RS Aggarwal 2018 Solutions for Class 8 Math Chapter 13 - Time And Work

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Page No 172:

Question 1:

Rajan can do a piece of work in 24 days while Amit can do it in 30 days. In how many days can they complete it, if they work together?

Answer:

$Work done by Rajan in 1 day = \frac{1}{24} Work done by Amit in 1 day = \frac{1}{30} Work done by Amit and Rajan together in 1 day = \frac{1}{24} + \frac{1}{30} = \frac{54}{720} = \frac{3}{40} ∴ They can complete the work in \frac{40}{3} days, i . e ., 13 \frac{1}{3} days if they work together .$

Page No 172:

Question 2:

Ravi can do a piece of work in 15 hours while Raman can do it in 12 hours. How long will both take to do it, working together?

Answer:

$Time taken by Ravi = 15 h Time taken by Raman = 12 h Work done per hour by Ravi = \frac{1}{15} Work done per hour by Raman = \frac{1}{12} Work done per hour by Ravi and Raman together = \frac{1}{15} + \frac{1}{12} = \frac{9}{60} = \frac{3}{20} ∴ Time taken by Ravi and Raman together to finish the work = \frac{20}{3} h = 6 \frac{2}{3} h$

Page No 172:

Question 3:

A and B, working together can finish a piece of work in 6 days, while A alone can do it in 9 days. How much time will B alone take to finish it?

Answer:

$Time taken by A and B to finish a piece of work = 6 days Work done per day by A and B = \frac{1}{6} Time taken by A alone = 9 days Work done per day by A alone = \frac{1}{9} Work done per day by B = (work done by A and B) - (work done by A) = \frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18} ∴ B alone will take 18 days to complete the work .$

Page No 172:

Question 4:

Two motor mechanics, Raju and Siraj, working together can overhaul a scooter in 6 hours. Raju alone can do the job in 15 hours. In how many hours can Siraj alone do it?

Answer:

$Time taken by Raju = 15 h Work done by Raju in 1 h = \frac{1}{15} Time taken by Raju and Siraj working together = 6 h Work done by Raju and Siraj in 1 h = \frac{1}{6} Work done by Siraj in 1 h = (work done by Raju and Siraj) - (work done by Raju) = \frac{1}{6} - \frac{1}{15} = \frac{5 - 2}{30} = \frac{3}{30} = \frac{1}{10} ∴ Siraj will take 10 h to overhaul the scooter by himself .$

Page No 172:

Question 5:

A, B and C can do a piece of work in 10 days, 12 days and 15 days respectively. How long will they take to finish it if they work together?

Answer:

$Time taken by A to complete the work = 10 days Time taken by B to complete the work = 12 days Time taken by C to complete the work = 15 days Work done per day by A = \frac{1}{10} Work done per day by B = \frac{1}{12} Work done per day by C = \frac{1}{15} Total work done per day = \frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{6 + 5 + 4}{60} = \frac{15}{60} = \frac{1}{4} A, B and C will take 4 days to complete the work if they work together .$

Page No 172:

Question 6:

A can do a piece of work in 24 hours while B alone can do it in 16 hours. If A, B and C working together can finish it in 8 hours, in how many hours can C alone finish the work?

Answer:

$Time taken by A to complete the piece of work = 24 h Work done per hour by A = \frac{1}{24} Time taken by B to complete the work = 16 h Work done per hour by B = \frac{1}{16} Total time taken when A, B and C work together = 8 h Work done per hour by A, B and C = \frac{1}{8} Work done per hour by A, B and C = (work done per hour by A) + (work done per hour by B) + (work done per hour by C) (Work done per hour by C) = (work done per hour by A, B and C) - (work done per hour by A) - (work done per hour by B) = \frac{1}{8} - \frac{1}{24} - \frac{1}{16} = \frac{6 - 2 - 3}{48} = \frac{1}{48} Thus, C alone will take 48 h to complete the work .$

Page No 172:

Question 7:

A, B and C working together can finish 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:

$A can complete the work in 20 h . Work done per hour by A = \frac{1}{20} B can complete the work in 24 h . Work done per hour by B = \frac{1}{24} It takes 8 h to complete the work if A, B and C work together . Work done together per hour by A, B and C = \frac{1}{8} (Work done per hour by A, B and C) = (work done per hour by A) + (work done per hour by B) + (work done per hour by C) OR (Work done per hour by C) = (work done per hour by A, B and C) - (work done per hour by A) - (work done per hour by B) = \frac{1}{8} - \frac{1}{24} - \frac{1}{20} = \frac{1}{30} ∴ C alone will take 30 h to complete the work .$

Page No 173:

Question 8:

A and B can finish a piece of work in 16 days and 12 days respectively. A started the work and worked at it for 2 days. He was then joined by B. Find the total time taken to finish the work.

