Q1. A can do a piece of work in 10 days and B in 15 days. How long will they take together to finish it?
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$\Rightarrow$ A's 1 days work = $\dfrac{1}{10}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{15}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{3 + 2}{30} = \dfrac{5}{30}$
$\therefore$ A and B can together finish the work in = $\dfrac{30}{5} = 6$ days Answer
$\Rightarrow$ B's 1 days work = $\dfrac{1}{15}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{3 + 2}{30} = \dfrac{5}{30}$
$\therefore$ A and B can together finish the work in = $\dfrac{30}{5} = 6$ days Answer
Q2. A and B together can do a piece of work in $6\dfrac{2}{3}$ days, but B alone can do it in 10 days. How long will A take to do it alone?
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$\Rightarrow$ Given, A and B can finish a work in = $6\dfrac{2}{3} = \dfrac{20}{3}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{\dfrac{20}{3}} = \dfrac{3}{20}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{10}$
$\therefore$ A's 1 days work = $\dfrac{3}{20} - \dfrac{1}{10} = \dfrac{3 - 2}{20} = \dfrac{1}{20}$
So, A can alone finish the work in = $20$ days Answer
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{\dfrac{20}{3}} = \dfrac{3}{20}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{10}$
$\therefore$ A's 1 days work = $\dfrac{3}{20} - \dfrac{1}{10} = \dfrac{3 - 2}{20} = \dfrac{1}{20}$
So, A can alone finish the work in = $20$ days Answer
Q3. A can do a work in 15 days and B in 20 days. If they work together on it for 4 days, what fraction of the work will be left?
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$\Rightarrow$ A's 1 days work = $\dfrac{1}{15}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{20}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{15} + \dfrac{1}{20} = \dfrac{4 + 3}{60} = \dfrac{7}{60}$
$\Rightarrow (A + B)$'s 4 days work = $\dfrac{7}{60} \times 4 = \dfrac{7}{15}$
$\therefore$ Work left = 1 $- \dfrac{7}{15} = \dfrac{15 - 7}{15} = \dfrac{8}{15}$ Answer
$\Rightarrow$ B's 1 days work = $\dfrac{1}{20}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{15} + \dfrac{1}{20} = \dfrac{4 + 3}{60} = \dfrac{7}{60}$
$\Rightarrow (A + B)$'s 4 days work = $\dfrac{7}{60} \times 4 = \dfrac{7}{15}$
$\therefore$ Work left = 1 $- \dfrac{7}{15} = \dfrac{15 - 7}{15} = \dfrac{8}{15}$ Answer
Q4. A, B and C can do a piece of work in 6 days, 12 days and 24 days. In what time will they finish it together?
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$\Rightarrow$ A's 1 days work = $\dfrac{1}{6}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{12}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{24}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24}$
$\Rightarrow \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} = \dfrac{4 + 2 + 1}{24} = \dfrac{7}{24}$
$\therefore$ A, B and C's can together finish the work in = $\dfrac{24}{7} = 3\dfrac{3}{7}$ days Answer
$\Rightarrow$ B's 1 days work = $\dfrac{1}{12}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{24}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24}$
$\Rightarrow \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} = \dfrac{4 + 2 + 1}{24} = \dfrac{7}{24}$
$\therefore$ A, B and C's can together finish the work in = $\dfrac{24}{7} = 3\dfrac{3}{7}$ days Answer
Q5. A and B working together can mow a field in 56 days and with the help of C, they could have mowed it in 42 days. How long would C take by himself?
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$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{56}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{42}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{42} - \dfrac{1}{56} = \dfrac{4 - 3}{168} = \dfrac{1}{168}$
$\therefore$ C alone can finish the work in = 168 days Answer
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{42}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{42} - \dfrac{1}{56} = \dfrac{4 - 3}{168} = \dfrac{1}{168}$
$\therefore$ C alone can finish the work in = 168 days Answer
Q6. A can do a piece of work in 24 days, A and B can do it in 16 days and A, B and C in $10\dfrac{2}{3}$ days. In how many days can A and C do it?
