# Formula - Time & work

**TIME AND WORK**

In most of the problems on time and work, either of the following basic parameters are to be calculated :

**TIME :**

If A can do a piece of work in X days, then A’s one day’s work = ^{1}⁄_{x}th part of whole work.

If A’s one day’s work = ^{1}⁄_{x}th part of whole work, then A can finish the work in X days.

If A can do a piece of work in X days and B can do it in Y days then A and B working together will do the same work in \(\frac{xy}{x + y}\) days.

If A, B and C can do a work in X, Y and Z days respectively then all of them working together can finish the work in

\(\frac{xyz}{xy + yz + xz}\) days.**Example 1**

**1. A can do a piece of work in 5 days, and B can do it in 6 days. How long will they take if both work together?**

**Sol.** A’s 1 day’s work = ^{1}⁄_{5}th part of whole work and

B’s 1 day’s work = ^{1}⁄_{6}th part of whole work

∴ (A + B)’s one day’s work = ^{1}⁄_{5} + ^{1}⁄_{6} = ^{11}⁄_{30}th part of whole work. So, both together will finish the work in ^{30}⁄_{11} days = 2^{8}⁄_{11} days.

By Direct Formula :

A + B can do the work in \(\frac{5 \times 6}{5 + 6} \; days = \frac{30}{11} = 2\tfrac{8}{11} \; days\).

**Example 2**

**2. Two men, Vikas and Vishal, working separately can mow a field in 8 and 12 hours respectively. If they work in stretches of one hour alternately, Vikas beginning at 8 a.m, when will the mowing be finished?**

**Sol.** In the first hour, Vikas mows ^{1}⁄_{8} of the field.

In the second hour, Vishal mows ^{1}⁄_{12} of the field.

∴ In the first 2 hours, \(\left ( \frac{1}{8} + \frac{1}{12} = \frac{5}{24}\right )\) of the field is mown.

∴ In 8 hours, \(\frac{5}{24} \times 4 = \frac{5}{6}\) of the field is mown.

Now, (1 – ^{5}⁄_{6}) = ^{1}⁄_{6} of the field remains to be mown.

In the 9th hour, Vikas mows ^{1}⁄_{8} of the field.

Remaining work = ^{1}⁄_{6} – ^{1}⁄_{8} = ^{1}⁄_{24}

∴ Vishal will finish the remaining work in (^{1}⁄_{24} ÷ ^{1}⁄_{12}) or ^{1}⁄_{2} of an hour.

∴ The total time required is (8 + 1 + ^{1}⁄_{2}) or 9^{1}⁄_{2} hours.

Thus, the work will be finished at 8 + 9^{1}⁄_{2} = 17^{1}⁄_{2} or 5.30 pm.

**Example 3**

**3. A can do a piece of work in 36 days, B in 54 days and C in 72 days. All the three began the work together on the Dec. 15, 2000, but A left 8 days and B 12 days before the completion of the work. If C took the rest for a week then in how many days,the work was finished from the day it started ?**

**Sol.** Let the total time taken be x days.

According to the given condition

=> \(\frac{x ? 8}{36} + \frac{x ? 12}{54} + \frac{x}{72} = 1\)

=> \(\frac{6(x ? 8) + 4(x ? 12) + 3x}{216} = 1\)

=> \(\frac{6x ? 48 + 4x ? 48 + 3x}{216} = 1 \Rightarrow \frac{13x ? 96}{216} = 1\)

=> 13x – 96 = 216 => 13x = 216 + 96 = 312

=> x = ^{312}⁄_{13} = 24

Since, C takes the rest for a week, so the number of days in which the work was finished from one day it started = 31 i.e.on 14.01.2001.

**Example 4**

**4. A and B can do a certain piece of work in 8 days, B and C can do it in 12 days and C and A can do it in 24 days.How long would each take separately to do it ?**

**Sol.** (A + B)’s one days’s work = 1/18,

(A + C)’s one days’s work = 1/24,

(B + C)’s one days’s work = 1/12,

Now add up all three equations :

2(A + B +C)’s one days’s work = ^{1}⁄_{18} + ^{1}⁄_{24} + ^{1}⁄_{12} = ^{13}⁄_{72}

(A + B + C)’s one days’s work = ^{13}⁄_{144}

A’s one days’s work = (A + B + C)’s one days’s work

– (B + C)’s one days’s work = ^{13}⁄_{144} – ^{1}⁄_{12} = ^{1}⁄_{144}

Since A completes of the work in 1 day, he will complete 1 work in ^{144}⁄_{1} = 144 days.

By similar logic we can find that B needs days and C will require ^{144}⁄_{5} days.

If A and B together can do a piece of work in X days and A alone can do it in Y days, then B alone can do the work in \(\frac{xy}{y ? x}\) days.

