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.