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$\Rightarrow$ Given, A's 1 days work = $\dfrac{1}{24}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{16}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{10\dfrac{2}{3}} = \dfrac{1}{\dfrac{32}{3}} = \dfrac{3}{32}$
$\Rightarrow$ C's 1 days work = $\dfrac{3}{32} - \dfrac{1}{16} = \dfrac{3 - 2}{32} = \dfrac{1}{32}$
$\therefore (A + C)$'s 1 days work = $\dfrac{1}{24} + \dfrac{1}{32} = \dfrac{4 + 3}{96} = \dfrac{7}{96}$
$\Rightarrow$ So, A and C can together finish the work in = $\dfrac{96}{7} = 13\dfrac{5}{7}$ days Answer
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{16}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{10\dfrac{2}{3}} = \dfrac{1}{\dfrac{32}{3}} = \dfrac{3}{32}$
$\Rightarrow$ C's 1 days work = $\dfrac{3}{32} - \dfrac{1}{16} = \dfrac{3 - 2}{32} = \dfrac{1}{32}$
$\therefore (A + C)$'s 1 days work = $\dfrac{1}{24} + \dfrac{1}{32} = \dfrac{4 + 3}{96} = \dfrac{7}{96}$
$\Rightarrow$ So, A and C can together finish the work in = $\dfrac{96}{7} = 13\dfrac{5}{7}$ days Answer
Q7. A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work?
Show Answer
$\Rightarrow$ A's 1 days work = $\dfrac{1}{20}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{15}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{20} + \dfrac{1}{15} = \dfrac{3 + 4}{60} = \dfrac{7}{60}$
$\Rightarrow (A + B)$'s 6 days work = $\dfrac{7}{60} \times 6 = \dfrac{7}{10}$
$\therefore$ Remaining work = 1 $- \dfrac{7}{10} = \dfrac{10 - 7}{10} = \dfrac{3}{10}$
$\Rightarrow$ B will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{3}{10}}{\dfrac{1}{15}} = \dfrac{3 \times 15}{10 \times 1} = \dfrac{45}{10} = \dfrac{9}{2} = 4\dfrac{1}{2}$ days Answer
$\Rightarrow$ B's 1 days work = $\dfrac{1}{15}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{20} + \dfrac{1}{15} = \dfrac{3 + 4}{60} = \dfrac{7}{60}$
$\Rightarrow (A + B)$'s 6 days work = $\dfrac{7}{60} \times 6 = \dfrac{7}{10}$
$\therefore$ Remaining work = 1 $- \dfrac{7}{10} = \dfrac{10 - 7}{10} = \dfrac{3}{10}$
$\Rightarrow$ B will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{3}{10}}{\dfrac{1}{15}} = \dfrac{3 \times 15}{10 \times 1} = \dfrac{45}{10} = \dfrac{9}{2} = 4\dfrac{1}{2}$ days Answer
Q8. A can finish a piece of work in 15 days and B can do it in 10 days. They worked together for 2 days and then B goes away. In how many days will A finish the remaining work?
Show Answer
$\Rightarrow$ A's 1 days work = $\dfrac{1}{15}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{10}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{2 + 3}{30} = \dfrac{5}{30} = \dfrac{1}{6}$
$\Rightarrow (A + B)$'s 2 days work = $\dfrac{1}{6} \times 2 = \dfrac{1}{3}$
$\therefore$ Remaining work = 1 $- \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3}$
$\Rightarrow$ B will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{2}{3}}{\dfrac{1}{15}} = \dfrac{2 \times 15}{3} = 10$ days Answer
$\Rightarrow$ B's 1 days work = $\dfrac{1}{10}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{2 + 3}{30} = \dfrac{5}{30} = \dfrac{1}{6}$
$\Rightarrow (A + B)$'s 2 days work = $\dfrac{1}{6} \times 2 = \dfrac{1}{3}$
$\therefore$ Remaining work = 1 $- \dfrac{1}{3} = \dfrac{3 - 1}{3} = \dfrac{2}{3}$
$\Rightarrow$ B will finish the remaining work in:
$\Rightarrow \dfrac{\dfrac{2}{3}}{\dfrac{1}{15}} = \dfrac{2 \times 15}{3} = 10$ days Answer
Q9. A can do a piece of work in 10 days, B in 18 days, and A, B and C togehter in 4 days. In what time would C alone do it?
