Q16. Two pipes P and Q would fill an empty cistern in 24 minutes and 32 minutes respectively. Both the pipes being opened together, find when the first pipe must be turned off so that the empty cistern may be just filled in 16 minutes.
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$\Rightarrow$ Let P's fill time before being closed be = t
$\therefore$ P's t mins fill = $\dfrac{t}{24}$
$\Rightarrow$ Q's 1 min fill = $\dfrac{1}{32}$
$\Rightarrow$ Q's 16 mins fill = $\dfrac{1}{32} \times 16 = \dfrac{16}{32} = \dfrac{1}{2}$
$\Rightarrow$ let the completed job be = 1
So we've the equation:
$\Rightarrow \dfrac{t}{24} + \dfrac{1}{2} = 1$
$\Rightarrow \dfrac{t}{24} = 1 - \dfrac{1}{2}$
$\Rightarrow \dfrac{t}{24} = \dfrac{2 - 1}{2} = \dfrac{1}{2}$
$\therefore t = \dfrac{1}{2} \times 24 = 12$ mins. Answer
$\therefore$ P's t mins fill = $\dfrac{t}{24}$
$\Rightarrow$ Q's 1 min fill = $\dfrac{1}{32}$
$\Rightarrow$ Q's 16 mins fill = $\dfrac{1}{32} \times 16 = \dfrac{16}{32} = \dfrac{1}{2}$
$\Rightarrow$ let the completed job be = 1
So we've the equation:
$\Rightarrow \dfrac{t}{24} + \dfrac{1}{2} = 1$
$\Rightarrow \dfrac{t}{24} = 1 - \dfrac{1}{2}$
$\Rightarrow \dfrac{t}{24} = \dfrac{2 - 1}{2} = \dfrac{1}{2}$
$\therefore t = \dfrac{1}{2} \times 24 = 12$ mins. Answer
Q17. A can do a piece of work in 20 days and B in 30 days. How long will they take together to complete it?
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$\Rightarrow$ A's 1 days work = $\dfrac{1}{20}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{30}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{20} + \dfrac{1}{30} = \dfrac{3 + 2}{60} = \dfrac{5}{60} = \dfrac{1}{12}$
$\therefore$ A and B together will finish the work in = $\dfrac{12}{1} = 12$ days Answer
$\Rightarrow$ B's 1 days work = $\dfrac{1}{30}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{20} + \dfrac{1}{30} = \dfrac{3 + 2}{60} = \dfrac{5}{60} = \dfrac{1}{12}$
$\therefore$ A and B together will finish the work in = $\dfrac{12}{1} = 12$ days Answer
Q18. A and B can do a piece of work in $15\dfrac{1}{3}$ days, but A alone can do it in 20 days. How long will B take to do it alone?
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$\Rightarrow (A + B)$ can complete a work in = $13\dfrac{1}{3} = \dfrac{40}{3}$ days
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{3}{40}$
$\Rightarrow$ A's 1 days work = $\dfrac{1}{20}$
$\Rightarrow$ B's 1 days work = $\dfrac{3}{40} - \dfrac{1}{20} = \dfrac{3 - 2}{40} = \dfrac{1}{40}$
$\therefore$ B can complete the work in = 40 days Answer
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{3}{40}$
$\Rightarrow$ A's 1 days work = $\dfrac{1}{20}$
$\Rightarrow$ B's 1 days work = $\dfrac{3}{40} - \dfrac{1}{20} = \dfrac{3 - 2}{40} = \dfrac{1}{40}$
$\therefore$ B can complete the work in = 40 days Answer
Q19. A and B can do a piece of work in 20 days, B and C in 15 days, C and A in 12 days.
(i) How long will it take them to do the work together?
(ii) In what time can each finish it working alone?
