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Solved Examples

Example-1
If 9 men working 6 hours a day can do a work in 88 days. Then 6 men working 8 hours a day can do it in how many days?
Solution
We know –
(M1*D1*H1/W1) = (M2*D2*H2/W2)
So, (9*6*88/1) = (6*8*d/1); On solving, D = 99 days.
 
Example-2

If 34 men completed 2/5th of a work in 8 days working 9 hours a day. How many more man should be engaged to finish the rest of the work in 6 days working 9 hours a day?

Solution

(M1*D1*H1/W1) = (M2*D2*H2/W2)
So, (34*8*9/(2/5)) = (x*6*9/(3/5))
So x = 136 men
Number of men to be added to finish the work = 136 - 34 = 102 men

 
Example-3
If 5 women or 8 girls can do a work in 84 days. In how many days 10 women and 5 girls can do the same work?
Solution
Given that 5 women is equal to 8 girls to complete a work
So, 10 women = 16 girls.
Therefore 10 women + 5 girls = 16 girls + 5 girls = 21 girls.
8 girls can do a work in 84 days
Then 21 girls = (8*84/21) = 32 days.
Therefore 10 women and 5 girls can a work in 32 days
 
 
Example-4

Worker A takes 8 hours to do a job. Worker B takes 10hours to do the same job. How long it takes both A & B, working together but independently, to do the same job?

Solution

A's one hour work = 1/8.
B's one hour work = 1/10
(A + B)'s one hour work = 1/8 + 1/10 = 9/40
Both A & B can finish the work in 40/9 days
 

  
Example-5
A can finish a work in 18 days and B can do the same work in half the time taken by A. Then, working together, what part of the same work they can finish in a day?
Solution
Given that B alone can complete the same work in days = half the time taken by A = 9 days
A's one day work = 1/18
B's one day work = 1/9
(A + B)'s one day work = 1/18 + 1/9 = 1/6
 
  
Example-6

A is twice as good a workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work.

Solution

If A takes x days to do a work then B takes 2x days to do the same work
1/x + 1/2x = 1/18
3/2x = 1/18
x = 27 days.
Hence, A alone can finish the work in 27 days.
 

  
Example-7
A can do a certain work in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?
Solution
Ratio of time taken by A & B = 160 : 100 = 8 : 5
Suppose B alone takes x days to do the job.
Then, 8 : 5 : : 12 : x
8x = 5*12
x = 15/2 days.
  
 
Example-8

A can do a piece of work n 7 days of 9 hours each and B alone can do it in 6 days of 7 hours each. How long wi ll they take to do it working together 8 & 2/5 hours a day?

Solution

A can complete the work in (7*9) = 63 Hours
B can complete the work in (6*7) = 42 Hours
A's one hour's work = 1/63 and
B's one hour work = 1/42
(A + B)'s one hour work = 1/63 + 1/42 = 5/126
Therefore, both can finish the work in 126/5 hours.
Number of days of 8 & 2/5 hours each = (126*5/ (5*42)) = 3days
 

 
Example-9
A takes twice as much time as B or thrice as much time to finish a piece of work. Working together they can finish the work in 2 days. B can do the work alone in?
Solution
Suppose A, B and C take x, x/2 and x/3 hours respectively finish the work then
1/x + 2/x + 3/x = 1/2
6/x = 1/2; x = 12; So, B takes 6 hours to finish the work.
 
  
Example-10

X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z can do 1/3 of work in 13 days. Who will complete the work first?

Solution

Whole work will be done by X in 10*4 = 40 days.
Whole work will be done by Y in (40*100/40) = 100 days.
Whole work will be done by Z in (13*3) = 39 days
Therefore, Z will complete the work first.
 

 
Example-11
A and B undertake to do a piece of work for Rs 600.A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they can finish it in 3 days, Find the share of each?
Solution
C's one day's work = (1/3) - (1/6 + 1/8) = 1/24
Therefore, A : B : C = Ratio of their one day’s work = 1/6 : 1/8 : 1/24 = 4 : 3 : 1
A's share = Rs. (600*4/8) = 300
B's share = Rs. (600*3/8) = 225
C's share = Rs. [600 - (300 + 225)] = Rs 75
  
Example-12

A can do a piece of work in 80 days. He works at it for 10 days & then B alone finishes the remaining work in 42 days. In how much time will A and B, working together, finish the work?

Solution

Work done by A in 10 days = 10/80 = 1/8
Remaining work = (1 - (1/8)) = 7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7) = 48 days
Therefore, A's one day's work = 1/80
B’s one day's work = 1/48
(A + B)'s one day's work = 1/80 + 1/48 = 8/240 = 1/30
Hence, both will finish the work in 30 days.
 

  
Example-13
Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?
Solution
Number of pages typed by Ronald in one hour = 32/6 = 16/3
Number of pages typed by Elan in one hour = 40/5 = 8
Number of pages typed by both in one hour = ((16/3) + 8) = 40 /3
Time taken by both to type 110 pages = 110*3/40 = 8 hours.
 
