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Now, after having being there and facing it all as a student first and then as a facilitator (though I take good number of Quants classes for CAT aspirants, I would not liked to be called a tutor or a teacher. These titles should be reserved for people of great wisdom and learning) for over 4-5 years I felt I had a really good grip on things; and even then, there were many things I am still struggling with.
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Let The Process Begin
I’m starting the process by posting below an interesting formula of the topic- time and work . The aim is to initiate the process of collaborative learning. So anyone who wants to make a difference can post a comment with any explanation, formula, shortcut, trick, solved examples, etc. on the same topic (Time and Work for this post). You can also help by sharing the information of this platform with your friends or anyone who wants to help others and learn from others in the process.
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Time and Work Formula
I think the single most useful formula for the topic Time and Work is
N1H1D1E1W2 = N2H2D2E2W1
Where:
N1 and N2 = number of person
H1 and H2 = Hours worked by per person per day (assumed constant)
D1 and D2 = days
E1 and E2 = Efficiency
W1 and W2= Amount of work done
Consider this example to understand the applicability of this formula –
A piece of work can be done by 16 men in 8 days working 12 hours a day.How many men are needed to complete another work, which is three times the first one,in 24 days working 8 hours a day. The efficiency of the second group is half that of the first group?
Solution –
N1H1D1E1W2 = N2H2D2E2W1
16*12*8*1*3 = N2*8*24*0.5*1
N2 = (16*12*8*1*3)/ (8*24*0.5*1) = 48
So number of men required is 48.
Note – you can remove anything from formula is not given in the question. For example if the question would have been –
“A piece of work can be done by 16 men in 8 days working 12 hours a day.How many men are needed to complete another work, which is three times the first one,in 24 days working 8 hours a day.”
The applicable formula would have been –
N1H1D1W2 = N2H2D2W1
Since nothing is mentioned about efficiency, we remove it from both sides.
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I am Vineet Patawari - PGDM (IIM Indore), ACA, B.Com(H). My passion for Mathematics, specially Vedic Maths encouraged me to start QuickerMaths
I believe that if trained properly using powerful tools like Vedic Maths, the immense intellect of human mind can be ignited instantly - find out more
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December 4th, 2012 - 15:57
Two pipes A And B can separately fill a tank in 12 minutes and 15 minutes respectively. both the pipes are opened together but 4 minutes after the start, pipe A is turned off. How much time will it take to fill the tank?
December 1st, 2012 - 12:12
A,B and C can do a piece of work in 18 days, 27days, 36 days respectively. they start working together. after working for 4 days, A goes away and leaves 7 days before the work is finished Only C remains at work from beginning to end. In how many days was the whole work done?
December 1st, 2012 - 22:57
lets take total work to be done as 108 units
then, A’s efficiency = 6 units/day
B’s efficiency = 4 units/day
C’s efficiency = 3 units/day
together, their efficiency = 13 units/day
in first 4 days they completed 52 units
lest is 56 units
last 7 days only C worked so he completed 7 x 3 = 21 units
left work = 108 – (52 + 21)= 35 units
B and C took 35/7 = 5 days to completed this work
so, totally they took 4+5+7 = 16 days
November 29th, 2012 - 02:37
B and C can do a work in 12 days,C and A can do a work in 8 days. All three can do it in 6 days.A and B together can complete in ..?
November 26th, 2012 - 23:26
A,B and C can do a piece of work in 60,48 and 20 days respectively .A started the work .after 3 days B joined A and after 2 more days C joined them .in how many days will the work get completed?
November 23rd, 2012 - 20:20
. A can do a work in 8 days, B can do a work in 7 days, C can do a work in 6 days. A works on the first day, B works on the second day and C on the third day respectively that is they work on alternate days. When will they finish the work.(which day will they finish the work)
November 25th, 2012 - 18:26
Let the total work be the LCM of 6,7,8 which is 168 units.
A can do 21 units in 1 day, B can do 24 units in 1 day and C can do 28 units in one day.
