time & work pipe and cisterns

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TIME AND WORKIn most of the problems on time and work, either of the followingbasic 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 thX

part of whole work.

If A’s one day’s work = 1 thX

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 XYX Y+

days.

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

XYZ

XY YZ XZ+ + days.

Example 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 ?

Solution :

A’s 1 day’s work = 1 th5

part of whole work and

B’s 1 day’s work = 1 th6

part of whole work

\ (A + B)’s one day’s work = 1 1 115 6 30

+ = th part of whole work.

So, both together will finish the work in30 days11

82 days.11

=

By Direct Formula :

A + B can do the work in 5 65 6

´+

days 30 8211 11

= = days.

Example 2 :Two men, Vikas and Vishal, working separately can mow afield in 8 and 12 hours respectively. If they work in stretchesof one hour alternately, Vikas beginning at 8 a.m, when willthe mowing be finished?

Solution :

In the first hour, Vikas mows 18

of the field.

In the second hour, Vishal mows 112

of the field.

\ In the first 2 hours, 1 1 58 12 24

æ ö+ =ç ÷è ø of the field is mown.

\ In 8 hours, 5 5424 6

´ = of the field is mown.

Now, 5 11–6 6

æ ö =ç ÷è ø of the field remains to be mown.

In the 9th hour, Vikas mows 18

of the field.

Remaining work =1 1–6 8

124

=

\ Vishal will finish the remaining work in 1 124 12

æ ö¸ç ÷è ø

or 12

of an hour..

\ The total time required is 18 1

2æ ö+ +ç ÷è ø

or 219 hours.

Thus, the work will be finished at 1 18 9 172 2

+ = or 5.30 pm.

Example 3 :A can do a piece of work in 36 days, B in 54 days and C in 72days. All the three began the work together on the Dec. 15,2014, but A left 8 days and B left 12 days before thecompletion of the work. If C took the rest for a week then inhow many days, the work was finished from the day itstarted ?

Solution :Let the total time taken be x days.According to the given condition

Þx 8 x 12 x 136 54 72- -

+ + =

Þ6 (x 8) 4 (x 12) 3x 1

216- + - +

=

Þ6x 48 4x 48 3x 1

216- + - +

= Þ 13x 96 1

216-

=

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

Þ312

x 2413

= =

Since, C takes the rest for a week, so the number of days inwhich the work was finished from one day it started = 31 i.e.on 14.01.2015.

TIME AND WORK &PIPES AND CISTERNS 10

CHA

PTER

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Example 4 :A and B can do a certain piece of work in 8 days, B and Ccan do it in 12 days and C and A can do it in 24 days. Howlong would each take separately to do it ?

Solution :(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 1 1 13

18 24 12 72+ + =

(A + B + C)’s one days’s work = 13

144A’s one days’s work = (A + B + C)’s one days’s work

– (B + C)’s one days’s work = 13 1 1144 12 144

- =

Since A completes of the work in 1 day, he will complete 1

work in 1441

= 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 Aalone can do it in Y days, then B alone can do the work in

XYY – X

days.

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

Solution :

(A + B)’s 1 day’s work = 1 th6

part of the whole work.

A’s 1 day’s work = 1 th9

part of the whole work.

\ B’s 1 day’s work = 1 1 3 2 1 th6 9 18 18

-- = =

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

6 9 54 18 days9 – 6 3

´= =

A and B can do a work in ‘X’ and ‘Y’ days respectively. Theystarted the work together but A left ‘a’ days before completionof the work. Then, time taken to finish the work is

Y(X a)X Y

++

If ‘A’ is ‘a’ times efficient than B and A can finish a work inX days, then working together, they can finish the work in

aXa 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 = Z(a 1)

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 Btogether then the number of days taken by A and B working

together is given by [ xy] days.

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

Solution :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. x 3 x 16 12-

+ =

3x 6 1 x 6 days12

-Þ = Þ =

By Direct Formula:

Required time = 12(6 3) 6 days

12 6+

=+

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

Solution :Let B can do the work in x daysand A can do the work in 2x days

Then, 1 1 1x 2x 14

+ = (given)

3x 14 21 days2

Þ = ´ =

By Direct Formula :

Time taken by B = 114 1 21days2

æ ö+ =ç ÷è ø

If n men or m women can do a piece of work in X days, thenN men and M women together can finish the work in

nmX days.nM mN+

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

Solution :It is clear that 10 men = 12 women or 5 men = 6 womenÞ 15 men + 6 women = (18 + 6) i.e. 24 womenNow 12 women can complete the work in 10 days\ 24 women will do it in 5 days.

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By Direct Formula :

Required time = 10 12 10 5 days

10 6 12 15´ ´

=´ + ´

Example 9 :If 3 men or 4 women can reap a field in 43 days, how longwill 7 men and 5 women take to reap it ?

Solution :

3 men reap143 of the field in 1 day..

\ 1 men reaps 1

43 3´ of the field in 1 day..

4 women reap 143 of the field in 1 day..

\ 1 woman reaps 1

43 4´of the field in 1 day..

\ 7 men and 5 women reap 7 5

43 3 43 4æ ö+ç ÷´ ´è ø

= 1

12 of the

field in 1 day.\ 7 men and 5 women will reap the whole field in 12 days.Alternate method

Required number of days = 1

7 543 3 43 4

é ù+ê ú´ ´ë û

= 43 3 4

4 4 5 3´ ´

´ + ´ = 12 days.

Example 10 :If 12 men and 16 boys can do a piece of work in 5 days and13 men and 24 boys can do it in 4 days, how long will 7 menand 10 boys take to do it ?

Solution :12 men and 16 boys can do the work in 5 days .... (i)13 men and 24 boys can do the work in 4 days .... (ii)Now it is easy to see that if the no. of workers be multipliedby any number, the time must be divided by the same number(derived from : more worker less time).Hence multiplying the no. of workers in (i) and (ii) by 5 and4 respectively, we get 5 (12 men + 16 boys) can do the workin 5/5 = 1 day

4 (13 men + 24 boys) can do the work in 44

= 1 dayor, 5(12m + 16b) = 4 (13m + 24b)or, 60 m + 80 b = 52 m + 96 bor, 60 m – 52 m = 96 b – 80 bor, 8 m = 16 b\ 1 men = 2 boys.Thus, 12 men + 16 boys = 24 boys + 16 boys = 40 boysand 7 men + 10 boys = 14 boys + 10 boys = 24 boysThe question now becomes :“If 40 boys can do a piece of work in 6 days how long will 24boys take to do it ?”Using basic formulawe have,40 × 5 = 24 × D2

or, D2 = 40 5

24´

= 813 days

Example 11 :Two men and 7 boys can do a piece of work in 14 days. 3men and 8 boys can do it in 11 days. In how many days can8 men and 6 boys do a work 3 times as big as the first ?

Solution :2 men + 7 boys in 14 days Þ 28 men + 98 boys in 1 day3 men + 8 boys in 11 days Þ 33 men + 88 boys in 1 day\ 28 men + 98 boys = 33 men + 88 boys\ 2 boys º 1 manNow, 2 men + 7 boys = 11 boys; 8 men + 6 boys = 22 boysMore boys, fewer days; more work, more daysBoys Days Work 11 14 1

22 x 3

\ x

14 =

1112

× 31

\ Number of days = 21 days.

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

Solution :Here, the alone time of kaberi is related to the alone times ofother two persons, so assume the alone time of kaberia = x,

Then, alone time of Kanti = x2

and of Kalpana = x3

Kaberi's 1 day work + Kanti’s 1 day work + Kalpana's 1 daywork = combined 1 days work

Þ 1x

+ 1

x / 2 + 1

x / 3 = 11

Þ x = 6

\ Alone time for Kaberi = 6 days, for Kanti = 6/2 = 3 days,Kalpana = 6/3 = 2 days,

Example 13 :1 man or 2 women or 3 boys can do a work in 44 days. Thenin how many days will 1 man, 1 woman and 1 boy do thework?

Solution :Number of required days

= 1

1 1 144 1 44 2 44 3

+ +´ ´ ´

= 44 1 2 3

6 3 2´ ´ ´+ +

= 24 days

If ‘M1’ persons can do ‘W1’ works in ‘D1’ days and ‘M2’persons can do ‘W2’ works in ‘D2’ days thenM1D1W2 = M2D2W1If T1 and T2 are the working hours for the two groups thenM1D1W2T1 = M2D2W1T2Similarly,M1D1W2T1E1 = M2D2W1T2E2, where E1 and E2 are theefficiencies 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 :

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a, assuming the amount of work done by each of them inthe given time is the same, or they are identical.

A is K times as good a worker as B and takes X days lessthan B to finish the work. Then the amount of time required

by A and B working together is 2K XK 1

´

- days.

If A is n times as efficient then B, i.e. A has n times as much

capacity to do work as B, A will take 1n

of the time taken by

B to do the same amount of work.Example 14 :

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

Solution :This example has an extra variable ‘time’ (hrs a day), so the‘basic-formula’ can’t work in this case. An extended formulais being given :M1 D1 T1 W2 = M2 D2 T2 W1Here, 5 × 6 × 6 × 16 = 12 × D2 × 8 × 10

\ D2 = 5 6 6 1612 8 10´ ´ ´

´ ´ = 3 days.