Show Answer
$\Rightarrow$ A's 1 days work = $\dfrac{1}{10}$
$\Rightarrow$ B's 1 days work = $\dfrac{1}{18}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{10} + \dfrac{1}{18} = \dfrac{9 + 5}{90} = \dfrac{14}{90} = \dfrac{7}{45}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{4}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{4} - \dfrac{7}{45} = \dfrac{45 - 28}{180} = \dfrac{17}{180}$
$\therefore$ C alone can finish the work in = $\dfrac{180}{17} = 10\dfrac{10}{17}$ days Answer
$\Rightarrow$ B's 1 days work = $\dfrac{1}{18}$
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{10} + \dfrac{1}{18} = \dfrac{9 + 5}{90} = \dfrac{14}{90} = \dfrac{7}{45}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{4}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{4} - \dfrac{7}{45} = \dfrac{45 - 28}{180} = \dfrac{17}{180}$
$\therefore$ C alone can finish the work in = $\dfrac{180}{17} = 10\dfrac{10}{17}$ days Answer
Q10. A can do $\dfrac{1}{4}$ of a work in 5 days and B can do $\dfrac{1}{3}$ of the same work in 10 days. Find the number of days in which both working together will complete the work.
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$\Rightarrow$ A's 5 days work = $\dfrac{1}{4}$
$\therefore$ A's 1 days work = $\dfrac{1}{4 \times 5} = \dfrac{1}{20}$
$\Rightarrow$ B's 10 days work = $\dfrac{1}{3}$
$\therefore$ B's 1 days work = $\dfrac{1}{3 \times 10} = \dfrac{1}{30}$
$\Rightarrow (A + B)$'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
$\therefore$ A's 1 days work = $\dfrac{1}{4 \times 5} = \dfrac{1}{20}$
$\Rightarrow$ B's 10 days work = $\dfrac{1}{3}$
$\therefore$ B's 1 days work = $\dfrac{1}{3 \times 10} = \dfrac{1}{30}$
$\Rightarrow (A + B)$'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
Q11. One tap can fill a cistern in 3 hours and waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?
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$\Rightarrow$ Tap's 1 min fill = $\dfrac{1}{3}$
$\Rightarrow$ Waste pipe can empty in 1 min = $\dfrac{1}{5}$
$\therefore$ Time taken for cistern to be full while the tap and waste pipe are kept open:
$\Rightarrow \dfrac{1}{3} - \dfrac{1}{5} = \dfrac{5 - 3}{15} = \dfrac{2}{15}$
$\therefore \dfrac{15}{2} = 7\dfrac{1}{2}$ Hrs. Answer
$\Rightarrow$ Waste pipe can empty in 1 min = $\dfrac{1}{5}$
$\therefore$ Time taken for cistern to be full while the tap and waste pipe are kept open:
$\Rightarrow \dfrac{1}{3} - \dfrac{1}{5} = \dfrac{5 - 3}{15} = \dfrac{2}{15}$
$\therefore \dfrac{15}{2} = 7\dfrac{1}{2}$ Hrs. Answer
Q12. A and B can do a work in 8 days, B and C in 12 days, and A and C in 16 days. in what time could they do it, all working together?
Show Answer
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{8}$
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{12}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{16}$
$\therefore 2(A + B + C)$'s 1 days work = $\dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{16}$
$\Rightarrow \dfrac{6 + 4 + 3}{48} = \dfrac{13}{48}$
$\therefore$ A + B + C's 1 days work= $\dfrac{13}{48 \times 2} = \dfrac{13}{96}$
So, A, B and C can together finish the work in = $\dfrac{96}{13} = 7\dfrac{5}{13}$ days Answer
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{12}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{16}$
$\therefore 2(A + B + C)$'s 1 days work = $\dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{16}$
$\Rightarrow \dfrac{6 + 4 + 3}{48} = \dfrac{13}{48}$
$\therefore$ A + B + C's 1 days work= $\dfrac{13}{48 \times 2} = \dfrac{13}{96}$
So, A, B and C can together finish the work in = $\dfrac{96}{13} = 7\dfrac{5}{13}$ days Answer
Q13. A and B complete a piece of work in 24 days, B and C do the same work in 36 days, and A, B and C together finish it in 18 days. In how many days will:
(i) A alone can complete the work?
