There are three boats A, B and C, working together they carry 60 people in each trip. One day an early morning A carried 50 people in few trips alone. When it stopped carrying the passengers B and C started carrying the people together. It took a total of 10 trips to carry 300 people by A, B and C. It is known that each day on an average 300 people cross the river using only one of the 3 boats A, B and C. How many trips it would take to A to carry 150 passengers alone?
A. 15
B. 30
C. 25
D. 10
E. None of these
Answer: Option A
Solution(By Examveda Team)
Combined efficiency of all the three boats = 60 passengers /trip Now, consider option (A) 15 trips and 150 passengers means efficiency of A = 10 passengers per trip A's efficiency = 10 passengers per trip Then, (B + C) combined efficiency = 50 passengers per trip Since, combined efficiency is 60 so option (A) is correctThis Question Belongs to Arithmetic Ability >> Time And Work
Join The Discussion
Comments ( 2 )
Related Questions on Time and Work
A. 18 days
B. 24 days
C. 30 days
D. 40 days
Given, A+B+C=60.. or, B+C=60-A.....
A carried 50 people, so in each trip,50/x=A or, Ax=50...(i)
if A took x trips to carry 50 people,so (B+C) took (10-x) trips to carry 250 people. therefore, 250/(10-x)=(B+C)...
or, (10-x)*(B+C)=250
(10-x)*(60-A)=250
or, 600-60x-10A+Ax=250
or, 600-60x-10A+50) =250 (Ax=50 from eqn(i)
or, 6x+A=40......(ii)
Now, solving eqn, i & ii we get,
6*50/A+A=40.... (Ax=50 or, x=50/A)
or, A^2-40A+300=0...so, A=10 or 30, but A can't be 30, trips value will be in fraction. so, A=10,
therefore, trips needed=150/10=15 trips to carry 150 people.
(Kaiser Mostafiz
CUET, BIBM)
kindly give the full solution