Example 15 :A and B can do a work in 45 days and 40 days respectively.They began the work together, but A left after some timeand B finished the remaining work in 23 days. After howmany days did A leave ?

Solution :B finished the remaining work in 23 days.

\ Work done by B in 23 days = 2340 work

\ A + B do together 1 – 2340 =

1740 work

Now, A + B do 1 work in 40 4540 45

´+

= 40 45

85´

days

\ A + B do 1740 work in

40 4585´

× 1740 = 9 days.

Alternate method :If we ignore the intermediate steps, we can write a direct

formula as : 40 4540 45

´+

40 23

40-æ ö

ç ÷è ø

= 9 days.

Example 16 :Two friends take a piece of work for ` 960. One alone coulddo it in 12 days, the other in 16 days with the assistance ofan expert they finish it in 4 days. How much remunerationthe expert should get ?

Solution :

First friend’s 4 day’s work = 4

12 =

13 (Since, the work is

finished in 4 days, when expert assists )

Second friends’s 4 day’s work = 4

16 = 14

The expert’s 4 day’s work = 1 – 1 13 4

æ ö+ç ÷è ø

= 5

12Now, total wages of ` 960 is to be distributed among twofriends and the expert in proportion to the amount of workdone by each of them.So, 960 is to be divided in the proportion of13 :

14

: 5

12 or 4 : 3 : 5

\ Share of expert = 5

12 x 960 = ̀ 400

Hence, the expert should get ` 400.Example 17 :

A certain number of men can do a work in 60 days. If therewere 8 men more it could be finished in 10 days less. Howmany men are there ?

Solution :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 = 8 50

10´

= 40 men.

Alternate method :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 = 50 8

10´

= 40 men.

Example 18 :Two coal loading machines each working 12 hours per dayfor 8 days handles 9,000 tonnes of coal with an efficiencyof 90%. While 3 other coal loading machines at an efficiencyof 80 % set to handle 12,000 tonnes of coal in 6 days. Findhow many hours per day each should work.

Solution :

Here 1 1 1 1

1

N D R EW

´ ´ ´ = 2 2 2 2

2

N D R EW

´ ´ ´

N1 = 2, R1 = 12h/ day : N2 = 3, R2 = ?

E1 = 90

100 W1 = 9,000 ;

E2 = 80

100 W2 = 12, 000

Þ2 8 12 909,000 100´ ´ ´

´= 23 6 R 80

12,000 100´ ´ ´

´

Þ R2 = 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.

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Example 19 :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 16

=

B’s one day’s work 18

=

C’s one day’s work 1

12=

A’s share : B’s share : C’s share1 1 1: :6 8 12

=

Multiplying each ratio by the L.C.M. of their denominators,the ratios become 4 : 3 : 2

\ B’s share 1350 3 4509

´= =`

Example 20 :If 6 men working 8 hours a day earn ` 1680 per week, thenhow much will 9 men working 6 hours a day earn per week ?

Solution :6 m 8 hours ` 16809 m 6 hours ?

6 91680

8 6´ ´ = ̀ 1890

Alternate method :As earnings are proportional to the work done, we have

1 1 2 22

1 2 2

M D M D 6 8 9 6W Rs.1890

W W 1680 W´ ´

= Þ = Þ == ` 1890

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

Solution :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 5

13 4 12

- + = of the work

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

A’s share = ` 145 Rs.153

´ = = ` 15

B’s share = ̀ 1

45 Rs.11.254

´ = = ̀ 11.25

C’s share = ̀ 5

45 Rs.18.7512

´ = = ` 18.75

Example 22 :If 8 men, working 9 hours per day can build a wall 18 meterlong, 2 meters wide and 12 meters high in 10 days, howmany men will be required to build a wall 32 meters long,3 meters wide and 9 meters high by working 6 hours a day in8 days ?

Solution :This method is a substitute for the conventional methodand 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 menSo X’s counterpart = No. of men, given = 8.So write X = 8x.......Now see the direct and indirect variation or simply see bywhich operation more men will be required & by which fewer:

We have 32 3 9 10 9

X 8 30 men18 2 12 8 6

= ´ ´ ´ ´ ´ =

Example 23 :If 5 engines consume 6 tonnes of coal when each runs 9 hrsper day, how much coal will be needed for 8 engines, eachrunning 10 hrs. per day, it being given that 3 engines of theformer type consume as much as 4 engines of latter type ?

Solution :

We have X = 8 10 36 8 tons5 9 4

´ ´ ´ =

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 º 5 × 9 × 4/3 engine hoursand let X º 8 × 10 × 1 engine hours.

or X = 6 tons × 8 10 1

8 tons5 9 (4 / 3)

´ ´=

´ ´Example 24 :

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

Solution :Since the garrison is reinforced by 500 men therefore thenare (1500 + 500) or 2000 men now,since 60 – 25 = 35 days.Þ The provisions left would last 1500 men 35 daysÞ Provisions left would last 1 man 35 × 1500 daysÞ Provisions left would last 2000 men

35 × 1500 26.252000

= days

Alternate method:1500 × 60 = (1500 × 25) + (2000 × X)

90000 – 37500 = 2000XX = 26.25 days.

Example 25 :40 men can cut 60 trees is 8 hrs. If 8 men leaves the job howmany trees will be cut in 12 hours ?

Solution :40 men – working 8 hrs – cut 60 trees

or, 1 man – working 1 hr – cuts 60

40 8´ tress

Thus, 32 – working 12 hrs – cut 60 32 12

40 8´ ´

´ = 73 trees.

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Using basic concepts :M1 = 40, D1 = 8 (As days and hrs both denote time)W1 = 60 (cutting of trees is taken as work)M2 = 40 – 8 = 32, D2 = 12, W2 = ?Putting the values in the formulaM1 D1 W2 = M2 D2 W1We have , 40 × 8 × W2 = 32 × 12 × 60

or, W2 = 32 12 60

40 8´ ´

´ = 72 trees.

Example 26 :Two women, Ganga and Sarswati, working separately canmow a field in 8 and 12 hrs respectively, If they work instretches of one hour alternately, Ganga beginning at 9 a.m.,when will the mowing be finished ?

Solution :

In the first hour Ganga mows 18

of the field.

In the second hour Saraswati mows 1

12 of the fields.

\ In the first 2 hrs 1 1 58 12 24

æ ö+ =ç ÷è ø

of the field is mown.

\ In 8 hrs 524

× 4 = 56

of the field is mown...... (i)

Now, 516

æ ö-ç ÷è ø

= 16 of the remains to be mown. In the 9th

hour Ganga mows 18

of the field.

\ Saraswati will finish the mowing of 1 16 8

æ ö-ç ÷è ø

= 124

of the

field in 1 124 12

æ ö¸ç ÷è ø

or 12 of an hours.

\ the total time required is 18 12

æ ö+ +ç ÷è ø

or 912 hrs.

Thus, the work will be finish at 9 + 912

= 1812

or 612

p.m.

Example 27 :I can finish a work in 15 days at 8 hrs a day. You can finish

it in 623 days at 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 one hr a day.Then I can finish the work in 15 × 8 = 120 days and you can

finish the work in 203 × 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 :A can do a work in 6 days. B takes 8 days to complete it. Ctakes as long as A and B would take working together. Howlong will it take B and C to complete the work together ?

Solution :

(A + B) can do the work in 6 86 8

´+

= 247 days.

\ C takes 247 days to complete the work.

\ (B + C) takes

24 87

24 87

´

+ =

24 824 56

´+

= 225 days.

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

Solution :Here, first of all, let us see how work can be defined. It isobvious 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 callW (which = 3)Þ 20 × 10 = 3 ........... (i)Do not be worried about the numerical values on eitherside. The point is that logically this equation is consistentas the LHS indicates “Cowdays” and the RHS indicates“Acres”, both of which are correct ways of measuring workdone.Now the field is twice as large. Hence the new equation isÞ C × 8 = 6 ........... (ii)Just divide (ii) by (i) to get the answer.

8C200 =

63

Þ 8C = 2 × 200 Þ C = 4008 = 50 cows.

Hence, there were 50 cows in the second group.

PIPE AND CISTERNSThe same principle of Time and Work is employed to solve theproblems on Pipes and Cisterns. The only difference is that inthis 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) iscalled an inlet, if it fills it.Outlet : A pipe connected with a tank is called an outlet, if itempties it. If a pipe can fill a tank in x hours, then the part filled in 1

hour 1x

=

If a pipe can empty a tank in y hours, then the part of the

full tank emptied in 1 hour 1y

= .

If a pipe can fill a tank in x hours and another pipe canempty the full tank in y hours, then the net part filled in 1

hour, when both the pipes are opened 1 1–x y

æ ö= ç ÷

è ø.

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\ Time taken to fill the tank, when both the pipes are

opened xyy – 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 xyy 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 1 1 1–x y z

é ù= +ê ú

ë û

\ Time taken to fill the tank xyzyz xz – xy

=+

hours.