(ii) C alone can complete the work?
(iii) A and C together complete the work?
Show Answer
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{24}$
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{36}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{18}$
(i) A's 1 days work = $\dfrac{1}{18} - \dfrac{1}{36} = \dfrac{2 - 1}{36} = \dfrac{1}{36}$
$\therefore$ A alone can finish the work in = $\dfrac{36}{1} = 36$ days Ans.
(ii) C's 1 days work = $\dfrac{1}{18} - \dfrac{1}{24} = \dfrac{4 - 3}{72} = \dfrac{1}{72}$
$\therefore$ C alone can finish the work in = $\dfrac{72}{1} = 72$ days Ans.
(iii) A's 1 days work = $\dfrac{1}{36}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{72}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{36} + \dfrac{1}{72} = \dfrac{2 + 1}{72} = \dfrac{3}{72}$
$\therefore$ A and C can finish the work in = $\dfrac{72}{3} = 24$ days Answer
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{36}$
$\Rightarrow (A + B + C)$'s 1 days work = $\dfrac{1}{18}$
(i) A's 1 days work = $\dfrac{1}{18} - \dfrac{1}{36} = \dfrac{2 - 1}{36} = \dfrac{1}{36}$
$\therefore$ A alone can finish the work in = $\dfrac{36}{1} = 36$ days Ans.
(ii) C's 1 days work = $\dfrac{1}{18} - \dfrac{1}{24} = \dfrac{4 - 3}{72} = \dfrac{1}{72}$
$\therefore$ C alone can finish the work in = $\dfrac{72}{1} = 72$ days Ans.
(iii) A's 1 days work = $\dfrac{1}{36}$
$\Rightarrow$ C's 1 days work = $\dfrac{1}{72}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{36} + \dfrac{1}{72} = \dfrac{2 + 1}{72} = \dfrac{3}{72}$
$\therefore$ A and C can finish the work in = $\dfrac{72}{3} = 24$ days Answer
Q14. A and B can do a piece of work in 40 days, B and C in 30 days, and C and A in 24 days. Find:
(i) How long will it take them to do the work together?
(ii) In what time can each finish it working alone?
Show Answer
$\Rightarrow (A + B)$'s 1 days work = $\dfrac{1}{40}$
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{30}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{24}$
(i) $2(A + B + C)$'s 1 days work = $\dfrac{1}{40} + \dfrac{1}{30} + \dfrac{1}{24}$
$\Rightarrow \dfrac{3 + 4 + 5}{120} = \dfrac{12}{120} = \dfrac{1}{10}$
$\therefore$ A + B + C's 1 days work= $\dfrac{1}{10 \times 2} = \dfrac{1}{20}$
So, A, B and C can together finish the work in = $\dfrac{20}{1} = 20$ days Ans.
(ii) A's 1 days work = $\dfrac{1}{20} - \dfrac{1}{30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$
$\therefore$ A alone can finish the work in = $\dfrac{60}{1} = 60$ days Ans.
$\Rightarrow$ B's 1 days work = $\dfrac{1}{20} - \dfrac{1}{24} = \dfrac{6 - 5}{120} = \dfrac{1}{120}$
$\therefore$ B alone can finish the work in = $\dfrac{120}{1} = 120$ days Ans.
$\Rightarrow$ C's 1 days work = $\dfrac{1}{20} - \dfrac{1}{40} = \dfrac{2 - 1}{40} = \dfrac{1}{40}$
$\therefore$ A alone can finish the work in = $\dfrac{40}{1} = 40$ days Answer
$\Rightarrow (B + C)$'s 1 days work = $\dfrac{1}{30}$
$\Rightarrow (A + C)$'s 1 days work = $\dfrac{1}{24}$
(i) $2(A + B + C)$'s 1 days work = $\dfrac{1}{40} + \dfrac{1}{30} + \dfrac{1}{24}$
$\Rightarrow \dfrac{3 + 4 + 5}{120} = \dfrac{12}{120} = \dfrac{1}{10}$
$\therefore$ A + B + C's 1 days work= $\dfrac{1}{10 \times 2} = \dfrac{1}{20}$
So, A, B and C can together finish the work in = $\dfrac{20}{1} = 20$ days Ans.
