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A can do a piece of work in 25 days and B can do the same work in 20 days. They started the work together but $B$ left after 4 days and $A$ continued to work . In how many days will the entire work be completed?
Let total work = LCM (25, 20)= 100 units
A's efficiency $=\frac{100}{25}=4$ units $/$ day
B's efficiency $=\frac{100}{20}=5$ units $/$ day
Work done by both in 4 days $=(4+5) \times 4=$ 36 units
Remaining work to be done by A alone $=100-36$ $=64$ units
$\Rightarrow$ A will finish remaing work in $\frac{64}{4}=16$ days
$\Rightarrow$ Total time taken to finish the entire work $=4+16=20$ days
Anuj’s efficiency is 150% of Vinod efficiency. If Anuj can complete a piece of work in 10 days, then how much time (in days) will they both take to complete the same work?
Let efficiency of Vinod be 2 units/day
Hence, efficiency of Anuj = 2 × (150/100) = 3 units/day
Now, total work = 3 × 10 = 30 unit
So, Time taken by Anuj and Vinod to complete the work = 30/(3 + 2) = 30/5 = 6 days
A completes two-third part of a work in 16 days, and the remaining work is completed by B alone. The entire work is completed in 21 days. A and B together will complete $65 \%$ of the same work in:
Let total work be x
A completed 2/3rd part of a work = 16 days
Total work done by A in = 16×3/2 =24 days
B completed 1/3rd part of a work in = 5 days
Total work done by B in = 5×3 =15 days
L.C.M of 25 and 15 is 120
Required Time $=\frac{120}{13}\times\frac{65}{100}= 6$
$A$ can do a piece of work in 15 days, while $B$ can do the same work in 21 days. If they work together, in how many days will the work be completed?
Let total work = LCM (15, 21) = 105 units
A's 1 day work $=\frac{105}{15}=7$ units
B's 1 day work $=\frac{105}{21}=5$ units
Required time $=\frac{105}{7+5}=8 \frac{3}{4}$ days
30 men working 8 hours a day can complete a piece of work in 28 days. How many men will be required to complete the same work in 20 days working 10 $\frac{1}{2}$ hours per day?
$M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2$
$30 \times 28 \times 8 = x \times 20 \times \frac{21}{2}$
$ x = 32 $
$\mathrm{A}$ and $\mathrm{B}$ together can do a certain work in 20 days, $\mathrm{B}$ and $\mathrm{C}$ together can do it in 30 days, and $\mathrm{C}$ and $\mathrm{A}$ together can do it in 24 days. B alone will complete $\frac{2}{3}$ part of the same work is:
2(A+B+C)=15 unit
(A+B+C) = 15/2=7.5 units
B alone can do = (A+B+C)-(A+C)=7.5-5 = 2.5 units
B alone can do 2/3 rd part of the work in = 120/2.5×2/3=32days
A can do a work in 40 days and B can do the same work in 50 days. They worked together for 5 days and then B left the work. In how many days will $A$ finish of remaining work?
worked done by A and B for 5 days= 9 × 5 = 45
Rest of the work completed by A= (200 - 45)/5 = 31 days.
A can do a piece of work in 45 days. He works for 30 days and leaves. If B completes the remaining work in 5 days, then in how many days can B alone complete the entire work?
A can do a piece of work in 45 days.
A's 1 days work =1/45
A's 30 days work =1/45×30=2/3
The remaining work =1-2/3=1/3
B's complete 1/3 th of work in 5 days
Then, B's 1 day work =1/(3×5)=1/15
B can complete the whole work =15 days
Pipe A is a filling pipe, while $\mathrm{B}$ and $\mathrm{C}$ are emptying pipes. Pipe A alone can fill a tank in 10 hours and pipe $\mathrm{C}$ alone can empty the full tank in 24 hours. If all three pipes are opened together, the tank is completely filled in 40 hours. In how many hours can pipe $B$ alone empty two-third part of the tank?
The LCM of 10,24 , and 40 =120 liters.
Efficiency of B = A- B-C -(A-C) = 3 - (12-5) = - 4
Now Time taken by Pipe B to empty 2/3 rd of a tank $= \frac{2}{3\times4}\times120 = 20$
18 persons can complete a work in 15 days when they work 6 hours a day. The time (in days) taken by 12 persons working 10 hours a day to complete the same work, is:
18×15×6=12× Number of working days ×10
Number of working days =$\frac{18×15×6}{12×10}$
Number of working days =$\frac{27}{2}$
Number of working days =13$\frac{1}{2}$ days.
$\mathrm{A}$ and $\mathrm{B}$ can do a piece of work in 12 days and 20 days respectively. They both work together for 6 days. The remaining work is completed by $C$ alone in 12 days. In how many days will $A$ and $C$ together complete the $\frac{2}{3}$ part of the work?
