Question

There is a playground measuring $50m \times 30m$ . In one corner of the ground, a pit of dimension $5m \times 4m \times 1m$ is dug and the mud is spread all over the ground uniformly. What is the approximate height of the layer of mud spread?
A. $7mm$
B. $8mm$
C. $14mm$
D. $10mm$

Answer 1

Hint: In this question we are asked to find the height of the layer of mud. In order to proceed with this question we need to find the volume of mud that is dug out. Volume is basically the space inside any 3-D figure. The pit of cuboid in shape so volume of mud can be calculated by$length \times width \times height$. Since, the mud is spread on the playground the volume will remain the same. Length and width of the playground will remain the same, only height will change.

Complete step by step solution:
$ \Rightarrow Volume\;of\;mud = 5 \times 4 \times 1$
$ \Rightarrow 20{m^3}$
$ \Rightarrow Area\;of\;playground = 50 \times 30$
$ \Rightarrow 1500{m^2}$
Since, some are of the playground are dug out, only some area of the playground is left on which the mud can be spread out.
$ \Rightarrow Area\;of\;playground\;left = (50 \times 30) - (5 \times 4)$
$ \Rightarrow Area\;of\;playground = 1480$
\[ \Rightarrow Height\;raised \times Area\;of\;playground = Volume\;of\;mud\]
$ \Rightarrow Height\;raised = \dfrac{{Volume\;of\;mud}}{{Area\;of\;playground}}$
$ \Rightarrow Height\;raised = \dfrac{{20{m^3}}}{{1480{m^2}}}$
The terms should be in same unit to be operated, so we have to convert both the terms in mm.
$ \Rightarrow Height\;raised = \dfrac{{20 \times 1000}}{{1480}}mm$
$ \Rightarrow Height\;raised = 13.3333mm \approx 14mm$
This is the required answer.

So, the correct answer is Option C.

Note: While calculating volume, you need to take care of units of values. Units which are used in formulas should be the same. For example- if you measure height in inches then length should also be in inches. Volume can only be calculated of 3-D figures, not of 2-D figures. We also need to be careful while calculating any of the missing terms.