Question

There are three boats $B_1,B_{2}and{B}_3$ working together they carry 60 people in each trip. One day an early mottling $B_1$ carried 50 people in few trips alone. When it stopped carrying the passengers $B_{2}and{B}_3$ started carrying the people together. It took a total of 10 trips to carry 300 people by $B_1,B_{2}and{B}_3$. It is known that each day on an average 300 crosses the river using only one of the 3 boats $B_1,B_{2}and{B}_3$. How many trips would it take to$B_1$ to carry 150 passengers alone?

• A. 15
• B. 30
• C. 25
• D. 10

Answer 1

The correct option is A

15

Combined efficiency of all the three boats = 60 passenger/trip = $60\frac{P}{t}$. Now, consider option (a).
15 trips and 150 passengers means efficiency of $B_1=10\frac{P}{t}$ which means in carrying 50 passengers $B_1$ must has taken 5 trips. So the rest trips equal to 5 (10 - 5 = 5) in which $B_2$ and $B_3$ together carried remaining 250 (300 - 50) passengers.