Answer:

$Time taken by A to complete the work = 16 days Work done per day by A = \frac{1}{16} Time taken by B to complete the work = 12 days Work done per day by B = \frac{1}{12} Work done per day by A and B = \frac{1}{12} + \frac{1}{16} = \frac{4 + 3}{48} = \frac{7}{48} Work done by A in two days = \frac{2}{16} = \frac{1}{8} Work left = 1 - \frac{1}{8} = \frac{7}{8} A and B together can complete \frac{7}{48} of the work in 1 day . Then, time taken to complete \frac{7}{8} of the work = \frac{7}{8} \div \frac{7}{48} = \frac{7}{8} \times \frac{48}{7} = 6 days ∴ Total time taken = 6 + 2 = 8 days .$

Page No 173:

Question 9:

A can do a piece of work in 14 days while B can do it in 21 days. They began together and worked at it for 6 days. Then, A fell ill and B had to complete the remaining work alone. In how many days was the work completed?

Answer:

$Time taken by A to complete the work = 14 days Work done by A in one day = \frac{1}{14} Time taken by B to complete the work = 21 days Work done by B in one day = \frac{1}{21} Work done jointly by A and B in one day = \frac{1}{14} + \frac{1}{21} = \frac{3 + 2}{42} = \frac{5}{42} Work done by A and B in 6 days = \frac{5}{42} \times 6 = \frac{5}{7} Work left = 1 - \frac{5}{7} = \frac{2}{7} With B working alone, time required to complete the work = \frac{2}{7} \div \frac{1}{21} = \frac{2}{7} \times 21 = 2 \times 3 = 6 days So, the total time taken to complete the work = 6 + 6 = 12 days$

Page No 173:

Question 10:

A can do $\frac{2}{3}$ of a certain work in 16 days and B can do $\frac{1}{4}$ of the same work in 3 days. In how many days can both finish the work, working together?

Answer:

$A can do \frac{2}{3} work in 16 days So, w ork done by A in one day = \frac{2}{48} = \frac{1}{24} B can do \frac{1}{4} work in 3 days So, w ork done by B in one day = \frac{1}{12} Work done jointly by A and B in one day = \frac{1}{24} + \frac{1}{12} = \frac{1 + 2}{24} = \frac{3}{24} = \frac{1}{8} So, A and B together will take 8 days to complete the work .$

Page No 173:

Question 11:

A, B and C can do a piece of work in 15, 12 and 20 days respectively. They started the work together, but C left after 2 days. In how many days will the remaining work be completed by A and B?

Answer:

$Time taken by A = 15 days Time taken by B = 12 days Time taken by C = 20 days Work d by A in one day = \frac{1}{15} Work done by B in one day = \frac{1}{12} Work done by C in one day = \frac{1}{20} Work don e in one day by A, B and C together = \frac{1}{15} + \frac{1}{12} + \frac{1}{20} = \frac{4 + 5 + 3}{60} = \frac{12}{60} = \frac{1}{5} Work done by A, B and C together in 2 days = \frac{2}{5} Work remaining = 1 - \frac{2}{5} = \frac{3}{5} Work done by A and B in one day = \frac{1}{15} + \frac{1}{12} = \frac{9}{60} = \frac{3}{20} Time required by A and B to complete the remaining work together = \frac{3}{5} \div \frac{3}{20} = \frac{3}{5} \times \frac{20}{3} = 4 days$

Page No 173:

Question 12:

A and B can do a piece of work in 18 days; B and C can do it in 24 days while C and A can finish it in 36 days. In how many days can A, B, C finish it, if they all work together?

