Question
A swimming bath is 24 m long and 15 m broad. When a number of men dive into the bath, the height of the water rises by 1 cm. If the average amount of water displaced by one of the men be 0.1 cu.m, how many men are there in the bath ?
Answer: Option B
Volume of water displaced :
$$\eqalign{
& = \left( {24 \times 15 \times \frac{1}{{100}}} \right){m^3} \cr
& = \frac{{18}}{5}{m^3} \cr} $$
Volume of water displaced by 1 man = 0.1 m3
∴ Number of men :
$$\eqalign{
& = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr
& = \left( {\frac{{18}}{5} \times 10} \right) \cr
& = 36 \cr} $$
Was this answer helpful ?
Volume of water displaced :
$$\eqalign{
& = \left( {24 \times 15 \times \frac{1}{{100}}} \right){m^3} \cr
& = \frac{{18}}{5}{m^3} \cr} $$
Volume of water displaced by 1 man = 0.1 m3
∴ Number of men :
$$\eqalign{
& = \left( {\frac{{\frac{{18}}{5}}}{{0.1}}} \right) \cr
& = \left( {\frac{{18}}{5} \times 10} \right) \cr
& = 36 \cr} $$
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution