EXERCISE 11
Q1 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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 [Multiplying both side by 4/3]
⇒ 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
Q2 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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.
Q3 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q4 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q5 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q6 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
$=2+\dfrac{8}{5}$
$=\dfrac{10+8}{5}$
$=\dfrac{18}{5}=3\dfrac{3}{5}$
Q7 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
ALTERNATE METHOD
⇒A's one day work $=\dfrac{1}{6}$
⇒B's one day work $=\dfrac{1}{8}$
⇒C's one day work $=\dfrac{1}{4}$
Total work done$=\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{4}$
$=\dfrac{4+3+6}{24}=\dfrac{13}{24}$
∴Shares of money of A , B and C = Ratio of their one day work / Total work × Total money
⇒A's share $\dfrac{\frac{1}{6}}{\frac{13}{24}}\times 780$
$=\dfrac{4}{13}\times 780$ = 240
⇒B's share $\dfrac{\frac{1}{8}}{\frac{13}{24}}\times 780$
$=\dfrac{3}{13}\times 780$ = 180
⇒C's share $\dfrac{\frac{1}{4}}{\frac{13}{24}}\times 780$
$=\dfrac{6}{13}\times 780$ = 360
A→240 , B→180 , C→360
Q8 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q9 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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)
$=\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{10}$
$=\dfrac{1+1+2}{20}=\dfrac{4}{20}=\dfrac{1}{5}$
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
Q10 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q11 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q12 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q13 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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
Q14 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper
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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