Math, asked by paswandharmendra9470, 4 months ago

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

Answered by stranger9799
1

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.

Answered by ItsGalaxy
2

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.

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