Answer:

$Time needed by A and B to finish the work = 18 days Time needed by B and C to finish the work = 24 days Time needed by C and A to finish the work = 36 days Work done by A and B in one day = \frac{1}{18} Work done by B and C in one day = \frac{1}{24} Work done by C and A in one day = \frac{1}{36} 2 × Work done by A, B and C in one day = \frac{1}{18} + \frac{1}{24} + \frac{1}{36} = \frac{4 + 3 + 2}{72} = \frac{9}{72} = \frac{1}{8} ∴ Work done by A, B and C in one day = \frac{1}{16} So, A, B and C working together will t ake 16 days to complete the work .$

Page No 173:

Question 13:

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

Answer:

$(A + B) can complete the work in 12 days . (B + C) can complete the work in 15 days . (C + A) can complete the work in 20 days . (A + B)' s 1 day work = \frac{1}{12} (B + C)' s 1 day work = \frac{1}{15} (C + A)' s 1 day work = \frac{1}{20} 2 (A + B + C)' s 1 day work = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} = \frac{5 + 4 + 3}{60} = \frac{12}{60} = \frac{1}{5} (A + B + C)' s 1 day work = \frac{1}{10} A' s 1 day work = \{(A + B + C)' s 1 day work\} - \{(B + C)' s 1 day work\} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30} A will take 30 days to complete the work, if he works alone .$

Page No 173:

Question 14:

Pipes A and B can fill an empty tank in 10 hours and 15 hours respectively. If both are opened together in the empty tank, how much time will they take to fill it completely?

Answer:

$A can fill a tank in 10 hours . B can fill a tank in 15 hours . Pipe A fills \frac{1}{10} of the tank in one hour. Pipe B fills \frac{1}{15} of the tank in one hour. Part of tank filled by pipes A and B together = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} Thus, pipes A and B require 6 hours to fill the tank .$

Page No 173:

Question 15:

Pipe A can fill an empty tank in 5 hours while pipe B can empty the full tank in 6 hours. If both are opened at the same time in the empty tank, how much time will they take to fill it up completely?

Answer:

$Pipe A can fill a tank in 5 hours . Pipe B can empty a full tank in 6 hours . Pipe A fills \frac{1}{5} of the tank in one hour. Pipe B empties \frac{1}{6} of the tank in one hour. Part of the tank filled in one hour using both pipes A and B = \frac{1}{5} - \frac{1}{6} = \frac{6 - 5}{30} = \frac{1}{30} It takes \frac{30}{1} or 30 hours to fill the tank completely .$

Page No 173:

Question 16:

Three taps A, B and C can fill an overhead tank in 6 hours, 8 hours and 12 hours respectively. How long would the three taps take to fill the empty tank, if all of them are opened together?

Answer:

$Time taken by tap A to fill the tank = 6 hours Time taken by tap B to fill the tank = 8 hours Time taken by tap C to fill the tank = 12 hours A fills \frac{1}{6} of the tank in one hour. B fills \frac{1}{8} of the tank in one hour. C fills \frac{1}{12} of the tank in one hour. Part of the tank filled in one hour using all the three pipes = \frac{1}{6} + \frac{1}{8} + \frac{1}{12} = \frac{4 + 3 + 2}{24} = \frac{9}{24} Time taken by A, B and C together to fill the tank = \frac{24}{9} = \frac{8}{3} = 2 \frac{2}{3} hours$

Page No 173:

Question 17:

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

Answer:

$Inlet A can fill the cistern in 12 minutes . Inlet B can fill the cistern in 15 minutes . Outlet C empties the filled cistern in 10 minutes . Part of the cistern filled by inlet A in one minute = \frac{1}{12} Part of the cistern filled by inlet B in one minute = \frac{1}{15} Part of the cistern emptied by outlet C in one minute = - \frac{1}{10} (water flows out from C and empties the cistern) Part of the cistern filled in one minute with A, B and C working together = \frac{1}{12} + \frac{1}{15} - \frac{1}{10} = \frac{5 + 4 - 6}{60} = \frac{3}{60} = \frac{1}{20} The time required to fill the cistern with all inlets, A, B and C, open is 20 minutes .$

Page No 173:

Question 18:

A pipe can fill a cistern in 9 hours. Due to a leak in its bottom, the cistern fills up in 10 hours. If the cistern is full, in how much time will it be emptied by the leak?

Answer:

$A pipe can fill a cistern in 9 hours . Part of the cistern filled by the pipe in one hour = \frac{1}{9} Let the leak empty the cistern in x hours . Part of the cistern emptied by the leak in one hour = - \frac{1}{x} (The leak drains out the water) Considering the leak, t he tank is filled in 10 hours . Part of the tank filled in one hour = \frac{1}{10} Therefore, \frac{1}{9} - \frac{1}{x} = \frac{1}{10} or, \frac{1}{x} = \frac{1}{9} - \frac{1}{10} = \frac{10 - 9}{90} = \frac{1}{90} x = 90$

The leak will empty the filled cistern in 90 hours.

Page No 173:

Question 19:

Pipe A can fill a cistern in 6 hours and pipe B can fill it in 8 hours. Both the pipes are opened and after two hours, pipe A is closed. How much time will B take to fill the remaining part of the tank?