On the first day, A works and does 21 units of the work.Similarly, next day B works, does 24 units and so on.At the end of the third day,21+24+28= 73 units of the work is done….So in 6 days, 146 units would be done….now on the 7th day, A would work and work out 21 more units.which becomes 146+21=167….Now only 1 unit of work is left to be done….B would work on the 8th day only for 1/24 of the day i.e. 1 hr…
Hence,Total time for finishing the work is 7 days and 1 hour
January 4th, 2013 - 14:38
Heyy…pravesh..
can u enlighten on …how the total work is LCM of 6,7,8…
like how u came to know…logic behind it..??
Thanx & Regards
October 5th, 2012 - 11:46
12 men can completed a work with in 9 days. after 3 days of the start the work, 6 men joined them and two men were replaced. how many days will they take to complete the remaining work? plese tell me answer with solution
November 18th, 2012 - 00:57
1 day 12 men can do 1/9 th of total work
1 day 1 man can do (1/9)/12 of total work=1/108
let x be the number of days they take to complete remaining work(18 people)
3*1/9 + x * (12+6)*(1/108)=1
=> x=4
November 26th, 2012 - 17:42
12men:6days=16 (12+6-2): Xdays
12*6/16=72/6=4.5days
September 28th, 2012 - 21:25
there are 250men and 150 women in a committee, if all will work they will complete 12 units per day, if all men work they will complete 15 units per day, how many units will women complete per day?
September 12th, 2012 - 19:02
R and S undertake a piece of work for Rs.9630/-. R and S can do a work alone in 10 and 15 days respectively. they work together for 5 days and then S leaves.
1] in how many days will the remaining work be completed by R?
2]find the share of R?
hey the answer to this problem given by ravi is wrong and that of kaush is right
(1)
one day work of R= 1/10
one day work of S= 1/15
one day work of (R+S)= 1/10+1/15=5/30=1/6
=
=
work completed after 5 days =1/6*5=5/6
work left=1-5/6=1/6
now R takes 10/6 = 5/3 days to complete the remaining work
(2)
S who can complete the work in 15 days works for 5 days that means he completed 1/3 of the work.. so the rest 2/3 is done by R hence his share will be 2/3 * 9630 = 6420
August 15th, 2012 - 07:37
sir please help….
A machine P can print one lakh book in 8 hrs,machine Q can print the same number of books in 10 hrs while machine R can print them in 12 hrs.all the machine are started at 9 a.m .while machine P is closed at 11 a.m .and remaining two machines complete work.Approximately at what time will the work(to print one lakh books)be finished?
August 19th, 2012 - 18:51
11:15 am
October 2nd, 2012 - 23:55
answer is 1:00 it is from Rs aggarwal
July 28th, 2012 - 13:17
Of the three taps X,Y and Z, Y takes 2 times as much time as X and Z together,while X takes 20mins more then Y and Z together to fill a tank. If three taps together fill the tank in 24mins,then how long does X take to fill the tank individually?
Thanks..
August 19th, 2012 - 19:03
Let tap X fills the tank in x min, Tank Y in y min and Tank Z in z min.
As it is given all the taps fill the tank so considering all are inflow pipes.
Framing the equations we get
y = 2(x+z) = 2x + 2z
x = 20 + y+z
therefore x = 20 +2x+2z
x + 2z + 20
August 19th, 2012 - 19:04
so x+2z+20 = 0;
so either x or z should be an outflow pipe….
July 16th, 2012 - 17:08
hi this site have good info.
July 11th, 2012 - 15:44
Hi Vineet Sir,
Time & work so far has been a tough topic, after coming across your informative blog, the perception has definitely changed. Kindly help me with the two questions mentioned below:
1. 24 men can complete a work in 16 days. 32 women can complete the same work in 24 days. 16 men & 16 women started working & worked for 12 days. How many men are added to complete the work in remaining two days?
2. 3 pipes A, B, C are attcahed to a cistern. A can fill in 20 mins & B can fill in 30 mins while empties in 15 mins. if A, B, C are kep[t alternatively for 1 min each. How soon the cistern will fill?
