Concept: Time and Work

Example: A can do a piece of work in 6 days, and B can do it in 12 days. What time will they require to do if working together?

Solution:
Let’s solve this question using the unitary method.

Assuming the total work to be done = 1 unit.

Efficiency of A = $\frac{1}{6}$

Efficiency of B = $\frac{1}{12}$

Combined efficiency of A and B = $\frac{1}{6}+\frac{1}{12}$ = $\frac{1}{4}$

Time required by A and B working together to finish the work = $\frac{1}{1/4}$ = 4 days

Alternately,

Number of days taken together = $\frac{6\times 12}{6+12}$ = $\frac{72}{18}$ = 4 days

Example: If A can do complete 2/3rd of the work in 10 days, how long will it take to complete the whole work?

Solution:
In 10 days $\frac{2}{3}$^rd work is done.

In 1 day work done is $\frac{2}{30}$th = $\frac{1}{15}$^th

Hence, to complete the entire work, it will take 15 days.

Example: A can do a piece of work in 6 days, B can do it in 12 days whereas C can do the same work in 4 days. What time will they require to do if working together?

Solution:
Let’s solve this question using the unitary method.

Assuming the total work to be done = 1 unit.

Efficiency of A = $\frac{1}{6}$

Efficiency of B = $\frac{1}{12}$

Efficiency of C = $\frac{1}{4}$

Combined efficiency of A and B = $\frac{1}{6}+\frac{1}{12}+\frac{1}{4}$ = $\frac{1}{2}$

Time required by A and B working together to finish the work = $\frac{1}{1/2}$ = 2 days

Example: A can do a piece of work in 20 days, and B can do it in 30 days. What time will they require to do if working together?

Solution:
Let’s solve this question using the LCM method.

Assuming the total work to be done = LCM(20, 30) = 60 units.

Efficiency of A = $\frac{\mathrm{Work\; done}}{\mathrm{Time\; taken}}$ = $\frac{60}{20}$ = 3 units/day

Efficiency of B = $\frac{60}{30}$ = 2 units/day

Combined efficiency of A and B = 3 + 2 = 5 units/day

Time required by A and B working together to finish the work = $\frac{\mathrm{Work\; to\; be\; done}}{\mathrm{combined\; efficiency}}$ = $\frac{60}{5}$ = 12 days

If you solve the same question using unitary method, then also the answer will be 12 days. In such questions the amount of work assumed does not change the answer.

Example: A can do a piece of work in 20 days, B can do it in 30 days whereas C can do the same work in 12 days. What time will they require to do if working together?

Solution:
Let’s solve this question using the LCM method.

Assuming the total work to be done = LCM(20, 30, 12) = 60 units.

Efficiency of A = $\frac{60}{20}$ = 3

Efficiency of B = $\frac{60}{30}$ = 2

Efficiency of C = $\frac{60}{12}$ = 5

Combined efficiency of A and B = 3 + 2 + 5 = 10

Time required by A and B working together to finish the work = 60/10 = 6 days

While using LCM method, we can avoid fractions and hence it may seem easier of the two approaches. But, it is important to be comfortable with both the approaches as some questions can be solved better using the Unitary method.

Example: A and B together can do a piece of work in 6 days. A alone can do it in 10 days, what time will B require to do it working alone?

Solution:
Now let us solve this question using the LCM method.

Here, let us assume the total work to be done = LCM(6 and 10) = 30

Work done by A and B in 1 day = $\frac{30}{6}$ = 5 units/day

Work done by A in one day = $\frac{30}{10}$ = 3 units/day

∴ Work done by B in 1 day = 5 – 3 = 2 units/day

∴ B alone can work it full in $\frac{30}{2}$ = 15 days.

Example: A and B can do a piece of work in 18 days, B and C in 24 days, A and C in 36 days. In what time can they do it all working together?

Solution:
(A + B)’s 1 day’s work = $\frac{1}{18}$

(B + C)’s 1 day’s work = $\frac{1}{24}$

(C + A)’s 1 day’s work = $\frac{1}{36}$,

Adding, we have 2 (A + B + C)’s one day’s work = $\frac{1}{18}+\frac{1}{24}+\frac{1}{36}$ = $\frac{1}{8}$

∴ (A + B + C)’s one day’s work = $\frac{1}{2}\times \frac{1}{8}$ = $\frac{1}{16}$

Hence, all will do the work in 16 days

Ans. 16 days

Example: A can do a piece of work in 40 days. The rate of working for A and B is in the ratio 3: 5. How many days will both of them together take to complete the whole work?

Solution:
Ratio of efficiency of A and B is 3: 5.

∴ The ratio of time taken for A and B alone will be 5: 3

Hence,

$\frac{40}{T_B}=\frac{5}{3}$

⇒ T_B = 24 days

∴ No. of days A and B will take to do the whole work = $\frac{40\times 24}{40+24}=\frac{40\times 24}{64}$ = 15 days

Example: A and B can do a piece of work in 40 days. Their skills of doing the work is in the ratio 8: 5. How many days will A take, if engaged alone?

