Question Detail

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.

Rs. 850
Rs. 860
Rs. 876
Rs. 886

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}

1. The radii of two cones are in ratio 2:1, their volumes are equal. Find the ratio of their heights.

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}

2. The perimeter of one face of a cube is 20 cm. Its volume will be:

\begin{aligned} 125 cm^3 \end{aligned}
\begin{aligned} 400 cm^3 \end{aligned}
\begin{aligned} 250 cm^3 \end{aligned}
\begin{aligned} 625 cm^3 \end{aligned}

Answer: Option A

Explanation:

Edge of cude = 20/4 = 5 cm

Volume = a*a*a = 5*5*5 = 125 cm cube

3. The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km square. The height of mountain is :

2.3 km
2.4 km
2.5 km
2.6 km

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}

4. How many cubes of 10 cm edge can be put in a cubical box of 1 m edge.

10000 cubes
1000 cubes
100 cubes
50 cubes

Answer: Option B

Explanation:

\begin{aligned}
\text{Number of cubes =}\frac{100*100*100}{10*10*10} \\
= 1000
\end{aligned}

Note: 1 m = 100 cm

5. A hollow spherical metallic ball has an external diameter 6 cm and is 1/2 cm thick. The volume of metal used in the metal is:

\begin{aligned} 47\frac{1}{5} cm^3 \end{aligned}
\begin{aligned} 47\frac{3}{5} cm^3 \end{aligned}
\begin{aligned} 47\frac{7}{5} cm^3 \end{aligned}
\begin{aligned} 47\frac{9}{5} cm^3 \end{aligned}

Answer: Option B

Explanation:

Please note we are talking about "Hollow" ball. Do not ignore this word in this type of question in a hurry to solve this question.
If we are given with external radius and thickness, we can get the internal radius by subtracting them. Then the volume of metal can be obtained by its formula as,

External radius = 3 cm,
Internal radius = (3-0.5) cm = 2.5 cm

\begin{aligned}
\text{Volume of sphere =}\frac{4}{3}\pi r^3 \\
= \frac{4}{3}*\frac{22}{7}*[3^2 - 2.5^2]cm^3 \\
= \frac{4}{3}*\frac{22}{7}*\frac{91}{8}cm^3 \\
= \frac{143}{3} cm^3 \\
= 47\frac{2}{3}cm^3
\end{aligned}