Rd Sharma 2019 2020 for Class 8 Maths Chapter 11 - Time And Work

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Rd Sharma 2019 2020 Solutions for Class 8 Maths Chapter 11 Time And Work are provided here with simple step-by-step explanations. These solutions for Time And Work are extremely popular among Class 8 students for Maths Time And Work Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 2020 Book of Class 8 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 2020 Solutions. All Rd Sharma 2019 2020 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 11.10:

Question 1:

Rakesh can do a piece of work in 20 days. How much work can he do in 4 days?

Answer:

$It is given that Rakesh can do a piece of work in 20 days . ∴ Rakesh' s 1 day's work = \frac{1}{20} ∴ Rakesh' s work for 4 days = \frac{4}{20} = \frac{1}{5} Thus, he can do \frac{1}{5} th of the work in 4 days .$

Page No 11.10:

Question 2:

Rohan can paint $\frac{1}{3}$ of a painting in 6 days. How many days will he take to complete the painting?

Answer:

$Rohan can paint \frac{1}{3} rd of a painting in 6 days . ∴ Time taken by Rohan to complete the painting = (6 \times 3) days = 18 days .$

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Question 3:

Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together?

Answer:

$Time taken by Anil to do the work = 5 days Time taken by Ankur to do the work = 4 days ∴ Work done by Anil in 1 day = \frac{1}{5} Work done by Ankur in 1 day = \frac{1}{4} ∴ Work done by Anil and Ankur in one day = \frac{1}{5} + \frac{1}{4} = \frac{4 + 5}{20} = \frac{9}{20} Thus, Anil and Ankur can do the work in \frac{20}{9} days i . e . 2 \frac{2}{9} days .$

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Question 4:

Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone?

Answer:

$Time taken by Mohan to do the work = 9 hours Time taken by Mohan and Sohan to do the work = 4 hours ∴ Work done by Mohan = \frac{1}{9} Work done by Mohan and Sohan = \frac{1}{4} ∴ Work done by Sohan = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} Thus, Sohan can do the work in \frac{36}{5} hours i . e . 7 \frac{1}{5} hours .$

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Question 5:

Sita can finish typing a 100 page document in 9 hours. Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together?

Answer:

$Time taken by Sita to do the work = 9 hours Time taken by Mita to do the work = 6 hours Time taken by Ri ta to do the work = 12 hours Now, Work done by Sita = \frac{1}{9} Work done by Mita = \frac{1}{6} Work done by Rita = \frac{1}{12} ∴ Work done by them together = \frac{1}{9} + \frac{1}{6} + \frac{1}{12} = \frac{4 + 6 + 3}{36} = \frac{13}{36} Thus, together they can do the work in \frac{36}{13} hours .$

Page No 11.10:

Question 6:

A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work?

Answer:

$Time taken by A to do the work = 20 hours Time taken by B to do the work = 24 hours Time taken by (A + B + C) to do the work = 8 hours Now, Work done by A = \frac{1}{20} Work done by B = \frac{1}{24} Work done by (A + B + C) = \frac{1}{8} ∴ Work done by C = \frac{1}{8} - (\frac{1}{20} + \frac{1}{24}) = \frac{1}{8} - (\frac{6}{120} + \frac{5}{120}) = \frac{1}{8} - (\frac{11}{120}) = \frac{15 - 11}{120} = \frac{4}{120} = \frac{1}{30} Thus, C can do the work in 30 hours .$

Page No 11.10:

Question 7:

A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together?

Answer:

$Time taken by (A + B) to do the work = 18 days Time taken by (B + C) to do the work = 24 days Time taken by (A + C) to do the work = 36 days Now, Work done by (A + B) = \frac{1}{18} Work done by (B + C) = \frac{1}{24} Work done by (A + C) = \frac{1}{36} ∴ Work done together = (A + B) + (B + C) + (A + C) = \frac{1}{18} + \frac{1}{24} + \frac{1}{36} = \frac{4 + 3 + 2}{72} = \frac{9}{72} = \frac{1}{8} ∴ Work done together = 2 (A + B + C) = \frac{1}{8} ∴ Work done by (A + B + C) = \frac{1}{16} Thus, together they can finish the work in 16 days .$

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Question 8:

A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work?

Answer:

$Time taken by (A + B) to do the work = 12 days Time taken by (B + C) to do the work = 15 days Time taken by (A + C) to do the work = 20 days Now, Work done by (A + B) = \frac{1}{12} Work done by (B + C) = \frac{1}{15} Work done by (A + C) = \frac{1}{20} ∴ Work done together = (A + B) + (B + C) + (A + C) = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} = \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5} ∴ Work done together = 2 (A + B + C) = \frac{1}{5} ∴ Work done by (A + B + C) = \frac{1}{10} ∴ Work done by A alone = (A + B + C) - (B + C) = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} Thus, A alone can do the work in 30 days .$

Page No 11.10:

Question 9:

A, B and C can reap a field in $15 \frac{3}{4}$ days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it?