A pipe can fill a tank in x hrs. Due to a leak in the bottom itis filled in y hrs. If the tank is full, the time taken by the leak

to empty the tank xyy – x

= hrs.

A cistern has a leak which can empty it in X hours. A pipewhich admits Y litres of water per hour into the cistern isturned on and now the cistern is emptied in Z hours. Then

the capacity of the cistern is 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 pipesare running together. If the largest pipe alone will fill it in Pminutes and the amount of water flowing in by each pipe isproportional to the square of its diameter, then the time inwhich the cistern will be filled by the three pipes is

2

2 2 2PZ

X Y Z

é ùê ú

+ +ê úë û minutes.

If one filling pipe A is n times faster and takes X minutes lesstime than the other filling pipe B, then the time they will taketo fill a cistern, if both the pipes are opened together, is

2nX

(n 1)

é ùê ú

-ê úë û minutes. A will fill the cistern in

Xn 1

æ öç ÷è ø-

minutes

and B will take to fill the cistern nXn 1

æ öç ÷è ø-

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 minutesmore or less than t and the second fill pipe B along takes Yminutes more or less than t minutes, then t is given by

[t xy]= minutes.Example 30 :

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 muchtime will it be emptied by the leak?

Solution :Part of the capacity of the cistern emptied by the leak in one

hour = 421

71

61

=÷øö

çèæ - of the cistern.

The whole cistern will be emptied in 42 hours.

Example 31 :Three pipes A, B and C can fill a cistern in 6 hrs. Afterworking together for 2 hrs, C is closed and A and B fill thecistern in 8 hrs. Then find the time in which the cistern canbe filled by pipe C.

Solution :

A + B + C can fill in 1 hr 16

= of cistern.

A + B + C can fill in 2 hrs 2 16 3

= = of cistern.

Remaining part 1 21–3 3

æ ö= =ç ÷è ø is filled by A + B in 8 hrs.

\ (A + B) can fill the cistern in 8 32´ = 12 hrs.

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

12 612 – 6

´ hours = 12 hours.

Example 32 :Pipe A can fill a tank in 20 hours while pipe B alone can fill itin 30 hours and pipe C can empty the full tank in 40 hours.If all the pipes are opened together, how much time will beneeded to make the tank full ?

Solution :By direct formula,

The tank will be fill in 20 30 40

30 40 20 40 – 20 30´ ´

=´ + ´ ´

120 117

7 7= = hrs.

Example 33 :Three pipes A, B and C can fill a tank in 6 minutes, 8 minutesand 12 minutes, respectively. The pipe C is closed 6 minutesbefore the tank is filled. In what time will the tank be full ?

Solution :Let it takes t minutes to completely fill the tank.

Now, 112

6t8t

6t

=-

++

or 124

12t2t3t4=

-++

or 9t – 12 = 24or 9t = 36 Þ t = 4 min.

Example 34 :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 ?

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

Then, 10 15 x 12

10 15 10x 15x´ ´ =

´ + +or, 150x = 150 × 12 + 25x × 12

8


or – 150x = 1800 \ x = – 12–ve sign shows that the third pipe is a waste pipe whichvacates the tank in 12 hrs.

Example 35 :4 pipes can fill a reservoir in 15, 20, 30 and 60 hoursrespectively. The first was opened at 6 am, second at 7 amthird at 8 am and fourth at 9 am. When will the reservoir befull ?

Solution :Let the time be t hours after 6 am.

\1 (t 1) (t 2) (t 3)t 1

15 20 30 60- - -

´ + + + =

\ 4t + 3 (t – 1) + 2 (t – 2) + (t – 3) = 60Þ t = 7 hoursÞ It is filled at 1 pm

Example 36 :A and B can fill a cistern in 7.5 minutes and 5 minutes re-spectively and C can carry off 14 litres per minute. If thecistern is already full and all the three pipes are opened,then it is emptied in 1 hour. How many litres can it hold ?

Solution :If the capacity is L litres, water filled in 1 hour = Waterremoved in 1 hour.

L LL 60 60 14 601 572

+ ´ + ´ = ´

\2LL 60 12L 14 6015

+ ´ + = ´ Þ L + 8L + 12L = 14 × 60

Þ 21L = 14 × 60 or L = 40 litres.So the capacity of the cistern is 40 litres.

Example 37:A cistern can be filled by two taps A and B in 25 minutesand 30 minutes respectively can be emptied by a third in 15minutes. If all the taps are turned on at the same moment,what part of the cistern will remain unfilled at the end of 100minutes ?

Solution :

We have 1 1 1 125 30 15 150

+ - = part filled in 1 minute

Hence, 1

1 100150

æ ö- ç ÷è ø = 1/3rd of the tank is unfilled after

100 minutes.Example 38 :

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

Solution :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 39 :A cistern can be filled by two pipes filling separately in 12and 16 min. respectively. Both pipes are opened togetherfor a certain time but being clogged, only 7/8 of the fullquantity of water flows through the former and only 5/6through the latter pipe. The obstructions, however, beingsuddenly removed, the cistern is filled in 3 min. from thatmoment. How long was it before the full flow began?

Solution :

Both the pipes A and B can fill 1 1 712 16 48

+ = of the

cistern in one minute, when their is no obstruction.With obstruction, both the pipes can fill1 7 1 5 7 5 1

12 8 16 6 96 96 8´ + ´ = + = of the cistern in one minute.

Let the obstructions were suddenly removed after x min-utes.

\ With obstruction, x8

of the cistern could be filled in

x minutes and so the remaining x 8 x18 8

-- = of the

cistern was filled without obstruction in 3 minutes, i.e. In

one minute, 8 x24- of the cistern was filled.

8 x 716 2x 7

24 48-

Þ = Þ - = Þ x = 4.5

9


1. A does a work in 10 days and B does the same work in 15days. In how many days they together will do the samework ?(a) 5 days (b) 6 days(c) 8 days (d) 9 days

2. A man can do a piece of work in 10 days but with theassistance of his son, the work is done in 8 days. In howmany days, his son alone can do the same piece of work?(a) 15 days (b) 22 days(c) 30 days (d) 40 days

3. A can finish a work in 18 days and B can do the same workin half the time taken by A. Then, working together, whatpart of the same work they can finish in a day?

(a)16 (b)

19

(c)25 (d)

27

4. George takes 8 hours to copy a 50 page manuscript whileSonia can copy the same manuscript in 6 hours. How manyhours would it take them to copy a 100 page manuscript, ifthey work together ?

(a)667 (b) 9

(c)597 (d) 14

5. A can do a piece of work in 25 days and B in 20 days. Theywork together for 5 days and then A goes away. In howmany days will B finish the remaining work ?(a) 17 days (b) 11 days(c) 10 days (d) None of these

6. A man is twice as fast as a woman. Together the man andthe woman do the piece of work in 8 days. In how manydays each will do the work if engaged alone?(a) man-14 days, woman-28 days(b) man-12 days, woman-24 days(c) man-10 days, woman-20 days(d) None of these

7. A and B can do a job in 16 days and 12 days respectively. 4days before finishing the job, A joins B. B has started thework alone. Find how many days B has worked alone?(a) 6 days (b) 4 days(c) 5 days (d) 7 days

8. A contractor undertakes to built a walls in 50 days. Heemploys 50 peoples for the same. However after 25 days hefinds that only 40% of the work is complete. How manymore man need to be employed to complete the work intime?(a) 25 (b) 30(c) 35 (d) 20

9. A is 30% more efficient than B. How much time will they,working together, take to complete a job which A alonecould have done in 23 days?(a) 11 days (b) 13 days

(c)320

17 days (d) None of these

10. A and B can finish a work in 10 days while B and C can doit in 18 days. A started the work, worked for 5 days, then Bworked for 10 days and the remaining work was finished byC in 15 days. In how many days could C alone have finishedthe whole work?(a) 30 days (b) 15 days(c) 45 days (d) 24 days

11. 24 men working 8 hours a day can finish a work in 10 days.Working at the rate of 10 hours a day, the number of menrequired to finish the same work in 6 days is :(a) 30 (b) 32(c) 34 (d) 36

12. 12 men complete a work in 18 days. Six days after they hadstarted working, 4 men joined them. How many days will allof them take to complete the remaining work?(a) 10 days (b) 12 days(c) 15 days (d) 9 days

13. A tyre has two punctures. The first puncture along wouldhave made the tyre flat in 9 minutes and the second alonewould have done it in 6 minutes. If air leaks out at a constantrate, how long does it take both the punctures together tomake it flat ?