Thanks & Regards
July 12th, 2012 - 05:54
Q1. 1 men work per day=1/(24*16)
1 woman work per day = 1/(24*32)
Therefore 16 men work for 12 days= (16*12)/(24*16)=1/2 work
16 women work for 12 days = (16*12)/(24*32)=1/4 work
Total work completed in 12 days = 1/2+1/4=3/4 work
Work left to be completed = 1-3/4 =1/4 work
It has to be completed by men in 2 days.
Let no. of men be x
Therefore,similar to the above process
x men work in 2 days should be 1/4
Equation is
x*2/(24*16)=1/4
Solving we get x=48 men
July 12th, 2012 - 05:55
Hi payal,
Solution to the 1st question.
Q1. 1 men work per day=1/(24*16)
1 woman work per day = 1/(24*32)
Therefore 16 men work for 12 days= (16*12)/(24*16)=1/2 work
16 women work for 12 days = (16*12)/(24*32)=1/4 work
Total work completed in 12 days = 1/2+1/4=3/4 work
Work left to be completed = 1-3/4 =1/4 work
It has to be completed by men in 2 days.
Let no. of men be x
Therefore,similar to the above process
x men work in 2 days should be 1/4
Equation is
x*2/(24*16)=1/4
Solving we get x=48 men
July 12th, 2012 - 14:34
Thank you Sir.
One more question as mentioned below:
A & B together can do a piece of work in 12 days, which B & C together can do in 16 days. After A has been working at it for 5 days & B for 7 days, C finishes it in 13 days. In how many days C alone will do the work?
July 12th, 2012 - 06:05
2nd Question:
A=20, B=30, C=15
LCM of 20,30,15 is 60
Let total units of water be 60.
A can fill 60/20= 3 units per min
B can fill 60/30= 2 units per min
C can empty 60/15= 4 units per min
In 1st min water filled is 3 units
In 2nd min water filled is 2 units
In 3rd min water emptied is 4 units
It means amount of water filled in total 3 min is= 3+2-4=1 units
simply we can say,
1 unit is filled in 3 min
Therefore 60 unit will be filled in= 60*3= 180 min or 3 hr
July 15th, 2012 - 02:05
Hello Payal, This is Chitranjan. The solution to your last question is given below:
A+B=12, (A+B) one day work=1/12
B+C=16, (B+C) one day work=1/16
From question,
=5 days A’s work+7 days B’s work+13 days C’s= 1 work
=5 days A’s work+ (5+2) days B’s work+ (11+2) days C’s= 1 work
=(5 days A’s work+ 5 days B’s work)+(2 days B’s work+ 2 days C’s work) +
11 days C’s= 1 work
=5 days (A+B) work+2 days (B+C) work + 11 days C’s work= 1 work
=5*1/12 + 2*1/16+ 11 days C’s work= 1
Solving we get,
11 days C’s work= 11/24
1 day C’s work= 11/(11*24)
1 day C’s work= 1/24
Therefore C can complete the work in 24 days.
If you have any problem you can ask me.
Thank You
July 15th, 2012 - 15:43
Hi Chitranjan Sir. U simplied the problem so well..Thank you so much!
August 19th, 2012 - 19:07
It will take 60 min for the cistern to fill
June 25th, 2012 - 11:52
A group of five people can do a work in certain number of days.if more people join the group,they take 12 days less to do the the same work.in how many days a group of three people can do the work? can anyone pls explain how to solve this problem..
August 19th, 2012 - 18:58
Data Insufficient
June 17th, 2012 - 15:36
plz….. can anyone explain about the theorem 2(a+b+c)
wat the theorem means
July 15th, 2012 - 02:09
It means twice the sum of three variables.
a,b,c can be any objects. ADD IT AND DOUBLE IT
July 19th, 2012 - 08:08
2(a+b+c)/ab+bc+ca
June 8th, 2012 - 12:10
i am confused with this question that ” A can a work in 20 days , B can do same work in 40 days and C can same work in 60 days if they started to work together and A leaves after 5 days and B leaves before 5 days of completion of work then in how many days whole work was finished ?” please sir give shortut and suitable solution .
