Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 13 - Time And Work

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Page No 172:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

Answer:

(c)

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

Page No 175:

Answer:

(b) 12 minutes

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

Page No 175:

Answer:

(c) 16 hours

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

Page No 175:

Answer:

(d) 14 hours

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

Page No 175:

Answer:

(b) 7 hours 30 minutes

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

Page No 176:

Answer:

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

Page No 176:

Answer:

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

Page No 176:

Answer:

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

Page No 176:

Answer:

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

Page No 176:

Answer:

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

Page No 176:

Answer:

(b) 3:2

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

Page No 176:

Answer:

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

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

Page No 176:

Answer:

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

Page No 176:

Answer:

(b) 14 hours

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

Page No 176:

Answer:

(c) 36 hours

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

Page No 176:

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 $\frac{20}{3} = 6 \frac{2}{3} days$

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