**Example 5**

**5. A and B together can do a piece of work in 6 days and A alone can do it in 9 days. In how many days can B alone do it?**

**Sol.** (A + B)’s 1 day’s work = ^{1}⁄_{6}th part of the whole work.

A’s 1 day’s work = ^{1}⁄_{9}th part of the whole work.

∴ B’s 1 day’s work = ^{1}⁄_{6} – ^{1}⁄_{9} = ^{(3 – 2)}⁄_{18} = ^{1}⁄_{18}th

part of the whole work.

∴ B alone can do the work in 18 days.

By Direct Formula :

B alone can do the whole work in

\(\frac{6 \times 9}{9 ? 6} = \frac{54}{3} = 18 \; days\)A and B can do a work in ‘X’ and ‘Y’ days respectively.They started the work together but A left ‘a’ days before completion of the work. Then, time taken to finish the work is \(\frac{Y(X + a)}{X + Y}\)

If ‘A’ is ‘a’ times efficient than B and A can finish a work in X days, then working together, they can finish the work in \(\frac{aX}{a + 1}\) days.

If A is ‘a’ times efficient than B and working together they finish a work in Z days then, time taken by A = \(\frac{Z\left ( a + 1 \right )}{a}\) days. and time taken by B = Z(a + 1) days.

If A working alone takes ‘x’ days more than A and B together,and B working along takes ‘y’ days more than A and B together then the number of days taken by A and B working together is given by \(\left [ \sqrt{XY} \right ]\) days.

**Example 6**

**6. A and B can do alone a job in 6 days and 12 days.They began the work together but 3 days before the completion of job, A leaves off. In how many days will the work be completed?**

(a) 6 days

(b) 4 days

(c) 5 days

(d) 7 days

**Sol.(a)** Let work will be completed in x days. Then,work done by A in (x – 3) days + work done by B in x days = 1

i.e. \(\frac{x ? 3}{6} + \frac{x}{12} = 1\)

=> \(\frac{3x ? 6}{12} = 1 \Rightarrow x = 6\) days

By Direct Formula :

Required time = \(\frac{12(6 + 3)}{12 + 6} = 6\) days

**Example 7**

**7. A is half good a workman as B and together they finish a job in 14 days. In how many days working alone will B finish the job.**

(a) 20 days

(b) 21 days

(c) 22 days

(d) None of these

**Sol.(b)** Let B can do the work in x days

and A can do the work in 2x days

Then, \(\frac{1}{x} + \frac{1}{2x} = \frac{1}{14}\) (given)

=> x = ^{3}⁄_{2} × 14 = 21 days

If n men or m women can do a piece of work in X days, then N men and M women together can finish the work in \(\frac{nmX}{nM + mN}\) days.

**Example 8**

**8. 10 men can finish a piece of work in 10 days,where as it takes 12 women to finish it in 10 days. If 15 men and 6 women undertake to complete the work, how many days they will take to complete it?**

(a) 7 days

(b) 5 days

(c) 4 days

(d) 6 days

**Sol.(b)** It is clear that 10 men = 12 women or 5 men = 6 women

=> 15 men + 6 women = (18 + 6) i.e. 24 women

Now 12 women can complete the work in 10 days

∴ 24 women will do it in 5 days.

By Direct Formula :

Required time = \(\frac{10 \times 12 \times 10}{10 \times 6 + 12 \times 15} = 5\) days.

**Example 9**

**9. 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 ?**

**Sol.**

**First Method :** 3 men reap ^{1}⁄_{43} of the field in 1 day.

∴ 1 men reaps ^{1}⁄_{(43 × 3)} of the field in 1 day.

4 women reap ^{1}⁄_{43} of the field in 1 day.

∴ 1 men reaps ^{1}⁄_{(43 × 4)} of the field in 1 day.

∴ 7 men and 5 women reap \(\left ( \frac{7}{43 \times 3} + \frac{5}{43 \times 4}\right ) = \frac{1}{12}\) of the field in 1 day.

∴ 7 men and 5 women will reap the whole field in 12 days.

**Second Method :**

3 men = 4 women

∴ 1 man = ^{4}⁄_{3} ∴ 7 men = ^{28}⁄_{3}

∴ 7 men + 5 women = ^{28}⁄_{3} + 5 = ^{43}⁄_{3} women now, the question becomes :

If 4 women can a field in 43 days, how long will ^{43}⁄_{3} women take to reap it ?

The basic-formula gives

\(4 \times 43 = \frac{43}{3} \times D_{2}\) or \(D_{2} = \frac{4 \times 43 \times 3}{43}\) = 12 days.**Examination method**

Required number of days = \(\frac{1}{\left [ \frac{7}{43 \times 3} + \frac{5}{43 \times 4}\right ]}\)

= \(\frac{43 \times 3 \times 4}{4 \times 4 + 5 \times 3} = 12\) days.

