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A is able to do a piece of work in 10 days and B can do the same work in 15 days. If they can work together for four days, what is the fraction of work left?
It is given that A can do a piece of work in 10 days and B can do the same work in 15 days.
Let the total work be LCM of (10, 15) = 30 units
Amount of work done by A in one day = $$\ \frac{\ 30}{10}$$ = 3 units
Amount of work done by B in one day = $$\ \frac{\ 30}{15}$$ = 2 units
Amount of work done by A and B together in one day = 3+2=5 units
So, together will complete 5*4 units i.e 20 of work in four days.
The amount of work left out = 30-20=10 units
The fraction of work left out = $$\ \frac{\ 10}{30}$$ = $$\ \frac{\ 1}{3}$$
D is the correct answer.
A, B and C can do a work in 6, 8 and 12 days respectively. If they do the work together and earn Rs. 2700, what is the share of C in that amount ?
shares are divided in the ratio of their efficiency.
Let us assume that a work of 24units is to be done
A does 4 units of work every day
B does 3 units of work everyday
C does 2 units of work everyday.
So shares will be divided in the ratio 4:3:2
So C's share will be $$\frac{2}{9}\cdot2700$$ = 600 Rs
Three pipes A, B and C can fill a tank in 12 hours. All the pipes started working together and after 3 hours, C is closed. If A and B can fill the remaining part in 10 hours, then the number of hours taken by C alone to fill the tank is .
Let efficiencies of A,B,C be a,b,c respectively.
Given a+b+c = $$\frac{W}{12}$$ => Total work = 12(a+b+c)
They worked together for 3 hours. So work done in 3 hours = 3(a+b+c)
Remaining= 9(a+b+c) which is filled by a and b only in 10 hours
So $$a+b\ =\frac{\left(9\left(a+b+c\right)\right)}{10}$$
On solving a+b=9c
So W= 12(9c+c)=120c
Time taken by C alone = $$\frac{120c}{c}$$=120 hours
A can do a piece of work in 25 days and B in 20 days. They work together for 5 days and then A goes away. In how many days will B finish the remaining work ?
Let 100 units of work be there.
$$\eta_A=\ \ \frac{\ 100}{25}=4units\ per\ day$$
$$\eta_B=\ \ \frac{\ 100}{20}=5units\ per\ day$$
Amount of work done when they work together for 5 days = $$5\times\ \left(\eta_A+\eta_B\right)\ =\ 5\times\ \left(5+4\right)\ =\ 45\ units$$
Remaining work = 100-45 = 55 units
Time taken by B = $$\ \frac{\ 55}{5}=11\ days$$
A and B together can complete a task in 20 days. B and C together can complete the same task in 30 days. A and C together can complete the same task in 40 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone ?
Let a, b, c be the amount of work done by A, B, C in 1 day and the total work be W units
(a+b) 20 = (b+c) 30 = (a+c) 40 = W
(a+b) 2 = (b+c) 3 = (a+c) 4
2a+2b = 3b+3c
b = 2a-3c
2a+2b = 4a+4c
2b = 2a+4c
2(2a-3c) = 2a+4c
2a=10c
a=5c
Number of days taken by A to complete the work = $$\ \frac{\ W}{a}$$
Number of days taken by C to complete the work = $$\ \frac{\ W}{c}$$
Required ratio = $$\ \frac{\ \ \frac{\ w}{a}}{\ \frac{\ w}{c}}$$
= 1:5
D is the correct answer.
A certain number of people were supposed to complete a work in 24 days. The work, however, took 32 days, since 9 people were absent throughout. How many people were supposed to be working originally?
Let the total work be W units and the number of people employed to do it were 'x'
24*x = 32*(x-9)
3x = 4(x-9)
x=36
$$\therefore$$ Number of people who were supposed to work originally = 36
C is the correct answer.
A pump can be operated both for filling a tank and for emptying it. The capacity of tank is 2400 $$m^3$$ .The emptying capacity of the pump is 10 $$m^3$$ per min higher than its filling capacity. Consequently, the pump needs 8 min less to empty the tank than to fill it. Find the filling capacity of the pump.
