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S.chand mathematics solution class 8 chapter 11 time and work

 EXERCISE 11


Question 1

A can do \dfrac{3}{4} of the work in 12 days . In how many days can he complete \dfrac{1}{8}th of the work ?

Sol :

⇒A can do \dfrac{3}{4} of work in 12 days

⇒ A can do 1 work in 12\times \dfrac{4}{3} days And  [Mutiplying both side by 3/4]

⇒ A can do \dfrac{1}{8} work in 12\times \dfrac{4}{3}\times \dfrac{1}{8} days [Multiplying both side by 1/8]

=4\times \dfrac{1}{2} days

= 2 days

 


Question 2

Tanu can do a piece of work in 24 days and Manisha can do it in 30 days . How long will they take to do the work working together ?

Sol :

⇒Tanu can do a piece of work in 24 days

⇒So Tanu’s one day work =\dfrac{1}{24}

⇒Manisha can do the same work in 30 days

⇒So Manisha’s one day work =\dfrac{1}{30}

⇒We know that

⇒Tanu and Manisha’s one day work = Tanu’s one day work + Manisha’s one day work

=\dfrac{1}{24}+\dfrac{1}{30} [Taking LCM]

=\dfrac {5 + 4}{120}

= \dfrac{9}{120}

= \dfrac{3}{40}

⇒Tanu and Manisha’s one day work = \dfrac{3}{40}

⇒Total time required for completing work \dfrac{1}{\text{one day work}}=\dfrac{1}{\dfrac{3}{40}}

⇒So Tanu and Manisha can complete the same work in = \dfrac{40}{3}=13\dfrac{1}{3} days.

 


Question 3

A and B can polish the floors of a building in 18 days . A alone can do \dfrac{1}{6} of this job in 4 days . in how many days can B alone polish the floors ?

Sol :

⇒A alone complete whole work in =\dfrac{4}{\dfrac{1}{6}}=4\times\dfrac{6}{1} = 24 days

⇒A’s one day work =\dfrac{1}{24}

⇒(A and B)’s one day work =\dfrac{1}{18}

⇒Also , (A+B)’s one day work = A’s one day work + B’s one day work

\dfrac{1}{18}=\dfrac{1}{24}+\dfrac{1}{x}

\dfrac{4-3}{72}=\dfrac{1}{x}

\dfrac{1}{72}=\dfrac{1}{x}

⇒x = 72

72 days

 


Question 4

A and B can weave a certain number of baskets in 12 days and 15 days respectively . They work together for 5 days and then B leaves . In how many days will A finish the rest of the work ?

Sol :

⇒A’s one day work =\dfrac{1}{12}

⇒B’s one day work =\dfrac{1}{15}

⇒(A + B)’s one day work = A’s one day work + B’s one day work

=\dfrac{1}{12}+\dfrac{1}{15}

=\dfrac{5+4}{60}

=\dfrac{9}{60}

=\dfrac{3}{20}

⇒(A + B)’s one day work =\dfrac{3}{20}

⇒They work together for 5 days =5\times \dfrac{3}{20}= \dfrac{3}{4}

⇒Remaining work =1-\dfrac{3}{4}

=\dfrac{4-3}{4}=\dfrac{1}{4}

⇒A can complete all work in 12 days . So , remaining work can be done in =\dfrac{1}{4}\times 12 days

= 3 days

 


Question 5

A , B and C working together take 8 min . to address a pile of envelopes . A and B together would take 10 min ; A and C together would take 15 min . How long would each take working alone ?

Sol :

⇒(A + B + C)’s one unit work =\dfrac{1}{8}

⇒(A + B)’s one unit work =\dfrac{1}{10}

⇒(A + C)’s one unit work =\dfrac{1}{15}

⇒C’s one unit work = (A + B + C)’s one unit work  – (A + B)’s one unit work

=\dfrac{1}{8}-\dfrac{1}{10}

=\dfrac{5-4}{40}=\dfrac{1}{40}

∴C alone take 40 min

⇒B’s one unit work = (A + B + C)’s one unit work  – (A + C)’s one unit work

=\dfrac{1}{8}-\dfrac{1}{15}

=\dfrac{15-8}{120}=\dfrac{7}{120}

∴B alone take \dfrac{120}{7} min

⇒A’s one unit work = (A + B + C)’s one unit work  – B’s one unit work – C’s one unit work

=\dfrac{1}{8}-\dfrac{7}{120}-\dfrac{1}{40}

=\dfrac{15-7-3}{120}=\dfrac{5}{120}=\dfrac{1}{24}

∴A alone take 24 min

A→24 min ; B→\dfrac{120}{7} min ; C→40 min

 


Question 6

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 .