**Example 10**

**10. If 12 men and 16 boys can do a piece of work in 5 days and 13 men and 24 boys can do it in 4 days, how long will 7 men and 10 boys take to do it ?**

**Sol.**

12 men and 16 boys can do the work in 5 days……. (1)

13 men and 24 boys can do the work in 4 days…….. (2)

Now it is easy to see that if the no. of workers be multiplied by any number, the time must be divided by the same number(derived from : more worker less time).

Hence multiplying the no. of workers in (1) and (2) by 5 and

4 respectively, we get 5 (12 men + 16 boys) can do the work in 5/5 = 1 day

4 (13 men + 24 boys) can do the work in 4/4= 1 day

or, 5(12m + 16b) = 4 (13m + 24b)

or, 60 m + 80 b = 52 m + 96 b

or,60 m – 52 m = 96 b – 80 b

or,8 m= 16 b

∴ 1 men = 2 boys.

Thus, 12 men + 16 boys = 24 boys + 16 boys = 40 boys and 7 men + 10 boys = 14 boys + 10 boys = 24 boys

The question now becomes :

“If 40 boys can do a piece of work in 6 days how long will 24 boys take to do it ?”Using basic formula, we have

40 × 5 = 24 × D_{2}

or, D_{2} = \(\frac{40 \times 5}{24} = 8\tfrac{1}{3}\) days.

**Example 11**

**11. Two men and 7 boys can do a piece of work in 14 days. 3 men and 8 boys can do it in 11 days. In how many days can 8 men and 6 boys do a work 3 times as big as the first ?**

**Sol.** 2 men + 7 boys in 14 days => 28 men + 98 boys in 1 day

3 men + 8 boys in 11 days => 33 men + 88 boys in 1 day

∴ 28 men + 98 boys = 33 men + 88 boys

∴ 2 boys ≡ 1 man

Now, 2 men + 7 boys = 11 boys;8 men + 6 boys = 22 boys

More boys, fewer days; more work, more days

∴ \(\frac{x}{14} = \frac{11}{12} \times \frac{3}{1}\)

Number of days = 21 days.

**Example 12**

**12. Kaberi takes twice as much time as Kanti and thrice as much as Kalpana to finish a place of work. They together finish the work in one day. Find the time taken by each of them to finish the work.**

**Sol.** Here, the alone time of kaberi is related to the alone times of other two persons, so assume the alone time of kaberia = x,

Then, alone time of kanti = x/2 and of kalpana = x/3

Kaberi’s 1 day work + kanti’s 1 day work + kalpana’s 1 daywork = combined 1 days work

=> \(\frac{1}{x} + \frac{1}{x/2} + \frac{1}{x/3} = \frac{1}{1} \Rightarrow x = 6\)

∴ Alone time for kaberi = 6 days, for kanti = 6/2 = 3 days,kalpana = 6/3 = 2 days,

**Example 13**

**13. 1 man or 2 women or 3 boys can do a work in 44 days. Then in how many days will 1 man, 1 woman and 1 boy do the work?**

**Sol.** Number of required days

= \(\frac{1}{\frac{1}{44 \times 1} + \frac{1}{44 \times 2} + \frac{1}{44 \times 3}} = \frac{44 \times 1 \times 2 \times 3}{6 + 3 + 2} = 24\) days.

If ‘M_{1}’ persons can do ‘W_{1}’ works in ‘D_{1}’ days and ‘M_{2}’persons can do ‘W_{2}’ works in ‘D_{2}’ days then

M_{1}D_{1}W_{2} = M_{2}D_{2}W_{1}

If T_{1} and T_{2} are the working hours for the two groups then

M_{1}D_{1}W_{2}T_{1} = M_{2}D_{2}W_{1}T_{2}

Similarly,

M_{1}D_{1}W_{2}T_{1}E_{1} = M_{2}D_{2}W_{1}T_{2}E_{2}, where E_{1} and E_{2} are the efficiencies of the two groups.

If the number of men to do a job is changed in the ratio a : b,then the time required to do the work will be in the ratio b : a,assuming the amount of work done by each of them in the given time is the same, or they are identical.

A is K times as good a worker as B and takes X days less than B to finish the work. Then the amount of time required by A and B working together is \(\frac{K \times X}{k^{2} ? 1}\) days.

If A is n times as efficient as B, i.e. A has n times as much capacity to do work as B, A will take 1/n of the time taken by B to do the same amount of work.

**Example 14**

**14. 5 men prepare 10 toys in 6 days working 6 hrs a day.Then in how many days can 12 men prepare 16 toys working 8 hrs a day ?**

**Sol.** This example has an extra variable ‘time’ (hrs a day), so the ‘basic-formula’ can’t work in this case. An extended formula is being given :

M_{1}D_{1}W_{2}T_{1} = M_{2}D_{2}W_{1}T_{2}

Here, 5 × 6 × 6 × 16 = 12 × D_{2} × 8 × 10

∴ D_{2} = \(\frac{5 \times 6 \times 6 \times 16}{12 \times 8 \times 10} = 3\) days.