(a)112 minutes (b)

132 minutes

(c)335 minutes (d)

144 minutes

14. A man, a woman or a boy can do a job in 20 days, 30 days or60 days respectively. How many boys must assist 2 menand 8 women to do the work in 2 days?(a) 15 boys (b) 8 boys(c) 10 boys (d) None of these

15. A can do 50% more work as B can do in the same time. Balone can do a piece of work in 20 hours. A, with help of B,can finish the same work in how many hours ?(a) 12 (b) 8

(c) 1133

(d) 152

16. A machine P can print one lakh books in 8 hours, machineQ can print the same number of books in 10 hours whilemachine R can print them in 12 hours. All the machines arestarted at 9 a.m. while machine P is closed at 11 a.m. and theremaining two machines complete the work. Approximatelyat what time will the work be finished?(a) 11 : 30 am (b) 12 noon(c) 12 : 30 pm (d) 1 pm

10


17. A can do a piece of work in 10 days, while B alone can do itin 15 days. They work together for 5 days and the rest ofthe work is done by C in 2 days. If they get ` 450 for thewhole work, how should they divide the money ?(a) ` 225, ̀ 150, ̀ 75 (b) ` 250, ̀ 100, ̀ 100(c) ` 200, ̀ 150, ̀ 100 (d) ` 175, ̀ 175, ̀ 100

18. A alone would take 8 days more to complete the job than ifboth A and B would together. If B worked alone, he took

142

days more to complete the job than A and B worked

together. What time would they take if both A and B workedtogether ?(a) 7 days (b) 5 days(c) 4 days (d) 6 days

19. 10 men and 15 women together can complete a work in 6days. It takes 100 days for one man alone to complete thesame work. How many days will be required for one womanalone to complete the same work ?(a) 90 (b) 125(c) 145 (d) None of these

20. After working for 8 days, Anil finds that only 13

of the work

has been done. He employs Rakesh who is 60% efficient asAnil. How many more days will Anil take to complete thejob ?(a) 15 days (b) 12 days(c) 10 days (d) 8 days

21. A sum of ` 25 was paid for a work which A can do in 32days, B in 20 days, B and C in 12 days and D in 24 days.How much did C receive if all the four work together ?

(a) ` 143 (b) `

163

(c) ` 153 (d) `

173

22. A and B can do a job in 15 days and 10 days, respectively.They began the work together but A leaves after some daysand B finished the remaining job in 5 days. After how manydays did A leave?(a) 2 days (b) 3 days(c) 1 day (d) None of these

23. Mr. Suresh is on tour and he has ` 360 for his expenses. Ifhe exceeds his tour by 4 days he must cut down dailyexpenses by ̀ 3. The number of days of Mr. Suresh’s tourprogramme is :(a) 20 days (b) 24 days(c) 40 days (d) 42 days

24. A can do a job in 3 days less time than B. A works at it alonefor 4 days and then B takes over and completes it. Ifaltogether 14 days were required to finish the job, then inhow many days would each of them take alone to finish it?(a) 17 days, 20 days (b) 12 days, 15 days(c) 13 days, 16 days (d) None of these

25. Two workers A and B working together completed a job in 5days. If A worked twice as efficiently as he actually did and

B worked 13 as efficiently as he actually did, the work would

have completed in 3 days. Find the time for A to completethe job alone.

(a) 164

days (b) 354

days

(c) 5 days (d) None of these26. 12 men can complete a piece of work in 4 days, while 15

women can complete the same work in 4 days. 6 men startworking on the job and after working for 2 days, all of themstopped working. How many women should be put on thejob to complete the remaining work, If it is to be completedin 3 days ?(a) 15 (b) 18(c) 22 (d) Data inadequate

27. If 6 men and 8 boys can do a piece of work in 10 days while26 men and 48 boys can do the same in 2 days, the timetaken by 15 men and 20 boys in doing the same type ofwork will be:(a) 4 days (b) 5 days(c) 6 days (d) 7 days

28. A contract is to be completed in 46 days and 117 men wereset to work, each working 8 hours a day. After 33 days, 4/7of the work is completed. How many additional men may beemployed so that the work may be completed in time, eachman now working 9 hours a day ?(a) 80 (b) 81(c) 82 (d) 83

29. Ramesh is twice as good a workman as Sunil and finishes apiece of work in 3 hours less than Sunil. In how many hoursthey together could finish the same piece of work ?

(a)312 (b) 2

(c)321 (d) None of these

30. One hundred men in 10 days do a third of a piece of work.The work is then required to be completed in another 13days. On the next day (the eleventh day) 50 more men areemployed, and on the day after that, another 50. How manymen must be discharged at the end of the 18th day so thatthe rest of the men, working for the remaining time, will justcomplete the work ?(a) 100 (b) 105(c) 110 (d) 115

31. If 12 men or 15 women or 18 boys can do a piece of work in15 days of 8 hours each, find how many men assisted by 5women and 6 boys will finish the same work in 16 days of 9hours each.(a) 6 men (b) 2 men(c) 8 men (d) 4 men

32. The work done by a man, a woman and a child is in the ratioof 3 : 2 : 1. There are 20 men, 30 women and 36 children in afactory. Their weekly wages amount to ` 780, which isdivided in the ratio of work done by the men, women andchildren. What will be the wages of 15 men, 21 women and30 children for 2 weeks ?(a) ` 585 (b) ` 292.5(c) ` 1170 (d) ` 900

11


33. 2 men and 3 boys can do a piece of work in 10 days while3 men and 2 boys can do the same work in 8 days. In howmany days can 2 men and 1 boy to the work ?

(a)1122

days (b)1112

days

(c)1

152

days (d)1

132

days

34. A can do a certain job in 12 days. B is 60% more efficientthan A. How many days B alone take to do the same job ?

(a)172

(b) 11

(c)182

(d) 8

35. 12 men and 16 boys can do a piece of work in 5 days, 13 menand 24 boys can do it in 4 days. Then the ratio of daily workdone by a man to that of a boy is(a) 2 : 1 (b) 3 : 1(c) 3 : 2 (d) 5 : 4

36. x is 3 times as faster as y and is able to complete the work in40 days less than y. Then the time in which they can completethe work together ?(a) 15 days (b) 10 days

(c)172

days (d) 5 days

37. Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipeC in 30 hours. If all the pipes are open, in how many hourswill the tank be filled ?(a) 2 (b) 2.5(c) 3 (d) 3.5

38. Pipe A and B running together can fill a cistern in 6 minutes.If B takes 5 minutes more than A to fill the cistern then thetimes in which A and B will fill the cistern separately will be,respectively:(a) 15 min, 20 min (b) 15 min, 10 min(c) 10 min, 15 min (d) 25 min, 20 min

39. Pipes A and B can fill a tank in 5 and 6 hours respectively.Pipe C can empty it in 12 hours. If all the three pipes areopened together, then the tank will be filled in :

(a)131 hours17 (b)

82 hours11

(c)93 hours

17(d)

14 hours2

40. Two taps can fill a tank in 12 and 18 minutes respectively.Both are kept open for 2 minutes and the first is turned off.In how many minutes more will the tank be filled ?(a) 15 min. (b) 20 min.(c) 11 min. (d) 13 min.

41. One fill pipe A is 3 times faster than second fill pipe B andtakes 10 minutes less time to fill a cistern than B takes. Findwhen the cistern will be full if fill pipe B is only opened.(a) 20 min (b) 18 min(c) 15 min (d) 10 min

42. Two pipes can fill a cistern in 14 and 16 hours respectively.The pipes are opened simultaneously and it is found that

due to leakage in the bottom, 32 minutes extra are taken forthe cistern to be filled up. If the cistern is full, in what timewould the leak empty it ?(a) 110 hr (b) 112 hr(c) 115 hr (d) 100 hr

43. Two pipes A and B can fill a cistern in 10 and 15 minutesrespectively. Both fill pipes are opened together, but at theend of 3 minutes, ‘B’ is turned off. How much time will thecistern take to fill ?(a) 6 min (b) 8 min(c) 10 min (d) 12 min

44. A cistern has two taps which fill it in 12 minutes and 15minutes respectively. There is also a waste pipe in the cistern.When all the three are opened, the empty cistern is full in 20minutes. How long will the waste pipe take to empty the fullcistern ?(a) 10 min (b) 12 min(c) 15 min (d) None of these

45. Two pipes A and B can fill a tank in 15 and 12 hours

respectively. Pipe B alone is kept open for 34

of time and

both pipes are kept open for remaining time. In how manyhours, the tank will be full ?(a) 18 h (b) 20 h(c) 10 h (d) 13.5 h

46. A pipe can fill a tank in 15 minutes and another one in 10minutes. A third pipe can empty the tank in 5 minutes. Thefirst two pipes are kept open for 4 minutes in the beginningand then the third pipe is also opened. In what time will thetank be empited ?(a) 35 min (b) 15 min(c) 20 min (d) Cannot be emptied

47. Two fill pipes A and B can fill a cistern in 12 and 16 minutesrespectively. Both fill pipes are opened together, but 4minutes before the cistern is full, one pipe A is closed. Howmuch time will the cistern take to fill ?

(a)19 min.7 (b)

13 min.3

(c) 5 min. (d) 3 min.48. Two fill taps A and B can separately fill a cistern in 45 and 40

minutes respectively. They started to fill a cistern togetherbut tap A is turned off after few minutes and tap B fills therest part of cistern in 23 minutes. After how many minutes,was tap A turned-off ?(a) 9 min (b) 10 min(c) 12 min (d) None of these

49. Three fill pipes A, B and C can fill separately a cistern in 3,4 and 6 minutes respectively. A was opened first. After 1minute, B was opened and after 2 minutes from the start ofA, C was also opened. Find the time when the cistern will befull ?