Solution:
Skill ratio of A and B is 8: 5.

∴ Work done by A is $\frac{8}{13}$ of the whole which he does in 40 days.

∴ No. of days A will take to do the whole work = $40\times \frac{13}{8}$ days = 65 days

Ans. 65 days

Example: A works twice as fast as B and thrice as fast as C. If C can finish the work alone in 11 days, In how many days can A, B and C together finish the work?

Solution:
efficiency of A = 2B and also efficiency of A = 3C

i.e. $B=\frac{1}{2}A,$$C=\frac{1}{3}A$

∴ efficiency of A + B + C = $A+\frac{1}{2}A+\frac{1}{3}A=\frac{11A}{6}$

Now C can finish a job in 11 days and since A = 3C

∴ efficiency of A = 3/11, and thus combined efficiency of all 3 is $\frac{11}{6}\times \frac{3}{11}=\frac{1}{2}$

Hence, all three together will take 2 days.

Example: A can do a work in 30 days. He worked for 5 days and the remaining work was finished by B in 15 days. How long does it take for B alone to finish the work.

Solution:
Work done by A in 5 days = $\frac{1}{30}$ × 5 = $\frac{1}{6}$

Remaining work = 1 - $\frac{1}{6}$ = $\frac{5}{6}$

Now 5/6 of the work is done by B in 15 days.

∴ Whole work is done by B in 15 × $\frac{6}{5}$ = 18 days

Example: A can do a work in 30 days. He worked for 5 days and the remaining work was finished by B in 15 days. How long does it take for B alone to finish the work.

Solution:
Let the time taken by B alone = B days.

Let us use the unitary method here. Hence, the total work to be done = 1 unit.

⇒ 1 = Work done in first 5 days + Work done in next 15 days.

⇒ 1 = e_A × 5 + e_B × 15

⇒ 1 = $\frac{1}{30}$ × 5 + $\frac{1}{\mathrm{B}}$ × 15

⇒ 1 = $\frac{1}{6}$ + $\frac{15}{\mathrm{B}}$

⇒ $\frac{5}{6}$ = $\frac{15}{\mathrm{B}}$

⇒ B = 18 days

Example: A can do a work in 20 days. He worked for 5 days and the remaining work was finished by A and B together in 9 days. How long does it take for B alone to finish the work.

Solution:
Let the time taken by B alone = B days.

Let us use the unitary method here. Hence, the total work to be done = 1 unit.

⇒ 1 = Work done by A + Work done by B.

⇒ 1 = e_A × 14 + e_B × 9

⇒ 1 = $\frac{1}{20}$ × 14 + $\frac{1}{\mathrm{B}}$ × 9

⇒ 1 = $\frac{7}{10}$ + $\frac{9}{\mathrm{B}}$

⇒ $\frac{3}{10}$ = $\frac{9}{\mathrm{B}}$

⇒ B = 30 days

Solution:
(A + B)’s 6 days work = $6\times \left(\frac{1}{20}+\frac{1}{15}\right)$ = $6\times \frac{7}{60}$ = $\frac{7}{10}$

Remaining work = 1 - $\frac{7}{10}$ = $\frac{3}{10}$

(A + C) can finish $\frac{3}{10}$ of the work in 4 days

(A + C)’s one day’s work = $\frac{3}{40}$ days

∴ C’s 1 day work = $\frac{3}{40}$ - $\frac{1}{20}$ = $\frac{1}{40}$

Hence, C alone can finish the work in 40 days.

Ans. 40 days

Example: A, B and C can do a work in 36, 54 and 72 days respectively. They started the work together but A left 8 days before the completion of the work while B left 12 days before the completion. The number of days for which C worked.

Solution:
(A + B + C)’s 1 day work = $\frac{1}{36}+\frac{1}{54}+\frac{1}{72}$ = $\frac{6+4+3}{216}=\frac{13}{216}$

A’s 8 days work = 8/36 = 2/9

B’s 12 days work =12/54 = 2/9

Total work if A and B had not left the work = 1 + 2/9 + 2/9 = 13/9

All the three do 13/9 of work in days = $\frac{13}{9}\times \frac{216}{13}$ = 24 days

∴ C worked for 24 days

Example: A can complete a work in 20 days while B can complete the same work in 30 days. If they complete the task together and receive Rs. 100 for it, what is the share of A?

Solution:
Ratio of number of days for A and B = 20: 30 = 2: 3

∴ Ratio of efficiency of A and B = 3 : 2

Ratio of payment for A and B = ratio of their efficiency = 3 : 2

A’s share = 3/5 × 100 = Rs. 60; B’s share = Rs. 40.

Example: A man can do a work in 10 days. With the help of a boy he can do the same work in 6 days. If they get Rs. 50 for that work, what is the share of that boy?

Solution:
Let the time taken by boy to complete the work alone = B

⇒ $\frac{1}{6}=\frac{1}{10}+\frac{1}{B}$

The boy can do the work in $\frac{10\times 6}{10-6}$ = 15 days

Ratio of number of days for man and boy = 10 : 15

∴ Ratio of efficiency of man and boy = 15 : 10 = 3 : 2

Since, both of them together complete the work in 6 days,

Ratio of payment for man and boy = ratio of their efficiency = 3 : 2

Man’s share = 3/5 × 50 = Rs. 30; Boy’s share = Rs. 20.