**Example 15**

**15. A and B can do a work in 45 days respectively.They began the work together, but A left after some time and B finished the remaining work in 23 days.After how many days did A leave ?**

**Sol.** B works alone for 23 days.

∴ Work done by B in 23 days = ^{23}⁄_{40} work

∴ A + B to together 1 – ^{23}⁄_{40} = ^{17}⁄_{40} work.

Now, A + B do 1 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

**Examination method :**

If we ignore the intermediate steps, we can write a direct formula as : \(\frac{40 \times 45}{40 + 45}\left ( \frac{40 ? 23}{40} \right ) = 9\) days.

**Example 16**

**16. Two friends take a piece of work for 960.One alone could do it in 12 days, the other in 16 days with the assistance of an expert they finish it in 4 days. How much remuneration the expert should get ?**

**Sol.** First friend’s 4 day’s work = 4/12 = 1/3 (Since, the work is finished in 4 days, when expert assists )

Second friends’s 4 day’s work = 4/16 = 1/4

The expert’s 4 day’s work = 1 – (^{1}⁄_{3} + ^{1}⁄_{4}) = ^{5}⁄_{12}

Now, total wages of 960 is to be distributed among two friends and the expert in proportion to the amount of work done by each of them.

So, 960 is to be divided in the proportion of

^{1}⁄_{3} : ^{1}⁄_{4} : ^{5}⁄_{12} or 4 : 3 : 5

∴ Share of expert = ^{5}⁄_{12} × 960 =400

Hence, the expert should get 400.

**Example 17**

**17. A certain number of men can do a work in 60 days. If there were 8 men more it could be finished in 10 days less. How many men are there ?**

**Sol.** Let there be x men originally.

(x + 8) men can finish the work in (60 –10) = 50 days.

Now, 8 men can do in 50 days what x men do in 10 days, then

by basic formula we have

∴ x = \(\frac{8 \times 50}{10} = 40 \; men\)

**Examination method (1) :**

We have :

x men to the work in 60 days and (x + 8) men do th work in

(60 – 10 = ) 50 days.

Then by “basic formula”, 60x = 50(x + 8)

∴ x = \(\frac{8 \times 50}{10} = 40 \; men\)

**Examination method (2) :** There exists a relationship :

Original number of workers

= \(\frac{No. \; of \; more \; worker \; \times \; Number \; of\; days \; taken \; by \; the \; second \; group}{No. \; of \; less \; days}\)

= \(\frac{8 \times \left ( 60 ? 10 \right )}{10} = \frac{8 \times 50}{10} = 40 \; man\)

**Example 18**

**18. Two coal loading machines each working 12 hours per day for 8 days handles 9,000 tonnes of coal with an efficiency of 90%. While 3 other coal loading machines at an efficiency of 80 % set to handle 12,000 tonnes of coal in 6 days.Find how many hours per day each should work.**

**
Sol.** Here \(\frac{N_{1} \times D_{1} \times R_{1} \times E_{1}}{W_{1}} = \frac{N_{2} \times D_{2} \times R_{2} \times E_{2}}{W_{2}}\)

N_{1} = R_{1} = 12 h/ day : N_{2} = 3R_{2} = ?

E_{1} = ^{90}⁄_{100}W_{1} = 9,000;

E_{2} = ^{80}⁄_{100}W_{2} = 12,000

=> \(\frac{2 \times 8 \times 12 \times 90}{9,000 \times 100} = \frac{3 \times 6 \times R_{2} \times 80}{12000 \times 100}\)

=> R_{2} = 16 h / day.

∴ Each machine should work 16 h/ day.

**WORK AND WAGES**

Wages are distributed in proportion to the work done and in indirect proportion to the time taken by the individual.

**Example 19**

**19. A, B and C can do a work in 6, 8 and 12 days respectively. Doing that work together they get an amount of1350. What is the share of B in that amount?**

**Sol.** 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 L.C.M. of their denominators,the ratios become 4 : 3 : 2

∴ B’s share = \(\frac{1350 \times 3}{9} =450\)

**Example 20**

**20. If 6 men working 8 hours a day earn 1680 per week, then how much will 9 men working 6 hours a day earn per week ?**

**Sol.**

6 m 8 hours 1680

9 m 6 hours ?

**Alternate method :**

As earnings are proportional to the work done, we have

**Example 21**

**21. A can do a piece of work in 15 days and B in 20 days. They finished the work with the assistance of C in 5 days and got 45 as their wages, find the share for each in the wages.**

**Sol.** A did in 5 days 1/3 of the work,

B did in 5 days 1/4 of the work.

C did in 5 days 1 – 1/3 + 1/4 = 5/12 of the work

Since A, B, C did in 5 days 1/3, 1/4, 5/12 of the work respectively.