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$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{20}$
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{15}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{12}$
(i) $2(A + B + C)$'s 1 days work = $\dfrac{1}{20} + \dfrac{1}{15} + \dfrac{1}{12}$
$\Rightarrow \dfrac{3 + 4 + 5}{60} = \dfrac{12}{60} = \dfrac{1}{5}$
$\therefore$ A + B + C's 1 days work= $\dfrac{1}{5 \times 2} = \dfrac{1}{10}$
So, A, B and C can together finish the work in = $\dfrac{10}{1} = 10$ days Ans.
(ii) A's 1 days work = $(A + B + C) - (B + C)$
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3 - 2}{30} = \dfrac{1}{30}$
$\therefore$ A alone can finish the work in = $\dfrac{30}{1} = 30$ days Ans.
$\Rightarrow$ B's 1 days work = $(A + B + C) - (A + C)$
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{12} = \dfrac{6 - 5}{60} = \dfrac{1}{60}$
$\therefore$ B alone can finish the work in = $\dfrac{60}{1} = 60$ days Ans.
$\Rightarrow$ C's 1 days work = $(A + B + C) - (A + B)$
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{20} = \dfrac{2 - 1}{20} = \dfrac{1}{20}$
$\therefore$ C alone can finish the work in = $\dfrac{20}{1} = 20$ days Answer
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{15}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{12}$
(i) $2(A + B + C)$'s 1 days work = $\dfrac{1}{20} + \dfrac{1}{15} + \dfrac{1}{12}$
$\Rightarrow \dfrac{3 + 4 + 5}{60} = \dfrac{12}{60} = \dfrac{1}{5}$
$\therefore$ A + B + C's 1 days work= $\dfrac{1}{5 \times 2} = \dfrac{1}{10}$
So, A, B and C can together finish the work in = $\dfrac{10}{1} = 10$ days Ans.
(ii) A's 1 days work = $(A + B + C) - (B + C)$
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3 - 2}{30} = \dfrac{1}{30}$
$\therefore$ A alone can finish the work in = $\dfrac{30}{1} = 30$ days Ans.
$\Rightarrow$ B's 1 days work = $(A + B + C) - (A + C)$
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{12} = \dfrac{6 - 5}{60} = \dfrac{1}{60}$
$\therefore$ B alone can finish the work in = $\dfrac{60}{1} = 60$ days Ans.
$\Rightarrow$ C's 1 days work = $(A + B + C) - (A + B)$
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{20} = \dfrac{2 - 1}{20} = \dfrac{1}{20}$
$\therefore$ C alone can finish the work in = $\dfrac{20}{1} = 20$ days Answer
Q20. A tap can fill a cistern in 10 minutes, while the other can empty it in 15 minutes. The cistern being empty and if both the taps are opened together, in how many minutes will the cistern be full?
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$\Rightarrow$ Tap A can fill the cistern in = 10 mins
$\Rightarrow$ Tap B can empty the cistern in = 15 mins
$\Rightarrow$ Tap A can fill the cistern in 1 min = $\dfrac{1}{10}$
$\Rightarrow$ Tap B can empty the cistern in 1 min = $\dfrac{1}{15}$
$\Rightarrow$ Tap A and B's 1 minute fill:
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3 - 2}{30} = \dfrac{1}{30}$ Mins.
$\therefore$ Tap A and B can together fill the cistern in = 30 mins Answer
$\Rightarrow$ Tap B can empty the cistern in = 15 mins
$\Rightarrow$ Tap A can fill the cistern in 1 min = $\dfrac{1}{10}$
$\Rightarrow$ Tap B can empty the cistern in 1 min = $\dfrac{1}{15}$
$\Rightarrow$ Tap A and B's 1 minute fill:
$\Rightarrow \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3 - 2}{30} = \dfrac{1}{30}$ Mins.
$\therefore$ Tap A and B can together fill the cistern in = 30 mins Answer
Q21. A can do a piece of work in 40 days. He works at it for 8 days and then B finishes his remaining work in 16 days. How long will they take to finish the work together?