Answer:

$Time taken by (A + B + C) to do the work = 15 \frac{3}{4} days = \frac{63}{4} days Time taken by (B + C + D) to do the work = 14 days Time taken by (C + D + A) to do the work = 18 days Time taken by (D + A + B) to do the work = 21 days Now, Work done by (A + B + C) = \frac{4}{63} Work done by (B + C + D) = \frac{1}{14} Work done by (C + D + A) = \frac{1}{18} Work done by (D + A + B) = \frac{1}{21} ∴ Work done by working together = (A + B + C) + (B + C + D) + (C + A + D) + (D + A + B) = \frac{4}{63} + \frac{1}{14} + \frac{1}{18} + \frac{1}{21} = \frac{4}{63} + (\frac{9 + 7 + 6}{126}) = \frac{4}{63} + \frac{22}{126} = \frac{4}{63} + \frac{11}{63} = \frac{15}{63} ∴ Work done by working together = 3 (A + B + C + D) = \frac{15}{63} ∴ Work done by (A + B + C + D) = \frac{15}{63 \times 3} = \frac{5}{63} Thus, together they can do the work in \frac{63}{5} days or 12 \frac{3}{5} days .$

Page No 11.10:

Question 10:

A and B can polish the floors of a building in 10 days. A alone can do $\frac{1}{4}$ th of it in 12 days. In how many days can B alone polish the floor?

Answer:

$It is given that A and B can polish the floors of the building in 10 days . ∴ Work done by (A + B) in 1 day = \frac{1}{10} Now, A alone can do \frac{1}{4} th of the work in 12 days . ∴ Time taken by A alone to do the complete work = (4 \times 12) = 48 days \Rightarrow Work done by A in 1 day = \frac{1}{48} Now, work done by B in 1 day = Work done by (A + B) in 1 day - Work done by A in 1 day = \frac{1}{10} - \frac{1}{48} = \frac{24 - 5}{240} = \frac{19}{240} Thus, B alone can polish the floor in \frac{240}{19} days or 12 \frac{12}{19} days .$

Page No 11.10:

Question 11:

A and B can finish a work in 20 days. A alone can do $\frac{1}{5}$ th of the work in 12 days. In how many days can B alone do it?

Answer:

$It is given that A and B can finish the work in 20 days . ∴ Work done by (A + B) in 1 day = \frac{1}{20} Now, A alone can do \frac{1}{5} th of the work in 12 days . ∴ Time taken by A alone to complete the work = (5 \times 12) = 60 days \Rightarrow Work done by A in 1 day = \frac{1}{60} Now, work done by B in 1 day = Work done by (A + B) in 1 day work - Work done by A in 1 day = \frac{1}{20} - \frac{1}{60} = \frac{3 - 1}{60} = \frac{2}{60} Thus, B alone can polish the floor in \frac{60}{2} days or 30 days .$

Page No 11.10:

Question 12:

A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work?

Answer:

$It is given that A can finish the work in 20 days and B can finish the same work in 15 days . ∴ Work done by A in 1 day = \frac{1}{20} Work done by B in 1 day = \frac{1}{15} ∴ Work done by (A + B) in 1 day = \frac{1}{20} + \frac{1}{15} = \frac{3 + 4}{60} = \frac{7}{60} ∴ Work done by (A + B) in 2 days = \frac{14}{60} = \frac{7}{30} Remaining work = 1 - \frac{7}{30} = \frac{23}{30} It is given that the remaining work is done by B . ∵ Complete work is done by B in 15 days . ∴ \frac{23}{30} of the work will be done by B in (15 \times \frac{23}{30}) days or \frac{23}{2} days or 11 \frac{1}{2} days . Thus, the remaining work is done by B in 11 \frac{1}{2} days .$

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Question 13:

A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work?

Answer:

$It is given that A can finish the work in 40 days and B can finish the same work in 45 days . ∴ Work done by A in 1 day = \frac{1}{40} Work done by B in 1 day = \frac{1}{45} ∴ Work done by (A + B) in 1 day = \frac{1}{40} + \frac{1}{45} = \frac{9 + 8}{360} = \frac{17}{360} ∴ Work done by (A + B) in 10 day s = 10 \times \frac{17}{360} = \frac{17}{36} Remaining work = 1 - \frac{17}{36} = \frac{19}{36} It is given that the remaining work is done by B . Complete work is done by B in 45 days . ∴ \frac{19}{36} of the work will be done by B in (45 \times \frac{19}{36}) days or 23 \frac{3}{4} days . Thus, the remaining work is done by B in 23 \frac{3}{4} days .$

Page No 11.10:

Question 14:

Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll?