Example: A, B and C can do a work in 6, 8 and 12 days respectively. Doing that work together they get an amount of Rs. 1350. What is the share of B in that amount?

Solution:
A’s one day’s work = 1/6
B’s one day’s work = 1/8
C’s one day’s work = 1/12

A’s share: B’s share: C’s share =1/6 : 1/8 : 1/12

Multiplying each ratio by the LCM of their denominators, the ratios become 4 : 3 : 2

∴ B’s share = 1350/9 × 3 = Rs. 450

Alternately,

A’s share : B’s share : C’s share

= B’s time × C’s time: A’s time × C’s time: A’s time × B’s time

= 96 : 72 : 48 = 4 : 3 : 2

∴ B’s share = 1350/9 × 3 = Rs. 450

Solution:
Work completes the whole work in 10 days.

Hence, work done by A in 10 days = 10/20 = ½

Remaining half of the work will be done by them together.

Ratio of number of days for A and B = 20 : 30 = 2: 3

∴ Ratio of efficiency of A and B = 3 : 2

Hence, of the remaining half of the work A would do 3/5th of the work.

So, total work done by A = 1/2 + 3/5 × 1/2 = 4/5

Hence, payment received by A would be 4/5th of total payment = 4/5 × 100 = Rs. 80

Example: A, B and C contract a work for Rs. 550. Together, A and B are supposed to do 7/11 of the work. How much does C get?

Solution:
A + B did 7/11 work and C did (1 - 7/11) = 4/11 work.

∴ (A + B)’s share: C’s share = 7/11 : 4/11 = 7 ∶ 4

∴ C’s share = 4/11 × 550 = Rs. 200.

Example: A and B undertook to do a piece of work for Rs. 4500. A alone could do the work in 8 days and B in 12 days. With the assistance of C, they finished the work in 4 days. How should the money be divided?

Solution:
A in 4 days does 4 × 1/8 of the work i.e. 1/2 of the work. B in 4 days does 4 × 1/12 of the work i.e. 1/3 of the work. C in 4 days does {1 - (1/2 + 1/3)} of the work i.e. 1/6 of the work.

A’s share: B’s share: C’s share =1/2 : 1/3 : 1/6 = 3 : 2 : 1

∴ A’s share = Rs. 3/6 × 4500 = Rs. 2250

B’s share = Rs. 2/6 × 4500 = Rs. 1500

C’s share = Rs. 1/6 × 4500 = Rs. 750

Example: If 3 men with 4 boys can earn Rs. 210 in 7 days, and 11 men with 13 boys can earn Rs. 830 in 8 days, in what time will 7 men with 9 boys earn Rs. 1100?

Solution:
Let a man’s one day wage be M and a boy’s one day wage B

∴ 3 M + 4 B = Rs. 210/7 and 11 M + 13 B = Rs. 830/8

Solving the two equations, we get :

Wage of a man for a day’s work = Rs. 5 and wage of a boy for a day’s work = Rs. 3.75

∴ Combined wages of 7 men + 9 boys for a day’s work.

= 35 + 135/4 = Rs. 275/4

Hence, number of days required = $\frac{1100}{\frac{275}{4}}=\frac{1100\times 4}{275}$ = 16 days
 

Example: A, B and C can do a piece of work in 20, 15 and 12 days respectively. A is assisted by B on one day and by C in the next day alternatively. How long the work would take to finish?

Solution:
Here we have a cycle of 2 days: In each cycle A and B work on first day whereas A and C work on second day.

(A + B)’s 1st day work = 1/20 + 1/15 = 7/60

(A + C)’s 2nd day work = 1/20 + 1/12 = 8/60

∴ In 1 cycle work done = 7/60 + 8/60 = 1/4

∴ Number of cycles required = $\frac{1}{1/4}$ = 4 cycles

∴ Whole work is done in 2 × 4 = 8 days

Example: Two labourers Ramesh and Suresh can do a piece of work independently in 4 days and 6 days respectively. If they are working on alternate days and Suresh starts. How long the work will take to be finished?

Solution:
Let us use the LCM method here. Work done = LCM(4, 6) = 12 units

e_R = 12/4 = 3

e_S = 12/6 = 2

Here we have a cycle of 2 days: Suresh works on first day and Ramesh works on second day.

Two days work done by Ramesh and Suresh = 3 + 2 = 5 units

∴ 5 units of work is done in each cycle.

⇒ Number of cycles required = 12/5 = $2\frac{2}{5}$ cycles

When the number of cycles is not an integer, we need to break them into complete and incomplete cycles

We have 2 complete cycles
Time = 2 × 2 = 4 days
Work done = 2 × 5 = 10 units.
Work left = 12 – 10 = 2 units

In the last incomplete cycle
Suresh works on first day (i.e. 5th day from the start) and he can complete 2 units of work in 1 day.
Hence, entire work would be completed on 5th day itself.