August 19th, 2012 - 19:14
21 days.
Work completed in 5 days = 5*[1/20 + 1/40 + 1/60] = 55/120
Remaining work = 1-(55/120) = 65/120 will be completed in x days.
As B left 5 days earlier to work completion B works for x-5 days and c work for x days to complete the remaining work 65/120
(x-5)(1/40) + x*(1/60) = 65/120. solving this we get x = 16. Total days = 5 (all three worked together) + x
total days =5+16 =21
June 6th, 2012 - 13:16
I am very, very confused with “Time & work” math plz explain me clearly. Example: If A can do a piece of work within 15 day & the same work B takes to complete 20 day, the question is how many days they will take if they work together ? Isn’t it ? A+B 1/15+1/20= 30 ??
June 25th, 2012 - 01:10
here in your e.g–
first step,1).. take lcm of 15 and 20 i.e 60.
2)so let total work = 60 units.
3)A takes 15 days to complete it so it means dat he completes 4 units per day..
4)similarily, B completes 3 units per day.
5)So combined they complete 7 units per day..(4+3 units)
6)So total work was 60,and work done per day if they are working together is 7 units..
7)So in 8 days they will complete 56 units
8)on 9th day remaining work is 4 units ,and they complete 7 units per day ,so 4/7th of the 9th day would be taken..
9)So they would take 8 and 4/7 days..
July 12th, 2012 - 14:48
it is much easier if u use d formula as x*y/ x+y
15*20/15+20=8.57
June 4th, 2012 - 09:26
VINEET SIR PLZZZ HELP….
20 men take 23 days to complete a job when all of them work together
(a)if all of 20 men start working together but one man leaves after every day of work , then find the approximate percentage of work completed when the last man leaves
(b)if all the 20 men start working together and one man joins additionally after every day of work , then fid the time taken to complete the work(in days)
June 6th, 2012 - 13:40
Let there be 23*20 = 460 units of work assuming each man does 1 unit of work each day.
First day 20 men working will do 20 units of work.
Second day 19 men working will do 19 units of work.
Third day 18 men working will do 18 units of work.
….and so on.
hence, total work done = 20 + 19 + 18 +…..3 + 2 + 1 = (20*21)/2 = 210
% of work done = (210/460)*100 = 45.65%
Using the same approach you can do the second problem….answer will be 16.55 days
June 9th, 2012 - 21:47
thank u sir
June 18th, 2012 - 10:40
am not getting thye exact ansr of b) question which is ansr 16.ss. plz send the ansr to my mail plz sir
May 13th, 2012 - 12:03
Sir pls help me.
8 boys can do a piece of work in 33 days.2 men can complete the same work in 33 days.in hw many days will 3 boys and 2men together can compelete the same work?
May 13th, 2012 - 21:00
8 Boys = 33 days
days
1 boy = (33*
So, 3 boys = (33*8)/3 days =88 days
similarly, 2 men = 33 days
1 man = (33* 2) days
Now req to find out 3 Boy+2men = ? days
as done above,
work done by 3 Boy+2men in one day = 1/88 +1/33 =1/11(1/8+1/3)= 1/11(11/24)= 1/24 days
Hence 3 Boy+2men can complete the task in 24 days
May 9th, 2012 - 17:05
A and B separately can do a piece of Work in 9 hours and 18 hours resptly.they started work together but before 3 days from the work being complete , A left the work, How many days will take to complete work”
May 10th, 2012 - 11:26
data is insufficient and ans cant b determine as hrs is given and days and no relation between them
May 13th, 2012 - 21:20
Hey, i think the question ll be like : A and B separately can do a piece of Work in 9 days and 18 days resptly.they started work together but before 3 days from the work being complete , A left the work, How many days will take to complete work”
Soln:
A completes in 9 days 1 task
So, in 1 Day = 1/9 task
Similarly,
B completes in 18 days 1 task
So, in 1 Day = 1/18 task
So, task completed by A+B in 1 day = 1/9+1/18 = 1/6 day.
