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A is 50% as efficient as B. C does half the work done by A and B together. If C alone does the work in 40 days, then in how many days will A, B and C together finish the work?

Problem: Couldn't relate work and time from data given for C to A and B.

My attempt:

Let B take 'b' days to complete a work W. Then A will take 2*b days to complete the same work (w).

Given that c's work = (W(A)+W(B))/2 = 0.75*W

C can do W in 40 days. Hence 0.75W can be done by C in 30 days.

After this, how do i relate the work done by A and B in 1 day to C's 1 day work so that i can find the combined 1 day's work for all three?

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Why not think of $A$ and $B$ as "one person" who is twice as efficient as $C$? (The information that $A$ is half as efficient as $B$ is not needed in this problem.) – David Mitra Sep 20 '12 at 20:27
@DavidMitra: Good observation. – copper.hat Sep 21 '12 at 17:05
up vote 1 down vote accepted

You might think of "work" as equivalent to "distance", and "efficiency" as "speed". You could restate the problem in those terms, but with the twist that it's how fast three people cover the same total distance between them (equivalent to having two people start at opposite ends of a distance, travelling at different speeds, and asking where they'll meet, but now we're adding this third person travelling along the same line).

Now, you solve this by reducing the number of variables. A, B and C are all unknowns, but are related:

$$A = .5B \equiv 2A = B$$ $$C = \dfrac{A+B}2 \equiv 2C = A+B$$

We can thus replace any of the letters with the other half of the equation in which the term(s) appear, reducing the scope of the problem to fewer variables. But, what's the problem?

Well, consider the body of work to be done, W. It has an unknown number of "units" of work to be performed, presumably in any order by anybody. C can do all of this work W in 40 days, so in one day, C does $W/40$ the work, or equivalently, 40 days times C's working pace equals the volume of work W.

$$C=W/40 \equiv W=40C$$

Now, we are asked how fast A, B and C could finish W; combining all three of their working paces, how many days does it take?

$$(A+B+C)X = W$$

Solve for X, by substituting A, B, and C for equivalent statements in terms of W:

$$(A+B+C)X = W \\ (2C+C)X = W\ \ \ \text{substitute 2C for A+B } \\ 3CX = W\ \ \ \text{combine like terms} \\ \dfrac{3WX}{40} = W\ \ \ \text{replace C with W/40 and arrange} \\ 3WX = 40W\ \ \ \text{multiply both sides by 40} \\ X = \dfrac{40W}{3W}\ \ \ \text{divide both sides by 3W} \\ X = \dfrac{40}{3}\ \ \ \text{W cancels out} \\ X = 13.\bar{3}\ \ \ \text{calculator work} \\ $$

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Hint: if B takes $b$ days to finish, he does $\frac 1b$ per day. Then A does $\frac 1{2b}$ per day. Then how much of the task does C do per day in terms of $b$. You are given that C does $\frac 1{40}$ per day, so....

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Nice hint, i got it! – Karan Sep 21 '12 at 13:48

Let $W$ denote the work that needs to be done. C works at a rate of $r_C = \frac{W}{40}$. We are given that $r_A = \frac{1}{2} r_B$, and $r_C = \frac{1}{2}(r_A+r_B)$. Hence $r_C = \frac{3}{4} r_B$, from which we get $r_B = \frac{W}{30}$. Then $r_A = \frac{W}{60}$.

Consequently, if all work together, completing work $W$ will take (mouse over to reveal answer):

$\frac{W}{r_A+r_B+r_C} = \frac{1}{\frac{1}{30}+ \frac{1}{40}+ \frac{1}{60}} = \frac{40}{3}$ days.

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Assume $B$ can do $4$ units per day. As $A$ is $50\%$ as efficient as $B$, so $A$ can do $2$ units per day. $C$ does half the work done by $A$ and $B$ together. So, $C$ can do $(4+2)/2$ = $3$ units per day. $C$ alone can complete the job in $40$ day. So, the total work is $120$ units. Now, $A$, $B$ and $C$ can do $4+2+3 = 9$ units per day. So, number of days required is $120/9 = 40/3$ days.

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Eficiency of A=e,B=2e,C=1.5 e Now 40*1.5e=1 e=1/60 Say x is the time required to finish the work by A+B+C x*(2e+e+1.5e)=1 x=40/3

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