∴ It takes 5 days to complete the whole work.

In such questions it is important to consider who is starting the work. In the previous example, if Ramesh had started the work, the answer would have been $4\frac{2}{3}$ days.

Example: A and B can do a piece of work separately in 10 days and 12 days respectively. If they work on alternate days. If A is starting, when will the work get finished?

Solution:
Let us use the LCM method here. Work done = LCM(10, 12) = 60 units

e_A = 60/10 = 6

e_B = 60/12 = 5

Here we have a cycle of 2 days: A works on first day and B works on second day.

Two days work done by A and B = 6 + 5 = 11 units

∴ 11 units of work is done in each cycle.

⇒ Number of cycles required = 60/11 = $5\frac{5}{11}$ cycles

Since the number of cycles is not an integer, we need to break them into complete and incomplete cycles

We have 5 complete cycles
Time = 5 × 2 = 10 days
Work done = 5 × 11 = 55 units.
Work left = 60 – 55 = 5 units

In the last incomplete cycle
A works on the first day (i.e. 11th day from the start) and he can complete 6 units of work in 1 day but the work remaining is only 5 units. That means the work would be complete before the end of 11th day.
The time taken to complete the remaining work = 5/6

∴ It takes 10 5/6 days to complete the whole work.

Example: A and B can do a piece of work separately in 8 days and 12 days respectively. If they work on alternate days. If B is starting, when will the work get finished?

Solution:
Let us use the LCM method here. Work done = LCM(8, 12) = 24 units

e_A = 24/8 = 3

e_B = 24/12 = 2

Here we have a cycle of 2 days: A works on first day and B works on second day.

Two days work done by A and B = 3 + 2 = 5 units

∴ 5 units of work is done in each cycle.

⇒ Number of cycles required = 24/5 = $4\frac{4}{5}$ cycles

Since the number of cycles is not an integer, we need to break them into complete and incomplete cycles

We have 4 complete cycles
Time = 4 × 2 = 8 days
Work done = 4 × 5 = 20 units.
Work left = 24 – 20 = 4 units

In the last incomplete cycle
B works on the first day (i.e. 9th day from the start) and he can complete 2 units of work in 1 day but the work remaining is 4 units. That means 2 units of work will be left after the 9th day.
A works on the second day (i.e. 10th day from the start) and he can complete 3 units of work in 1 day but the work remaining is only 2 units. The time taken to complete the remaining work = 2/3

∴ It takes 92/3 days to complete the whole work.

Example: If 3 men can complete a piece of work in 10 days, how long will it take 5 to complete the same piece of work?

Solution:
We have, M₁ × D₁ = M₂ × D₂

3 × 10 = 5 × D₂

∴ D₂ = 6 days

Example: If 3 men working for 10 hours every day can complete a piece of work in 12 days, how long will it take 5 men working for 9 hours every day to complete the same piece of work?

Solution:
We have, M₁ × H₁ × D₁ = M₂ × H₂ × D₂

3 × 10 × 12 = 5 × 9 × D₂

∴ D₂ = 8 days

Solution:
M₁ × D₁ × H₁ = M₂ × D₂ × H₂

∴ 15 × 14 × 12 = 21 × x × 10

⇒ x = $\frac{15\times 14\times 12}{21\times 10}$ = 12 days

Example: If 3 men working for 10 hours every day can complete a piece of work in 12 days, how long will it take 5 men working for 9 hours every day to complete thrice the amount of work?

Solution:
We have, $\frac{M_1\times H_1\times D_1}{W_1}=\frac{M_2\times H_2\times D_2}{W_2}$

$\frac{3\times 10\times 12}{W}=\frac{5\times 9\times D_2}{3W}$ (W is the amount of work done by first group of people)

$\frac{3\times 10\times 12}{1}=\frac{5\times 9\times D_2}{3}$

∴ D₂ = 24 days

Solution:
We know that

$\frac{M_1\times H_1\times D_1}{W_1}=\frac{M_2\times H_2\times D_2}{W_2}$

⇒ (30 × 8 × 24)/1 = (18 × 10 × X)/2

⇒ x = $\frac{30\times 8\times 24\times 2}{18\times 10\times 1}$ = 64 day

Example: It took 72 men working 8 hours a day to construct a wall 250 yards long, 24 ft. high and 6 ft. wide. How many men should be engaged to fulfill a contract for constructing a wall 275 yards long, 2.5 ft. high and 9 ft. wide if they are to work 9 hours a day and finish the work in the same time as the past.

Solution:
Let the required number of men be x to finish the work in the same time (say d, days)

Then compound proportion

Men Hours Length Height Width Days
72       8     250 × 3    24        6       D
X        9      275 × 3    25        9       D
 

$\frac{M_1\times H_1\times D_1}{W_1}=\frac{M_2\times H_2\times D_2}{W_2}$

⇒ 72 × 8 × 275 × 3 × 25 × 9 = X × 9 × 250 × 3 × 24 × 6 × d

⇒ x = $\frac{72\times 8\times 275\times 3\times 25\times 9\times d}{9\times 250\times 3\times 24\times 6\times d}$

x = 110 men

Example: If 3 men or 4 women can reap a field in 43 days, how long will 7 men and 5 women take to reap it?

