A and B start at the same time to reach the same destination. B travelled at \(\frac{5}{7}\) of A's speed and reached the destination 1 hour 20 minutes after 4. What was the time taken by B to reach the destination?

The Correct answer is Option 1. 

Key PointsLet the speed of A be v v and the speed of B be 5/7 v 7/5 ​ v (since B travels at 5/7 ​ of A's speed).

Let the time taken by A to reach the destination be t t hours. Therefore, the time taken by B to reach the destination can be expressed as:

⇒ Time taken by B = Distance / Speed of B = d/ 5/ 7 v =  7d /5 v  ​

⇒ Since distance d d can also be expressed in terms of A's speed and time:  d=vt

⇒ Substituting this into the equation for B's time:

⇒Time taken by B= 7vt / 5v ​ = 7t/5 ​

⇒ According to the problem, B reaches the destination 1 hour 20 minutes after A. Converting 1 hour 20 minutes to hours gives:

⇒ 1 hour 20 minutes = 4 / 3 hours 

⇒ Thus, we can set up the equation: 7t/5 = t + 4/3 ​

⇒ Now, let's solve for  t:

⇒ Multiply through by 15 to eliminate the fractions:

⇒ 15 ⋅ 7t/ 5 = 15 t + 15 ⋅ 4/ 3 

⇒ 21t=15t+20

⇒ Rearranging gives: 21 t − 15 t = 20

⇒ 6 t = 20

⇒ t = 20/6 = 10/3 hour = 3 hour 20 min  

Time taken by B= 7t/5 = 70 / 15 = 14 / 3  hours=4 hours 40 minutes Thus, the time taken by B to reach the destination is: 1) 4 hours 40 minutes.