Sol :

⇒A’s one day work =\dfrac{1}{6}

⇒B’s one day work =\dfrac{1}{4}

⇒A’s 2 days work =2\times \dfrac{1}{6}=\dfrac{1}{3}

⇒Remaining work =1-\dfrac{1}{3}

=\dfrac{3-1}{3}=\dfrac{2}{3}..(i)

⇒(A+B)’s one day work = A’s one day work + B’s one day work

=\dfrac{1}{6}+\dfrac{1}{4}

=\dfrac{2+3}{12}=\dfrac{5}{12}

or

⇒Total time required to complete work together=\dfrac{1}{\text{one day work}} =\dfrac{1}{\dfrac{5}{12}}=\dfrac{12}{5}

⇒Time to complete the remaining work together =\dfrac{2}{3}\times \dfrac{12}{5} =\dfrac{8}{5}..(ii)

⇒Total time required to complete work = A’s alone working time + (A+B)’s working together time

=2+\dfrac{8}{5}

=\dfrac{10+8}{5}

 


Question 7

780 is paid for a task which A can do in 6 days , B in 8 days and C in 4 days . If all work together , how much money should each receive ?

Sol :

⇒A’s one day work =\dfrac{1}{6}

⇒B’s one day work =\dfrac{1}{8}

⇒C’s one day work =\dfrac{1}{4}

∴Ratio of shares of money of A , B and C = Ratio of their one day work

=\dfrac{1}{6}:\dfrac{1}{8}:\dfrac{1}{4} =\dfrac{1}{6}\times 24:=\dfrac{1}{8}\times 24 :\dfrac{1}{4}\times 24 [L.C.M of 6 , 8 and 24]

= 4 : 3 : 6

∴Sum of the ratios = 4 + 3 + 6 = 13

⇒A’s share =\dfrac{4}{13}\times 780 = 240

⇒B’s share =\dfrac{3}{13}\times 780 = 180

⇒C’s share =\dfrac{6}{13}\times 780 = 360

A→240 , B→180 , C→360

 


Question 8

X works thrice as fast as Y . If Y alone can complete a job in 18 days , find how ,many days will X and Y together take to complete the job .

Sol :

⇒Y’s one day work =\dfrac{1}{18}

A.T.Q

⇒3X=Y

X=\dfrac{Y}{3}

X=\dfrac{18}{3}

⇒X=6

⇒X’s one day work =\dfrac{1}{6}

⇒(X+Y)’s one day work = X’s one day work + Y’s one day work

=\dfrac{1}{6}+\dfrac{1}{18}

=\dfrac{3+1}{18}=\dfrac{4}{18}=\dfrac{2}{9}

⇒Total time required by (X+Y) to complete work =\dfrac{1}{\text{one day work}} =\dfrac{1}{\dfrac{2}{9}} =\dfrac{9}{2} or =4\dfrac{1}{2} days

 


Question 9

A can do a job in 20 days , B in 30 days and C in 60 days . If A is helped by B and C on every third day , then in how many days will the job be finished ?

Sol :

⇒A’s one day work =\dfrac{1}{20}..(i)

⇒(A+B+C)’s one day work = A’s one day work + B’s one day work + C’s one day work

=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{60}

=\dfrac{3+2+1}{60}

=\dfrac{6}{60}=\dfrac{1}{10}..(i)

⇒According to question, A work starting two days alone and on every third day all together work .

Total time to complete work = 1st day + 2nd day + 3rd day

1 and 2 day equal to (i) and 3 day equal to (ii)

=\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}

=\dfrac{1+1+2}{20}=\dfrac{4}{20}=\dfrac{1}{5}

Which means in 3 days 1/5 work is completed

\text{3 days}=\dfrac{1}{5}\text{ work}

3\times 5 = 1 \text{ work}

= 15 days to complete whole work

 


Question 10

Sonia can copy 50 pages in 10 hours . Priya and Sonia can copy 300 pages in 40 hours . In how much time can Priya copy 30 pages ?