Solution:
Let’s assume efficiency of each man = m and each woman = w units/day.

Work done by 3 men in 43 days = m × 3 × 43 …(1)

Work done by 4 women in 43 days = w × 4 × 43 …(2)

Since, they are doing the same amount of work, (1) = (2)

⇒ m × 3 × 43 = w × 4 × 43

∴ 3m = 4w …(3)

This means, work done by 3 men in a day is same as work done by 4 women in 1 day.

Now, work done by 7 men and 5 women in d days = (7m + 5w) × d …(4)

This is same as work done by 3 men or 4 women in 43 days, hence we can equate (4) with either of (1) or (2)

Let’s day (4) = (1)

⇒ (7m + 5w) × d = m × 3 × 43

⇒ (7m + 5 × 3m/4) × d = m × 3 × 43 (w = 3m/4 from (3))

⇒ 43m/4 × d = m × 3 × 43

⇒ d = 12 days

Example: 3 women and 4 men can reap a field in 15 days, while 2 women and 5 men can do it in 20 days. How long will 5 women and 2 men take to reap it?

Solution:
Let’s assume efficiency of each woman = w and each man = m units/day.

Work done by 3 women and 4 men in 15 days = (3w + 4m) × 15 …(1)
Work done by 2 women and 5 men in 20 days = (2w + 5m) × 20 …(2)
Work done by 5 women and 2 men in d days = (5w + 2m) × d …(3)

Since, they are doing the same amount of work, (1) = (2) = (3)

Solving (1) = (2)
⇒ (3w + 4m) × 15 = (2w + 5m) × 20
∴ w = 8m …(4)

This means, work done by a woman in a day is same as work done by 8 men in 1 day.

Now, solving (1) = (3)
⇒ (3w + 4m) × 15 = (5w + 2m) × d (substituting w = 8m)
⇒ (28m) × 15 = (42m) × d
⇒ d = 10 days

Example: I can finish a work in 15 days working 8 hrs a day. You can finish it in 6 2/3 days working 9 hrs a day. Find in how many days we can finish it working together 10 hrs a day.

Solution:
First suppose each of us works for only 1 hr a day

Then I can finish the work in 15 × 8 = 120 days and you can finish the work in 20/3×9=60 days

Now we together can finish the work in (120 × 60)/(120 + 60) = 40 days

But, here we are given that we do the work 10 hrs a day. Then, clearly we can finish the work in 4 days.

Solution:
(A + B) can do the work in (6×8)/(6+8) = 24/7 days

∴ C takes 24/7 days to complete the work

∴ (B + C) take $\frac{\frac{24}{7}\times 8}{\frac{24}{7}+8}$ = $\frac{24\times 8}{24+56}$ = $2\frac{2}{5}$ days

Example: A and B can do a work in 45 days and 40 days respectively. They began the work together, but A left after sometime and B finished the remaining work in 23 days. After how many days did A leave?

Solution:
B worked alone for 23 days.

∴ Work done by B in 23 days = 23/40 work

∴ A + B do together 1 - 23/40 = 17/40 work

Now, A + B do the work in $\frac{40\times 45}{40+45}$ = $\frac{40\times 45}{85}$ days

∴ A + B do 17/40 work in $\frac{40\times 45}{85}\times \frac{17}{40}$ = 9 days

Example: A cistern is filled by pipe A in 10 minutes and can be leaked out by a waste pipe B in 12 minutes. If both the pipes are opened, in what time the cistern is full?

Solution:
The portion of the cistern is full in one minute when both the pipes are opened = 1/10 - 1/12 = (6-5)/60 = 1/60

∴ The cistern shall be full in 60 minutes.

Solution: Work done in 1 hour by the filling pipe = 1/9
Work done in 1 hour by the leak and the filling pipe = 1/10
∴ Work done by the leak in 1 hour = 1/9 - 1/10 = 1/90
Hence, the leak can empty it in 90 hours.

Solution:
In the first hour pipe A fill 1/h of the tank.

In the second hour pipe B fills 1/6 of the tank.

∴ In the first 2 hours (1/4 + 1/6) or 5/12 of the tank is full or in 4 hours 2 × 5/12 i.e. 5/6 of the tank is full.

Now 1/6 of the tank remains to be filled which pipe A can fill in 4 × 1/6 i.e. 2/3 hour.

∴ the total time = (4 + 2/3) hour or 4 2/3 hours

Example: A bath can be filled by cold water pipe in 10 minutes and by hot water pipe in 15 minutes. A person leaves the bath room after turning on both pipes simultaneously and returns at the moment when the bath should be full. Findings, however, that the waste pipe has been open, he now closes it. In 4 minutes more the bath is full. In what time would the waste pipe empty it?

Solution:
In one minute the cold and hot water can fill (1/10 + 1/15) = 1/6 of the bath. Thus, the bath should have been full in 6 minutes.

The waste pipe is closed, and in 4 minutes more the bath is full.

