Manoj travels from City A to City B. If Manoj drives his car at 3/5 of his normal speed, then he reaches City B 10 minutes late. Find the time (in minutes) that Manoj would have taken to travel from City A to City B if he drove at his normal speed.

Correct option is D

Given:
Let normal time taken = x minutes
Reduced speed = $$\frac 35$$​of normal speed
Time difference = 10 minutes
Formula Used:
Time = Distance / Speed
If speed is reduced, time increases in inverse ratio.
Solution:
If speed is $$\frac 35$$ of normal, then time taken will be$$\frac 53$$ of normal time.
So, increased time = $$\frac 53$$​ × x

Difference in time = $$\frac 53$$​x - x = 10
​$$\Rightarrow \frac{5x - 3x}{3} = 10\\\Rightarrow \frac{2x}{3} = 10\\\Rightarrow x = \frac{10 \times 3}{2} = 15\\$$​​
Therefore, Manoj would have taken 15 minutes at his normal speed. 
Alternate Solution:
As we know, time is inversely proportional to speed.
Let the usual speed of the person be 5x
Speed of the person after decreasing = 3x
Ratio of usual speed to after decreasing = 5x : 3x
Ratio of usual time to after decreasing = 3x : 5x
5x – 3x = 10
=> x = 5
3x = 5 × 3 =15 min