(a)12 min9 (b)

14 min2

(c)33 min4

(d) None of these

12


50. A tap can fill a tank in 16 minutes and another can empty itin 8 minutes. If the tank is already 1/2 full and both the tapsare opened together, will the tank be filled or emptied? Howlong will it take before the tank is either filled or emptiedcompletely as the case may be ?(a) Emptied; 16 min (b) Filled; 8 min(c) Emptied; 8 min (d) Filled; 12 min

51. A pump can be operated both for filling a tank and foremptying it. The capacity of tank is 2400 m3. The emptyingcapacity of the pump is 10 m3 per minute higher than itsfilling capacity. Consequently, the pump needs 8 minutesless to empty the tank to fill it. Find the filling capacity ofpump.(a) 50 m3/min (b) 60 m3/min(c) 58 m3/min (d) None of these

52. A cistern has three pipes, A, B and C. The pipes A and B canfill it in 4 and 5 hours respectively and C can empty it in 2hours. If the pipes are opened in order at 1, 2 and 3 a.m.respectively, when will the cistern be empty ?(a) 3 p.m. (b) 4 p.m.(c) 5 p.m. (d) 6 p.m.

53. A tank is filled in 5 hours by three pipes A, B and C. Thepipe C is twice as fast as B and B is twice as fast as A. Howmuch time will pipe A alone take to fill the tank ?(a) 20 hrs (b) 25 hrs(c) 35 hrs (d) Cannot be determind

54. Two pipes A and B can fill a tank in 15 hours and 20 hoursrespectively while a third pipe C can empty the full tank in25 hours. All the three pipes are opened in the begining.After 10 hours, C is closed. In how much time, will the tankbe full ?(a) 12 hrs (b) 13 hrs(c) 16 hrs (d) 18 hrs

55. Three taps A, B and C can fill a tank in 12, 15 and 20 hoursrespectively. If A is open all the time and B and C are openfor one hour each alternately, then the tank will be full in :

(a) 6 hrs. (b)2

6 hrs.3

(c) 7 hrs. (d)17 hrs.2

56. Two pipes can fill a tank in 20 and 24 minutes respectivelyand a waste pipe can empty 3 gallons per minute. All thethree pipes working together can fill the tank in 15 minutes.The capacity of the tank is :(a) 60 gallons (b) 100 gallons(c) 120 gallons (d) 180 gallons

57. A hot pipe takes 3 minutes longer to fill a tank than the coldpipe. Together they take 6 minutes 40 seconds. Time takenby the cold pipe alone to fill the tank is :(a) 6 minutes (b) 18 minutes(c) 9 minutes (d) 12 minutes

58. Water flows at 3 metres per sec through a pipe of radius 4cm. How many hours will it take to fill a tank 40 metres long,30 metres broad and 8 metres deep, if the pipe remains full?(a) 176.6 hours (b) 120 hour(c) 135.5 hours (d) None of these

59. 4 pipes each of 3 cm diameter are to be replaced by a singlepipe discharging the same quantity of water. What shouldbe the diameter of the single pipe, if the speed of water isthe same.(a) 2 cm (b) 4 cm(c) 6 cm (d) 8 cm

60. A ship 55 kms. from the shore springs a leak which admits 2tones of water in 6 min ; 80 tones would suffer to sink her,but the pumps can throw out 12 tones an hour. Find theaverage rate of sailing that she may just reach the shore asshe begins to sink.(a) 5.5 km/h (b) 6.5 km/h(c) 7.5 km/h (d) 8.5 km/h

61. Two pipes A and B can fill a tank in 24 minutes and 32minutes respectively. If both the pipes are openedsimultaneously, after how much time should B be closedso that the tank is full in 18 minutes?(a) 6 min. (b) 8 min.(c) 12 min. (d) 14 min.

62. A can build up a wall in 8 days while B can break it in 3 days.A has worked for 4 days and then B joined to work with Afor another 2 days only. In how many days will A alonebuild up the remaining part of wall?

(a)1133 days (b)

173 days

(c)163 days (d) 7 days

63. A group of men decided to do a job in 4 days. But since20 men dropped out every day, the job completed at theend of the 7th day. How many men were there at thebeginning?(a) 240 (b) 140(c) 280 (d) 150

64. One man and six women working together can do a jobin 10 days. The same job is done by two men in 'p' daysand by eight women in p + 5 days. By what percentageis the efficiency of a man greater than that of a woman?(a) 300% (b) 500%(c) 600% (d) 700%

65. The total number of men, women and children working ina factory is 18. They earn ` 4000 in a day. If the sum ofthe wages of all men, all women and all children is in theratio of 18 : 10 : 12 and if the wages of an individual man,woman and child is in the ratio 6 : 5 : 3, then how mucha woman earn in a day?(a) ` 400 (b) ` 250(c) ` 150 (d) ` 120

66. A can do a job in 3 days less time than B. A works at it alonefor 4 days and then B takes over and completes it. Ifaltogether 14 days were required to finish the job, then inhow many days would each of them take alone to finish it?(a) 17 days, 20 days (b) 12 days, 15 days(c) 13 days, 16 days (d) None of these

13


67. 3 small pumps and a large pump are filling a tank. Each of thethree small pumps works at 2/3rd the rate of the large pump.If all 4 pumps work at the same time, they should fill the tankin what fraction of the time that it would have taken the largepump alone?(a) 4/7 (b) 1/3(c) 2/3 (d) 3/4

68. The Bubna dam has four inlets. Through the first threeinlets, the dam can be filled in 12 minutes; through thesecond, the third and the fourth inlet, it can be filled in 15minutes; and through the first and the fourth inlet, in 20minutes. How much time will it take all the four inlets to fillup the dam?(a) 8 min (b) 10 min(c) 12 min (d) None of these

69. Seventy-five men are employed to lay down a railway linein 3 months. Due to certain emergency conditions, the workwas to be finished in 18 days. How many more men shouldbe employed to complete the work in the desired time ?(a) 300 (b) 325(c) 350 (d) 375

70. A, B and C together can do a piece of work in 40 days. Afterworking with B and C for 16 days, A leaves and then B andC complete the remaining work in 40 days more. A alonecould do the work in(a) 80 days (b) 90 days(c) 100 days (d) 120 days

71. Three pipes A, B and C can fill a tank in 6 hours. Afterworking it together for 2 hours, C is closed and A and B canfill the remaining part in 7 hours. The number of hours takenby C alone to fill the tank is(a) 10 (b) 12(c) 14 (d) 16

72. Pratibha is thrice as efficient as Sonia and is therefore ableto finish a piece of work in 60 days less than Sonia. Pratibhaand Sonia can individually complete the work respectivelyin(a) 30, 60 days (b) 60, 90 days(c) 30, 90 days (d) 40, 120 days

73. A can do a certain work in the same time in which B and Ctogether can do it. If A and B together could do it in 10 daysand C alone in 50 days, then B alone could do it in(a) 15 days (b) 20 days(c) 25 days (d) 30 days

74. A can do a piece of work in 6 days. B can do the same workin 15 days. How long would both of them take to do thesame work ?(a) 2 days (b) 4 days(c) 6 days (d) 8 days

75. 12 men construct 1.5 km of road in 7 days. 28 men willconstruct 12 km of roads in(a) 20 days (b) 24 days(c) 28 days (d) 38 days

76. X and Y can do a piece of work in 30 days. They worktogether for 6 days and then X quits and Y finishes the workin 32 more days. In how many days can Y do the piece ofwork alone?(a) 30 days (b) 32 days(c) 34 days (d) 40 days

77. 40 men can finish a piece of work in 60 days. After somedays, 10 men leave the work so that the work is finished in70 days. The number of days after which 10 men left thework is(a) 20 days (b) 25 days(c) 30 days (d) 40 days

14


1. (b) A’s 1 day’s work = 1

10 and B’s 1 day’s work =

115

\ (A + B)’s 1 day’s work = 1 1 1 .

10 15 6æ ö+ =ç ÷è ø

So, both together will finish the work in 6 days.

2. (d) (Man + Son)’s one day’s work = 18

Man’s one day’s work = 1

10

Þ Son’s one day’s work = 1 1 18 10 40

- =

\ Son can do it in 40 days.

3. (a) A’s 1 day’s work = 1

18 and B’s 1 day’s work =

1 .9

\ (A + B)’s 1 day’s work = 1 1 1 .18 9 6

æ ö+ =ç ÷è ø

4. (a) In an hour, George and Sonia together can copy1 1 76 8 24

+ = of a 50-page manuscript.

i.e. In an hour they together can copy 748

of the

100-page manuscript.i.e. They together can copy a 100-page manuscript in487

hours, i.e. 667

hours.

5. (b) (A + B)’s 5 days’ work

= 5 1 1 45 925 20 100 20

æ ö+ = =ç ÷è ø

Remaining work = 2011

2091 =÷

øö

çèæ -

1120

of the work would be finished by B in

.days11

2012011

=

6. (b) Let the man alone do the work in x days.Then, the woman alone do the work in 2x days.

Their one day’s work = 18 th part of whole work

i.e. 1 1 1x 2x 8

+ =

x 12 daysÞ =\ man takes 12 days and woman 2x = 24 days.

