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I came across a question which says:

3 members of a family A, B and C work together to get all household chores done. The time it takes them to do the work together is six hours less than A would have taken working alone, one hour less than B would have taken alone, and half the time C would have taken working alone. How long did it take them to do these chores working together ?

My solution follows:

Let time taken by A, B and C to do the work be $A$, $B$ and $C$ hours respectively. Their speeds are $\frac{1}{A}$, $\frac{1}{B}$ and $\frac{1}{C}$ respectively. Hence the time taken working together should be $$\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{BC+AC+AB}{ABC} = X$$

Next, according to question,

$$X = A-6 = B-1 = \frac{C}{2}$$

Taking the last two equations, we get:

$$2B-2 = C$$

But if I substitute $C$ in the value of $X$, I would get a very long value which will still remain unsolvable ...

A thorough Google Search revealed no helpful answers.

Any hints or suggestions would be greatly appreciated.

Thank you.

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  • $\begingroup$ Why $(1/A)+(1/B)+(1/C)$? $\endgroup$ Jan 11 '14 at 13:51
  • $\begingroup$ @GerryMyerson Should it be not ? I thought adding the rates usually gave the time? $\endgroup$ Jan 11 '14 at 13:53
  • $\begingroup$ If it takes each of them 1 hour, will it take them 3 hours working together? $\endgroup$ Jan 11 '14 at 13:55
  • $\begingroup$ @GerryMyerson Hmmm... I see you are right. Silly me. Can you please explain how should I solve this question (a hint/ suggestion something like that)? Thanks a lot. $\endgroup$ Jan 11 '14 at 13:57
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The correct equation to start with here is

$${1\over X}={1\over A}+{1\over B}+{1\over C}$$

Everything else you did looks correct, so see if you can take it from here.

It's worth noting what's wrong in writing $X={1\over A}+{1\over B}+{1\over C}$. Each of $A$, $B$, $C$, and $X$, as you've defined them, have dimensions of time, assumed in hours. This means that $1/A$ etc. have dimensions of "per hour." You yourself called it "speed." Just as you can't (or shouldn't) add apples and oranges, you can't equate "hours" with "per hours." In working problems of this type, it often helps to keep track of the dimensions that various quantities have, and make sure the dimensions match up whenever you're adding or equating them.

Added at OP's request: Since $X$ is what you want to know, it's best here to convert your equation(s) $X=A-6=B-1=C/2$ into $A=X+6$, $B=X+1$, and $C=2X$, and then change the original equation to

$${1\over X}={1\over X+6}+{1\over X+1}+{1\over2X}$$

Bring the $1/2X$ over the left hand side, which gives

$${1\over 2X}={1\over X+6}+{1\over X+1}$$

From this you should wind up with a quadratic in $X$. Can you take it from here?

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  • $\begingroup$ oh, i see . it's indeed a lot of trouble when everything is messed up. BTW, you say that after using the above equation the question should be solvable. But I am not able to solve it still :( Please help. $\endgroup$ Jan 11 '14 at 14:22
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Let $x$ be the hours to have the work done when all three work together. Therefore, A would take $(x+6)$ hours, B would take $(x+1)$ hours and C $2x$ hours

$$\frac{1}{1/(x+6)+1/(x+1) +1/2x}=x$$

Rearrange the equation, one would have: $$\frac{(x+6)(x+1)2x}{((x+1)2x+(x+6)2x+(x+6)(x+1))}=x$$ Cancel x on both sides and cross multiply, one would have $$2(x+6)(x+1)=(x+1)2x+(x+6)2x+(x+6)(x+1)$$ Then, $$(x+6)(x+1)=(x+1)2x+(x+6)2x$$ Expand each expression, and combine similar items, one would have a quadratic equation: $$3x^2+7x-6=0$$ Solve the equation, one would find $x=-3$ or $\frac 23$

$\frac 23$hours is the answer.

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  • $\begingroup$ Hi George, welcome to MSE! Hope to see you around here more. Also, please do use MathJax to format your posts. I edited your post as an example so that it is more clear. $\endgroup$ Jan 20 '18 at 14:05
  • $\begingroup$ Thanks, Gaurang! It looks much better :) $\endgroup$
    – George Guo
    Jan 20 '18 at 22:51
  • $\begingroup$ You're welcome :) $\endgroup$ Jan 21 '18 at 2:13

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