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?





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.


4 Answers 4


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!


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:

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


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

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

  • $\begingroup$ Could you please have a look at this question : math.stackexchange.com/q/2567879/394202 $\endgroup$
    – Soumee
    Dec 15, 2017 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. $\endgroup$ Dec 15, 2017 at 22:47

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%.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.