If a right circular cone of height 24 cm has a volume of 1232 cm cube, then the area of its curved surface is :

Answer: Option B

Explanation:

Area of the wet surface =
2[lb+bh+hl] - lb = 2 [bh+hl] + lb
= 2[(4*1.25+6*1.25)]+6*4 = 49 m square

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 of the largest cone = Volume of the cone with diameter of base 7 and height 7 cm

\begin{aligned}
\text{Volume of cone =}\frac{1}{3}\pi r^2h \\
= \frac{1}{3}*\frac{22}{7}*3.5*3.5*7 \\
= \frac{269.5}{3}cm^3 \\
= 89.8 cm^3
\end{aligned}

Note: radius is taken as 3.5, as diameter is 7 cm

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}