A swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is:

Answer: Option C

Explanation:

\begin{aligned}
Volume = \pi r^2h \\
Volume = \left(\frac{22}{7}*1*1*14\right)m^3 \\
= 44 m^3
\end{aligned}

Answer: Option C

Explanation:

We will first calculate the Surface area of cube, then we will calculate the quantity of paint required to get answer.
Here we go,

\begin{aligned}
\text{Surface area =}6a^2 \\
= 6 * 8^2 = 384 \text{sq feet} \\
\text{Quantity required =}\frac{384}{16} \\
= 24 kg\\
\text{Cost of painting =} 36.50*24 \\
= Rs. 876
\end{aligned}

Answer: Option B

Explanation:

In this type of question, first we will calculate the volume of water displaces then will multiply with the density of water.
Volume of water displaced = 3*2*0.01 = 0.06 m cube

Mass of Man = Volume of water displaced * Density of water

= 0.06 * 1000 = 60 kg

Answer: Option C

Explanation:

In this question we need to calculate the diagonal of cuboid,
which is =
\begin{aligned}
\sqrt{l^2+b^2+h^2} \\
= \sqrt{8^2+6^2+2^2} \\
= \sqrt{104} \\
= 2\sqrt{26}
\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}