A and B can do a piece of work in 12 days. B and C can do it in 15 days; A and C can do it 20 days. Who among these will take the least time if put to do it alone?
Answers
Given:-
- A+B=12 days
- B+C=15 days
- A+C=20 days
To Find:-
- Who will take least time to do it alone
Solution:-
- At first calculate the total units of days which is done by A B C. To calculate the unit we have to find out the lcm]
Lcm of 12 , 15 , 20=3×2×2×5=60 units
3[12, 15, 20]
2[4 , 5 , 20]
2[2, 5, 10]
5[1, 5 , 5]
1, 1, 1
=>Total units of days=60
=>A+B+B+C+A+C=1/12+1/15+1/20
=>2A+2B+2C=5+4+3/60
=>A+B+C=12/120
=>A+B+C=120/12=10 units
=>Now, calculate units here separately
=>A+B=12
=>A+B=60/12=5units
=>B+C=15
=>B+C=60/15=4units
=>A+C=20
=>A+C=60/20=3 units
=>Now calculate the work alone by A B and C
=>A+B+C-(A+B)
=>10-5=5 units (for C)
=>A+B+C-(B+C)
=>10-4=6 units ( for A)
=>A+B+C-(A+C)
=>10-3=7 units (for B)
It's clearly mentioned above that C take least time to do it alone
Hence,
- C will take least time.
- A & B together do 1/12 th work per day.
- B & C does 1/15 th work per day
- C & D does 1/20 th work per day
Now add all of these.
⇒ (A+B)+(B+C)+(C+D) = 1/12 + 1/15 + 1/20
⇒ 2(A+B+C) = 1/12 + 7/60
⇒ 2(A+B+C) = 12/60
⇒ (A+B+C) = 1/10
So per day work of A&B&C is 1/10.
Now find per day work A , B , C.
A's per day work :-
( A+B+C ) - ( B+C )
⇒ 1/10 - 1/15
⇒ 1/30
B's per day work :-
( A+B+C ) - ( A+C )
⇒ 1/10 - 1/20
⇒ 1/20
C's per day work :-
( A+B+C ) - ( A+B )
⇒ 1/10 - 1/12
⇒ 1/60
As B is most efficient . It will do work in 20 days .
So B takes least time if they work alone .