Sol :

⇒Sonia in 10 hours can copy 50 pages

∴In 1 hour sonia can copy \dfrac{50}{10}=\dfrac{5}{1} pages

⇒Priya and Sonia together in 40 hours can copy 300 pages

∴Priya and Sonia together in 1 hours can copy \dfrac{300}{40}=\dfrac{30}{4} pages

⇒Priya’s 1 hour work = Priya and Sonia together in 1 hours work – Sonia’s 1 hour work

=\dfrac{30}{4}-\dfrac{5}{1}

=\dfrac{30-20}{4}=\dfrac{10}{4}=\dfrac{5}{2}

⇒Number of hours required to complete the work =\dfrac{\text{Work to be done}}{\text{one day work}}

=\dfrac{30}{\dfrac{5}{2}}

=5\times\dfrac{2}{5}

= 12 hours

 


Question 11

A tap can fill a cistern in 15 minutes and another can empty it in 18 minutes . Find in how many minutes the tank will be filled up ?

Sol :

⇒Tank filled by tap in one hour =\dfrac{1}{15} part

⇒Tank empty by tap in one hour =\dfrac{1}{18} part

⇒Tank filled in one hour when both taps are open =\dfrac{1}{15}-\dfrac{1}{18}

=\dfrac{6-5}{90}=\dfrac{1}{90}

= 90 mins the tank will be filled up

 


Question 12

Two taps can fill a cistern in 15 hours and 20 hours respectively . A third tap can empty it in 30 hours . How long will they take to fill the cistern if all the taps are opened ?

Sol :

⇒Tank filled by first tap in one hour =\dfrac{1}{15} part

⇒Tank filled by second tap in one hour =\dfrac{1}{20} part

⇒Tank empty by third tap in one hour =\dfrac{1}{30} part

⇒Tank filled in one hour when all three taps are open =\dfrac{1}{15}+\dfrac{1}{20}-\dfrac{1}{30}

=\dfrac{4+3-2}{60}=\dfrac{5}{60}=\dfrac{1}{12}

= 12 hours take the tank to fill up

 


Question 13

Two taps running together can fill a bath in 4 minutes , which is filled by one of the taps by itself in 7 minutes . How long would it take if the other pipe is running by itself ?

Sol :

⇒Bathtub filled by (1st + 2nd) taps in one hour =\dfrac{1}{4} part

⇒Bathtub filled by first tap in one hour =\dfrac{1}{7} part

⇒Bathtub filled by (1st + 2nd) taps in one hour = time taken by 1st tap in one hour + time taken by 2nd tap in one hour

\dfrac{1}{4}=\dfrac{1}{7}+x

\dfrac{1}{4}-\dfrac{1}{7}=x

\dfrac{7-4}{28}=x

x=\dfrac{3}{28}

Time taken by Second tap =\dfrac{28}{3}=9\dfrac{1}{3} minutes

 


Question 14

A cistern can be filled by 3 taps A, B and C when turned on separately in 12 min , 10 min and 15 min respectively . If all are turned on together for 2\dfrac{2}{3} minutes and if B and C are then turned off , how much time will A alone take to fill the cistern ?

Sol :

⇒Time taken when all 3 taps are open for a unit time = Tap A open for unit time + Tap B open for unit time + Tap C open for unit time

=\dfrac{1}{12}+\dfrac{1}{10}+\dfrac{1}{15}

=\dfrac{5+6+4}{60}=\dfrac{15}{60}=\dfrac{1}{4}

A.T.Q

⇒All are turned for 2\dfrac{2}{3} minutes =\dfrac{1}{4}\times 2\dfrac{2}{3} =\dfrac{1}{4}\times \dfrac{8}{3}=\dfrac{2}{3}

⇒Remaining cistern to be filled =1-\dfrac{2}{3} =\dfrac{3-2}{3}=\dfrac{1}{3}

⇒Tap A alone fill the remaining cistern [1/3 of 12min] =\dfrac{1}{3}\times 12

= 4 minutes

 


 

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