Let the filling capacity of the pump be x $$m^3$$/min
Then the emptying capacity of the pump = (x+10) $$m^3$$/min
The time required for filling the tank = $$\ \frac{\ 2400}{x}$$
The time required for emptying the tank = $$\ \frac{\ 2400}{x+10}$$
The pump needs 8 min less to empty the tank than to fill it.
$$\ \frac{\ 2400}{x}$$ - $$\ \frac{\ 2400}{x+10}$$ = 8
$$\ \frac{\ 300}{x}$$ - $$\ \frac{\ 300}{x+10}$$ = 1
300(x+10) - 300(x) = x(x+10)
$$x^2+10x-3000$$ = 0
(x+60)(x-50) = 0
x = 50 or -60
Since x cannot be negative, x = 50
Filling capacity of the pump = 50 $$m^3$$/min
C is the correct answer.
8 men and 4 women together can complete a piece of work in 6 days. The work done by a man in one day is double the work done by a woman in one day. If 8 men and 4 women started working and after 2 days 4 men left and 4 new women joined, in how many more days will the work be completed?
Let the work done by a woman in one day = x units
The work done by a man in one day = 2x units
8 men and 4 women together can complete a piece of work in 6 days.
Total work = (8*2x+4*x)6 = 120x units
Amount of work done by 8 men and 4 women in 2 days = (8*2x+ 4*x)2 = 40x units
The remaining 80x units will be done by 8 women and 4 men
Number of days taken by 8 women to complete 80x units of work = $$\ \frac{\ 80x}{16x}$$
=5 days
A is the correct answer.
B is twice efficient as A and A can do a piece of work in 15 days. A started the work and after a few days B joined him. They completed the work in 11 days, from the starting. For how many days did they work together?
A can do a piece of work in 15 days.
B is twice as efficient as A.
So B can do the same piece of work in 7.5 days.
Assume the total work to be 15 units
A does 1 unit per day and B does 2 units per day.
For the first x days, A worked alone and hence work done per day is
1 * x = x units.
For the next (11 - x) days, A and B worked together. In 1 day, they will do (1 + 2) = 3 units.
So in 11 - x days, they will do 33 - 3x units.
33 - 3x + x = 15
2x = 18 and x = 9.
Thus A worked alone for 9 days the A and B worked together for 2 days.
Four men and three women can do a job in 6 days. When 5 men and 6 women work on the same job, the work gets completed in 4 days. How long will 2 women and 3 men take to do the job?
(4M+ 3W)6 = (5M+6W)4 = W
4M = 6W
2M = 3W
let x be the number of days taken by 2 women and 3 men take to do the job
(4M+3W)6=(2W+3M)x
9W*6=6.5W*x
x=8.3 days
C is the correct answer.
Ram completes 60% of a task in 15 days and then takes the help of Rahim and Rachel. Rahim is 50% as efficient as Ram is and Rachel is 50% as efficient as Rahim is. In how many more days will they complete the work?
Let the total work be 25x units
Ram completes 60% of the task in 15 days.
Ram's efficiency = x units per day
Rahim is 50% as efficient as Ram is and Rachel is 50% as efficient as Rahim is
Efficiency of Rahim = 0.5x units per day
Efficiency of Rahim = 0.25x units per day
Ram will come 15x units in 15 days.
The remaining 10x units of work will be done by Ram, Rahim, Rachel.
=$$\ \frac{10x}{x+0.5x+0.25x}$$
= $$\ \frac{10}{1.75}$$
=$$\ \frac{1000}{175}$$
=$$\frac{40}{7}$$
C is the correct answer.
A and B can do a piece of work in 21 and 24 days respectively. They start the work together and after some days A leaves the work and B completes the remaining work in 9 days. After how many days did A leave?
Let the total work be LCM of (21, 24) = 168 units.
Amount of work done by A on 1 day = $$\ \frac{\ 168}{21}$$ = 8 units
Amount of work done by B on 1 day = $$\ \frac{\ 168}{24}$$ = 7 units
Let p be the number of days for which they worked together.
In P days, the amount of work done by A and B = 15p
Remaining work = 168x-15p, which is completed by B in 9 days.
168x-15p=9*7
p=7
So A left after 7 days.
B is the correct answer.
Amit can do a work in 12 days and Sagar in 15 days. If they work on it together for 4 days, then the fraction of the work that is left is:
Let the total work be LCM of (12, 15) = 60 units.
Amount of work done by Amit on one day = $$\ \frac{\ 60}{12}$$ = 5 units
Amount of work done by Sagar on one day = $$\ \frac{\ 60}{15}$$ = 4 units
Amount of work completed by them on 4 days = (5+4)*4 = 36 units
The amount of work left out = 60-36 = 24 units.
Fraction of work left out = $$\ \frac{\ 24}{60}$$
=$$\ \frac{\ 2}{5}$$
C is the correct answer.