Answer:

$Pipe A can fill a cistern in 6 hours . Pipe B can fill a cistern in 8 hours . Part of the cistern filled by pipe A in one hour = \frac{1}{6} Part of the cistern filled by pipe B in one hour = \frac{1}{8} Part of the cistern filled by pipes A and B in one hour = \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24} Part of the cistern filled by pipes A and B in 2 hours = \frac{7}{24} \times 2 = \frac{7}{12} Part of the tank empty after 2 hours = 1 - \frac{7}{12} = \frac{5}{12} Time taken by pipe B to fill the remaining tank = \frac{5}{12} \div \frac{1}{8} = \frac{5}{12} \times 8 = \frac{10}{3} = 3 \frac{1}{3} hours$

Page No 174:

Question 1:

Tick (✓) the correct answer:
A alone can do a piece of work in 10 days and B alone can do it in 15 days. In how many days will A and B together do the same work?
(a) 5 days
(b) 6 days
(c) 8 days
(d) 9 days

Answer:

(b) 6 days

$A can do a work in 10 days . A' s 1 day work = \frac{1}{10} B can do a work in 15 days . B' s 1 day work = \frac{1}{15} (A + B)' s 1 day work = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6} A and B together will take 6 days to complete the work .$

Page No 174:

Question 2:

Tick (✓) the correct answer:
A man can do a piece of work in 5 days. He and his son working together can finish it in 3 days. In how many days can the son do it alone?
(a) $6 \frac{1}{2}$ days
(b) 7 days
(c) $7 \frac{1}{2}$ days
(d) 8 days

Answer:

(c) $7 \frac{1}{2} d a y s$

$A man can do a work in 5 days . The man' s 1 day work = \frac{1}{5} The man and the son can do the work in 3 days . The man and his son' s 1 day work = \frac{1}{3} Let the son' s 1 day work be \frac{1}{x} . Therefore, \frac{1}{3} = \frac{1}{5} + \frac{1}{x} or, \frac{1}{x} = \frac{1}{3} - \frac{1}{5} = \frac{5 - 3}{15} = \frac{2}{15} x = \frac{15}{2} = 7 \frac{1}{2} days$

Page No 174:

Question 3:

Tick (✓) the correct answer:
A can do a job in 16 days and B can do the same job in 12 days. With the help of C, they can finish the job in 6 days only. Then, C alone can finish it in
(a) 34 days
(b) 22 days
(c) 36 dyas
(d) 48 days

Answer:

(d) 48 days

$A can do a job in 16 days . B can do the job in 12 days . Suppose C can do the job in x days . A' s 1 day work = \frac{1}{16} B' s 1 day work = \frac{1}{12} C' s 1 day work = \frac{1}{x} A, B and C together can complete the work in 6 days . (A + B + C)' s 1 day work = \frac{1}{6} Therefore, \frac{1}{6} = \frac{1}{16} + \frac{1}{12} + \frac{1}{x} \Rightarrow \frac{1}{x} = \frac{1}{6} - \frac{1}{16} - \frac{1}{12} = \frac{8 - 3 - 4}{48} = \frac{1}{48} x = 48 Therefore, C alone can complete the job in 48 days .$

Page No 174:

Question 4:

Tick (✓) the correct answer:
To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?
(a) 30 days
(b) 35 days
(c) 40 days
(d) 45 days

Answer:

(a) 30 days

$L e t B t a k e x d a y s t o c o m p l e t e t h e w o r k . T h e n A t a k e s (x + \frac{50}{100} x) = 1.5 x A' s 1 d a y' s w o r k = \frac{1}{1.5 x} = \frac{2}{3 x} B' s 1 d a y' s w o r k = \frac{1}{x} (A + B) t a k e s 18 d a y s t o c o m p l e t e t h e w o r k . (A + B)' s 1 d a y' s n e t w o r k = \frac{1}{18} o r \frac{1}{18} = \frac{2}{3 x} + \frac{1}{x} \Rightarrow \frac{1}{18} = \frac{5}{3 x} B y c r o s s - m u l t i p l i c a t i o n, w e g e t : x = 30 d a y s ∴ B a l o n e w i l l t a k e 30 d a y s t o c o m p l e t e t h e w o r k .$

Page No 174:

Question 5:

Tick (✓) the correct answer:
A works twice as fast as B. If both of them can together finish a piece of work in 12 days, then B alone can do it in
(a) 24 days
(b) 27 days
(c) 36 days
(d) 48 days