Hence, it is clear that during the first 6 min. The waste pipe emptied as much as the other two can together fill in 4 minutes.

i.e. 4/6 or 2/3 of the bath.

∴ Waste pipe would empty the whole in 6 × 3/2 or 9 min.

Example: Three pipes A, B and C can fill a cistern in 6 hours. After working at it together for 2 hours, C is closed and A and B fill it in 7 hours more. How many hours will C alone take to fill the cistern?

Solution:
(A + B + C)’s work for 2 hours = 2 × 1/6 = 1/3

Remaining portion unfilled =1-1/3=2/3

In 7 hours (A + B)’s can fill = 2/3 of the cistern.

∴ In 1 hour (A + B)’s can fill = 1/7 × 2/3 of the cistern

= 2/21 of the cistern

But (A + B + C)’s 1 hour’s work = 1/6

∴ C’s one hour work = 1/6 - 2/21 = 1/14 of the cistern.

∴ C can fill it alone in 14 hours

Solution:
When both pipes are working, the portion of the cistern full in 1 hr = 1/12

And the leak can empty 1/8 of the cistern in 1 hour.

∴ Work done by a filling pipe in 1 hour = 1/8 - 1/12 = 1/24

So, the cistern is full by the filling pipe in 24 hours.

∴ Capacity of the cistern = water admitted by the pipe for 24 hours at the rate of 6 litres per minute = 24 × 60 × 6 = 8640 litres.

Example: Two pipes can fill a tank in 8 and 12 hours respectively whereas an escape pipe can empty it in 6 hours. If the three pipes are open at 1 pm, 2 pm and 3 pm respectively at what time will the tank be filled?

Solution:
By 3 pm, the water filled by two pipes = 2 hours work done by pipe A + 1 hour work done by pipe B

= 2 × 1/8 + 1 × 1/12 = 1/4 + 1/12 = 1/3

After 3 pm, all the three pipes are working and the work done by all the three pipes in one hour = 1/8 + 1/12 - 1/6 = $\frac{\mathrm{(3+2-4)}}{24}$ = 1/24

After 3 pm only 2/3 of tank is to be filled

Time taken in filling 2/3 of the tank = 24 × 2/3 = 16 hours

∴ 16 hours after 3 PM = 7 AM

Example: A pipe can fill a cistern in 8 hours and another in 10 hours. Both are opened at the same time. If the second pipe is closed 2 hours before the cistern is full, find in what time, the cistern will be full?

Solution:
Work done by A and B in one hour = 1/8 + 1/10 = (5+4)/40 = 9/40

Had the second pipe not been closed, the work done by the pipes = 1 + 2 hours work done by the second pipe 1 + 2 × 1/10 = 6/5 of the cistern

∴ Time taken to fill 6/5 of the cistern at the rate 9/40 per hour.

= 6/5 ÷ 9/40 = 6/5 × 40/9 = 16/3 hours or 5 hours 20 minutes.

Ans. 5$\frac{1}{3}$ hours

Example: Two pipes X and Y can fill a cistern in 24 minutes and 32 minutes respectively. If both the pipes are opened together, then after how much time should Y be closed so that the cistern is full in 18 minutes?

Solution:
Let pipe Y be closed after x minutes.

Then, $x\left(\frac{1}{24}+\frac{1}{32}\right)+\left(18-x\right)\frac{1}{24}$ = 1

⇒ $\frac{7x}{96}+\frac{18-x}{24}$ = 1 or 7x + 72 - 4x = 96

∴ 3x = 24 or x = 8 minutes.

Example: Find the time required for A, B and C to fill a tank separately if A and B together fill the tank in 6 hours, B and C together in 10 hours and A and C together in 7.5 hours.

Solution:
A + B fill the tank in 6 hours, B + C fill the tank in 10 hours. A + C fill the tank in 7.5 hours.

∴ 2 (A + B + C) can fill up $\frac{1}{6}+\frac{1}{10}+\frac{1}{7.5}$ of the tank in 1 hour i.e. $\frac{1}{6}+\frac{1}{10}+\frac{2}{15}=\frac{5+3+4}{30}=\frac{2}{5}$ of the tank in 1 hour, or (A + B + C) fill up the tank in 5 hours

A ≡ [(A + B + C) – (B + C)] = 1/5 - 1/10 = 1/10 of the tank in 1 hour.

⇒ A can fill the tank in 10 hours.

B ≡ [(A + B + C) – (A + C)] = 1/5 - 2/15 = 1/15 of the tank in 1 hour.

C ≡ [(A + B + C) – (A + B)] =1/5 - 1/6 = 1/30 of the tank in 1 hour.

⇒ C can fill the tank in 30 hours.

Example: A pipe fills 1/3 rd of a tank in 10 min. In what time would it fill 5/6th of the tank?

Solution:
$\frac{t_1}{a_1}=\frac{t_2}{a_2}$, where t_1 and t_2 are the respective time and a_1, a_2 are the parts filled.

or, $\frac{10}{1/3}=\frac{t_2}{5∕6}$ ⇒ $t_2=10\times 3\times \frac{5}{6}$= 25 min

Example: Two pipes A and B can fill up a tank in 6 minutes and 7 minutes respectively. If they are opened for 1 minute alternately, A beginning, in how much will they fill up the tank?