A’s share = 45 × ^{1}⁄_{3} =15

B’s share = 45 × ^{1}⁄_{4} =11.25

C’s share = 45 × ^{5}⁄_{12} =18.75

**Example 33**

**22. If 8 men, working 9 hours per day can build a wall 18 meter long, 2 meters wide and 12 meters high in 10 days, how many men will be required to build a wall 32 meters long, 3 meters wide and 9 meters high by working 6 hours a day in 8 days ?**

**Sol.** This method is a substitute for the conventional method and can be safely employed for most of the problems.

Step 1 : Assume the thing to be found as ‘X’

Step 2 : In the first place look for X’s counterpart.

e.g. in the above example, X = no. of men

So X’s counterpart = No. of men, given = 8.

So write X = 8x…….

Now see the direct and indirect variation or simply see by which operation more men will be required & by which fewer:

We have X = \(8 \times \frac{32}{18} \times \frac{3}{2} \times \frac{9}{12} \times \frac{10}{8} \times \frac{9}{6} = 30\) men

**Example 23**

**23. If 5 engines consume 6 tonnes of coal when each runs 9 hours per day,how much coal will be needed for 8 engines, each running 10 hours. per day,it being given that 3 engines of the former type consume as much as 4 engines of latter type ?**

Sol.We have X = \(6 \times \frac{8}{5} \times \frac{10}{9} \times \frac{3}{4} = 8\) tons

Explanation : (1) More engines more coal ( > 1)

(2) More time, more coal ( > 1)

(3) Latter consumes less coal than former ( < 1).

In case of men working we have more time, less men (< 1) but here we have more time, more coal ( > 1).

Here let W = 6 tonnes ≡ 5 × 9 × 4/3 engine hours and let X ≡ 8 × 10 × 1 engine hours.

or X = 6 tons × \(\frac{8 \times 10 \times 1}{5 \times 9 \times \left ( \frac{4}{3} \right )}\) = 8 tons

**Example 24**

**24. A garrison of 1500 men is provisioned for 60 days. After 25 days the garrison is reinforced by 500 men, how long will the remaining provisions last ?**

**Sol.** Since the garrison is reinforced by 500 men therefore then are (1500 + 500) or 2000 men now,

since 60 – 25 = 35 days.

=> The provisions left would last 1500 men 35 days

=> Provisions left would last 1 man 35 × 1500 days

=> Provisions left would last 2000 men

35 × ^{1500}⁄_{2000} = 6.25 days

By work equivalence method

1500 × 60 = (1500 × 25) + (2000 × X)

Solve to get X = 26.25 days.

**Example 25**

**25. 40 men can cut 60 trees is 8 hours. If 8 men leaves the job how many trees will be cut in 12 hours ?**

**Sol.** 40 men – working 8 hours – cut 60 trees

or, 1 man – working 1 hour – cuts \(\frac{60}{40 \times 8}\) trees.

Thus, 32 – working 12 hours – cut \(\frac{60 \times 32 \times 12}{40 \times 8} = 73\) trees.

**Using basic concepts :**

M_{1} = 40, D_{1} = 8 (As days and hrs both denote time)

W_{1} = 60 (cutting of trees is taken as work)

M_{2} = 40 – 8 = 32, D_{2} = 12, W_{2} = ?

Putting the values in the formula

M_{1} D_{1} W_{2} = M_{2} D_{2} W_{1}

We have , 40 × 8 × W_{2} = 32 × 12 × 60

or, W_{2} = \(\frac{60 \times 32 \times 12}{40 \times 8} = 72\) trees.

**Example 26**

**26. Two women, Ganga and Sarswati, working separately can mow a field in 8 and 12 hours respectively, If they work in stretches of one hour alternately, Ganga beginning at 9 a.m., when will the mowing be finished ?**

**Sol.** In the first hour Ganga mows 1/8 of the field.

In the second hour Saraswati mows 1/12 of the fields.

∴ in the first 2 hrs \(\left ( \frac{1}{8} + \frac{1}{12} = \frac{5}{24}\right )\) of the field is mown.

∴ in 8 hours ^{5}⁄_{24} × 4 = ^{5}⁄_{6} of the field is mown. —–(1)

Now, (1 – ^{5}⁄_{6}) = ^{1}⁄_{6} of the remains to be mown. In the 9^{th} hour Ganga mows ^{1}⁄_{8} of the field.

∴ Saraswati will finish the mowing of (^{1}⁄_{6} – ^{1}⁄_{8}) = ^{1}⁄_{24} of the field in (^{1}⁄_{24} ÷ ^{1}⁄_{12}) or ^{1}⁄_{2} of an hours.

∴ the total time required is (8 + 1 + ^{1}⁄_{2}) or 9^{1}⁄_{2} hrs.

Thus, the work will be finish at 9 + 9^{1}⁄_{2} = 18^{1}⁄_{2} or 6^{1}⁄_{2} p.m.

