Bhanu travels from City A to City B. If Bhanu drives his car at$$\frac 14$$​ of his normal speed, then he reaches City B 51 minutes late. Find the time (in minutes) that Bhanu would have taken to travel from City A to City B if he drove at his normal speed.

Correct option is B

Given:

Bhanu drives at ​$$\frac 14$$​ of his normal speed.

Bhanu is 51 minutes late when driving at this reduced speed.

We need to find the time it would take Bhanu to travel from City A to City B if he drove at his normal speed.

Concept Used:

The time taken to travel is inversely proportional to the speed, meaning that if speed decreases, time increases.

Solution: 

Let T be the time Bhanu would take at his normal speed and S be Bhanu’s normal speed.

When Bhanu drives at  $$\frac{1}{4}$$​ of his normal speed, the time taken will increase by a factor of  $$\frac{4}1$$​​

The time taken at reduced speed is: $$4 \times T$$​​

The difference in time is 28 minutes:

$$4\times T - T = 51$$​​

​T = $$\frac{51}{3 } = 17$$ min

Thus, the time Bhanu would have taken to travel from City A to City B at his normal speed is 17 minutes 

Alternate Solution: ​

For constant distance; 

Speed $$\propto \frac1{Time}$$

Speed ratio = 4    :     1

Time ratio =   1   :     4
4 - 1 = 51 min.

3 unit = 51 min.

1 unit = 17 min. 

time taken by normal speed = 3 × 7 = 21 min.