Running at the same constant rate, 6 identical machines can produce a total of 180 bottles per hour. How many bottles could 15 such machines produce in 30 minutes?
Machines are identical. So, each machine produces 180/6 = 30 bottles/hr
15 such machines will produce 450 bottles in an hour.
So, in 30 mins 15 machines will produce 225 bottles.
It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B alone takes 8 hours to fill the same tank. A, B and another pump C all together fill the tank in 2 hours. How long would pump C take, used alone, to fill the tank?
Assuming tank volume to be 48 litres,
Flow rate through pump A = 48/6 = 8 litres/hr
Flow rate through pump B = 48/8 = 6 litres/hr
Flow rate through pumps A, B, C together = 48/2 = 24 litres/hr
So, flow rate of pump C alone = 24 - (8+6) = 10 litres/hr
$$\therefore\ $$ Time taken by C alone to fill the tank = 48/10 = 4.8 hrs
A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am, pump A is started. At what time will the swimming pool be filled if pump B is started at 10 am?
Let pool volume be 30 litres.
Flow rate through A = 30/3 = 10 litres/hr
Flow rate through B = 30/6 = 5 litres/hr
A is opened at 9:00am and then B is opened at 10:00 am. So, A alone fills the tank for an hour to 10 litres.
After 10:00am both are opened simultaneously.
Remaining 20 litres of the pool are filled at a combined flow rate of 15 litres/hr which takes 80 min to get filled.
$$\therefore\ $$ The pool is filled 80 mins after 10:00am i.e. 11:20am
Stuart, Jack and Leo are colleagues working in a plant. Stuart and Jack can do a work in 10 days, Jack and Leo can do the same work in 15 days while Stuart and Leo can do it in 12 days. All of them started the work together. After two days, Leo was shifted to some other work. How many days will Stuart and Jack take to finish the rest of the work?
Let the total work be 120 units.
Let efficiencies of Stuart = s , Jack = j , Leo = l.
We know, Efficiency x Time = Work done.
Forming 3 eqns.:
s + j = 120 units/ 10 days = 12 units/day
j + l = 120 units/ 15 days = 8 units/day
s + l = 120 units/ 10 days = 10 units/day
On solving the above 3 eqns. , we get
s = 7 units/day , l = 3 units/day , j = 5 units/day
All 3 start together but Leo leaves after 2 days.
(15 x 2) + (12 x n) = 120
=> n = 7.5 days
$$\therefore\ $$ To finish rest of the work, Stuart and Jack take 7.5 days.
A contract is to be completed in 50 days and 105 men were set to work, each working 8 hours a day. After 25 days, $$\frac{2}{5}th$$ of the work is finished. How many additional men be employed so that the work may be completed on time, each man now working 9 hours a day?
It is given that in the initial 25 days, the work done by 105 men was $$(2/5)^{th}$$ of the total work.
Thus, the total work left is $$(3/5)^{th}$$ of the total work.
MDH/W is constant
so we get
$$\frac{105\times\ 25\times\ 8}{\frac{2W}{5}}=\frac{M\times\ 25\times\ 9}{\frac{3W}{5}}$$
We get M =140
So additional men = 140-105 = 35
A can built up a structure in 8 days and B can break it in 3 days. A has worked for 4 days and then B joined to work with A for another 2 days only. In how many days will A alone build up the remaining part of the structure?
Let the total work be 24 units
So we can say work done by A will be +3 units / day
and work done by B will be -8 units/day
Now as per the given condition :
work done by A in 4 days = 3(4) 12 units
Now, both worked for 2 days:
so work done will be (3-8)(2) = -10 units
Now the net work will be 12-10 = 2 units
Now remaining work 24-2 =22 units
Therefore A will complete this in 22/3 = $$7\frac{1}{3}$$ days
If 5 spiders can catch five flies in five minutes. How many flies can hundred spiders catch in 100 minutes?
The rate at which spiders catch the flies remains the same. Hence multiplying the time frame and the number of spiders relative to the information provided gives the solution.
Distributing 5 spiders as a group. There are 20 groups and each group catches 5 spiders in 5 minutes ad hence 100 spiders are caught in 5 minutes and in 100 minutes.
100*(100)/5 = 2000 flies.
$$\frac{\left(M1\cdot D1\cdot\right)}{W1}\ =\frac{\left(M2\cdot D2\right)}{W2}$$
$$\frac{\left(5\cdot5\right)}{5}\ =\frac{\left(100\cdot100\right)}{W2}$$
=> W= 2000
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