7. (c) A’s one day’s work = 1 th work

16

B’s one day’s work 1 th work

12=

Let B has worked alone = x days. Then,A’s amount of work + B’s amount of work = 1

1 14 (x 4) 116 12

æ ö æ öÞ + + =ç ÷ ç ÷è ø è ø

1 x 4 31 x 12 44 12 4

+Þ + = Þ = ´ -

x 5 daysÞ =8. (a) 50 men complete 0.4 work in 25 days.

Applying the work rule, 122211 wdmwdm ´´=´´we have,

4.025m6.02550 2 ´´=´´

or m2 = men754.025

6.02550=

´´´

Number of additional men required = (75 – 50) = 259. (b) Ratio of times taken by A and B = 100 : 130 = 10 : 13.

Suppose B takes x days to do the work.

1 (b) 11 (b) 21 (b) 31 (b) 41 (c) 51 (a) 61 (b) 71 (c)2 (d) 12 (d) 22 (b) 32 (c) 42 (b) 52 (c) 62 (b) 72 (c)3 (a) 13 (c) 23 (a) 33 (a) 43 (b) 53 (c) 63 (b) 73 (c)4 (a) 14 (b) 24 (b) 34 (a) 44 (a) 54 (a) 64 (b) 74 (c)5 (b) 15 (b) 25 (a) 35 (a) 45 (c) 55 (c) 65 (b) 75 (b)6 (b) 16 (d) 26 (a) 36 (a) 46 (c) 56 (c) 66 (b) 76 (d)7 (c) 17 (a) 27 (a) 37 (c) 47 (a) 57 (d) 67 (b) 77 (c)8 (a) 18 (d) 28 (b) 38 (c) 48 (a) 58 (a) 68 (b)9 (b) 19 (d) 29 (a) 39 (c) 49 (a) 59 (c) 69 (a)

10 (c) 20 (c) 30 (c) 40 (d) 50 (c) 60 (a) 70 (c)

ANSWER KEY

15


Then, 10 : 13 : : 23 : x 23 13 299x x .10 10´æ öÞ = Þ =ç ÷

è ø

A’s 1 day’s work = 1 ;23

B’s 1 days work = 10 .299

(A + B)’s 1 day’s work = 1 10 23 1 .23 299 299 13

æ ö+ = =ç ÷è ø

\ A and B together can complete the job in 13 days.Alternate Method:

A and B together complete work in 1.3×23

=13 days1.3+1

10. (c) Let C completes the work in x days.

Work done by (A + B) in 1 day = 101

Work done by (B +C) in 1 day = 181

A’s 5 days’ work + B’s 10 days’ work + C’s 15 days’work = 1or (A + B)’s 5 days’ work + (B + C)’s 5 days’ work

+ C’s 10 days’ work = 1

or 5 5 10 110 18 x

+ + = or x = 45 days

11. (b) 1 1 1m d t´ ´ = 2 2 2m d t´ ´24 × 10 × 8 = m2 × 6 × 10

Þ men32106

81024m2 =´

´´=

12. (d) In 1 day, work done by 12 men = 1

18

In 6 days, work done by 12 men = 31

186

=

Remaining work = 32

Now, 122211 wdmwdm ´´=´´

or 1d16321812 2 ´´=´´

or days916

2184d2 =´´

=

13. (c) 1 minute’s work of both the punctures = 1 1 5 .9 6 18

æ ö+ =ç ÷è ø

So, both the punctures will make the tyre flat in18 33 min.5 5

=

14. (b) Man’s two day’s work = 12 th work20

´ 1 th work

10=

Woman’s two days’s work1 12 th work th work30 15

= ´ =

Boy’s two day’s work 12 th60

= ´ work = 1 th work30

Now, let 2 men, 8 women and x boys can completework in 2 days. Then ,2 men’s work +8 women’s work + x boy’s work =1

1 1 12 8 x 110 15 30

æ ö æ ö æ ö+ + =ç ÷ ç ÷ ç ÷è ø è ø è ø

1 8x 1 305 15

æ öÞ = - - ´ç ÷è ø

Þ x = 8 boys

15. (b) B alone can do a work in 20 hours.

\ A alone can do 32

of the work in 20 hours.

i.e., A alone can do the same work in 403 hours

\ (A + B)’s one hour’s work 3 1 5 140 20 40 8

= + = =

Þ A and B together can finish the whole work in8 hours.

16. (d) (P + Q + R)’s 1 hour’s work = 1 1 1 37 .8 10 12 120

æ ö+ + =ç ÷è ø

Work done by P, Q and R in 2 hours = 37 372 .

120 60æ ö´ =ç ÷è ø

Remaining work = 37 231 .60 60

æ ö- =ç ÷è ø

(Q + R)’s 1 hour’s work = 1 1 11 .

10 12 60æ ö+ =ç ÷è ø

Now, 1160

work is done by Q and R in 1 hour..

So, 2360

work will be done by Q and R in

60 23 2311 60 11

æ ö´ =ç ÷è ø

hours » 2 hours.

So, the work will be finished approximately 2 hoursafter 11 a.m., i.e., around 1 p.m.

17. (a) Work done by A and B in 5 days = 655

151

101

=´÷øö

çèæ +

Work remaining = 61

651 =-

\ C alone can do the work in 6 × 2 = 12 days

Ratio of their share work = 1:2:3122:

155:

105

=

Share of wages = ` 225, ̀ 150, ̀ 75.18. (d) Let if both A and B work together, they take x days.

\ (A + B)’s 1 days’s work 1 th workx

= .

A’s 1 day’s work 1 th work

x 8=

+.

16


B’s 1 day’s work 1 th workx 9 / 2

=+

.

Now, 1 2 1

x 8 2x 9 x+ =

+ +

x(2x 9 2x 16) (x 8)(2x 9)Þ + + + = + +Þ 4x2 + 25x = 2x2 + 25x + 72Þ x2 = 36 Þ x = 6 daysAlternate Method:

A and B together finish the work in 98× =6 days2

19. (d) 1 man’s 1 day’s work = 1 .

100

(10 men + 15 women)’s 1 day’s work = 1.6

15 women’s 1 day’s work

= 1 10 1 1 1 .6 100 6 10 15

æ ö æ ö- = - =ç ÷ ç ÷è ø è ø

\ 1 woman’s 1 day’s work = 1 .

225\ 1 woman alone can complete the work in 225 days.

20. (c) In 8 days, Anil does 1 rd work3

= .

\ in 1 day, he does 1 th work24

= .

\ Rakesh’s one day’s work = 60% of 124

= 1 th work40

.

Remaining work 1 213 3

= - =

(Anil and Rakesh)’s one day’s work

= 1 124 40

+1 th work

15=

Now, 1 th15

work is done by them in one day..

\ 23 rd work is done by them in

215 10days3

´ =

21. (b) A’s one day’s work = 132

B's one day’s work = 120

(B + C)'s one day’s work = 1

12

\ C's one day’s work = 1 1 1

12 20 30- =

D's one day’s work = 124

\ (A + B + C + D)’s one day’s work

= 1 1 1 132 20 30 24

+ + + 75 120 80 100

2400+ + +

=

= 375 15 52400 96 32

= =

\ Out of 532

of work done, 130

of the work is done

by C.Þ Out of ̀ 25 paid for the work, C will receive

` 1/30 25,5 / 32

´ i.e. 1 32 25,30 5

´ ´ i.e. ̀ 163

22. (b) A’s one day’s work = 1 th work

15.

B’s one day’s work 1 th work

10= .

(A + B)’s one day’s work 1 1 1 th15 10 6

= + = work.

Let A left after x days.

\ (A + B)’s x days’ work x th work6

= .

Remaining work x 6 x1 th6 6

-= - = work.

Now, in 5 days, work done by 6 xB th work

6-

= .

\ In 1 day work done by B 6 x th work30-

=

and 6 x 130 10-

=

\ x 3 days=23. (a) Let Suresh undertakes a tour of x days.

Then, expenses for each day = x

360

Now, 3x

3604x

360-=

+

or 1 1360

x x 4æ ö-ç ÷è ø+

= 3

or 0480x4x2 =-+ or x = – 24 or x = 20Since, 24x -¹ we have x = 20

24. (b) Let B can finish the work in x days.Then A can finish the work in (x – 3) days.

B’s one day’s work 1 th workx

=

A’s one day’s work 1 thx 3

=-

work

A’s 4 days’ work = 4 th work

x 3-

17


Remaining work 4 x 71 th workx 3 x 3

-= - =

- -The remaining work done by B in 14 – 4 = 10 days.

Now, in 10 days, work done by x 7Bx 3

-=

- th work

\ In 1 day, work done by B = 1 x 7 th10 x 3

-æ öç ÷-è ø

work

and 1 x 7 1

10 x 3 x-æ ö =ç ÷-è ø

x 15daysÞ =\ B will finish in 15 days and A will finish in 12 days

25. (a) (A + B)’s one day’s work 1 th work5

=

Let A can do job in x days. Then,

A’s one day’s work 1 th workx

=

and B’s one day’s work 1 1–5 x

=x 5 th work5x-

=

Now , (2A)’s work + 13æ öç ÷è ø

B’s work = 13

rd work

2 1 x 5 1x 3 5x 3

-æ öÞ + =ç ÷è ø

25 1x 6 days4 4

Þ = =

26. (a) 1 man’s 1 day’s work = 1 ;48

1 woman’s 1 day’s work = 1 .60

6 men’s 2 day’s work = 6 12 .48 4

æ ö´ =ç ÷è ø

Remaining work = 1 31 .4 4

æ ö- =ç ÷è ø

Now, 160

work is done in 1 day by 1 woman.