**Example 27**

**27. I can finish a work in 15 days at 8 hours a day.You can finish it in 6 ^{2}⁄_{3} days at 9 hrs a day.Find in how many days we can finish it working together 10 hrs a day.**

**Sol.** First suppose each of us works for only one 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.

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

**Example 28**

**28. A can do a work in 6 days. B takes 8 days to complete it. C takes as long as A and B would take working together. How long will it take B and C to complete the work together ?**

**Sol.** (A + B) can do the work in \(\frac{6 \times 8}{6 + 8} = \frac{24}{7}\) days.

∴ C takes ^{24}⁄_{7} days to complete the work.

∴ (B + C) takes \(\frac{\frac{24}{7} \times 8}{\frac{24}{7} + 8} = \frac{24 \times 8}{24 + 56} = 2\tfrac{2}{5}\) days.

**Example 29**

**29. A group of 20 cows can graze a field 3 acresin size in 10 days. How many cows can graze a field twice as large in 8 days ?**

**Sol.** Here, first of all, let us see how work can be defined. It is obvious that work can be measured as “acres grazed”.

In the first case, there were 20 cows in the group.

They had to work for 10 days to do the work which we call W (which = 3)

=> 20 × 10 = 3……….. (1)

Do not be worried about the numerical values on either side.

The point is that logically this equation is consistent as the LHS indicates “Cowdays” and the RHS indicates “Acres”,both of which are correct ways of measuring work done.

Now the field is twice as large. Hence the new equation is

=> C × 8 = 6……….. (2)

Just divide (2) by (1) to get the answer.

\(\frac{8C}{200} = \frac{6}{3}\)=> 8C = 2 × 200 => C = ^{400}⁄_{8} = 50 Cows.

Hence, there were 50 cows in the second group.

**Example 30**

**30. In how many days can the work be completedby Aand B together ?
I. A alone can complete the work in 8 days.
II. If A alone works for 5 days and B alone works for 6 days, the work gets completed.
III. B alone can complete the work in 16 days.**

(a) I and II only

(b) II and III only

(c) Any two of the three

(d) II and either I or III

**Sol.(c)** I. A can complete the job in 8 days. So, A’s 1 days’ work = ^{1}⁄_{8}

II. A works for 5 days, B works for 6 days and the work is completed.

III. B can complete the job in 16 days. So B’s 1 days’s work = ^{1}⁄_{16}

I and III : (A + B)’s 1 days’ work = \(\left ( \frac{1}{8} + \frac{1}{16}\right ) = \frac{3}{16}\)

∴ Both can finish the work in 16/3 days.

II and III : Suppose A takes x days to finish the work

Then, \(\frac{5}{x} + \frac{6}{16} = 1 \Rightarrow \frac{5}{x} = \left ( 1 ? \frac{3}{8} \right ) = \frac{5}{8} \Rightarrow x = 8\)

∴ (A + B)’s 1 days’ work = \(\left ( \frac{1}{8} + \frac{1}{16}\right ) = \frac{3}{16}\)

∴ Both can finish it in 16/3 days.

I and II : A’ 1 day’s work = 1/8.

Suppose B takes x days to finish the work

Then from II, \(\left ( 5 \times \frac{1}{8} + 6 \times \frac{1}{x} = 1 \right )\)

=> \(\frac{6}{x} = \left ( 1 ? \frac{5}{8} \right ) = \frac{3}{8} \Rightarrow x = \left ( \frac{8 \times 6}{3} \right ) = 16\)

∴ (A + B)’s1 days’ work = (^{1}⁄_{8} + ^{1}⁄_{16}) = ^{3}⁄_{16}

∴ Both can finish it in 16/3 days.

**Example 31**

**31. In how many days can the work be done by 9 men and 15 women ?**

**I. 6 men and 5 women can complete the work in 6 days**

**II. 3 men and 4 women can complete the work in 10 days**

**III. 18 men and 15 women can complete the work in 2 days.**

(a) III only

(b) All I,II and III

(c) Any two of the three

(d) Any of the three

**Sol.(c)** Clearly, any two of the three will give two equations in x and y, which can be solved simultaneously.

For example I and II together give

\(\left ( 6x + 5y = \frac{1}{6}, 3x + 4y = \frac{1}{10} \right )\)**Example 32**

**32. 8 men and 14 women are working together in a field. After working for 3 days, 5 men and 8 women leave the work. How many more days will be required to complete the work ?**

**I. 19 men and 12 women together can complete the work in 18 days.**

**II. 16 men complete two-third of the work in 16 days**

**III. In a day, the work done by three men is equal to the work done by four women.**

(a) I only

(b) II only

(c) III only

(d) I or II or III

**Sol.(d)** Clearly, I only gives the answer

Similarly, II only gives the answer

And, III only gives the answer

**PIPES AND CISTERNS**

The same principle of Time and Work is employed to solve the problems on Pipes and Cisterns. The only difference is that in this case, the work done is in terms of filling or emptying a cistern(tank) and the time taken is the time taken by a pipe or a leak(crack) to fill or empty a cistern respectively.