  
Example-14

A and B can do a work in 12 days, B and C in 15 days, C and A in 20 days. If A, B and C work together, they will complete the work in how many days?

Solution

(A + B)'s one day's work = 1/12;
(B + C)'s one day's work = 1/15;
(A + C)'s one day's work = 1/20;
Adding we get 2(A + B + C)'s one day's work = 1/12 + 1/15 + 1/20 = 12/60 = 1/5
(A + B + C)'s one day work = 1/10
So, A, B, and C together can complete the work in 10 days.
 

  
Example-15
A and B can do a work in 8 days, B and C c an do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in how many days?
Solution
(A + B + C)'s one day's work = 1/6;
(A + B)'s one day's work = 1/8;
(B + C)'s one day's work = 1/12;
(A + C)'s one day's work
 = 2(A + B + C)'s one day's work - ((A + B)'s one day’s work + (B + C)'s one day work)
 = (2/6) - (1/8 + 1/12) = (1/3) - (5/24) = 3/24 = 1/8
So, A and C together will do the work in 8 days.
 
  
Example-16

A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in how many days?

Solution

(A + B)'s one day's work = 1/10;
C's one day's work = 1/50
(A + B + C)'s one day's work = (1/10 + 1/50) = 6/50 = 3/25
Also, A's one day's work = (B + C)’s one day's work
Therefore, 2*(A's one day's work) = 3/25
A's one day's work = 3/50
B's one day’s work = (1/10 - 3/50) = 2/50 = 1/25
B alone could complete the work in 25 days
 

  
Example-17
A is thrice as good a workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
Solution
Ratio of times taken by A and B = 1 : 3
If difference of time is 2 days, B takes 3 days
If difference of time is 60 days, B takes (3*60/2) = 90 days
So, A takes 30 days to do the work = 1/90
A's one day's work = 1/30;
B's one day's work = 1/90;
(A + B)'s one day's work = 1/30 + 1/90 = 4/90 = 2/45
Therefore, A & B together can do the work in 45/2 days
 
 
Example-18

A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. In how much time will A & B, working together, finish the work?

Solution

Work Done by A n 10 days = 10/80 = 1/8
Remaining work = 1 - 1/8 = 7/8
Now 7/8 work is done by B in 42 days
Whole work will be done by B in 42*8/7 = 48 days
A's one day's work = 1/80 and B's one day's work = 1/48
(A + B)'s one day's work = 1/80 + 1/48 = 8/240 = 1/30
Hence both will finish the work in 30 days
 

  
Example-19
Two pipes A & B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
Solution
Part filled by A in 1 hour = 1/36
Part filled by B in 1 hour = 1/45;
Part filled by (A + B)'s in 1 hour = 1/36 + 1/45 = 9/180 = 1/20
Hence, both the pipes together will fill the tank in 20 hours.
 
 
Example-20

Two pipes can fill a tank in 10 hours & 12 hours respectively. While 3rd pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?

Solution

Net part filled in 1 hour = 1/10 + 1/12 - 1/20 = 8/60 = 2/15
The tank be filled in 15/2hours = 7 hrs 30 min
 

  
Example-21
A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?
Solution
Net part filled in 1 hour = 1/4 - 1/9 = 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours.
 
 
Example-22

If two pipes function simultaneously, the reservoir will be filled in 12 days. One pipe fills the reservoir 10 hours faster than the other. How many hours does it take the second pipe to fill the reservoir.

Solution

Let the reservoir be filled by the 1st pipe in x hours.
The second pipe will fill it in (x + 10) hours
1/x + (1/(x + 10)) = 1/12
(2x + 10)/((x)*(x + 10)) = 1/12
x = 20
So, the s econd pipe will take 30 hours to fill the reservoir.
 

  
Example-23
A cistern has two taps which fill it in 12 min and 15 min respectively. There is also a waste pipe in the cistern. When all the three are opened, the empty cistern is full in 20 min. How long will the waste pipe take to empty the full cistern?
Solution
Work done by a waste pipe in 1 min
 = 1/20 - (1/12 + 1/15) = -1/10 (- ve means emptying)
 
  
Example-24

A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

Solution

Time taken by one tap to fill the half of the tank = 3 hours
Part filled by the four taps in 1 hour = 4/6 = 2/3
Remaining part = 1 - 1/2 = 1/2
Therefore, 2/3 : 1/2 : : 1 : x
Or x = (1/2)*1*(3/2) = 3/4 hours.
I.e. 45 min
So, total time taken = 3 hrs 45 min.
 