Solution:
Portion of tank fill up by A in 1 minute = 1/6

∴ Portion of tank filled up by A and B together in first two minutes = $\frac{1}{6}+\frac{1}{7}=\frac{13}{42}$

∴ Portion of tank filled up by A and B together in first six minutes = $\frac{13}{42}\times 3=\frac{13}{14}$

∴ Remaining portion of tank =1 - 13/14 = 1/14

Now, after 6 minutes, it is the turn of pipe A

∵ Pipe A fills 1/6 portion of tank in = 1 minute

∴ Pipe A fills 1/14 portion of tank in = $\frac{1}{14}\times \frac{6}{1}=\frac{3}{7}$ minute

∴ Total time taken to fill up the tank = $6\frac{3}{7}$ minutes.

Example: There is a leak in the bottom of a cistern. When the cistern had no leak, it was filled in 2 1/2 hours. It now takes half an hour longer. If the cistern is full of water, how long would it take in leaking itself empty, in case the water leaks out at double the rate after half the cistern becomes empty?

Solution:
When there was no leak:

The cistern was filled in $2\frac{1}{2}$ hours i.e. 5/2 hours

∴ In 1 hour, 2/5 of the cistern was filled.

When there was leak:

The cistern was filled in 3 hours

∴ In 1 hour, 1/3 of the cistern was filled.

∴ In 1 hour, $\frac{2}{5}-\frac{1}{3}=\frac{1}{15}$ of the cistern is emptied

i.e., 1 cistern is emptied in = 15 hours

∴ 1/2 cistern is emptied in = 15/2 hours

After this, the cistern starts getting emptied at the rate of 2/15 cistern in 1 hour.

i.e., 1 cistern is emptied in = $\frac{15}{2}$ hrs

∴ $\frac{1}{2}$ (remaining half) cistern is emptied in = $\frac{15}{2}\times \frac{1}{2}=\frac{15}{4}$ hrs.

Hence, the system will leak itself empty in = $\frac{15}{2}$ + $\frac{15}{4}$ = $\frac{45}{4}$ = 11 hours 15 minutes

Example: Two taps can fill a cistern separately in 60 and 80 minutes respectively. To fill the empty cistern, both the taps are kept open for 20 minutes and then the second tap is turned off. Find the additional time required to fill the cistern .

Solution:
∵ The first tap can fill the cistern in 60 minutes

∴ The first tap in 1 minute can fill = $\frac{1}{60}$ cistern

∵ The second tap can fill the cistern in 80 minutes

∴ The second tap in 1 minute can fill = $\frac{1}{80}$ cistern

∴ Both the taps in 1 minute can fill = $\left(\frac{1}{60}+\frac{1}{80}\right)$ cistern = $\frac{4+3}{240}=\frac{7}{240}$ cistern

∴ Both the taps in 20 minutes can fill = $\frac{7}{240}\times 20=\frac{7}{12}$ cistern

∴ Remaining (unfilled) portion of the cistern = 1 - $\frac{7}{12}$ = $\frac{5}{12}$ cistern

This portion is to be filled by the first tap.

∵ The first tap fills 1 cistern in = 60 minutes

∴ The first tap will fill $\frac{5}{12}$ cistern in = 60 × $\frac{2}{12}$ minutes = 25 minutes more

Example: A pipe can fill a bath in 20 minutes and another can fill it in 30 minutes. A person opens both the pipes alternately. When the bath should have been full, he finds that the waste pipe was open. He then closes the waste pipe and in 3 minutes more, the bath is full. In what time, would the waste pipe empty it?

Solution:
In the first minute, A fills $\frac{1}{20}$ bath

In the second minute, B fills $\frac{1}{30}$ bath

∴ In the first two minutes, bath filled = $\frac{1}{20}$ + $\frac{1}{30}$ = 1/12

∴ The whole bath should have been filled in 2 × 12 = 24 minutes

In the last three minute, when the leak is closed:

∴ In three minutes bath filled = $3\left(\frac{1}{12}+\frac{1}{20}\right)=\frac{1}{4}$

∵ The waste pipe empties $\frac{1}{4}$ bath in = 24 minutes

∴ The waste pipe empties 1 bath in = 4 × 24 minutes = 96 minutes

Solution:
First tap in 1 minute fills = $\frac{1}{60}$ cistern

Second tap in 1 minute fills $\frac{1}{75}$ cistern

Let the third tap can empty the cistern in x minutes.

∴ Third tap in 1 minute empties = 1/x cistern

∴ Three taps fill in 1 minute = $\left(\frac{1}{60}+\frac{1}{75}-\frac{1}{x}\right)$ cistern

∴ Three taps fill in 50 minutes = $\left(\frac{1}{60}+\frac{1}{75}-\frac{1}{x}\right)50$ cistern

According to the question $\left(\frac{1}{60}+\frac{1}{75}-\frac{1}{x}\right)50=1$

or $\frac{1}{60}+\frac{1}{75}-\frac{1}{x}=\frac{1}{50}$

or $\frac{1}{x}$ = $\frac{1}{60}+\frac{1}{75}-\frac{1}{50}$ = $\frac{5+4-6}{300}=\frac{1}{100}$

or x = 100

Hence, the third pipe can empty the cistern in 100 minutes.

