Q31. A, B and C working together, can do a piece of work in 18 days. A could do it alone in 72 days. After working together for 6 days , A is fallen ill. How long will B and C together take to finish the remaining work?
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$\Rightarrow$ (A + B + C)'s 1 days work = $\dfrac{1}{18}$
$\Rightarrow$ (A + B + C)'s 6 days work = $\dfrac{1}{18} \times 6 = \dfrac{1}{3}$
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{72}$
$\therefore$ (B + C) 1 day's work:
$\Rightarrow \dfrac{1}{18} - \dfrac{1}{72} = \dfrac{4 - 1}{72} = \dfrac{3}{72} = \dfrac{1}{24}$ days
$\Rightarrow$ Remaining work = 1 $- \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3}$
$\therefore$ (B + C) will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{2}{3}}{\dfrac{1}{24}} = \dfrac{2 \times 24}{3} = 2 \times 8 = 16$ days Answer
$\Rightarrow$ (A + B + C)'s 6 days work = $\dfrac{1}{18} \times 6 = \dfrac{1}{3}$
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{72}$
$\therefore$ (B + C) 1 day's work:
$\Rightarrow \dfrac{1}{18} - \dfrac{1}{72} = \dfrac{4 - 1}{72} = \dfrac{3}{72} = \dfrac{1}{24}$ days
$\Rightarrow$ Remaining work = 1 $- \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3}$
$\therefore$ (B + C) will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{2}{3}}{\dfrac{1}{24}} = \dfrac{2 \times 24}{3} = 2 \times 8 = 16$ days Answer
Q32. For doing a certain work, A's ability is equal to the joint abilities of B and C. If A and B can it in 2 days and C alone in 10 days, in what time could B alone do it?
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$\Rightarrow$ Given: A's work = (B + C)'s work
$\Rightarrow$ (A + B)'s 1 day's work = $\dfrac{1}{2}$
$\Rightarrow$ C's 1 day's work = $\dfrac{1}{10}$
$\Rightarrow$ Let B's 1 day's work be = $\dfrac{1}{x}$
$\therefore$ A's 1 day's work = $\dfrac{1}{x} + \dfrac{1}{10}$
$\Rightarrow$ So, we have the equation = A + B = (A + B)'s 1 days work
$\Rightarrow (\dfrac{1}{x} + \dfrac{1}{10}) + \dfrac{1}{x} = \dfrac{1}{2}$
$\Rightarrow (\dfrac{1}{x} + \dfrac{1}{x}) + \dfrac{1}{10} = \dfrac{1}{2}$
$\Rightarrow \dfrac{2}{x} + \dfrac{1}{10} = \dfrac{1}{2}$
$\Rightarrow \dfrac{2}{x} = \dfrac{1}{2} - \dfrac{1}{10}$
$\Rightarrow \dfrac{2}{x} = \dfrac{5 - 1}{10} = \dfrac{4}{10}$
$\Rightarrow 2 \times 10 = 4x$
$\Rightarrow 4x = 20$
$\Rightarrow x = \dfrac{20}{4} = 5$ days Answer
$\Rightarrow$ (A + B)'s 1 day's work = $\dfrac{1}{2}$
$\Rightarrow$ C's 1 day's work = $\dfrac{1}{10}$
$\Rightarrow$ Let B's 1 day's work be = $\dfrac{1}{x}$
$\therefore$ A's 1 day's work = $\dfrac{1}{x} + \dfrac{1}{10}$
$\Rightarrow$ So, we have the equation = A + B = (A + B)'s 1 days work
$\Rightarrow (\dfrac{1}{x} + \dfrac{1}{10}) + \dfrac{1}{x} = \dfrac{1}{2}$
$\Rightarrow (\dfrac{1}{x} + \dfrac{1}{x}) + \dfrac{1}{10} = \dfrac{1}{2}$
$\Rightarrow \dfrac{2}{x} + \dfrac{1}{10} = \dfrac{1}{2}$
$\Rightarrow \dfrac{2}{x} = \dfrac{1}{2} - \dfrac{1}{10}$
$\Rightarrow \dfrac{2}{x} = \dfrac{5 - 1}{10} = \dfrac{4}{10}$
$\Rightarrow 2 \times 10 = 4x$
$\Rightarrow 4x = 20$
$\Rightarrow x = \dfrac{20}{4} = 5$ days Answer
Q33. A completes $\dfrac{3}{5}$ of a piece of work in 36 days and then gets the assistance of B. They work together for 6 days, when B leaves and A finishes the remaining work in 9 days. How long B would have taken to do the work alone?