So, 34

work will be done in 3 days by

3 1604 3

æ ö´ ´ç ÷è ø

= 15 women.

27. (a) Let 1 man’s 1 day’s work = x and1 boy’s 1 day’s work = y.

Then, 6x + 8y = 1

10 and 26x + 48y =

1 .2

Solving these two equations, we get :1 1x and y .

100 200= =

\ (15 men + 20 boys)’s 1 day’s work

= 15 20 1 .100 200 4

æ ö+ =ç ÷è ø

\ 15 men and 20 boys can do the work in 4 days.28. (b) Let x additional men employed.

117 men were supposed to finish the whole work in46 × 8 = 368 hours.

But 117 men completed 47

of the work in 33 × 8

= 264 hours\ 117 men could complete the work in 462 hours.

Now (117 + x) men are supposed to do 37

of the work,

working 9 hours a day, in 13 × 9 = 117 hours, so as tofinish the work in time.i.e. (117 + x) men are supposed to complete the whole

work in 71173

´ = 273 hours.

\ (117 + x) × 273 = 117 × 462Þ (117 + x) × 7 = 3 × 462Þ x + 117 = 3 × 66 = 198 Þ x = 81\ Required number of additional men to finish thework in time = 81.

29. (b) Let Sunil finishes the job in x hours.

Then, Ramesh will finish the job in 2x hours.

We have, 6x32xx =Þ=-

Therefore, Sunil finishes the job in 6 hours and Rameshin 3 hours.

Work done by both of them in 1 hour = 21

31

61

=+

They together finish the piece of work in 2 hours.30. (c) Suppose that X men must be discharged at the end of

the 18th day.100 × 10 + 150 × 1 + 200 × 7 + (200 – X) × 5 = 100 × 305X = 550 Þ X = 110 men

31. (b) Given 12 men º 15 women º 18 boys\ 1 Man º 1.5 boys, 1 woman = 6/5 boys.Now, 5W + 6B = 12B.Required answer is calculated as follows :Total no. of boys reqd. = 18 × [(15/16) × (8/9)]

= 15 boysThe number of boys already present = 12.Hence, 3 boys more required.But 3 boys = 2 men.So, 2 men are required.

32. (c) Men Women ChildrenWork 3 2 1Numbers 20 30 36Ratio of wages = (3 × 20) : (2 × 20) : (1 × 36) = 5 : 5 : 3

Total wages of men = 5 780 300

13´ =`

\ Wages of a man = ̀ 15Similarly, wages of woman = ̀ 10and wages of child = ` 5

18


Total waves of 15 men, 21 women and 30 children= 15 × 15 + 21 × 10 + 30 × 5 = 585

Total wages for 2 weeks = ̀ 117033. (a) Let 1 man’s 1 days’ work= x & 1 boy’s 1 day’s work = y

Then, 2x + 3y = 1

10 and 3x + 2y =

18

Solving, we get : 7x

200= and

1y100

=

\ (2 men + 1 boy)’s 1 day’s work

= 7 1 16 2

2 1200 100 200 25

æ ö´ + ´ = =ç ÷è ø

So, 2 men and 1 boy together can finish the work in

1122

days.

34. (a) Ratio of time taken by A and B = 160 : 100 = 8 : 5

Suppose, B alone takes x days to do the jobthen, 8 : 5 : : 12 : x8x = 5 × 12

x = 5 12 17

8 2´

= days.

35. (a) Let 1 man’s 1 days work = x1 boy’s 1 day’s work = y

12x + 16y = 15

13x + 24y = 14

Solving these two equation we get,

1 1,100 200

x y= =

Required ratio = 2 : 136. (a) If x complete a work in x days. y will do the same task

in 3x days.3x – x = 40Þ x = 20y will finish the task in 60 days(x + y)’s 1 days work

= 1 1 120 60 15

+ =

Both of them will complete the work in 15 days.

37. (c) Part filled by (A + B + C) in 1 hour = 1 1 1 15 10 30 3

æ ö+ + =ç ÷è ø

.

\ All the three pipes together will fill the tank in3 hours.

38. (c) Let pipe A fills the cistern in x minutes.Therefore, pipe B will fill the cistern in (x + 5) minutes.

Now, 61

5x1

x1

=+

+ Þ x = 10

Thus, the pipes A and B can fill the cistern in10 minutes and 15 minutes, respectively

39. (c) Net part filled in 1 hour = 1 1 1 17–5 6 12 60

æ ö+ =ç ÷è ø

.

\ The tank will be full in 6017 hrs i.e.,

9317 hrs.

40. (d) Part filled by first tap in one min 1 th12

=

Part filled by second tap in one min 1 th18

=

Now, 1 12 unfilled part = 112 18

é ù+ +ê úë û

13unfilled part = th18

Þ

Q 1 th18

part of tank is filled by second tap in 1min.

\ 13 th18

part of tank is filled by second tap in 1 min.

1318 min18

= ´ = 13 min.

41. (c) Let B can fill the cistern in x min. Then,

then A can fill the cistern in x min3

Given xx 10 x 15min3

- = Þ =

42. (b) Cistern filled by both pipes in one hour

=1 1 15 th

14 16 112+ =

\ Both pipes filled the cistern in 112 hrs15

.

Now, due to leakage both pipes filled the cistern in

112 32 8hrs.15 60

+ =

\ Due to leakage, filled part in one hour 18

=

\ part of cistern emptied, due to leakage in one hour

15 1 1 th112 8 112

= - =

\ In 112 hr, the leakage would empty the cistern.

43. (b) In one min, (A + B) fill the cistern 1 1 1 th10 15 6

= + =

In 3 min, (A + B) fill the cistern = 3 1 th6 2

=

Remaining part 1 112 2

= - =

Q 1 th

10 part filled by A in one min.

19


\ 12

nd part filled by A in 110 5min2

´ = .

\ Total time = 3 + 5 = 8 min.44. (a) Work done by the waste pipe in 1 minutes

= 1 1 1 120 12 15 10

æ ö- + = -ç ÷è ø

[–ve sign means emptying]

\ Waste pipe will empty the full cistern in 10 minutes.45. (c) Let the required time be x hours, then

1 3 1 3 1 3x x x x x 1

12 4 15 4 12 4æ ö æ ö æ ö+ - + - =ç ÷ ç ÷ ç ÷è ø è ø è ø

Þx x x 1

16 60 48+ + =

Þ x = 10 hours46. (c) Proportion of the volume of the tank filled by both the

pipes in 4 min = ÷øö

çèæ +

101

1514 =

32

rd of the tank.

Volume of the tank filled by all the pipes working

together = 30

151

101

151 -

=-+

i.e. 301 tank is emptied in 1 min.

\ 32

rd of the tank can be emptied in 3302´ = 20 min

47. (a) Let cistern will be full in x min. Then,part filled by B in x min + part filled by A in (x – 4)min = 1

x x 4 116 12

-Þ + =

64 1x 9 hours.7 7

Þ = =

48. (a) Let A was turned off after x min. Then,cistern filled by A in x min + cistern filled byB in (x + 23) min = 1

x x 23 145 40

+Þ + =

17x 207 360 x 9 min.Þ + = Þ =49. (a) Let cistern will be full in x min. Then,

part filled by A in x min + part filled by B in (x – 1) min+ part filled by C in (x – 2)min = 1

x x 1 x 2 13 4 6

- -Þ + + =

19 19x 19 x 2 min9 9

Þ = Þ = =

50. (c) If both the pumps are opened together, then the tankwill be emptied because the working efficiency of pumpempting is more than that of the pump filling it. Thus in1 min net proportion of the volume of tank filled

= 161

161

81

=÷øö

çèæ -

or the tank will be emptied in 16 min

Þ 21

tank will be emptied in 8 min.

51. (a) Let the filling capacity of pump be x m3/min.Then, emptying capacity of pump = (x + 10) m3 /min.

\ 2400 2400 8

x x 10- =

+Þ x2 + 10 x – 3000 = 0Þ (x – 50)(x + 60) = 0 Þ x = 50 m3/min.

52. (c) Hint : Let the time be t hours after 1 a.m.

\ ( ) ( )t 1 t 2t 0

4 5 2- -

+ - =

1 14 5 4 5t t t

+ - = -

Þ t = 1616 hours from 1 a.m is 5 p.m

53. (c) Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take x2

and x4

hours

respectively to fill the tank.

1 2 4 1 7 1 x 35 hrs.x x x 5 x 5

\ + + = Þ = Þ =

54. (a) Part filled in 10 hours = 1 1 1 2310 .15 20 25 30

æ ö+ - =ç ÷è ø

Remaining part = 23 7130 30

æ ö- =ç ÷è ø

.