So task can be completd by A+B in 6 days.
Now, its mentioned that A left the work before 3 days from the work being complete. So, we can assume that if both worked together, work can be completed in 6 days,
Hence they both worked for 3 days.So half of the work is remaining
B completes in 18 days 1 task
So, in 1 Day = 1/18 task
So 1/2 task = 9 days…
July 27th, 2012 - 10:08
I guess, the above solution is wrong.
B would work 1/18th of work per day
so, the last 3 days, B alone had worked, making it 3/18=1/6th work in it
The remaining 5/6work has to be completed by A and B together
A and B together complete 1/6 work per day.
A and B together hence would work for 5 days.
thus the total time taken is 5+3=8 days
April 17th, 2012 - 20:22
How to solve this using the trick?
X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
April 17th, 2012 - 20:28
THE WORK LASTED FOR 6 DAYS ………………..
May 31st, 2012 - 15:04
Xs 1 day work 1/20
xs 4 days work 4/20
remaining worik 1-4/20=6/5
(x+y)s 1 days work = 1/20+1/12=2/15
3/15 work is done by X+y in 1 day
1 work done by X+Y in 15/2
6/5 work done by X+Y in 15/2*6/5=9 days
May 4th, 2012 - 17:08
x-20
y-12
let total work 60 lcm of 20 & 12
then x will do in 1 day work 3
& y 1 day work in 5
then x worked for 4 days so work done by x in 4 days is 3*4 12 work done
now 48 work remain & x & y will do work together so they will do 8 work in 1 day
nw 48 will be done in 6 days
so all work will be done in 10 days ok
ye samjhane k liye lamba likha hai varna 2 step me ho jata hai.
April 16th, 2012 - 16:48
I want to know why in the above formula mentioned W2 comes on left hand side and w1 comes on right hand side of the equation
April 16th, 2012 - 18:53
ITS A SIGN CONVENTION IN TIME and WORK FORMULA . CONSIDER THIS
M1D1T1E1W2 = M2D2E2T2W1 where M- No Of Men , D- No Of Days , T- Time taken to complete the work, E- Efficiency Of men doing Work, W- Amount Of work done by the man .
HUN TAAN HO GEYA CLEAR BIBA ………………….
June 5th, 2012 - 19:16
Refer the following Points:
The fundamental rules on variation also apply in Time and Work.
(i) Work and men are directly proportional to each other ie, if the work increases, the no. of men required to do it, also increases, if the work is to be completed in the same number of days.
(ii) Men and days are inversely proportional, ie, if the number of men increases, the number of days required to complete the same work decreases and vice versa.
(iii) Work and days are directly proportional, ie, if the work increases, the number of days required also increases, if the work is to be completed with the same number of men and vice versa.
April 13th, 2012 - 02:21
16 men can complete a work in 12 days. 24 children can complete the same work in 18 days. 12 men and 8 children started working and after 8 days 3 more children joined them. how many days will they now take to complete the remaining work?
August 19th, 2012 - 19:18
4 days..
April 12th, 2012 - 11:08
lllllllllllllllllllllllllllllllllllllllsnnsnnsnsnsnsns
April 8th, 2012 - 17:52
very gud work osam site far better than d other sites that claim 2 b d best for quick maths
April 4th, 2012 - 21:36
Dear Vineet Sir Please provide Solution to this Time and Work Problem ::-
A and B separately can do a piece of Work in 8 hours and 10 hours resptly.If they work for one hour alternately with A in the beginning then in how many hours will the Work be Completed ?