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$\Rightarrow$ A's 1 days work = $\dfrac{1}{40}$
$\Rightarrow$ A's 8 days work = $\dfrac{1}{40} \times 8 = \dfrac{1}{5}$
$\therefore$ Remaining work = 1 $- \dfrac{1}{5} = \dfrac{5 - 1}{5} = \dfrac{4}{5}$
$\Rightarrow$ B finishes $\dfrac{4}{5}$ of work in 16 days.
$\therefore$ B will finish the whole work in = $\dfrac{16}{\dfrac{4}{5}} = \dfrac{16 \times 5}{4} = 4 \times 5 = 20$
$\Rightarrow$ B's 1 day work = $\dfrac{1}{20}$
$\therefore (A + B)$'s 1 day work:
$\Rightarrow \dfrac{1}{40} + \dfrac{1}{20} = \dfrac{1 + 2}{40} = \dfrac{3}{40}$
$\therefore (A + B)$ will together finish the work in = $\dfrac{40}{3} = 13\dfrac{1}{3}$ days Answer
$\Rightarrow$ A's 8 days work = $\dfrac{1}{40} \times 8 = \dfrac{1}{5}$
$\therefore$ Remaining work = 1 $- \dfrac{1}{5} = \dfrac{5 - 1}{5} = \dfrac{4}{5}$
$\Rightarrow$ B finishes $\dfrac{4}{5}$ of work in 16 days.
$\therefore$ B will finish the whole work in = $\dfrac{16}{\dfrac{4}{5}} = \dfrac{16 \times 5}{4} = 4 \times 5 = 20$
$\Rightarrow$ B's 1 day work = $\dfrac{1}{20}$
$\therefore (A + B)$'s 1 day work:
$\Rightarrow \dfrac{1}{40} + \dfrac{1}{20} = \dfrac{1 + 2}{40} = \dfrac{3}{40}$
$\therefore (A + B)$ will together finish the work in = $\dfrac{40}{3} = 13\dfrac{1}{3}$ days Answer
Q22. A man, a woman and a boy can do a piece of work in 9, 12 and 18 days respectively. What fraction of work would remain unfinished, after 2 men, 3 women and 4 boys had been working at it for 1 day?
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$\Rightarrow$ 1 man's 1 day work = $\dfrac{1}{9}$
$\therefore$ 2 men's 1 day work = $\dfrac{1}{9} \times 2 = \dfrac{2}{9}$
$\Rightarrow$ 1 woman's 1 day work = $\dfrac{1}{12}$
$\therefore$ 3 women's 1 day work = $\dfrac{1}{12} \times 3 = \dfrac{1}{4}$
$\Rightarrow$ 1 boy's 1 day work = $\dfrac{1}{18}$
$\therefore$ 4 boy's 1 day work = $\dfrac{1}{18} \times 4 = \dfrac{2}{9}$
$\Rightarrow$ 2 men's, 3 women's and 4 boy's 1 day's work:
$\Rightarrow \dfrac{2}{9} + \dfrac{1}{4} + \dfrac{2}{9} = \dfrac{8 + 9 + 8}{36} = \dfrac{25}{36}$
$\therefore$ Remaining work = 1 $- \dfrac{25}{36} = \dfrac{36 - 25}{36} = \dfrac{11}{36}$ Answer
$\therefore$ 2 men's 1 day work = $\dfrac{1}{9} \times 2 = \dfrac{2}{9}$
$\Rightarrow$ 1 woman's 1 day work = $\dfrac{1}{12}$
$\therefore$ 3 women's 1 day work = $\dfrac{1}{12} \times 3 = \dfrac{1}{4}$
$\Rightarrow$ 1 boy's 1 day work = $\dfrac{1}{18}$
$\therefore$ 4 boy's 1 day work = $\dfrac{1}{18} \times 4 = \dfrac{2}{9}$
$\Rightarrow$ 2 men's, 3 women's and 4 boy's 1 day's work:
$\Rightarrow \dfrac{2}{9} + \dfrac{1}{4} + \dfrac{2}{9} = \dfrac{8 + 9 + 8}{36} = \dfrac{25}{36}$
$\therefore$ Remaining work = 1 $- \dfrac{25}{36} = \dfrac{36 - 25}{36} = \dfrac{11}{36}$ Answer
Q23. A can do a piece of work in 9 days, B in 11 days. If they work together for 4 days, what fraction of the work will be left unfinished?
