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A, B and C need a certain unique time to do a certain work. C needs 1 hours less than A to complete the work. Working together, they require 30 minutes to complete 50% of the job. The work also gets completed if A and B start working together and A leaves after 1 hour and B works for a further 3 hours. How much work does C do per hour?

(A)16.66%

(B)33.33%

(C)50%

(D)66.66%

My attempt:

Let the total work be 100 units.

Let the work done by A,B and C be a units/hour,b units/hour,c units/hour respectively.

Let the time taken by A alone to complete the work be t hours.

ATQ: \begin{align*} (a+b+c) \cdot \frac{1}{2} & =50 \tag{1}\\ (a+b) \cdot 1+b \cdot 3 & =100 \tag{2}\\ c \cdot (t-1)& =100 \tag{3}\\ at & =100 \tag{4} \end{align*} Please help me solve these equations. When I am solving it is getting cumbersome.

Also if someone tells us some other way of solving, that would be helpful as well. Thanks.

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Let $A$, $B$ and $C$ be the rates measured in an amount of work per hour $\left(\frac{work}{hour}\right)$ at which workers $A$, $B$ and $C$ accomplish work. If $C$ is the amount of work worker $C$ can do in one hour and $1$ ($100\%$) is the amount of work of the entire job, then $t$ is the time it takes him to bring it to completion: $t=\frac{1}{C}\cdot\frac{\text{work}}{\text{work/hour}}$. We also know that it takes worker $A$ one hour longer to complete the job than it takes worker $C$: $A(t+1)=1 \implies A\left(\frac{1}{C}+1\right)=1$. Another condition that we have is that three of them together require $30$ (or $\frac{1}{2}$ of an hour, don't forget that time is masured in hours here) minutes to complete $50\%$ of the job: $\frac{1}{2}(A+B+C)=\frac{1}{2} \implies (A+B+C)=1$. It is aslo stated that the work also gets completed if worker $A$ and worker $B$ start working together and worker $A$ leaves after $1$ hour and worker $B$ works for a further $3$ hours: $1\cdot(A+B)+3B=1 \implies A+4B=1$. We're all set to go now. Our conditions:

$$ A\left(\frac{1}{C}+1\right)=1\\ A+B+C=1\\ A+4B=1 $$


Solution:

$$ A\cdot\left(\frac{1}{C}+1\right)=1 \implies A=\frac{C}{1+C},\\ A+4B=1 \implies B=\frac{1-A}{4} \implies B=\frac{1}{4(1+C)},\\ A+B+C=1 \implies \frac{C}{1+C}+\frac{1}{4(1+C)}+C=1 \implies C^2+C-1=0 \implies \\ C_{1}=-\frac{3}{2}\text{ (discarded)}\\ C_{2}=\frac{1}{2}=\frac{1\text{ work}}{2\text{ hour}}≡50\% $$

Answer: $C$ does $50\%$ of work per hour.

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  • $\begingroup$ Could you please have a look at this question : math.stackexchange.com/q/2567879/394202 $\endgroup$ – Soumee Dec 15 '17 at 17:59
  • $\begingroup$ I did. Rohan's solution is pretty good. I would be writing pretty much the same thing. You won't be able to do it better than that. The algebra just gets really hairy. $\endgroup$ – Michael Rybkin Dec 15 '17 at 22:47
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Let us assume, they take $a, b,$ and $c$ hours to finish the job individually, respectively. Note the following points:

  • They complete half the job in half an hour $\implies$ full job is completed in a hour. Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

  • Also, A takes 1 hour more time than C. So, $a= c+1\implies \frac{1}{a}=\frac{1}{c+1}$

  • If A works for a hour, and B works for four hours, then $\frac{1}{a}+\frac{4}{b}=1 \implies \frac{1}{b}=\frac{1-\frac{1}{a}}{4}=\frac{1-\frac{1}{c+1}}{4}=\frac{c}{4+4c}$.

Now, solving for $c$ gives us the quadratic: $3c^2-4c-4=0 \implies c=2, c= -\frac{2}{3}$. As, $c >0$, we have, the rate of work C does $=0.5$.


Solving by your method: $a =\frac{100}{t}, b = \frac{100-a}{4}=\frac{25t-25}{t}$. Then, we get, $$a +b + c =100 \implies 100+25t-25 + ct = 100t \implies t = \frac{75}{75-c}$$

Putting this in Eq.(3), we get, $$\frac{100}{c}=\frac{c}{75-c}\equiv 7500-100c = c^2 \equiv c^2+100c-7500=0 \implies c = \frac{1}{2}, -\frac{3}{2}$$

Rejecting the negative solution, we have $c =\frac12$, the same answer as obtained earlier.

No mistake in your approach!

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Let the total work be 1, or 100%.

Let the efficiency of A,B and C be a,b and c(percentage of work be done in an hour)


(1) $\frac{1}{c}$ + 1 = $\frac{1}{a}$

(2) $(a+b+c)\cdot\frac{1}{2}$ = $\frac{1}{2}$

(3) $(a+b)\cdot1 + b\cdot3 = 1$


from (3) you get

a + 4b = 1

b = $\frac{1-a}{4}$


put it into (2)

a + $\frac{1-a}{4}$ + c = 1

4a + 1 - a + 4c = 4

a = $\frac{3-4c}{3}$


put it into (1)

$\frac{1}{c} + 1 = \frac{3}{3-4c}$

$\frac{1+c}{c} = \frac{3}{3-4c}$

$4c^2 + 4c - 3 = 0$

$4c^2 + 4c + 1 = 4$

$(2c + 1)^2 = 4$

2c + 1 = 2 (c can not be negative)

c = 0.5

The final answer is C 50%.

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Ok I'm going to check just options so if you are looking for equations this is not your answer but question can be solved in this way too...

Let's take option B , i.e C does 33.33 % work in one hours That means for 100% it requires 3 hours which means A requires 4 hours for 100% work to be done.

That means per hour A does 25% work Now ATQ for they all take 1 hour together to complete 100% work this means B will do 100-25-33.33 = 41.66% of work in 1 hour

Now verify this with another statement i.e A+4B=100 which shows that B is far more than required

Thus C should be more.

Now check for 50% work by C in 1 hour and everything fits perfectly.

PS: this method may seem cumbersome but it just require basic mathematics and no paper and pen is needed. Thus saves lot of time.

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