There are bricks with 24 cm x 12 cm x 8 cm dimensions. Find the total number of bricks required to construct a wall 24 m long, 8 m high and 60 m thick with 10% of wall filled with mortar.

Answer: Option B

Explanation:

Let the radius of the base be r km. Then,
\begin{aligned}
\pi r^2 = 1.54 \\
r^2 = \frac{1.54*7}{22} = 0.49\\
= 0.7 km \\
\text{Now l=2.5 km, r = 0.7 km} \\
h = \sqrt{2.5^2 - 0.7^2} km \\
=\sqrt{6.25 - 0.49}\\
=\sqrt{5.76} km \\
= 2.4 km
\end{aligned}

Answer: Option A

Explanation:

Let the edges be a and b of two cubes, then

\begin{aligned}
\frac{a^3}{b^3} = \frac{27}{1} \\
=> \left( \frac{a}{b} \right)^3 = \left( \frac{3}{1} \right)^3 \\
\frac{a}{b}=\frac{3}{1} \\
=> a:b = 3:1
\end{aligned}

Answer: Option C

Explanation:

So as per question,
\begin{aligned}
\text{Volume of wall} = (2400 * 800 * 60 ) cm^3 \\
\text{Volume of bricks} = \text { 90% of volume of wall } \\
= [ \frac{90}{100} * 2400 * 800 * 60 ] cm^3 \\
\text{ Volume of 1 brick = } (24 * 12 * 8) cm^3 \\
\text{Number of Bricks required} \\
= \frac{ (\frac{90}{100}) * (2400 * 800 * 60)}{24 * 12 * 8} \\
= 45000
\end{aligned}

Answer: Option C

Explanation:

Volume will be length * breadth * height, but in this case two heights are given so we will take average,

\begin{aligned}
Volume = \left(12*9*\left(\frac{1+4}{2}\right)\right)m^3 \\
12*9*2.5 m^3 = 270 m^3
\end{aligned}

Answer: Option D

Explanation:

We will first subtract the cone volume from wood volume to get the wood wasted.
Then we can calculate its percentage.

\begin{aligned}
\text{Sphere Volume =}\frac{4}{3}\pi r^3 \\
\text{Cone Volume =}\frac{1}{3}\pi r^2h\\

\text{Volume of wood wasted =}\\
\left(\frac{4}{3}\pi *9*9*9\right)-\left(\frac{1}{3}\pi *9*9*9\right) \\
= \pi *9*9*9 cm^3 \\
\text{Required Percentage =} \\
\frac{\pi *9*9*9}{\frac{4}{3}\pi *9*9*9}*100 \% \\
= \frac{3}{4}*100 \% \\
= 75\%
\end{aligned}