Total work done in 6 days = 6$\times$(5+3) =48
remaining work = 60 - 48 = 12
Efficiency of C = 12/12 = 1
2/3 rd part of the work = 60$\times$2/3 = 40
Required days = 40/6= $6 \frac{2}{3}$ days
$X$ and $Y$ together can do a piece of work in 8 days. If $X$ alone can do the same work in 40 days, then in how many days will $Y$ do the work alone?
efficiency of y = 5-1 = 4
Required days $=\frac{40}{4} = 10 $
A can do $33 \frac{1}{3} \%$ of a work in 10 days and B can do $66 \frac{2}{3} \%$ of the same work in 8 days. Both worked together for 8 days. C alone completed the remaining work in 3 days. $A$ and $C$ together will complete $\frac{5}{6}$ part of the original work in:
A does 1/3 part of the work in = 10 days
A can do whole work alone in = (10×3) days = 30 days ⋯(1)
B does 2/3 part of the work in = 8 days
B will do the whole work alone in = (8×3)/2 days =12 days ⋯(2)
A and B worked for 8 days=(2+5)×8 Units =56 units
Now, the remaining work = (60 - 56) = 4 units
The efficiency of C=4/3 units
The total efficiency of A and C=2+(4/3) =10/3 units
A and C have to do 5/6 of the original work=(5/6)×60=50 units
work to be done by A and C =50/(10/3)=15 days
$\text{F}$ and $\text{M}$ together can do a piece of work in 8 days. $\text{F}$ alone can do the same work in 12 days. In how many days can $\text{M}$ alone do the same work?
Let the total work be $ = 24$ unit , (LCM of 8 and 12)
Efficiency of $F = \frac{24}{12} = 2$ unit
Efficiency of $F+M = \frac{24}{8} = 3$ unit
Then Efficiency of $M = 3-2 = 1$ unit
Time taken by $M$ to do $24$ unit of work $ = \frac{24}{1} = 24$ days
A, B and C can do a piece of work in 10,12 and 15 days, respectively. In how many days can B do the work, if he is assisted by $A$ and $C$ together on alternate days?
Let total work $= 60$ unit
Efficiency of A $= \frac{60}{10}=6$
Efficiency of B $= \frac{60}{12}=5$
Efficiency of C $= \frac{60}{15}=4$
On the first day, B is assisted by A and on the Second day, B is assisted by C and this continues
So in two days $(6+5) + (5+4) = 20$ units of work get completed.
To complete 60 units of work, It will take $2\times3 = 6$ days
A can do $33 \frac{1}{3} \%$ of a work in 10 days and B can do $66 \frac{2}{3} \%$ of the same work in 8 days. Both worked together for 8 days. C alone completed the remaining work in 3 days. $A$ and $C$ together will complete $\frac{5}{6}$ part of the original work in how many days:
A can do $33\frac{1}{3}\%$ work in 10 days
So he will complete the total work in $(10\times \frac{3}{1})= 30$ days, (As $33\frac{1}{3}\% = \frac{1}{3}$ )
B can do $66\frac{2}{3}\%$ work in 8 days
So he will complete the total work in $(8\times\frac{3}{2}) = 12$ days, (As $66\frac{2}{3}\%= \frac{2}{3}$)
LCM of $30$ and $12= 60$ unit
Efficiency of A $= \frac{60}{30} = 2$ unit
Efficiency of B $=\frac{60}{12} = 5$ unit
In 8 days they both will complete = $8\times (5+2) = 56$ unit
Remaining work $=60-56 = 4$ unit
C completed $4$ unit work in $3$ days
so Efficiency of C $= \frac{4}{3}$
A and C will complete $\frac{5}{6}$ of the total work in
$\Rightarrow \frac{5}{6}\times \frac{60}{2+\frac{4}{3}} = \frac{5\times60\times3}{10\times6}$
$\Rightarrow 15$ Days
Mohan and Ramesh together can complete a work in 15 days. Mohan is 50% more efficient than Ramesh. Find the time taken by Mohan alone to complete the work.
Let the Efficiency of Ramesh is $2$ unit
then Efficiency of Mohan will be $3$ unit, (As $50\% = \frac{1}{2}$)
So Total work $ = 15\times (2+3) = 75$ unit
Time taken by Mohan to complete the work $= \frac{75}{3} = 25$ days
$\frac{4}{11}$ part of a bucket is filled in 5 minutes. Find the time taken to fill the remaining bucket.
Given that, $\frac{4}{11}$ part is filled
So,
Remaining $1-\frac{4}{11} = \frac{7}{11}$
$\frac{4}{11}$ part is filled in $5$ minutes
so, $\frac{1}{11}$ part will be filled in $\frac{5}{4}$ minutes
To fill $\frac{7}{11}$ part, It will take -
$7\times \frac{5}{4} = \frac{35}{4}$ = 8 minutes 45 seconds
$A$ can do a piece of work in 18 days. $A$ and $B$ working together can do the same work in 8 days, while $B$ and $C$ working together can do the same work in 6 days. In how many days can $A, B$ and $C$ working together do the same work?
As we need efficiencies of A, B and C to find the number of days taken to complete the total work, We can use the data of A and B+C to get the required efficiencies
Let the total work $= 18$ unit, (As the LCM of $6$ and $18 = 18$)
Efficiency of $A= \frac{18}{18} = 1$ unit
Efficiency of $B$ and $C=\frac{18}{6} = 3$ unit
So A,B and C will complete the total work in
$\frac{18}{3+1} = \frac{18}{4} = 4\frac{1}{2}$ days
$X$ and $Y$ together can do a piece of work in 8 days. If $X$ alone can do the same work in 40 days, then in how many days will $Y$ do the work alone?
Let the total work be 40 unit , then
Efficiency of X $= \frac{40}{40} = 1$
And Efficiency of X+Y $= \frac{40}{8} = 5$
So Efficiency of Y $=5-1=4$
Time taken by Y to complete total work $=\frac{40}{4}=10$ Days