**Inlet :** A pipe connected with a tank (or a cistern or a reservoir) is called an inlet, if it fills it.

**Outlet :** A pipe connected with a tank is called an outlet, if it empties it.

If a pipe can fill a tank in x hours, then the part filled in 1 hour = 1/x

If a pipe can empty a tank in y hours, then the part of the full tank emptied in 1 hour 1/y.

If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, then the net part filled in 1 hour, when both the pipes are opened = (^{1}⁄_{x} - ^{1}⁄_{y}).

∴ Time taken to fill the tank, when both the pipes are opened = \(\frac{xy}{y - x}\).

If a pipe can fill a tank in x hours and another can fill the same tank in y hours, then time taken to fill the tank = \(\frac{xy}{y + x}\),

when both the pipes are opened

If a pipe fills a tank in x hours and another fills the same tank is y hours, but a third one empties the full tank in z hours,and all of them are opened together,then net part filled in 1 hr = \(\left [ \frac{1}{x} + \frac{1}{y} - \frac{1}{z}\right ]\)

∴ Time taken to fill the tank = \(\frac{xyz}{yz + xz - xy}\) hour.

A pipe can fill a tank in x hrs. Due to a leak in the bottom it is filled in y hrs. If the tank is full, the time taken by the leak to empty the tank = \(\frac{xy}{y - x}\) hrs.

A cistern has a leak which can empty it in X hours. A pipe which admits Y litres of water per hour into the cistern is turned on and now the cistern is emptied in Z hours. Then the capacity of the cistern is \(\frac{x + y + z}{z - x}\) litres.

A cistern is filled by three pipes whose diameters are X cm.,Y cm. and Z cm. respectively (where X < Y < Z). Three pipes are running together. If the largest pipe alone will fill it in P minutes and the amount of water flowing in by each pipe is proportional to the square of its diameter, then the time in which the cistern will be filled by the three pipes is => \(\left [ \frac{PZ^{2}}{X^{2} + Y^{2} + Z^{2}} \right ]\) minutes.

If one filling pipe A is n times faster and takes X minutes less time than the other filling pipe B, then the time they will take to fill a cistern, if both the pipes are opened together,is \(\left [ \frac{nX}{\left ( n^{2} - 1 \right )} \right ]\) minutes. A will fill the cistern in \(\left [ \frac{X}{\left ( n - 1 \right )} \right ]\) minutes and B will take to fill the cistern \(\left [ \frac{nX}{\left ( n - 1 \right )} \right ]\) minutes.

Here, A is the faster filling pipe and B is the slower one.

Two filling pipes A and B opened together can fill a cistern in t minutes. If the first filling pipe A alone takes X minutes more or less than t and the second fill pipe B along takes Y minutes more or less than t minutes, then t is given by [t = \(\sqrt{xy}\)] minutes.

**Example 1**

**1. A pipe can fill a cistern in 6 hours. Due to a leak in its bottom, it is filled in 7 hours. When the cistern is full, in how much time will it be emptied by the leak?**

(a) 42 hours

(b) 40 hours

(c) 43 hours

(d) 45 hours

**Sol.(a)** Part of the capacity of the cistern emptied by the leak in one hour = (^{1}⁄_{6} – ^{1}⁄_{7}) = ^{1}⁄_{42} of the cistern.

The whole cistern will be emptied in 42 hours.

**Example 2**

**2. Three pipes A, Band C can fill a cistern in 6hrs. After working together for 2 hrs, C is closed and A and B fill the cistern in 8 hrs. Then find the time in which the cistern can be filled by pipe C.**

**Sol.** A + B + C can fill in 1 hr 1/6 of cistern.

A + B + C can fill in 2 hrs = 2/6 = 1/3 of cistern.

Remaining part = (1 – 1/3) = 2/3 is filled by A + B in 8 hrs.

∴ (A + B) can fill the cistern in \(\frac{8 \times 3}{2}\) = 12 hrs.

Since (A + B + C) can fill the cistern in 6 hrs.

∴ C = (A + B + C) – (A + B) can fill the cistern in

\(\frac{12 \times 6}{12 ? 6}\) hours = 12 hours.**Example 3**

**3. Pipe A can fill a tank in 20 hours while pipe B alone can fill it in 30 hours and pipe C can empty the full tank in 40 hours. If all the pipes are opened together, how much time will be needed to make the tank full?**

**Sol.** By direct formula,The tank will be fill in

The tank will be fill in \(\frac{20 \times 30 \times 40}{30 \times 40 + 20 \times 40 ? 20 \times 30}\)

= ^{120}⁄_{7} = 17^{1}⁄_{7} hrs.

**Example 4**

**4. Three pipes A, B and C can fill a tank in 6 minutes, 8 minutes and 12 minutes, respectively. The pipe C is closed 6 minutes before the tank is filled. In what time will the tank be full ?**

(a) 4 min

(b) 6 min

(c) 5 min

(d) Data inadequate

**Sol.(a)** Let it takes t minutes to completely fill the tank.

Now, \(\frac{t}{6} + \frac{t}{8} + \frac{t ? 6}{12} = 1\)

or \(\frac{4t + 3t + 2t ? 12}{24} = 1\)

or 9t – 12 = 24

or 9t = 36 => t = 4 min.