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$\Rightarrow$ A's 6 day's work = $\dfrac{3}{5}$
$\Rightarrow$ A's 1 day's work = $\dfrac{3}{5 \times 36} = \dfrac{3}{180} = \dfrac{1}{60}$
Now total no. of days in which A finished the work = $36 + 6 + 9 = 51$ days
$\Rightarrow$ We have, A's 1 days work = $\dfrac{1}{60}$
$\therefore$ A's 51 day's work = $\dfrac{1}{60} \times 51 = \dfrac{51}{60} = \dfrac{17}{20}$
$\Rightarrow$ And, B's 6 day's work = 1 $- \dfrac{17}{20} = \dfrac{20 - 17}{20} = \dfrac{3}{20}$
$\therefore$ B's 1 day's work = $\dfrac{3}{20 \times 6} = \dfrac{3}{120} = \dfrac{1}{40}$
$\Rightarrow$ Hence, B can alone finish the work in = $\dfrac{40}{1} = 40$ days Answer
$\Rightarrow$ A's 1 day's work = $\dfrac{3}{5 \times 36} = \dfrac{3}{180} = \dfrac{1}{60}$
Now total no. of days in which A finished the work = $36 + 6 + 9 = 51$ days
$\Rightarrow$ We have, A's 1 days work = $\dfrac{1}{60}$
$\therefore$ A's 51 day's work = $\dfrac{1}{60} \times 51 = \dfrac{51}{60} = \dfrac{17}{20}$
$\Rightarrow$ And, B's 6 day's work = 1 $- \dfrac{17}{20} = \dfrac{20 - 17}{20} = \dfrac{3}{20}$
$\therefore$ B's 1 day's work = $\dfrac{3}{20 \times 6} = \dfrac{3}{120} = \dfrac{1}{40}$
$\Rightarrow$ Hence, B can alone finish the work in = $\dfrac{40}{1} = 40$ days Answer
Q34. John can do a piece of work in 20 days. Ram is found to be 25$\%$ less efficient than John. Find, in how many days will:
(i) Ram can do the work alone?
(ii) John and Ram do the same work together?
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(i) John's 1 day's work = $\dfrac{1}{20}$
$\Rightarrow$ Ram's 1 day's work = $\dfrac{1}{20} - 25\% \text{ of } \dfrac{1}{20}$
$\Rightarrow \dfrac{1}{20} - \dfrac{25}{100} \times \dfrac{1}{20}$
$\Rightarrow \dfrac{1}{20} - \dfrac{5}{100} \times \dfrac{1}{4}$
$\Rightarrow \dfrac{1}{20} - \dfrac{1}{20} \times \dfrac{1}{4}$
$\Rightarrow \dfrac{1}{20} - \dfrac{1}{80} = \dfrac{4 - 1}{80} = \dfrac{3}{80}$
$\therefore$ Hence, Ram can alone do the work in = $\dfrac{80}{3} = 26\dfrac{2}{3}$ days Answer
(ii) John's 1 day's work = $\dfrac{1}{20}$
$\Rightarrow$ Ram's 1 day's work = $\dfrac{3}{80}$
$\Rightarrow$ (John + Ram)'s 1 day's work:
$\Rightarrow \dfrac{1}{20} + \dfrac{3}{80} = \dfrac{4 + 3}{80} = \dfrac{7}{80}$
$\therefore$ Hence, (John + Ram) can together do the work in = $\dfrac{80}{7} = 11\dfrac{3}{7}$ days Answer
$\Rightarrow$ Ram's 1 day's work = $\dfrac{1}{20} - 25\% \text{ of } \dfrac{1}{20}$
$\Rightarrow \dfrac{1}{20} - \dfrac{25}{100} \times \dfrac{1}{20}$
$\Rightarrow \dfrac{1}{20} - \dfrac{5}{100} \times \dfrac{1}{4}$
$\Rightarrow \dfrac{1}{20} - \dfrac{1}{20} \times \dfrac{1}{4}$
$\Rightarrow \dfrac{1}{20} - \dfrac{1}{80} = \dfrac{4 - 1}{80} = \dfrac{3}{80}$
$\therefore$ Hence, Ram can alone do the work in = $\dfrac{80}{3} = 26\dfrac{2}{3}$ days Answer
(ii) John's 1 day's work = $\dfrac{1}{20}$
$\Rightarrow$ Ram's 1 day's work = $\dfrac{3}{80}$