(A + B)'s 1 hour’s work = 1 1 7

15 20 60æ ö+ =ç ÷è ø

.

7 7: ::1:x60 30 or

7 60x 1 2 hours.30 7

æ ö= ´ ´ =ç ÷è ø

\ The tank will be full in (10 + 2) hrs = 12 hrs.

55. (c) (A + B)’s 1 hour’s work = 1 1 9 3

12 15 60 20æ ö+ = =ç ÷è ø

(A + C)'s 1 hour’s work = 1 1 8 2

12 20 60 15æ ö+ = =ç ÷è ø

Part filled in 2 hrs = 3 2 17

20 15 60æ ö+ =ç ÷è ø

Part filled in 6 hrs = 17 17360 20

æ ö´ =ç ÷è ø

Remaining part = 17 3120 20

æ ö- =ç ÷è ø

Now, it is the turn of A and B and 320 part is filled by

20


A and B in 1 hour.\ Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

56. (c) Work done by the waste pipe in 1 minute

= 1 1 1 1 11 1 .

15 20 24 15 120 40æ ö æ ö- + = - = -ç ÷ ç ÷è ø è ø

[–ve sign means emptying]

\ Volume of 140 part = 3 gallons.

Volume of whole = (3 × 40) gallons = 120 gallons.57. (d) Pipe 1 (Hot) ® 3 + X, X ® Pipe 2 (cold)

Together X(X 3) 26 min.2X 3 3

+=

+

= X(X 3) 2 206 min.2X 3 3 3

+= =

+40X + 60 = 3X (X + 3)Þ 40X + 60 = 3X2 + 9XÞ 3X2 – 31X – 60 = 0Þ X = 12 minutes

58. (a) Radius of the pipe (r) = 4 cm = 0.04 meterVolume of water flowing out per sec= pr2 × rate of flow

= 222 0.04 3 cu meters 0.0151cubic m7

´ ´ =

Time taken to fill the tank = 40 × 30 × 8

0.0151 sec

= 40 30 8 1 hours 176.6 hours

0.01 3600´ ´

´ =

59. (c) Let h be the length of water column discharged in 1hour or 1 minute.Volume discharged by the 4 pipes = Volume dischargedby the single pipe.4 × p × (1.5)2 × h = p × (r)2 × h\ r2 = 9 \ r = 3, Diameter = 6 cm.

60. (a) Rate of admission of water

= 26 tonnes / min =

13 tonnes/ min

Rate of pumping out of water

= 1260 tonnes/min =

15 tonnes/ min.

Rate of accumulation = 2 126 60

æ ö-ç ÷è ø

tonnes / min.

Time to accumulate 80 tonnes of water

= Amount of waterAccumulation rate

= 80

1 13 5

æ ö-ç ÷è ø

= 600 min. = 10 hours

\ Average sailing rate so as avoid sinking

= Distance

Time =

5510 km/ h = 5.5 km/h

61. (b) Let B be closed after x minutes. Then, part filled by(A + B) in x min. + part filled by A in (18 – x) min = 1.

\1 1 1

(18 )24 32 24

x xæ ö+ + - ´ç ÷è ø = 1

or,7 18 196 24

x x-+ = or, 7x + 4(18 – x) = 96

or, 3x = 24 \ x = 8.So, B should be closed after 8 min.Direct Formula:

Pipe B should be closed after 181 32

24æ ö- ´ç ÷è ø

= 8 min.

62. (b) A’s one day’s work 1 th work8

=

B’s one day’s work 1 rd work3

=

\ A’s 4 day’s work = 1 14 nd work8 2

´ =

\ In next two days, total wall 1 1 12 22 8 3

æ ö æ ö= + -ç ÷ ç ÷è ø è ø

1 th wall

12=

Remaining wall 1 111 th

12 12= - =

Now, 1 th8

wall is built up by A in one day..

\ 11 th12

wall is built up by A in 11812

´ = 17 days3

.

63. (b) Go through option140 × 4 = (140 + 120 + 100 + ... + 20)560 = 560Alternatively: Let n be the initial number of workerthenn × 4 = n + (n – 20) + (n – 40) + ... + (n – 120)4n = 7n – 420Þ 3n = 420Þ n = 140 workers

64. (b) Let the work (in units) done by a man and a womanin one day be M and W respectively. Total work (inunits) = 10(M + 6W) = 10M + 60W

Þ10 60 10 60

8 2M W M W

W M+ +

- = 5

5 304M WW M

- = 52

On putting MW = x, we get

5 30 54 2x

x- = Þ x = 6 or

MW = 6

\ The efficiency of a man is greater than that ofa woman by 500%.

65. (b) Ratio of number of men, women and children

21


= 18 10 12: :6 5 3

= 3x : 2x : 4x

\ (3x + 2x + 4x) = 18\ x = 2Therefore, number of women = 4

Share of all women = 10 400040

´ = ` 1000

(Q 18 + 10 + 12 = 40)

\ Share of each woman = 1000

4 ` 250

66. (b) Let B can finish the work in x days.Then A can finish the work in (x – 3) days.

B’s one day’s work 1 th workx

=

A’s one day’s work 1 thx 3

=-

work

A’s 4 days’ work = 4 th work

x 3-

Remaining work 4 x 71 th work

x 3 x 3-

= - =- -

The remaining work done by B in 14 – 4 = 10 days.

Now, in 10 days, work done by x 7Bx 3

-=

- th work

\ In 1 day, work done by B = 1 x 7 th

10 x 3-æ ö

ç ÷-è ø work

and 1 x 7 1

10 x 3 x-æ ö =ç ÷-è ø

x 15daysÞ =\ B will finish in 15 days and A will finish in 12 days

67. (b) Suppose large pump takes t hours to fill a tank

\ 1 hour work of large pump fills 1t

= part

1 hour work of each small pump fills 1 23t

= ´

1 hour work of all 4 pumps fill1 23

3t t= + ´ 3

t=

Therefore, 3t

part is filled by all 4 pumps in 1 hour

\ Whole tank would be filled in 13 3t t h´ = this is

1/3 of the time taken by large pump i.e., t hour68. (b) Let the inlets be A, B, C and D.

A + B + C = 8.33 %B + C + D = 6.66%A + D = 5%

Thus 2A + 2B + 2C + 2D = 20%and A + B + C + D = 10%® 10 minutes would be required to fill the tankcompletely.

69. (a) More the no. of men less time they take to completework.Let x men are added.

75 1875 90x

=+

(Inverse Proportion)

75 175 5x

=+

375 – 75 = xx = 300

70. (c) (A + B + C)'s 1 day's work = th1

40æ öç ÷è ø part of whole

work

(A + B + C)'s 16 days work = 16 240 5

= of whole work

(B + C) completes remaining work in 40 days (B + C)

completes th3

5æ öç ÷è ø part of work in 40 days.

(B + C) completes whole work in 40 5 200

3 3´

= days.

1 1 1 1A B C 40

+ + =

1 3 1A 200 40

+ =

1 1 1 4A 40 200 200

= - =

1 1A 50

=

A alone can complete whole work in 50 days.

71. (c)1 1 1 1A B C 6

+ + =

(A + B + C) can do 2 16 3

= part of work in 2 days.

Remaining work = 1 2

13 3

- =

In one hour (A + B) can do 2

3 7´ part of work

1 1 1 1C 6 B C

æ ö= - +ç ÷è ø

1 1 2 3C 6 21 42

= - =

C = 14 hours72. (c) Let Pratibha can finish the work in x days then, Sonia

can finish the same work in 3x daysAccording to question3x – x = 602x = 60 Þ x = 30Pratibha and Sonia can individually complete the workin 30 days and 90 days respectively.

73. (c) (A + B)’s 1 day’s work 1 ;

10= C’s 1day’s work

150

=

22


(A + B + C)’s 1 day’s work 1 1 6 3

10 50 10 25æ ö= + = =ç ÷è ø

...(1)

Also, A’s 1 day’s work = (B + C)’s 1 day’s work ...(2)

From (1) and (2), we get : 2 × (A’s 1 day’s work) 325

=

Þ A’s 1 day’s work 350

=

\ B’s 1 day’s work 1 3 2 1–

10 50 50 25æ ö= = =ç ÷è ø

So, B alone could do the work in 25 days.

74. (c) A’s 1 day’s work 1

10= and B’s 1 day’s work

115

=

\ (A + B)’s 1 day’s work 1 1 1

10 15 6æ ö= + =ç ÷è ø

So both together will finish the work in 6 days.75. (b) Let the required number of days be x.

Then, more men, more km (Direct proportion)more days, more km (Direct proportion)menMen 12 : 28

::1.5 :12Days 7 : x

üýþ

\ 12 × 7 × 12 = 28 × x × 1.512 7 12 24

28 1.5x ´ ´

= =´

76. (d) (x + y)’s 6 days’ work = 1 1630 5

æ ö´ =ç ÷è ø

.

Remaining work = 1 415 5

æ ö- =ç ÷è ø

Now, 45

work is done by y in 32 days.

Whole work will be done by y in 5324

æ ö´ç ÷è ø

= 40 days.

23


This chapter is taken from :

time & work pipe and cisterns

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