April 5th, 2012 - 21:31
Let there be total 40 (LCM of 8 & 10) units of work. So A does 5 units of work per hour and B does 4 units of work per hour. So in 2 hours (first hr A, 2nd hr B) do 9 units of work. Hence, in 8 hours they do 36 units of work. Then it’s A’s turn and he does the remaining 4 units of work in 4/5 of an hour = 48 minutes. Hence total 8 hours 48 minutes will be taken to complete the work
April 5th, 2012 - 23:13
THANK U SIR . I ALSO WANT TO SHARE AN IMP RELATION USED 4 SOLVING TIME and WORK PROBLEMS . that is IF M1 PERSONS DO W1 WORK IN DI DAYS WORKING T1 HOURS A DAY and M2 PERSONS DO W2 WORK WORKING T2 HOURS A DAY then there is relation in TIME and WORK ACCORDING TO WHICH :;
M1*D1*T1*W2 = M2*D2*T2*W1 .
April 4th, 2012 - 21:35
Dear Vineet Sir Please provide Solution to this Time and Work Problem ::-
A and B separately can do a piece of Work in 8 hours and 10 hours resptly.If they work for one hour alternately with A in the beginning then in how many hours will the Work be Completed ?
June 6th, 2012 - 13:42
Assume there are 40 units of total work (LCM of 8,10)
So A does 40/8 = 5 units of work per hour
B does 40/10 = 4 units of work per hour.
So total hours in which 40 units will be completed is 8 hours 48 minutes
April 4th, 2012 - 15:50
a certain job was assigned to a group of man to do in 20 days. but 12 men didn’t turn up for the job and remaining man did the job in 32 days. what was original no of man?Pls Give me solution
April 27th, 2012 - 13:19
total no. of man=n
so12 man are turn up for job so man are left=n-12
n*20=(n-12)*32
20*n=32*n-384
n=384/12
n=32
so ans isoriginal no man is=32
March 30th, 2012 - 11:45
Hi….
i am Mohana,very much impressed with u r Site,i am also follow the veic maths tricks.
But i have doubt about calendars problems.
1.How it comes 5 odd days for 100 years.
2.The answer for the qs
Which of the following is not a leap year?
A. 700 B. 800
C. 1200 D. 2000
in the above qs ans is 700.
but reference to the below link it is a leap year.
REF:http://www.timeanddate.com/calendar/?year=700&country=35
Plz solve my problem…..
Thanking you sir…….
August 19th, 2012 - 19:28
Answer is 700
Refer the following link
http://leap-year-guide.blogspot.in/2011/03/leap-year-calculation.html
March 28th, 2012 - 17:40
There are 4 taps in a swimming pool.
First one alone fills the pool takes one day.
second tap alone fills the pool in 2 days.
Third one alone will take 3 days.
fourth one will fill the pool with just 6 hours.
How long it will take to fill the pool if the four taps are open together?
March 30th, 2012 - 11:58
Ans:
1st pipe=1 day=24 hours
2nd pipe=2 days=48hrs
3rd pipe=3days=72hrs
4th pipe=6 hrs
according to the information all alone fill it
1/24,1/48,1/72,1/6
then together 1/24+1/48+1/72+1/6=35/144
then all together in 144/35 hrs=4.11 hrs
April 27th, 2012 - 13:06
work &time sudy
March 7th, 2012 - 18:03
One day work of A=1/40
One day work of B=1/60
One day work of c=1/120
3 days work of A and One day work of B and One day work of C= 3(1/40)+1/60+1/120.
3 days work = 1/10.
Work done in =3*10.
=30 days.
August 19th, 2012 - 19:34
Let the total work is x.
If all of them work for 30 days together, work completed is completed is 1.5 x
For the work to complete it will take 30 days of A’s work and 10 days each of B and C’s work.
February 16th, 2012 - 06:56
Sr pls tale me name of a shortcut math
February 8th, 2012 - 10:43
A and B can do a piece of work in 16days. B and C can do it in 20 days. A and C can do it in 24days. In how many days they all will finish the same work?