**Example 5**

**5. If three taps are opened together,a tank is filled in 12 hrs. One of the taps can fill it in 10 hrs and another in 15 hrs. How does the third tap work ?**

**Sol.** We have to find the nature of the third tap, whether it is a filler or a waste pipe.Let it be a filler pipe which fills in x hrs.

Then, \(\frac{10 \times 15 \times x}{10 \times 15 + 10x + 15x}\) = 12

or, 150x = 150 × 12 + 25x × 12

or –150x = 1800

∴ x = –12

–ve sign shows that the third pipe is a waste pipe which vacates the tank in 12 hrs.

**Example 6**

**6. 4 pipes can fill a reservoir in 15, 20, 30 and 60 hours respectively. The first was opened at 6 am, second at 7 am third at 8 am and fourth at 9 am. When will the reservoir be full?**

**Sol.(1)** Let the time be t hours after 6 am.

∴ \(\frac{1}{15} \times t + \frac{\left ( t ? 1 \right )}{20} + \frac{\left ( t ? 2 \right )}{30} + \frac{\left ( t ? 3 \right )}{60} = 1\)

∴ 4t + 3 (t – 1) + 2 (t – 2) + (t– 3) = 60

∴ t = 7 hours

∴ It is filled at 1 pm

**Example 7**

**7. A and B can fill a cistern in 7.5 minutes and 5 minutes respectively and C can carry off 14 litres per minute. If the cistern is already full and all the three pipes are opened,then it is emptied in 1hour. How many litres can it hold ?**

**Sol.** If the capacity is L litres, water filled in 1 hour = Water removed in 1 hour.

∴ \(L + \frac{2L}{15} \times 60 + 12L = 14 \times 60\) => L + 8L + 12L = 14 × 60

=> 21L = 14 × 60 or L = 40 litres.

So the capacity of the cistern is 40 litres.

**Example 8**

**8. A cistern can be filled by two taps A and B in 25 minutes and 30 minutes respectively can be emptied by a third in 15 minutes. If all the taps are turned on at the same moment,what part of the cistern will remain unfilled at the end of 100 minutes ?**

**Sol.** We have \(\frac{1}{25} + \frac{1}{30} ? \frac{1}{15} = \frac{1}{150}\) part filled in 1 minute

Hence, 1 – 100(^{1}⁄_{150}) 1/3rd of the tank is unfilled after 100 minutes.

**Example 9**

**9. A barrel full of beer has 2 taps one midway,,which draw a litre in 6 minutes and the other at the bottom,which draws a litre in 4 minutes. The lower tap is lower normally used after the level of beer in the barrel is lower than midway.The capacity of the barrel is 36 litres. A new assistant opens the lower tap when the barrel is full and draws out some beer. As a result the lower tap has been used 24 minutes before the usual time. For how long was the beer drawn out by the new assistant ?**

**Sol.** The top tab is operational till 18 litres is drawn out.

∴ Time after which the lower tap is usually open

= 18 × 6 = 108 minutes

∴ Time after which it is open now = 108 – 24 = 84 minutes

∴ Litres drawn = 84/6 = 14 litres

∴ 18 – 14 = 4 litres were drawn by the new assistant.

∴ Time = 4 × 4 = 16 minutes

**Example 10**

**10. A cistern can be filled by two pipes filling separately in 12 and 16 min. respectively. Both pipes are opened together for a certain time but being clogged, only 7/8 of the full quantity of water flows through the former and only 5/6 through the latter pipe. The obstructions, however, being suddenly removed, the cistern is filled in 3 min. from that moment. How long was it before the full flow began?**

(a) 2.5 min

(b) 4.5 min

(c) 3.5 min

(d) 5.5 min

**Sol.(b)** Both the pipes A and B can fill \(\frac{1}{12} + \frac{1}{16} = \frac{7}{48}\) of the cistern in one minute, when their is no obstruction.

With obstruction, both the pipes can fill

\(\frac{1}{12} \times \frac{7}{8} + \frac{1}{16} \times \frac{5}{6} = \frac{7}{96} + \frac{5}{96} = \frac{1}{8}\) of the cistern in one minute.Let the obstructions were suddenly removed after x minutes.

∴ With obstruction, x/8 of the cistern could be filled in x minutes and so the remaining \(1 ? \frac{x}{8} = \frac{8 ? x}{8}\) of the cistern was filled without obstruction in 3 minutes, i.e. In one minute, \(\frac{8 ? x}{24}\) of the cistern was filled.

=> \(\frac{8 ? x}{24} = \frac{7}{48} \Rightarrow 16 ? 2x = 7 \Rightarrow x = 4.5\)