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$\Rightarrow$ A's 1 day's work = $\dfrac{1}{9}$
$\Rightarrow$ B's 1 day's work = $\dfrac{1}{11}$
$\Rightarrow (A + B)$'s 1 day work = $\dfrac{1}{9} + \dfrac{1}{11} = \dfrac{11 + 9}{99} = \dfrac{20}{99}$
$\Rightarrow (A + B)$'s 4 day's work = $\dfrac{20}{99} \times 4 = \dfrac{80}{99}$
$\therefore$ Remaining work = 1 $- \dfrac{80}{99} = \dfrac{99 - 80}{99} = \dfrac{19}{99}$ Answer
$\Rightarrow$ B's 1 day's work = $\dfrac{1}{11}$
$\Rightarrow (A + B)$'s 1 day work = $\dfrac{1}{9} + \dfrac{1}{11} = \dfrac{11 + 9}{99} = \dfrac{20}{99}$
$\Rightarrow (A + B)$'s 4 day's work = $\dfrac{20}{99} \times 4 = \dfrac{80}{99}$
$\therefore$ Remaining work = 1 $- \dfrac{80}{99} = \dfrac{99 - 80}{99} = \dfrac{19}{99}$ Answer
Q24. A does $\dfrac{3}{10}$ of a work in 6 days, he then call in B and they finish the work in 7 days more. How long would B take to do the work by himself?
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$\Rightarrow$ A's 6 day's work = $\dfrac{3}{10}$
$\Rightarrow$ A's 1 day's work = $\dfrac{3}{10 \times 6} = \dfrac{3}{60} = \dfrac{1}{20}$
$\Rightarrow$ Let B's time to to finish the work be = b
$\Rightarrow$ Because A works a total of 13 days, we have:
$\Rightarrow \dfrac{13}{20} + \dfrac{7}{b} = 1$
$\Rightarrow \dfrac{13b + (7)20}{20b} = 1$
$\Rightarrow 13b + 140 = 20b$
$\Rightarrow 140 = 20b - 13b$
$\Rightarrow 140 = 7b$
$\Rightarrow 7b = \dfrac{140}{7} = 20$
$\therefore$ B can alone finish the work in = 20 days Answer
$\Rightarrow$ A's 1 day's work = $\dfrac{3}{10 \times 6} = \dfrac{3}{60} = \dfrac{1}{20}$
$\Rightarrow$ Let B's time to to finish the work be = b
$\Rightarrow$ Because A works a total of 13 days, we have:
$\Rightarrow \dfrac{13}{20} + \dfrac{7}{b} = 1$
$\Rightarrow \dfrac{13b + (7)20}{20b} = 1$
$\Rightarrow 13b + 140 = 20b$
$\Rightarrow 140 = 20b - 13b$
$\Rightarrow 140 = 7b$
$\Rightarrow 7b = \dfrac{140}{7} = 20$
$\therefore$ B can alone finish the work in = 20 days Answer
Q25. Sanjay can do a piece of work in 10 days and Anju can do the same work in 8 days. Both work on it for 2 days and then Sanjay stops working. How long will Anju take to complete the remaining work?
