The cost of the paint is Rs. 36.50 per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint outside of a cube having 8 feet each side.

Answer: Option A

Explanation:

In this type of question, just equate the two volumes to get the answer as,
\begin{aligned}
\text{Volume of cylinder =}\pi r^2h\\
\text{Volume of sphere =} \frac{4}{3}\pi r^3\\
=> 12*\frac{4}{3}\pi r^3 = \pi r^2h \\
=> 12*\frac{4}{3}\pi r^3 = \pi *8*8*2 \\
=> r^3 = \frac{8*8*2*3}{12*4} \\
=> r^3 = 8 \\
=> r = 2 cm \\
=> \text{Diameter =}2*2 = 4 cm
\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 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 A

Explanation:

Let their radii be 2x, x and their heights be h and H resp.
Then,
\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
\frac{\frac{1}{3}*\pi *{(2x)}^2*h}{\frac{1}{3}*\pi *{x}^2*H} \\
=> \frac{h}{H} = \frac{1}{4} \\
=> h:H = 1:4
\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}