# Solved Examples

(M

_{1}*D

_{1}*H

_{1}/W

_{1}) = (M

_{2}*D

_{2}*H

_{2}/W

_{2})

So, (9*6*88/1) = (6*8*d/1); On solving, D = 99 days.

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?

(M_{1}*D_{1}*H_{1}/W_{1}) = (M_{2}*D_{2}*H_{2}/W_{2})

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

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

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?

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

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

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.

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.

Suppose B alone takes x days to do the job.

Then, 8 : 5 : : 12 : x

8x = 5*12

x = 15/2 days.

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?

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

1/x + 2/x + 3/x = 1/2

6/x = 1/2; x = 12; So, B takes 6 hours to finish the work.

X can do Â¼ 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?

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.

Therefore, A : B : C = Ratio of their one dayâ€™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

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?

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

Bâ€™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.

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.

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?

(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.

(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 dayâ€™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.

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?

(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)â€™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 dayâ€™s work = (1/10 - 3/50) = 2/50 = 1/25

B alone could complete the work in 25 days

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

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?

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

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.

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?

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

Therefore the cistern will be filled in 36/5 hours or 7.2 hours.

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.

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.

= 1/20 - (1/12 + 1/15) = -1/10 (- ve means emptying)

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?

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.

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.

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?

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

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.

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.

Part filled by A in 18 min = 18/24

Remaining part = 6/24 = 1/4

Â¼ can be filled by B in 32*1/4 = 8; Hence B must be closed after 8 min.

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.

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:

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.

Suppose the tank is filled in x minutes

Then, x/2(1/24 + 1/40) = 1

(x/2)*(1/15) = 1

x = 30 min.

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.

(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.

(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

A Booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m^{3}. The emptying capacity of the tank is 10 m^{3} 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?

Let, the filling capacity of the pump be x m^{3}/min

Then, emptying capacity of the pump = (x + 10) m^{3}/min.

So, 2400/x â€“ 2400/(x + 10) = 8; On solving x = 50.

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.

Two pipes A and B can fill a cistern in 37 Â½ 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:

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