x}=\dfrac{1}{{{x}^{2}}} \\
 & \Rightarrow \dfrac{{{V}_{n}}}{{{V}_{o}}}={{x}^{2}} \\
\end{align}$
Since $\dfrac{{{V}_{n}}}{{{V}_{o}}}=0.8$, we have
${{x}^{2}}=0.8$, which is the same as obtained above. Hence following in a similar way as done above, we get l = 5.367 and hence option [b] is correct.