Show Answer
$\Rightarrow$ Sanjay's 1 day's work = $\dfrac{1}{10}$
$\Rightarrow$ Anju's 1 day's work = $\dfrac{1}{8}$
$\Rightarrow (Sanjay + Anju)$'s 1 day's work = $\dfrac{1}{10} + \dfrac{1}{8} = \dfrac{4 + 5}{40} = \dfrac{9}{40}$
$\Rightarrow (Sanjay + Anju)$'s 2 day's work = $\dfrac{9}{40} \times 2 = \dfrac{9}{20}$
$\Rightarrow$ Remaining work = 1 $- \dfrac{9}{20} = \dfrac{20 - 9}{20} = \dfrac{11}{20}$
$\therefore$ Anju will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{11}{20}}{\dfrac{1}{8}} = \dfrac{11 \times 8}{20} = \dfrac{88}{20} = \dfrac{44}{10} = \dfrac{22}{5} = 4.4$ days Answer
$\Rightarrow$ Anju's 1 day's work = $\dfrac{1}{8}$
$\Rightarrow (Sanjay + Anju)$'s 1 day's work = $\dfrac{1}{10} + \dfrac{1}{8} = \dfrac{4 + 5}{40} = \dfrac{9}{40}$
$\Rightarrow (Sanjay + Anju)$'s 2 day's work = $\dfrac{9}{40} \times 2 = \dfrac{9}{20}$
$\Rightarrow$ Remaining work = 1 $- \dfrac{9}{20} = \dfrac{20 - 9}{20} = \dfrac{11}{20}$
$\therefore$ Anju will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{11}{20}}{\dfrac{1}{8}} = \dfrac{11 \times 8}{20} = \dfrac{88}{20} = \dfrac{44}{10} = \dfrac{22}{5} = 4.4$ days Answer
Q26. Three taps can fill a cistern in 30 minutes, 45 minutes and 54 minutes. The cistern being empty, all the three taps are kept open and then after 9 minutes, the third tap is closed. In how many more minutes will the cistern be full?
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$\Rightarrow$ Tap A's 1 minute fill = $\dfrac{1}{30}$
$\Rightarrow$ Tap B's 1 minute fill = $\dfrac{1}{45}$
$\Rightarrow$ Tap C's 1 minute fill = $\dfrac{1}{54}$
$\Rightarrow$ Tap (A + B + C)'s 1 minute fill:
$\Rightarrow \dfrac{1}{30} + \dfrac{1}{45} + \dfrac{1}{54}$
$\Rightarrow \dfrac{9 + 6 + 5}{270} = \dfrac{20}{270} = \dfrac{2}{27}$
$\therefore$ Tap (A + B + C) can fill the cistern in = $\dfrac{27}{2} = 13\dfrac{1}{2}$ minutes
$\Rightarrow$ Tap C's 9 minute fill = $\dfrac{54}{9} = 6$
$\therefore$ If Tap C is closed after 9 minutes the cistern will be filled in :
$\Rightarrow \dfrac{27}{2} - 6 = \dfrac{27 - 12}{2} = \dfrac{15}{2}$
$\therefore$ Required more time = $\dfrac{27}{2} - \dfrac{15}{2} = \dfrac{27 - 15}{2} = \dfrac{12}{2} = 6$ mins. Answer
$\Rightarrow$ Tap B's 1 minute fill = $\dfrac{1}{45}$
$\Rightarrow$ Tap C's 1 minute fill = $\dfrac{1}{54}$
$\Rightarrow$ Tap (A + B + C)'s 1 minute fill:
$\Rightarrow \dfrac{1}{30} + \dfrac{1}{45} + \dfrac{1}{54}$
$\Rightarrow \dfrac{9 + 6 + 5}{270} = \dfrac{20}{270} = \dfrac{2}{27}$
$\therefore$ Tap (A + B + C) can fill the cistern in = $\dfrac{27}{2} = 13\dfrac{1}{2}$ minutes
$\Rightarrow$ Tap C's 9 minute fill = $\dfrac{54}{9} = 6$
$\therefore$ If Tap C is closed after 9 minutes the cistern will be filled in :
$\Rightarrow \dfrac{27}{2} - 6 = \dfrac{27 - 12}{2} = \dfrac{15}{2}$
$\therefore$ Required more time = $\dfrac{27}{2} - \dfrac{15}{2} = \dfrac{27 - 15}{2} = \dfrac{12}{2} = 6$ mins. Answer
Q27. A can do $\dfrac{1}{3}$ of a piece of work in 6 days, while B can do $\dfrac{1}{2}$ of the same work in 12 days. In how many days can both do it together?
