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How many bricks, each measuring 25cm*11.25cm*6cm, will be needed to build a wall 8m*6m*22.5m

  • 6100
  • 6200
  • 6300
  • 6400

Answer: Option D

Explanation:

To solve this type of question, simply divide the volume of wall with the volume of brick to get the numbers of required bricks
So lets solve this

Number of bricks =
\begin{aligned}
\frac{\text{Volume of wall}}{\text{Volume of 1 brick}} \\
= \frac{800*600*22.5}{25*11.25*6} \\
= 6400
\end{aligned}

Similar Questions :

1. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

  • 4 cm
  • 6 cm
  • 8 cm
  • 10 cm

Answer: Option B

Explanation:

\begin{aligned}
\text{Curved surface area of sphere =}\\
\frac{4}\pi r^2 \\
\text{Surface area of cylinder =} \\
2\pi rh \\
=> \frac{4}\pi r^2 = 2\pi rh \\
=> r^2 = \frac{6*12}{2} \\
=> r^2 = 36 \\
=> r = 6
\end{aligned}

Note: Diameter of cylinder is 12 so radius is taken as 6.

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

  • \begin{aligned} 450 cm^2 \end{aligned}
  • \begin{aligned} 550 cm^2 \end{aligned}
  • \begin{aligned} 650 cm^2 \end{aligned}
  • \begin{aligned} 750 cm^2 \end{aligned}

Answer: Option B

Explanation:

Volume is given, we can calculate the radius from it, then by calculating slant height, we can get curved surface area.
\begin{aligned}
\frac{1}{3}*\pi *r^2*h = 1232 \\
\frac{1}{3}*\frac{22}{7}*r^2*24 = 1232 \\
r^2 = \frac{1232*7*3}{22*24} = 49 \\
r = 7 \\
\text{Now, r = 7cm and h = 24 cm } \\
l = \sqrt{r^2+h^2} \\
= \sqrt{7^2+24^2} = 25cm \\
\text{Curved surface area =}\pi rl\\
= \frac{22}{7}*7*25 = 550 cm^2
\end{aligned}

3. A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface.

  • 42 m sqaure
  • 49 m sqaure
  • 52 m sqaure
  • 64 m sqaure

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

4. A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weights 8g/cm cube, then find the weight of the pipe.

  • 3.696 kg
  • 3.686 kg
  • 2.696 kg
  • 2.686 kg

Answer: Option A

Explanation:

In this type of question, we need to subtract external radius and internal radius to get the answer using the volume formula as the pipe is hollow. Oh! line become a bit complicated, sorry for that, lets solve it.

External radius = 4 cm
Internal radius = 3 cm [because thickness of pipe is 1 cm]

\begin{aligned}
\text{Volume of iron =}\pi r^2h\\
= \frac{22}{7}*[4^2 - 3^2]*21 cm^3\\
= \frac{22}{7}*1*21 cm^3\\
= 462 cm^3 \\
\end{aligned}
Weight of iron = 462*8 = 3696 gm
= 3.696 kg

5. Two right circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.

  • \begin{aligned} \sqrt{3}:1 \end{aligned}
  • \begin{aligned} \sqrt{7}:1 \end{aligned}
  • \begin{aligned} \sqrt{2}:1 \end{aligned}
  • \begin{aligned} 2:1 \end{aligned}

Answer: Option C

Explanation:

Let their heights be h and 2h and radii be r and R respectively then.

\begin{aligned}
\pi r^2h = \pi R^2(2h) \\
=> \frac{r^2}{R^2} = \frac{2h}{h} \\
= \frac{2}{1} \\
=> \frac{r}{R} = \frac{\sqrt{2}}{1} \\
=> r:R = \sqrt{2}:1 \\
\end{aligned}

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