Example: Three taps empty a cistern in 3 hours. First tap alone can empty it in 6 hours and second tap alone can empty in 9 hours. How many hours would third tap alone take to empty the cistern?

Solution:
∵ 3 taps in 3 hours empty = 1 cistern

∴ 3 taps in 1 hour empty = $\frac{1}{3}$ cistern

∵ First tap in 6 hours empties = 1 cistern

∴ First tap in 1 hour empties $\frac{1}{6}$ cistern

∵ Second tap in 9 hour empties = 1 cistern

∴ Second tap in 1 hour empties = $\frac{1}{9}$ cistern

∴ Third tap in 1 hour empties = $\frac{1}{3}-\left(\frac{1}{6}+\frac{1}{9}\right)=\frac{1}{18}$ cistern

∴ Third tap alone will take 18 hours to empty the cistern.

Example: A supply of water lasts for 200 days if 10 gallons leak off every day, but only for 180 days if 16 gallons leak off daily. Find the total quantity of water in the supply.

Solution:
In the first case, 200 × 10 i.e. 2000 gallons leak off altogether.

In the second case, 180 × 16 i.e., 2880 leaks off altogether.

It is now clear that leakage in second case is 880 gallons more than in first case, hence the supply lasts for 20 days less.

∴ It is clear that 880 gallons are necessary for 20 days consumption i.e., $\frac{880}{20}$0 = 44 gallons are necessary for 1 day’s consumption.

Now, taking the first case, we find that supply lasts for 200 days.

∴ 200 × 44 = 8800 gallons are required for consumption in that time. Also, leakage in that time = 2000 gallons.

∴ Total quantity of water in supply = (8800 + 2000) gallons = 10,800 gallons

Example: A and B can do a piece of work in 20 days. They worked together for 15 days and then A left. Remaining work was finished by B in 6 more days. In how many days A alone can do?

Solution:
A works with B for 15 days and the remaining work B finishes in (21 – 15), i.e., 6 days

(A + B)’s 15 days work = $\frac{1}{20}\times 15=\frac{3}{4}$

Remaining 6 days B’s work = 1 - $\frac{3}{4}$ = $\frac{1}{4}$

B’s 1 day work = $\frac{1}{6}\times \frac{1}{4}=\frac{1}{24}$

A’s 1 day work = (A + B)’s 1 day work – B’s 1 day work = $\frac{1}{20}-\frac{1}{24}=\frac{6-5}{120}=\frac{1}{120}$ of the work

∴ A can finish the whole work by himself in 120 days.

Ans. 120 days

Example: Pipe A of circular cross-section can fill a tank in 12 hours. How long will it take for another pipe:
(a) whose cross-section is twice that of pipe A
(b) whose radius is twice that of pipe A
(c) in which water flows at thrice the rate as compared to A
(d) whose radius is twice that of A, but water flows at half the rate of A

Solution:
(a) Efficiency of a pipe ∝ cross-section
Since cross-section is double, efficiency will also be double. Hence, time taken will become half.
∴ this pipe will take 6 hours to fill the tank.

(b) Efficiency of a pipe ∝ r²
Since radius is double, efficiency will also be four times. Hence, time taken will become one-fourth.
∴ this pipe will take 3 hours to fill the tank.

(c) Efficiency of a pipe ∝ speed of the flow of water
Since flow rate is thrice, efficiency will also be thrice. Hence, time taken will become one-third.
∴ this pipe will take 4 hours to fill the tank.

(d) Efficiency of a pipe ∝ r² × (speed of the flow of water)
Since radius is double and flow rate is half, efficiency will also be 4 × $\frac{1}{2}$ = 2 times. Hence, time taken will become half.
∴ this pipe will take 6 hours to fill the tank.

Example: In what time would a cistern be filled by three pipes whose diameters are 1 cm, 1 $\frac{1}{3}$ cm, 2 cm, running together, when the largest alone will fill it in 61 minutes; the amount of water flowing in by each pipe being proportional to the square of its diameter?

Solution:
Work done by 2 cm pipe in 1 minute = $\frac{1}{61}$ unit.

∴ Work done by 1 cm pipe in 1 minute = $\left(\frac{1}{61}\times \frac{1}{4}\right)$ unit.

∴ Work done by 4/3 cm pipe in 1 minute = $\left(\frac{1}{61}\times \frac{1}{4}\times \frac{16}{9}\right)$ unit.

∴ Work done by all three in 1 minute = $\left(\frac{1}{61}\times \left(1+\frac{1}{4}+\frac{4}{9}\right)\right)$ = $\frac{1}{61}\times \frac{61}{36}$ = $\frac{1}{36}$

∴ The whole cistern will take 36 minutes to be full.