GRE Work/Rate Problems - Tips and Tricks you must know with examples
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1. If a man can do a piece of work in N days (or hours or any other unit of time), then the work done by him in one day will be \(\frac{1}{N}\) of the total work.
1. If A and B together finish a piece of work in 10 days & B alone can finish it in 20 days. In how many days can A alone finish the work ?
Sol. Let X and Y be the number of days required by A and B respectively. ⇒ By the standard formula, \(\frac{XY}{(X + Y)} = 10\) & \(Y = 20 ⇒ \frac{X \times 20 }{(X + 20)} = 10\) or X = 20 days
2. If A is twice as good a workman as B, then A will take half the time B takes to finish a piece of work.
2. Four men working together all day, can finish a piece of work in 11 days; but two of them having other engagements can work only one half–time and quarter time respectively. How long will it take them to complete the work ?
Sol. Each man will take 11 x 4 = 44 days to complete the work. If one man works half day/day he will take 44 x 2 = 88 days to finish the work. Similarly, a man working quarter \(\frac{day}{day}\) will take 44 x 4 = 176 days to finish the work. When these work together they will require \(\frac{1 }{\left[ \begin{array}{cc|r} (\frac{1}{44}) + (\frac{1}{44}) + (\frac{1}{88}) + (\frac{1}{176}) \end{array} \right] }= 16\) days.
3. If A and B can do a piece of work in X & Y days respectively while working alone, they will together take \(\frac{XY }{ (X + Y)}\) days to complete it.
3. 20 men can complete a piece of work in 10 days, but after every 4 days 5 men are called off, in what time will the work be finished ?
Total work = 20 x 10 = 200 monday
1. First 4 days' work = 20 x 4 = 80 md
2. Next 4 days' work = 15 x 4 = 60 md
3. Next 4 days' work = 10 x 4 = 40 md
4. Next 4 days' work = 5 x 4 = 20
\(Σ = 200\) md
Hence, days reqd = 4 + 4 + 4 + 4 = 16
4. If A, B, C can do a piece of work in X, Y, Z days respectively while working alone, they will together take \(\frac{XYZ }{ [XY + YZ + ZX]}\) days to finish it.
4. A vessel can be filled by one pipe A in 10 minutes, by a second pipe B in 15 minutes. It can be emptied by a waste pipe C in 9 minutes. In what time will the vessel be filled if all the three were turned on at once ?
Sol. We will follow exactly the same method as in time & work. The part of a vessel filled in 1 minute when all three are on \(= \frac{1}{10} + \frac{1}{15} – \frac{1}{9} = \frac{1}{18} ⇒\) Total vessel will be filled in 18 minutes
5. If an inlet pipe can fill a cistern in X hours, the part filled in 1 hour = \(\frac{1}{X}\)
5. Three pipes A, B and C can fill a cistern in 15, 20 and 30 min resp. They were all turned on at the same time. After 5 minutes the first two pipes were turned off. In what time will the cistern be filled ?
Sol. A, B and C can fill \((\frac{1}{15} + \frac{1}{20} + \frac{1}{30})\) or \(\frac{3}{20}\) of the cistern in 1 minute ⇒ A, B and C filled \((\frac{3}{20} x 5)\) or \(\frac{3}{4}\) of the cistern in 5 min. Now A and B are turned off \(⇒ (1 – \frac{3}{4})\) or \(\frac{1}{4}\) of the cistern wil be filled by C ⇒ C will fill \(\frac{1}{4}\) of the cistern in \((30 \times \frac{1}{4})\) or 7.5 minutes ⇒ The cistern will be filled in 7.5 + 5 or 12.5 min
6. If an inlet pipe can fill a tank in X hours and an outlet pipe empties the full tank in Y hours, then the net part filled in 1 hour when both the pipes are opened \(= (\frac{1}{X}) – (\frac{1}{Y}) ⇒\) In 1 hour, the part filled (or emptied) \(= \frac{1}{X }– \frac{1}{Y} ⇒ \) Time required to fill or empty the tank \(= \frac{XY }{ (X \sim Y)}\) hours. \(X \sim Y\) indicates \([X – Y]\) or \([Y – X]\), whichever is positive).
6. A cistern can be filled by two taps A and B in 12 minutes and 14 minutes respectively and can be emptied by a third in 8 minutes. If all the taps are turned on at the same moment, what part of the cistern will remain unfilled at the end of 7 minutes ?
Sol. We have \((\frac{7}{12}) + (\frac{7}{14}) – \frac{7}{8} = \frac{5}{24}\) part filled in 7 minutes. Hence \(1 – \frac{5}{24} = \frac{19}{24}\) th of the tank is unfilled.
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