Answer:

$Aasheesh can paint a doll in 20 minutes, and Chinki can do the same in 25 minutes . ∴ Work done by Aasheesh in 1 minute = \frac{1}{20} ∴ Work done by Chinki in 1 minute = \frac{1}{25} ∴ Work done by them together = \frac{1}{20} + \frac{1}{25} = \frac{5 + 4}{100} = \frac{9}{100} ∴ Work done by them in 5 minutes = 5 \times \frac{9}{100} = \frac{9}{20} Remaining work = 1 - \frac{9}{20} = \frac{11}{20} It is given that the remaining work is done by Aasheesh . The work done by Aasheesh in 20 minutes . ∴ \frac{11}{20} th work will be done by Aasheesh in (20 \times \frac{11}{20}) minutes or 11 minutes . Thus, the remaining work is done by Aasheesh in 11 minutes .$

Page No 11.10:

Question 15:

A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.

Answer:

$A can do a work in 6 days, and B can do the same work in 4 days . ∴ Work done by A in 2 days = \frac{2}{6} = \frac{1}{3} Remaining work = 1 - \frac{1}{3} = \frac{2}{3} ∴ Work done by (A + B) in 1 day = (\frac{1}{6} + \frac{1}{4}) = \frac{2 + 3}{12} = \frac{5}{12} ∵ \frac{5}{12} th work is done by A and B in 1 day . ∴ \frac{2}{3} rd work will be done by A and B in (\frac{12}{5} \times \frac{2}{3}) days or \frac{8}{5} days . ∴ Total time taken = (\frac{8}{5} + 2) days = \frac{18}{5} days = 3 \frac{3}{5} days$

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Question 16:

6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job?

Answer:

$It is given that 6 men can complete a job in 7 days . ∴ Time taken by 1 man to complete the same job = (6 \times 7) = 42 days ∴ Time taken by 21 men to complete the same job = \frac{42}{21} = 2 days$

Page No 11.10:

Question 17:

8 men can do a piece of work in 9 days. In how many days will 6 men do it?

Answer:

$It is given that 8 men can complete a job in 9 days . ∴ Time taken by 1 man to complete the same job = (8 \times 9) = 72 days ∴ Time taken by 6 men to complete the same job = \frac{72}{6} = 12 days$

Page No 11.10:

Question 18:

Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?

Answer:

$It is given that Reema weaves 35 baskets in 25 days . ∴ Time taken by her to weave 1 basket = \frac{25}{35} day ∴ Time taken by her to weave 55 baskets = (\frac{25}{35} \times 55) days = \frac{1375}{35} days = \frac{275}{7} days = 39 \frac{2}{7} days$

Page No 11.10:

Question 19:

Neha types 75 pages in 14 hours. How many pages will she type in 20 hours?

Answer:

$It is given that Neha types 75 pages in 14 hours . ∴ Number of pages typed by her in 1 hour = \frac{75}{14} ∴ Number of pages typed by her in 20 hours = (\frac{75}{14} \times 20) pages = \frac{1500}{14} pages = \frac{750}{7} pages = 107 \frac{1}{7} pages Thus, Neha types 107 \frac{1}{7} pages in 20 hours .$

Page No 11.10:

Question 20:

If 12 boys earn Rs 840 in 7 days, what will 15 boys earn in 6 days?

Answer:

$Earnings of 12 boys in 7 days = Rs . 840 ∴ Earnings of 12 boys in 1 day = Rs . \frac{840}{7} = Rs . 120 ∴ Earnings of 1 boy in 1 day = Rs . \frac{120}{12} = Rs . 10 ∴ Earnings of 1 boy in 6 days = Rs . (10 \times 6) = Rs . 60 ∴ Earnings of 15 boys in 6 days = Rs . (60 \times 15) = Rs . 900$

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Question 21:

If 25 men earn Rs 1000 in 10 days, how much will 15 men earn in 15 days?

Answer:

$Earnings of 25 men in 10 days = Rs . 1000 ∴ Earnings of 25 men in 1 day = Rs . \frac{1000}{10} = Rs . 100 ∴ Earning of 1 man in 1 day = Rs . \frac{100}{25} = Rs . 4 ∴ Earnings of 1 men in 15 days = Rs . (15 \times 4) = Rs . 60 ∴ Earnings of 15 men in 15 days = Rs . (60 \times 15) = Rs . 900$

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Question 22:

Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days?

