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A can do a piece of work four hours faster than B. They worked together for two hours and then the remaining part of the work was done by A in an hour. How many hours would B take to complete the job if he were to work alone?

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2 Answers 2

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let t be the time A takes in hours, then A does $\frac{1}{t}$ per hour. B does $\frac{1}{t+4}$ per hour. we know $\frac{3}{t}+\frac{2}{t+4}=1$ $\rightarrow 3+\frac{2t}{t+4}=t\rightarrow 3(t+4)+2t=t^2+4t\rightarrow t^2-t-12=0\rightarrow (t-4)(t+3)=0$ so $t=4$

therefore A takes 4 hours and B takes 8 hours.

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  • $\begingroup$ No, B does $\frac 1{t+4}$ per hour. $\endgroup$ Nov 21, 2013 at 19:02
  • $\begingroup$ there is a little error in the explanation above.Nevertheless,i have understood the method to be used to arrive at the answer. $\endgroup$
    – ayush
    Nov 21, 2013 at 19:20
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Hint: let $a$ be the fraction of the job that $A$ can do in one hour and $b$ be the fraction that $B$ can do in one hour. Write two equations that express the first two sentences of your problem. You have two simultaneous equations in two unknowns, so solve them. Your answer is then $\frac 1b$

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