Answer:

(c) 36 days

$Let A take x days to complete the work . Then B takes 2 x days to complete the work . A' s 1 day' s work = \frac{1}{x} B' s 1 day' s work = \frac{1}{2 x} A and B take 12 days to complete the work . Net work done by (A + B) in 1 day = \frac{1}{12} = \frac{1}{x} + \frac{1}{2 x} = \frac{3}{2 x} \Rightarrow 2 x = 36 \Rightarrow x = 18 A can complete the work by himself in 18 days . B will take 36 days, i . e ., twice as long as the time taken by A .$

Page No 174:

Question 6:

Tick (✓) the correct answer:
A alone can finish a piece of work in 10 days which B alone can do in 15 days. If they work together and finish it, then out of total wages of Rs 3000, A will get
(a) Rs 1200
(b) Rs 1500
(c) Rs 1800
(d) Rs 2000

Answer:

(c) Rs. 1800

Since the wage distribution will follow the work distribution ratio, we have:

Work done by A in 1 day $= \frac{1}{10}$
Work done by B in 1 day $= \frac{1}{15}$

Net work done by (A+B) in 1 day $= \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6}$

i.e., (A+B) will take 6 days to complete the work.

A's share of work in a day = $\frac{1}{10} \div \frac{1}{6} = \frac{1}{10} \times \frac{6}{1} = \frac{6}{10} = \frac{3}{5}$

∴ A's wage = $\frac{3}{5} \times 3000 = Rs 1800$

Page No 174:

Question 7:

Tick (✓) the correct answer:
The rates of working of A and B are in the ratio 3 : 4. The number of days taken by them to finish the work are in the ratio
(a) 3 : 4
(b) 9 : 16
(c) 4 : 3
(d) 16 : 9

Answer:

(c) 4:3

The number of days taken for working is the reciprocal of the rate of work.

$i . e ., number of days taken = \frac{1}{rate of work} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

Page No 174:

Question 8:

Tick (✓) the correct answer:
A and B together can do a piece of work in 12 days; B and C can do it in 20 days while C and A can do it in 15 days. A, B and C all working together can do it in
(a) 6 days
(b) 9 days
(c) 10 days
(d) $10 \frac{1}{2}$ days

Answer:

(c) 10 days

$(A + B) can do a work in 12 days . (B + C) can do a work in 20 days . (C + A) can do a work in 15 days . Now, we have : Work done by (A + B) in 1 day = \frac{1}{12} Work done by (B + C) in 1 day = \frac{1}{20} Work done by (C + A) in 1 day = \frac{1}{15} Net work done by 2 (A + B + C) = \frac{1}{12} + \frac{1}{20} + \frac{1}{15} = \frac{5 + 3 + 4}{60} = \frac{12}{60} = \frac{1}{5} Net work done by (A + B + C) in 1 day = \frac{1}{10} ∴ If A, B and C work together, they will complete the work in 10 days .$

Page No 174:

Question 9:

Tick (✓) the correct answer:
3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to do it?
(a) 6 days
(b) 5 days
(c) 4 days
(d) 3 days

Answer:

(c) 4 days

$Three men can complete the work in 12 days . Thus, one man can complete the work in 36 days . Rate of work done by one man in 1 day = \frac{1}{36} Similarly, rate of work done by one woman in 1 day = \frac{1}{5 \times 12} = \frac{1}{60} Now, six men will do \frac{6}{36}, i . e ., \frac{1}{6} unit of work in a day . Five women will do \frac{5}{60}, i, e ., \frac{1}{12} unit of work in a day . ∴ Total work done in 1 day = \frac{1}{6} + \frac{1}{12} = \frac{1}{4} unit Thus, six men and five women will take 4 days to complete the work .$

The work can be completed in 4 days.

Page No 174:

Question 10:

Tick (✓) the correct answer:
A can do a piece of work in 15 days. B is 50% more efficient than A. A can finish it in
(a) 10 days
(b) $7 \frac{1}{2}$ days
(c) 12 days
(d) $10 \frac{1}{2}$ days

Answer:

(a) 10 days

$Work done by A in 1 day = \frac{1}{15} B is 50 % more efficient than A . ∴ Work done by B in 1 day = \frac{150}{100} \times \frac{1}{15} = \frac{1}{10} Thus, B can complete the work in 10 days .$

Page No 175:

Question 11:

Tick (✓) the correct answer:
A does 20% less work than B. If A can finish a piece of work in $7 \frac{1}{2}$ hours, then B can finish it in
(a) 5 hours
(b) $5 \frac{1}{2}$ hours
(c) 6 hours
(d) $6 \frac{1}{2}$ hours

Answer:

(c) 6 hours

$Time taken by A to finish the piece of work = 7 \frac{1}{2} hours = \frac{15}{2} hours Work done by A in 1 hour = \frac{2}{15} Let B take x hours to finish the work . Work done by B in 1 hours = \frac{1}{x} A can work 20 % less than B, or A can do \frac{4}{5} of B' s work . Now, \frac{(\frac{4}{5})}{1} = \frac{(\frac{2}{15})}{(\frac{1}{x})} \Rightarrow \frac{4}{5} = \frac{2 x}{15} \Rightarrow x = \frac{15 \times 4}{5 \times 2} = 6 hours$

Page No 175:

Question 12:

Tick (✓) the correct answer:
A can do a piece of work in 20 days which B alone can do in 12 days. B worked at it for 9 days. A can finish the remaining work in
(a) 3 days
(b) 5 days
(c) 7 days
(d) 11 days

Answer:

(b) 5 days

$A can complete the work in 20 days . Work done by A in 1 day = \frac{1}{20} B can complete the work in 12 days . Work done by B in 1 day = \frac{1}{12} In 9 days, B completes \frac{9}{12}, i . e ., \frac{3}{4} of the work and leaves 1 - \frac{3}{4}, i . e ., \frac{1}{4} of the work undone . ∴ Time taken by A = \frac{1}{4} \div \frac{1}{20} = \frac{1}{4} \times 20 = 5 days$

Page No 175:

Question 13:

Tick (✓) the correct answer:
A can do a piece of work in 25 days, which B alone can do in 20 days. A started the work and was joined by B after 10 days. The work lasted for
(a) $12 \frac{1}{2}$ days
(b) 15 days
(c) $16 \frac{2}{3}$ days
(d) 14 days

Answer:

(c)

$A can do the piece of work in 25 days . Work done by A in 1 day = \frac{1}{25} B can do the same work in 20 days . Work done by B in 1 day = \frac{1}{20} A alone completes \frac{10}{25}, i, e ., \frac{2}{5} of the work in 10 days . Now, work remaining = 1 - \frac{2}{5} = \frac{3}{5} Work done by (A + B) in 1 day = \frac{1}{25} + \frac{1}{20} = \frac{9}{100} ∴ Time taken if they work together = \frac{3}{5} \div \frac{9}{100} = \frac{3}{5} \times \frac{100}{9} = \frac{20}{3} = 6 \frac{2}{3} days$

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Question 14:

Tick (✓) the correct answer:
Two pipes can fill a tank in 20 minutes and 30 minutes respectively. If both the pipes are opened simultaneously, then hte tank will be filled in
(a) 10 minutes
(b) 12 minutes
(c) 15 minutes
(d) 25 minutes

Answer:

(b) 12 minutes

$First pipe can fill a tank in 20 minutes . Second pipe can fill the tank in 30 minutes . Part of tank filled by the first pipe in one minute = \frac{1}{20} P art of tank filled by the second pipe in one minute \frac{1}{30} Part of tank filled by both pipes in one minute = \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12} Thus, it takes 12 minutes to fill the tank using both the pipes.$

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Question 15:

Tick (✓) the correct answer:
A tap can fill a cistern in 8 hours and another tap can empty the full cistern in 16 hours. If both the taps are open, the time taken to fill the cistern is
(a) $5 \frac{1}{3}$ hours
(b) 10 hours
(c) 16 hours
(d) 20 hours

Answer:

(c) 16 hours

$A tap can fill a cistern in 8 hours . Part of cistern filled in one hour = \frac{1}{8} A tap can empty the cistern in 16 hours . Part of cistern emptied in one hour = - \frac{1}{16} (negative sign shows that the cistern is being drained) ∴ Part of cistern filled in one hour = \frac{1}{8} - \frac{1}{16} = \frac{1}{16} Time required to fill the cistern = 16 hours$

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Question 16:

Tick (✓) the correct answer:
A pump can fill a tank in 2 hours. Due to a leak in the tank it takes $2 \frac{1}{3}$ hours to fill the tank. The leak can empty the full tank in
(a) $2 \frac{1}{3}$ hours
(b) 7 hours
(c) 8 hours
(d) 14 hours

Answer:

(d) 14 hours

$A pump can fill a tank in 2 hours . Part of the tank filled by the pump in one hour = \frac{1}{2} Suppose the leak empties a full tank in x hours . Part of the tank emptied by the leak in one hour = - \frac{1}{x} Part of tank filled in one hour = \frac{1}{2} - \frac{1}{x} = \frac{3}{7} (given) \frac{1}{x} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} x = 14 hours$

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Question 17:

Tick (✓) the correct answer:
Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be full?
(a) 7 hrs 15 min
(b) 7 hrs 30 min
(c) 7 hrs 45 min
(d) 8 hrs

Answer:

(b) 7 hours 30 minutes

$Part of the tank filled by the first pipe in one hour = \frac{1}{10} Part of the tank filled by the second pipe in one hour = \frac{1}{12} Part of the tank filled by the third pipe in one hour = \frac{- 1}{20} Part of the tank filled by three pipes in one hour = \frac{1}{10} + \frac{1}{12} - \frac{1}{20} = \frac{2}{15} Total time taken to fill the tank = \frac{15}{2} hrs = 7 hours 30 minutes$

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Question 1:

A can do a piece of work in 10 days while B alone can do it in 15 days. In how many days can both finish the same work?

Answer:

$A can do a piece of work in 10 days . A' s 1 day work = \frac{1}{10} B can do a piece of work in 15 days . B' s 1 day work = \frac{1}{15} (A + B)' s 1 day work = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} A and B working together can complete the work in 6 days .$

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Question 2:

A and B can do a piece of work in 15 days; B and C in 12 days; C and A in 20 days. How many days will be taken by A, B and C working together to finish the work?

Answer:

$(A + B) can do a work in 15 days . ∴ (A + B)' s 1 day work = \frac{1}{15} (B + C) can do a work in 12 days . ∴ (B + C)' s 1 day work = \frac{1}{12} (C + A) can do a work in 20 days . ∴ (C + A)' s 1 day work = \frac{1}{20} 2 (A + B + C)' s 1 day work = \frac{1}{15} + \frac{1}{12} + \frac{1}{20} = \frac{4 + 5 + 3}{60} = \frac{12}{60} = \frac{1}{5} (A + B + C)' s 1 day work = \frac{1}{10} A, B and C working together require 10 days to complete the work .$

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Question 3:

Tap A can fill a cistern in 8 hours and tap B can empty it in 12 hours. How long will it take to fill the cistern if both of them are opened together?

Answer:

$Tap A can fill a cistern in 8 hours . Part of cistern filled by Tap A in 1 hour = \frac{1}{8} Tap B empties the cistern in 12 hours . Part of cistern emptied by Tap B in 1 hour = - \frac{1}{12} (negative sign shows that tap B drains the tank) Part of cistern filled in one hour when both taps are opened together = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24} Therefore, it will take 24 hours to fill the cistern .$

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Question 4:

2 men or 3 women can do a piece of work in 16 days. In how many days can 4 men and 6 women do the same work?

Answer:

$Work of 2 men = Work of 3 women \Rightarrow Work of 1 man = \frac{3}{2} women Three women can do a piece of work in 16 days . As 4 men and 6 women = (4 \times \frac{3}{2}) women + 6 women = 6 women + 6 women = 12 women Also, 3 women can do the work in 16 days . So, work done by 3 women in one day = \frac{1}{16} ∴ Work done by 1 woman in one day = \frac{1}{48} \Rightarrow Work done by 12 women in one day = \frac{1}{4} Thus, 4 men and 6 women will complete the work in 4 days .$

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Question 5:

A pipe can fill a cistern in 9 hours. Due to a leak in its bottom, the cistern fills up in 10 hours. If the cistern is full, in how much time will it be emptied by the leak?

Answer:

$Time taken by the pipe to fill the cistern = 9 hours Part of the cistern filled in one hour = \frac{1}{9} Suppose the leak empties the full cistern in x hours . Part of the cistern emptied in one hour = - \frac{1}{x} (negative sign implies a leak) Time taken by the cistern to fill completely due to the leak = 10 hours Part of the cistern filled in one hour due to the leak = \frac{1}{10} ∴ \frac{1}{10} = \frac{1}{9} - \frac{1}{x} \Rightarrow \frac{1}{x} = \frac{1}{9} - \frac{1}{10} = \frac{1}{90} x = 90 hours Therefore, the leak will empty a full cistern in 90 hours .$

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Question 6:

Mark (✓) against the correct answer:
The rates of working of two tapes A and B are in the ratio 2 : 3. The ratio of the time taken by A and B respectively to fill a given cistern is
(a) 2 : 3
(b) 3 : 2
(c) 4 : 9
(d) 9 : 4

