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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}

Similar Questions :

1. Find the surface area of a 10cm*4cm*3cm brick.

  • 154 cm square
  • 156 cm square
  • 160 cm square
  • 164 cm square

Answer: Option D

Explanation:

Surface area of a cuboid = 2(lb+bh+hl) cm square
So,
Surface area of a brick = 2(10*4+4*3+3*10) cm square
= 2(82) cm square = 164 cm square

2. A cone of height 9 cm with diameter of its base 18 cm is carved out from a wooden solid sphere of radius 9 cm. The percentage of the wood wasted is :

  • 45%
  • 56%
  • 67%
  • 75%

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}

3. A circular well with a diameter of 2 meters, is dug to a depth of 14 meters. What is the volume of the earth dug out.

  • \begin{aligned} 40 m^3 \end{aligned}
  • \begin{aligned} 42 m^3 \end{aligned}
  • \begin{aligned} 44 m^3 \end{aligned}
  • \begin{aligned} 46 m^3 \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}

4. 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}

5. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :

  • \begin{aligned} 2:1 \end{aligned}
  • \begin{aligned} 1:\sqrt{2} \end{aligned}
  • \begin{aligned} \sqrt{2}:1 \end{aligned}
  • \begin{aligned} \sqrt{3}:1 \end{aligned}

Answer: Option C

Explanation:

Let the radius of hemisphere and cone be R,
Height of hemisphere H = R.
So the height of the cone = height of the hemisphere = R
Slant height of the cone
\begin{aligned}
= \sqrt{R^2+R^2} \\
= \sqrt{2}R \\
\frac{\text{Hemisphere Curved surface area}}{\text{Cone Curved surface area}} = \\
\frac{2\pi R^2}{\pi *R*\sqrt{2}R} \\
= \sqrt{2}:1
\end{aligned}

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