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$\Rightarrow$ A's 6 day's work = $\dfrac{1}{3}$
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{3 \times 6} = \dfrac{1}{18}$
$\Rightarrow$ B's 12 day's work = $\dfrac{1}{2}$
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{2 \times 12} = \dfrac{1}{24}$
$\therefore (A + B)$'s 1 day work:
$\Rightarrow \dfrac{1}{18} + \dfrac{1}{24} = \dfrac{4 + 3}{72} = \dfrac{7}{72}$
$\therefore (A + B)$ will together finish the work in = $\dfrac{72}{7} = 10\dfrac{2}{7}$ days Answer
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{3 \times 6} = \dfrac{1}{18}$
$\Rightarrow$ B's 12 day's work = $\dfrac{1}{2}$
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{2 \times 12} = \dfrac{1}{24}$
$\therefore (A + B)$'s 1 day work:
$\Rightarrow \dfrac{1}{18} + \dfrac{1}{24} = \dfrac{4 + 3}{72} = \dfrac{7}{72}$
$\therefore (A + B)$ will together finish the work in = $\dfrac{72}{7} = 10\dfrac{2}{7}$ days Answer
Q28. Two men undertake to do a piece of work for Rs.2800. One man can do it alone in 7 days, the other man in 6 days and with the assistance of a boy, they finnished the work in 3 days. How should the money be divided?
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$\Rightarrow$ Let 2 men be = A and B
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{7}$
$\Rightarrow$ A's 3 day's work = $\dfrac{1 \times 3}{7} = \dfrac{3}{7}$
$\Rightarrow$ B's 1 day's work = $\dfrac{1}{6}$
$\Rightarrow$ A's 3 day's work = $\dfrac{1 \times 3}{6} = \dfrac{1}{2}$
$\Rightarrow$ A's Share = $\dfrac{3}{7} \times 2800 = 3 \times 400 = Rs.1200$ Ans.
$\Rightarrow$ B's Share = $\dfrac{1}{2} \times 2800 = 1 \times 1400 = Rs.1400$ Ans.
$\Rightarrow$ Boy's share = $1400 + 1200 = 2600$
$\Rightarrow 2800 - 2600 = Rs.200$ Answer
$\Rightarrow$ A's 1 day's work = $\dfrac{1}{7}$
$\Rightarrow$ A's 3 day's work = $\dfrac{1 \times 3}{7} = \dfrac{3}{7}$
$\Rightarrow$ B's 1 day's work = $\dfrac{1}{6}$
$\Rightarrow$ A's 3 day's work = $\dfrac{1 \times 3}{6} = \dfrac{1}{2}$
$\Rightarrow$ A's Share = $\dfrac{3}{7} \times 2800 = 3 \times 400 = Rs.1200$ Ans.
$\Rightarrow$ B's Share = $\dfrac{1}{2} \times 2800 = 1 \times 1400 = Rs.1400$ Ans.
$\Rightarrow$ Boy's share = $1400 + 1200 = 2600$
$\Rightarrow 2800 - 2600 = Rs.200$ Answer
Q29. Ajit can do as much work in 2 days as Baljit can do in 3 days, and Baljit can do as much work in 4 days as Chandola in 5 days. A piece of work takes 20 days, if all work together. How long would Baljit take to do all the work by himself?
