A and B can do a piece of work in 5 days, B and C can do it in 7 days and A and
C can do it in 4 days. Who among them will take the least time if put to do
it alone ?
Answers
Answer:
Let the work be W.
A and B can do a piece of work in 5 days ∴ 5A+5B=W..........(1)
B and C can do it in 7 days ∴ 7B+7C=W......(2)
A and C can do it in 4 days ∴ 4A+4C=W.......(3)
Solving (1) (2) and (3)We get,
Speed of A= 28043work/day
Speed of B= 28013work/day
Speed of C= 28027work/day
Since speed of A is maximum among all three,
Hence A will take least time to complete the work.
Answer:
A will do it in the least time.
Step-by-step explanation:
Make fractions :
A and B = 1/5
B and C = 1/7
A and C = 1/4
add them = (1/5) + (1/7) + (1/4)
which makes 83/140 if you multiply to get the same denominator
next:
a b and c together do : 83/(2*140) = 83/240th part of the work in a day. And all three working together will complete the work in 280/83 = 3.373493976 days. Now to find out who is the most efficient.
C working alone will do (83/280)-(1/5) = (83–56)/280 = 27/280 = 1/10.37037037th part of the work in a day.
B working alone will do (83/280)-(1/4) = (83–70)/280 = 13/280 = 1/21.53846154th part of the work in a day.
A working alone will do (83/280)-(1/7) = (83–40)/280 = 43/280 = 1/6.511627907th part of the work in a day.
So A is the most efficient and he will do the job in 6.511627907 days.
Next comes C who will do the job in 10.37037037 days.
And the least efficient is B who will do the job in 21.53846154 days.