(ii) A's 1 days work = $\dfrac{1}{20} - \dfrac{1}{30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$
$\therefore$ A alone can finish the work in = $\dfrac{60}{1} = 60$ days Ans.
$\Rightarrow$ B's 1 days work = $\dfrac{1}{20} - \dfrac{1}{24} = \dfrac{6 - 5}{120} = \dfrac{1}{120}$
$\therefore$ B alone can finish the work in = $\dfrac{120}{1} = 120$ days Ans.
$\Rightarrow$ C's 1 days work = $\dfrac{1}{20} - \dfrac{1}{40} = \dfrac{2 - 1}{40} = \dfrac{1}{40}$
$\therefore$ A alone can finish the work in = $\dfrac{40}{1} = 40$ days Answer
Q15. A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. How long did the work last?
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$\Rightarrow$ Let no. of days work lasted be = $x$ days
$\Rightarrow$ Given A worked for 2 days
$\Rightarrow$ A's 1 days work = $\dfrac{1}{10}$
$\Rightarrow$ A's 2 days work = $\dfrac{1}{10} \times 2 = \dfrac{1}{5}$
Now, B worked for (x - 3) days
$\Rightarrow$ Given, B's 1 days work = $\dfrac{1}{12}$
$\Rightarrow$ B's (x - 3) days work = $\dfrac{1}{12} \times (x - 3)$
$\Rightarrow \dfrac{1}{12} \times (x - 3) = \dfrac{x}{12} - \dfrac{3}{12} = \dfrac{x}{12} - \dfrac{1}{4}$
And, C worked for $x$ days
$\Rightarrow$ C's 1 days work = $\dfrac{1}{15}$
$\therefore$ C's $x$ days work = $\dfrac{1}{15} \times x = \dfrac{x}{15}$
$\Rightarrow$ Now, let completed work be = 1
So, we've the equation:
$\Rightarrow \dfrac{1}{5} + \dfrac{x}{12} - \dfrac{1}{4} + \dfrac{x}{15} = 1$
$\Rightarrow \dfrac{12 + 5x - 15 + 4x}{60} = 1$
$\Rightarrow \dfrac{9x - 3}{60} = 1$
$\Rightarrow 9x - 3 = 60$
$\Rightarrow 9x = 60 + 3 = 63$
$\Rightarrow x = \dfrac{63}{9} = 7$ days Answer
$\Rightarrow$ Given A worked for 2 days
$\Rightarrow$ A's 1 days work = $\dfrac{1}{10}$
$\Rightarrow$ A's 2 days work = $\dfrac{1}{10} \times 2 = \dfrac{1}{5}$
Now, B worked for (x - 3) days
$\Rightarrow$ Given, B's 1 days work = $\dfrac{1}{12}$
$\Rightarrow$ B's (x - 3) days work = $\dfrac{1}{12} \times (x - 3)$
$\Rightarrow \dfrac{1}{12} \times (x - 3) = \dfrac{x}{12} - \dfrac{3}{12} = \dfrac{x}{12} - \dfrac{1}{4}$
And, C worked for $x$ days
$\Rightarrow$ C's 1 days work = $\dfrac{1}{15}$
$\therefore$ C's $x$ days work = $\dfrac{1}{15} \times x = \dfrac{x}{15}$
$\Rightarrow$ Now, let completed work be = 1
So, we've the equation:
$\Rightarrow \dfrac{1}{5} + \dfrac{x}{12} - \dfrac{1}{4} + \dfrac{x}{15} = 1$
$\Rightarrow \dfrac{12 + 5x - 15 + 4x}{60} = 1$
$\Rightarrow \dfrac{9x - 3}{60} = 1$
$\Rightarrow 9x - 3 = 60$
$\Rightarrow 9x = 60 + 3 = 63$
$\Rightarrow x = \dfrac{63}{9} = 7$ days Answer