$\Rightarrow$ (John + Ram)'s 1 day's work:
$\Rightarrow \dfrac{1}{20} + \dfrac{3}{80} = \dfrac{4 + 3}{80} = \dfrac{7}{80}$
$\therefore$ Hence, (John + Ram) can together do the work in = $\dfrac{80}{7} = 11\dfrac{3}{7}$ days Answer
Q35. Manisha is able to do 40$\%$ more work than Neetu in the same time. If Manisha can do a certain piece of work in 60 days, find in how days will:
(i) Neetu do the work alone?
(ii) Both do the work together?
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(i) Let Neetu's work be completed in = $x$ days
$\therefore$ Neetu's 1 day's work = $\dfrac{1}{x}$
$\Rightarrow$ Manisha being 40\% more efficient than Neetu means:
$\Rightarrow \dfrac{1}{x} + 40\% \text{ of } \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{40}{100} \times \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{8}{20} \times \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{2}{5} \times \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{2}{5x}$
$\Rightarrow \dfrac{5 + 2}{5x} = \dfrac{7}{5x}$
$\Rightarrow$ Given, Manisha's 1 day's work = $\dfrac{1}{60}$
$\Rightarrow$ On the fact that Neetu is 40\% less equivalent than Manisha, we have:
$\Rightarrow \dfrac{7}{5x} = \dfrac{1}{60}$
$\Rightarrow 7 \times 60 = 5x$
$\Rightarrow 5x = 420$
$\Rightarrow x = \dfrac{420}{5} = 84$
$\Rightarrow$ Hence, Neetu can finish the work in alone in 84 days. Answer
(ii) (Manisha + Neetu)'s 1 days work = $\dfrac{1}{60} + \dfrac{1}{84}$
$\Rightarrow \dfrac{7 + 5}{420} = \dfrac{12}{420} = \dfrac{2}{70} = \dfrac{1}{35}$
$\therefore$ Manisha and Neetu can together finish the work in = 35 days Answer
$\therefore$ Neetu's 1 day's work = $\dfrac{1}{x}$
$\Rightarrow$ Manisha being 40\% more efficient than Neetu means:
$\Rightarrow \dfrac{1}{x} + 40\% \text{ of } \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{40}{100} \times \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{8}{20} \times \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{2}{5} \times \dfrac{1}{x}$
$\Rightarrow \dfrac{1}{x} + \dfrac{2}{5x}$
$\Rightarrow \dfrac{5 + 2}{5x} = \dfrac{7}{5x}$
$\Rightarrow$ Given, Manisha's 1 day's work = $\dfrac{1}{60}$
$\Rightarrow$ On the fact that Neetu is 40\% less equivalent than Manisha, we have:
$\Rightarrow \dfrac{7}{5x} = \dfrac{1}{60}$
$\Rightarrow 7 \times 60 = 5x$
$\Rightarrow 5x = 420$
$\Rightarrow x = \dfrac{420}{5} = 84$
$\Rightarrow$ Hence, Neetu can finish the work in alone in 84 days. Answer
(ii) (Manisha + Neetu)'s 1 days work = $\dfrac{1}{60} + \dfrac{1}{84}$
$\Rightarrow \dfrac{7 + 5}{420} = \dfrac{12}{420} = \dfrac{2}{70} = \dfrac{1}{35}$
$\therefore$ Manisha and Neetu can together finish the work in = 35 days Answer
Q36. Three taps can fill a cistern in 36, 30 and 20 minutes. The cistern being empty, all the three taps are kept open and after 6 minutes the first tap is closed. Find in how many more minutes will the cistern be full?
