NCERT Exemplar Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 12.1 in Chapter 12 - Surface Areas and Volumes

Question 6 Surface Areas and Volumes - Exercise 12.1

A mason construction a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that \frac{1}{8} space is covered by the mortar. Then the number of bricks used to construct the wall is

The volume of the wall covered by mortar = \frac{1}{8} part

So, the volume covered by bricks of wall

= \left(1-\frac{1}{8}\right) volume of wall

=\frac{7}{8} volume of wall

Bricks (Cuboid)

l1 = 22.5 cm

b1 = 11.25 cm

h1 = 8.75 cm

Wall (Cuboid)

l = 270 cm

b = 300 cm

h = 350 cm

Let n be the number of bricks.

According to the question, we have

Volume of n bricks = \frac{7}{8} Volume of wall (Cuboid)

\Rightarrow \mathrm{n} \times \mathrm{l}_{1} \times \mathrm{b}_{1} \times \mathrm{h}_{1}=\frac{7}{8} l \times b \times h \\ \Rightarrow n=\frac{7 \times l \times b \times h}{8 \times l_{1} \times b_{1} \times h_{1}}=\frac{7 \times 270 \times 300 \times 350}{8 \times 22.5 \times 11.25 \times 8.75} \\ \Rightarrow n=\frac{7 \times 270 \times 300 \times 350 \times 10 \times 100 \times 100}{8 \times 225 \times 1125 \times 875}

⇒ n = 2 × 4 × 350 × 4 = 32 × 350 = 11200 bricks

Hence, right option is 11200.

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