Answer:

$If Ashu works for 8 hours daily, he can finish the work in 18 days . If he works for 1 hour daily, he can finish the work in (18 \times 8) days or 144 days . ∴ Number of hours he should work daily to finish the work in 12 days = \frac{144}{12} = 12$

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Question 23:

If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour?

Answer:

$Number of garlands made by 9 girls in 1 hour = \frac{135}{3} = 45 Number of garlands made by 1 girl in 1 hour = \frac{45}{9} = 5 ∴ Number of girls required to make 270 garlands in 1 hour = \frac{270}{5} = 54 .$

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Question 24:

A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together?

Answer:

$Time taken by the first tap to fill the cistern = 8 hours Time taken by the second tap to fill the cistern = 4 hours ∴ Work done by the first tap in 1 hour = \frac{1}{8} Work done by the second tap in 1 hour = \frac{1}{4} ∴ Work done by both the taps in 1 hour = \frac{1}{8} + \frac{1}{4} = \frac{1 + 2}{8} = \frac{3}{8} Thus, both the taps together will fill the cistern in \frac{8}{3} hours or 2 \frac{2}{3} hours .$

Page No 11.11:

Question 25:

Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and they B is turned off. How much time will A take to fill the remaining tank?

Answer:

$Pipe A can fill the tank in 10 hours, and pipe B can fill the tank in 15 hours . ∴ In 1 hour, A can fill \frac{1}{10} th part of the tank . In 1 hour, B can fill \frac{1}{15} th part of the tank . ∴ In 1 hour, A and B can fill (\frac{1}{10} + \frac{1}{15}) = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} th part of the tank ∴ In 4 hours, A and B can fill (\frac{1}{6} \times 4) = \frac{2}{3} rd part of the tank Remaining part of the tank = 1 - \frac{2}{3} = \frac{1}{3} Now, A can fill the tank in 10 hours . ∴ \frac{1}{3} rd part of the tank can be filled by A in (\frac{1}{3} \times 10) hours or \frac{10}{3} hours or 3 \frac{1}{3} hours .$

Page No 11.11:

Question 26:

A pipe can fill a cistern in 10 hours. Due to a leak in the bottom it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?

Answer:

$When there is no leakage, the pipe can fill the cistern in 10 hours . Thus, the pipe can fill \frac{1}{10} th part of the cistern in 1 hour . When there is leakage, the pipe can fill the cistern in 12 hours . Therefore, in case of leakage, the pipe can fill \frac{1}{12} th part of the cistern in 1 hour . Thus, in one hour, due to leakge, (\frac{1}{10} - \frac{1}{12}) th or \frac{1}{60} th part of the cistern is emptied . Hence, the cistern will be emptied by the leakage in 60 hours .$

Page No 11.11:

Question 27:

A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?

Answer:

$Time taken by tap A to fill the cistern = 12 hours Time taken by tap B to fill the cistern = 15 hours Let C be the outlet that can empty the cistern in 10 hours . Time taken by tap C to empty the cistern = 10 hours Now, Tap A fills \frac{1}{12} th part of the cistern in 1 hour . Tap B fills \frac{1}{15} th part of the cistern in 1 hour . Tap C empties out \frac{1}{10} th part of the cistern in 1 hour . Thus, in one hour, (\frac{1}{12} + \frac{1}{15} - \frac{1}{10}) th part of the cistern is filled . We have : \frac{1}{12} + \frac{1}{15} - \frac{1}{10} = \frac{10 + 8 - 12}{120} = \frac{6}{120} = \frac{1}{20} Thus, in 1 hour, \frac{1}{20} th part of the cistern is filled . Hence, the cistern will be filled completely in 20 hours if all the three taps are opened together .$

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Question 28:

A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern if both the tap and the pipe are opened together?

Answer:

$Time taken by the tap to fill the cistern = 4 hours ∴ Tap fills \frac{1}{4} th part of the cistern in 1 hour . Time taken by the pipe to empty the cistern = 6 hours ∴ Pipe empties out \frac{1}{6} th part of the cistern in 1 hour . Thus, in 1 hour, (\frac{1}{4} - \frac{1}{6}) th part of the cistern is filled . We have : \frac{1}{4} - \frac{1}{6} = \frac{6 - 4}{24} = \frac{2}{24} = \frac{1}{12} Thus, in 1 hour, \frac{1}{12} th part of the cistern is filled . Hence, the cistern will be filled in 12 hours .$

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