 
Example-25
A water tank is two - fifth full. Pipe A can fill a tank in 10 min. And B can empty it in 6 min. If both pipes are open, how long will it take to empty or fill the tank completely?
Solution
Clearly, pipe B is faster than A and So, the tank will be emptied.
Part to be emptied = 2/5.
Part emptied by (A + B) in 1 min = 1/6 - 1/10 = 1/15
Therefore, 1/15 : 2/5 : : 1 : x or x = ((2/5)*1*15) = 6 min.
So, the tank is emptied in 6 min.
 
 
Example-26

Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P & Q, having each turn together to fill the empty drum?

Solution

Let the capacity of P be x lit.
Then capacity of Q = x/3 lit
Capacity of the drum = 60x lit
Required number of turns = 60x/(x + (x/3)) = 60x * 3/4x = 45
 

 
Example-27
Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 32 min more to fill the cistern. When the cistern is full, in what ti me will the leak empty it?
Solution
Work done by the two pipes in 1 hour = 1/14 + 1/16 = 15/112
Time taken by these two pipes to fill the tank = 112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min + 32 min = 8 hours
Therefore, work done by (two pipes + leak) in 1 hr = 1/8
Work done by leak n 1 hour = 15/112 - 1/8 = 1/112
Leak will empty full cistern n 112 hours.
 
  
Example-28

Two pipes A & B can fill a tank in 24 min and 32 min respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min.

Solution

Part filled by A in 18 min = 18/24
Remaining part = 6/24 = 1/4
¼ can be filled by B in 32*1/4 = 8; Hence B must be closed after 8 min.
 

 
Example-29
Two pipes A & B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
Solution
Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x + 6 hours
1/x + 1/x + 6 = 1/4
Solving this we get x = 6; Hence, A takes 6 hours to fill the cistern separately.
 
  
Example-30

A tank is filled by 3 pipes with uniform flow. The first two pipes operating simultaneously fill the tan in the same time during whic h the tank is filled by the third pipe alone. The 2nd pipe fills the tank 5 hours faster than first pipe and 4 hours slower than third pipe. The time required by first pipe is:

Solution

Suppose, first pipe take x hours to fill the tank then
B & C will take (x - 5) and (x - 9) hours respectively.
Therefore, 1/x + 1/(x - 5) = 1/(x - 9)
On solving, x = 15; Hence, time required by first pipe is 15 hours.
 

 
Example-31
A large tanker can be filled by two pipes A & B in 60min and 40 min respectively. How many minutes will i t take to fill the tanker from empty state if B is used for half the time and A & B fill it together for the other half?
Solution
Part filled by (A + B) n 1 min = (1/60 + 1/40) = 1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24 + 1/40) = 1
(x/2)*(1/15) = 1
x = 30 min.
 
 
Example-32

Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A s opened first, in how many hours, the tank shall be full.

Solution

(A + B)'s 2 hours work when opened alternatively = 1/6 + 1/4 = 5/12
(A + B)'s 4 hours work when opened alternatively = 10/12 = 5/6
Remaining part = 1 - 5/6 = 1/6.
Now, it is A's turn and 1/6 part is filled by A in 1 hour.
So, total time taken to fill the tank = (4 + 1) = 5 hours.
 

  
Example-33
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternatively, the tank will be full in.
Solution
(A + B)'s 1 hour's work = 1/12 + 1/15 = 9/60 = 3/20
(A + C)'s 1 hour's work = 1/20 + 1/12 = 8/60 = 2/15
Part filled in 2 hours = 3/20 + 2/15 = 17/60
Part filled in 2 hours = 3/20 + 2/15 = 17/60
Part filled in 6 hours = 3*17/60 = 17/20
Remaining part = 1 - 17/20 = 3/20
Now, it is the turn of A & B and 3/20 part is filled by A & B in 1 hour.
Therefore, total time taken to fill the tank = 6 + 1 = 7 hours
 
 
Example-34

A Booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3. The emptying capacity of the tank is 10 m3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

Solution

Let, the filling capacity of the pump be x m3/min
Then, emptying capacity of the pump = (x + 10) m3/min.
So, 2400/x – 2400/(x + 10) = 8; On solving x = 50.
 

 
Example-35
A leak in the bottom of a tank can empty the full tan in 8 hr. An inlet pipe fills water at the rate of 6 litres a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hrs. How many litres does the cistern hold?
Solution
Work done by the inlet in 1 hr = 1/8 - 1/12 = 1/24
Work done by the inlet in 1 min = (1/24)*(1/60) = 1/1440
Therefore, Volume of 1/1440 part = 6 lit
Volume of whole = (1440*6) lit = 8640 lit.
 
 
Example-36

Two pipes A and B can fill a cistern in 37 ½ min and 45 minutes respectively. Both the pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after:

Solution

Let B be turned off after x min. T hen,
Part filled by (A + B) in x min + part filled by A in (30 - x) min = 1
Therefore, x(2/75 + 1/45) + (30 - x) (2/75) = 1
11x/225 + (60 - 2x)/75 = 1
11x + 180 - 6x = 225
x = 9.; So, B must be turned off after 9 minutes
 





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