February 24th, 2012 - 12:06
we can do by theorem 2(a+b+c)
a+b=16
b+c =20
a+c=24 16*20*24/ 16*20 + 20*24 + 24*16 = 320/29 or 11 1/29
March 9th, 2012 - 12:34
(a+b) 1 day work=1/16
(b+c ) 1 day work =1/20
(a+c) 1 day work=1/24
then add all these works
2(a+b+c)=(1/16+1/20+1/24)
solve the above eqn we can get the answer
February 7th, 2012 - 20:23
A, B and C can do a piece of work in 40, 60 and 120 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
Helo Sir,
Please help me to solve this question.
February 17th, 2012 - 04:38
Check out the 3rd questn and its anws. ….It will help u out.
http://www.indiabix.com/aptitude/time-and-work/
Question is of same format as u have writtn here!
rgrds: Frnd!
January 25th, 2012 - 20:34
i am attend the competitve exams, so i want shortcuts for maths for all types
January 20th, 2012 - 11:07
A can do a piece of work in 24 days,which B can do in 16 days.with the help of C they finished the work in 8 days.c alone will do the work in? please any body help me to solve this….
i am so much heck with this time and work problems……
January 20th, 2012 - 11:20
Simple way to do it is; assume that total work is LCM(24,16,8) = 48 units
So A does 48/24 = 2 units/day
B does 48/16 = 3 units/day
A+B+C does 48/8 = 6 units/day
Therefore C does 6-2-3 = 1 unit/day
Hence C alone will do it in 48/1 = 48 days
January 26th, 2012 - 11:49
thank u very much
sir……
January 26th, 2012 - 11:53
bt look last step..48/1= 28?
January 27th, 2012 - 14:13
thank u sir…..now i understood well….
July 9th, 2012 - 19:14
A’s work=1/24
B’s work=1/16
with the help of the C means(A+B+C)=1/8
C=1/8-(A+B)
C=1/8-(1/24+1/16)
C=1/48
So, C’s alone will do in 48 days
January 26th, 2012 - 11:27
thank you very much sir……thanks…
January 26th, 2012 - 21:08
Q. A can do a piece of work in 24 days,which B can do in 16 days.with the help of C they finished the work in 8 days.c alone will do the work in? please any body help me to solve this….
Sol: Ok let me give it a try…
As per quesn. A’s 1day’s work = 1/24; likewise B’s 1day’s work= 1/16 and let C’s 1day’s work= 1/C.
And working all of them together, they can finish the job in 8 days so 1 day work of all of them working together would be 1/8. Now equating 1 day work would yield the equation:
1/24+1/16+1/C=1/8
or, 1/C = 1/8-1/24-1/16
or, 1/C=(6-2-3)/48= 1/48
So, C’s 1 day’s work is 1/48 and hence C alone can do the job in 48 days.
N.B.: Attn. Padma Priya….I have solved the problem elaborately to understand the concept behind that and Vineet has just done a typing error; it would be 48/1=48 days.
Please follow Vineet’s handy and yet conceptual technique to attempt this types of problems quickly.
January 27th, 2012 - 13:57
thank u sir….
January 18th, 2012 - 17:00
please tell me Anupam Das, How to solv it… contact me my mail. ashish1.sethi@gmail.com
January 18th, 2012 - 20:45
@ASHISH
Please let me know which part u r asking about and what you want to solved by me….please be specific my friend.
December 5th, 2011 - 12:31
in 1 hr tap fill the tank 1/9 +1/6 -1/8 =11/72
in 5 hr ………..=55/72.
November 25th, 2011 - 19:31
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
December 1st, 2011 - 04:16
ans :: 4 days
Use Basic one as given above:
(6+8)*10=35*x
x=4days
April 27th, 2012 - 13:42
so 6m & 8b=10days
os that i can write in following form
3m & 4b =20 days so,
15m & 20b=4 days this is the ANS
November 3rd, 2011 - 16:31
R and S undertake a piece of work for Rs.9630/-. R and S can do a work alone in 10 and 15 days respectively. they work together for 5 days and then S leaves.
1] in how many days will the remaining work be completed by R?
2]find the share of R?
November 9th, 2011 - 18:44
1-R takes 5/3 days more to complete the task..