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$\Rightarrow$ Tap (A + B + C) can fill a cistern in = $\dfrac{1}{36} + \dfrac{1}{30} + \dfrac{1}{20}$
$\Rightarrow \dfrac{5 + 6 + 9}{180} = \dfrac{20}{180} = \dfrac{1}{9}$
$\therefore$ Three taps can together fill the cistern in = 9 minutes
$\Rightarrow$ Tap A's 6 minutes fill = $\dfrac{1}{36} \times 6 = \dfrac{1}{6}$
$\therefore$ Tap A's fill before being closed = 6 minutes
$\therefore$ Required more minutes = $9 - 6 = 3$ more minutes Answer
$\Rightarrow \dfrac{5 + 6 + 9}{180} = \dfrac{20}{180} = \dfrac{1}{9}$
$\therefore$ Three taps can together fill the cistern in = 9 minutes
$\Rightarrow$ Tap A's 6 minutes fill = $\dfrac{1}{36} \times 6 = \dfrac{1}{6}$
$\therefore$ Tap A's fill before being closed = 6 minutes
$\therefore$ Required more minutes = $9 - 6 = 3$ more minutes Answer
Q37. A takes twice as much as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. Find how long B would take it alone to finish the work?
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$\Rightarrow$ Let no. of days taken by B be = $x$
$\therefore$ B's 1 days work = $\dfrac{1}{x}$
$\Rightarrow$ Given, A works twice as much as B = 2x
$\therefore$ A's 1 day's work = $\dfrac{1}{2x}$
$\Rightarrow$ Given, C's work thrice as much as A
$\Rightarrow$ 3 = 2x = $\dfrac{2x}{3}$
$\therefore$ C's 1 day's work = $\dfrac{3}{2x}$
$\Rightarrow$ As such, (A + B + C)'s 1 days work = $\dfrac{1}{2x} + \dfrac{1}{x} + \dfrac{3}{2x}$
$\Rightarrow$ But given, (A + B + C)'s 1 days work = $\dfrac{1}{2}$
$\therefore \dfrac{1}{2x} + \dfrac{1}{x} + \dfrac{3}{2x} = \dfrac{1}{2}$
$\Rightarrow \dfrac{1 + 2 + 3}{2x} = \dfrac{1}{2}$
$\Rightarrow \dfrac{6}{2x} = \dfrac{1}{2}$
$\Rightarrow 6 \times 2 = 2x$
$\Rightarrow 12 = 2x$
$\therefore x = \dfrac{12}{2} = 6$ days Answer
$\therefore$ B's 1 days work = $\dfrac{1}{x}$
$\Rightarrow$ Given, A works twice as much as B = 2x
$\therefore$ A's 1 day's work = $\dfrac{1}{2x}$
$\Rightarrow$ Given, C's work thrice as much as A
$\Rightarrow$ 3 = 2x = $\dfrac{2x}{3}$
$\therefore$ C's 1 day's work = $\dfrac{3}{2x}$
$\Rightarrow$ As such, (A + B + C)'s 1 days work = $\dfrac{1}{2x} + \dfrac{1}{x} + \dfrac{3}{2x}$
$\Rightarrow$ But given, (A + B + C)'s 1 days work = $\dfrac{1}{2}$
$\therefore \dfrac{1}{2x} + \dfrac{1}{x} + \dfrac{3}{2x} = \dfrac{1}{2}$
$\Rightarrow \dfrac{1 + 2 + 3}{2x} = \dfrac{1}{2}$
$\Rightarrow \dfrac{6}{2x} = \dfrac{1}{2}$
$\Rightarrow 6 \times 2 = 2x$
$\Rightarrow 12 = 2x$
$\therefore x = \dfrac{12}{2} = 6$ days Answer