a) 30 minutes

b) 45 minutes

c) 60 minutes

d) 75 minutes

correct answer is: c) 60 minutes

#### Explanation

*Let,*

The usual speed of the man is ‘x’ m/min.*And,* he takes ‘y’ minutes to reach his destination.*Now,*

If the man increase his speed 25\% faster, then the man new speed is :

\{x+\left(x\times\large\frac{25}{100}\right)\} m/min

\rightarrow\left(x+\large\frac{25x}{100}\right) m/min

\rightarrow\large\frac{125x}{100} m/min*And,* if he reaches 48 minutes earlier, then the new time the man takes =\left(y-48\right) minutes to reach the destination.

*According to the question,*

x\times y=\large\frac{125x}{100}\normalsize\times\left(y-48\right) _{[Distance = speed × time]}

\rightarrow x\times y=\large\frac{5x}{4}\normalsize\times\left(y-48\right)

\rightarrow x\times y=\large\frac{5xy-240x}{4}

\rightarrow 4xy=5xy-240x

\rightarrow 4xy-5xy=-240x _{[interchange]}

\rightarrow -xy=-240x

\rightarrow y=\large\frac{-240x}{-x}

\therefore y=240 minutes

\therefore The man usually takes 240 minutes to reach his destination.*And,* the distance is =\left(x\times y\right) metre. _{[Distance = speed × time]}

\rightarrow\left(x\times240\right) metre.

\therefore 240x metre.

*Now,*

If he less his speed 20\%, then his new speed would be =

\{x-\left(x\times\large\frac{20}{100}\right)\} m/min

\rightarrow\left(x-\large\frac{20x}{100}\right) m/min

\rightarrow\large\left(\frac{100x-20x}{100}\right) m/min

\rightarrow\large\frac{80x}{100} m/min

\therefore\large\frac{4x}{5} m/min.

*So that,*

The man takes =\left(240x\div\large\frac{4x}{5}\right) minutes. _{[Time = Distance/Speed]}

\rightarrow\left(240x\times\large\frac{5}{4x}\right) minutes

\rightarrow\left(60\times5\right) minutes

\therefore 300 minutes.*So,* the man would be late by =\left(300-240\right) minutes

\rightarrow 60 minutes.**Ans:** The man would be late by 60 minutes, if he travels at 20\% less then his usual speed.