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$\Rightarrow$ Let Ajit's 2 day's work be = $2x$
$\Rightarrow$ So Ajit's 1 day's work be = $\dfrac{2x}{2} = x$
$\Rightarrow$ Baljit's 3 day's work be = $2x$
$\Rightarrow$ Baljit's 1 day's work be = $\dfrac{2x}{3}$
$\Rightarrow$ Baljit's 4 day's work be = $\dfrac{2x}{3} \times 4 = \dfrac{8x}{3}$
$\Rightarrow$ Chandola's 5 day's work be = $\dfrac{8x}{3}$
$\Rightarrow$ Chandola's 1 day's work be = $\dfrac{8x}{3 \times 5} = \dfrac{8x}{15}$
$\Rightarrow$ (Ajit + Baljit + Chandola)'s 1 day's work:
$\Rightarrow x + \dfrac{2x}{3} + \dfrac{8x}{15} = \dfrac{33x}{15}$
$\Rightarrow$ (Ajit + Baljit + Chandola)'s combined work for 20 days:
$\Rightarrow \dfrac{33x}{15} \times 20 = \dfrac{33x \times 4}{3} = 11x \times 4 = 44x$
$\Rightarrow$ We have, Baljit's 1 day's work = $\dfrac{2x}{3}$
$\Rightarrow$ As such, Baljit can do a peice of work in = $\dfrac{2x}{3} = \dfrac{3}{2x}$
$\therefore$ Baljit can alone finish the work in:
$\Rightarrow \dfrac{3}{2x} \times 44x = 3 \times 22 = 66$ days Answer
$\Rightarrow$ So Ajit's 1 day's work be = $\dfrac{2x}{2} = x$
$\Rightarrow$ Baljit's 3 day's work be = $2x$
$\Rightarrow$ Baljit's 1 day's work be = $\dfrac{2x}{3}$
$\Rightarrow$ Baljit's 4 day's work be = $\dfrac{2x}{3} \times 4 = \dfrac{8x}{3}$
$\Rightarrow$ Chandola's 5 day's work be = $\dfrac{8x}{3}$
$\Rightarrow$ Chandola's 1 day's work be = $\dfrac{8x}{3 \times 5} = \dfrac{8x}{15}$
$\Rightarrow$ (Ajit + Baljit + Chandola)'s 1 day's work:
$\Rightarrow x + \dfrac{2x}{3} + \dfrac{8x}{15} = \dfrac{33x}{15}$
$\Rightarrow$ (Ajit + Baljit + Chandola)'s combined work for 20 days:
$\Rightarrow \dfrac{33x}{15} \times 20 = \dfrac{33x \times 4}{3} = 11x \times 4 = 44x$
$\Rightarrow$ We have, Baljit's 1 day's work = $\dfrac{2x}{3}$
$\Rightarrow$ As such, Baljit can do a peice of work in = $\dfrac{2x}{3} = \dfrac{3}{2x}$
$\therefore$ Baljit can alone finish the work in:
$\Rightarrow \dfrac{3}{2x} \times 44x = 3 \times 22 = 66$ days Answer
Q30. A is thrice as good a workman as B and is therefore able to finish a piece of work in 60 days less than B. Find the time in which they can do it together.
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$\Rightarrow$ A is thrice as good a workman as B means that if A does a piece of work in 1 day then B does it in 3 days.
$\therefore$ If the difference is 2 days then A does the work in 1 day and B in 3 days. But the given difference is 60. As such, it means that A does the work in 30 days and B in 90.
$\Rightarrow$ Now, (A + B)'s 1 day's work:
$\Rightarrow \dfrac{1}{30} + \dfrac{1}{90} = \dfrac{3 + 1}{90} = \dfrac{4}{90} = \dfrac{2}{45}$
$\therefore$ (A + B) can together finish the work in:
$\Rightarrow \dfrac{45}{2} = \dfrac{45}{2} = 22\dfrac{1}{2}$ days Answer
$\therefore$ If the difference is 2 days then A does the work in 1 day and B in 3 days. But the given difference is 60. As such, it means that A does the work in 30 days and B in 90.
$\Rightarrow$ Now, (A + B)'s 1 day's work:
$\Rightarrow \dfrac{1}{30} + \dfrac{1}{90} = \dfrac{3 + 1}{90} = \dfrac{4}{90} = \dfrac{2}{45}$
$\therefore$ (A + B) can together finish the work in:
$\Rightarrow \dfrac{45}{2} = \dfrac{45}{2} = 22\dfrac{1}{2}$ days Answer