Answer:

(b) 3:2

$Rates at which taps A and B work = 2 : 3 The ratio of time taken by taps A and B to fill the cistern = \frac{1}{Rates at which taps A and B work} = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 3 : 2$

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Question 7:

Mark (✓) against the correct answer:
A can finish a piece of work in 12 hours while B can finish it in 15 hours. How long will both take to finish it, working together?
(a) 9 hours
(b) $6 \frac{2}{3}$ hours
(c) $6 \frac{3}{4}$ hours
(d) $8 \frac{1}{3}$ hours

Answer:

(b) $6 \frac{2}{3} h o u r s$

$A' s 1 hour work = \frac{1}{12} B' s 1 hour work = \frac{1}{15} (A + B)' s 1 hour work = \frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20} Time taken by A and B to complete the work together = \frac{20}{3} = 6 \frac{2}{3} hours$

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Question 8:

Mark (✓) against the correct answer:
A can do a piece of work in 14 days and B is 40% more efficient than A. In how many days can B finish it?
(a) 10 days
(b) $7 \frac{1}{2}$ days
(c) $5 \frac{1}{4}$ days
(d) $5 \frac{3}{5}$ days

Answer:

(a) 10 days
$A' s 1 day work = \frac{1}{14} B is 40 % more efficient than A . ∴ B' s 1 day work = \frac{140}{100} \times \frac{1}{14} = \frac{1}{10} B takes 10 days to complete the work .$

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Question 9:

Mark (✓) against the correct answer:
A pump can fill a cistern in 2 hours. Due to a leak in the tank it takes $2 \frac{1}{3}$ hours to fill it. The leak can empty the full tank in
(a) 7 hours
(b) 14 hours
(c) 8 hours
(d) 3 hours

Answer:

(b) 14 hours

$A pump fills a tank in 2 hours . Part of tank filled by the pump in one hour = \frac{1}{2} Let x hours be the time required for water to leak from the cistern . Part of tank drained by the leak in one hour = - \frac{1}{x} Time required to fill the leaking cistern = 2 \frac{1}{3} = \frac{7}{3} hours Part of tank stored with water in one hour = \frac{3}{7} \Rightarrow \frac{3}{7} = \frac{1}{2} - \frac{1}{x} \frac{1}{x} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} x = 14 hours$

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Question 10:

Mark (✓) against the correct answer:
A works twice as fast as B. If both of them can together finish a peice of work in 12 hours, then B alone can do it in
(a) 24 hours
(b) 27 hours
(c) 36 hours
(d) 18 hours

Answer:

(c) 36 hours

$Suppose B take s x hour s to complete the work . ∴ B' s 1 hour work = \frac{1}{x} A works twice as fast as B . ∴ A's 1 hour work = \frac{2}{x} \frac{1}{12} = \frac{1}{x} + \frac{2}{x} = \frac{3}{x} \Rightarrow x = 36 hours$

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Question 11:

Fill in the blanks.
(i) A tap can fill a tank in 6 hours. The part of the tank filled in 1 hour is .........
(ii) A and B working together can finish a piece of work in 6 hours while A alone can do it in 9 hours. B alone can do it in ......... hours.
(iii) A can do a work in 16 hours and B alone can do it in 24 hours. If A, B and C working together can finish it in 8 hours, then C alone can finish it in ......... hours.
(iv) If A's one day's work is $\frac{3}{20}$ , then A can finish the whole work in ......... days.

Answer:

(i) A tap can fill a tank in 6 hours. In 1 hour, $\frac{1}{6}$ of the tank is filled.

(ii) 18 hours
$(A + B)' s 1 hour work = \frac{1}{6} A' s 1 hour work = \frac{1}{9} B' s 1 hour work = \frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18} Thus, B takes 18 hours to finish the work .$

(iii) 48 hours
$A' s 1 hour work = \frac{1}{16} B' s 1 hour work = \frac{1}{24} C' s 1 hour work = \frac{1}{x} (A + B + C)' s 1 hour work = \frac{1}{8} Therefore, \frac{1}{x} = \frac{1}{8} - \frac{1}{16} - \frac{1}{24} = \frac{6 - 3 - 2}{48} = \frac{1}{48} or, x = 48 hours Thus, C alone takes 48 hours to complete the wor k .$

(iv) The time for completion is the reciprocal of the work done in one day. Therefore, A can complete the whole work in $\frac{20}{3} = 6 \frac{2}{3} days$

View NCERT Solutions for all chapters of Class 8