2-share of R in 5 days,,R work 1/2 work in 5 days
February 4th, 2012 - 16:21
Can u plz give me clarity in ur solution Kaush sir…..
March 31st, 2012 - 11:35
(1)
one day work of R= 1/10
one day work of S= 1/15
one day work of (R+S)= 1/10+1/15=5/30=1/6
=
=
work completed after 5 days =1/6*5=5/6
work left=1-5/6=1/6
the R take 6 day to complete the remaining work
(2)
share of R is rs 5633.55 or 5634
October 3rd, 2011 - 21:21
Sir that formula was really helpful, i had a doubt if they ask question like
“45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete remaining work”
give me a shortcut for this type of ques…?
pls do reply to my mail…
October 23rd, 2011 - 17:24
45 men complete work in 16 days
45 men’s work done in one day=1/16
45 men work 6 days after=6/16
remaining work=1-(6/16)=5/8
1 men work done in 1 day=1/16*1/45=1/720
75 men work in 1 day=75/720=5/48
so 5/8 work done in 48/5*5/8=6 days
March 31st, 2012 - 11:41
thank you for the solution
November 20th, 2012 - 23:29
sirsend me other tricks to my email id,that tricks is also helpful to me plz
September 13th, 2011 - 00:17
Hi Vineet and all!!! let me share what I have learnt from one of my mentor its pretty handy and conceptual………………..
You have given the formulae N1H1D1E1W2 = N2H2D2E2W1
Though its a pretty formula and can be remember easily but why bother our brain……and there’s a risk of getting it forget and applying it correctly…lets take your question….
Q.> A piece of work can be done by 16 men in 8 days working 12 hours a day.How many men are needed to complete another work, which is three times the first one,in 24 days working 8 hours a day. The efficiency of the second group is half that of the first group?
Solution: 16*(1/2)*(3/1)*(24/8)*(8/12)=48 in one line itself…
let me explain a bit.
First consider the variable that is required to be found in the question…here it is how many man needed…ok so take the original man first.
16
Now comes the efficiency which is 1/2 times the original men so this (1/2) efficiency will bring a negative impact (if I say those which are greater than 1 is positive impact and those which are less than 1 are negative impact and not just -ve/+ve literally) say on men needed means it will made more men to work so I am taking 1/2 and not 2/1..
Now comes the task which is thrice so the next factor would be (3/1)…a positive impact
now it becomes 16*(1/2)*(3/1)
Now the number of days which was earlier 8 and now its 24 that means we are giving more day this implies less men will be required that’s again a positive effect on number of men required
now its 16*(1/2)*(3/1)*(24/8)
And lastly, hours earlier 12 hours and now they are working 8 hours this means other things being equal this will bring an negative impact on number of men required, so I have written 8/12 and not 12/8
so, finally 16*(1/2)*(3/1)*(24/8)*(8/12)=48 in one line itself…
I have tried to explain the things as much as possible by writing…if some mistake crept in feel free to comment….
September 13th, 2011 - 00:26
and also stress that mine last equation is a bit different from that shared by Vineet (16*12*8*1*3)/ (8*24*0.5*1) = 48
mine: 16*(1/2)*(3/1)*(24/8)*(8/12)=48 [or,
(16*1*3*24*8)/(2*1*8*12)=48]
but it will always yield the same result. Looking forward for your views.
September 13th, 2011 - 11:02
Thanks for your efforts. Friends like make this blog a more lively and useful place!
April 28th, 2011 - 13:54
This is pretty amazing………..I m enjoying it……I will not read it for free so i request God to transfer ONE Day of my life into your account……
Pls Keep posting and increase our interest……..
September 13th, 2011 - 10:56
Thanks Dear for such a wonderful comment.
April 25th, 2011 - 11:50
thanks
December 16th, 2010 - 13:30
QUITE INTERESTING…USUAL BASICS OFTEN CAN BE SOLVED CALCULUS EASILY
October 18th, 2010 - 14:51
thanks for the post
December 16th, 2010 - 13:30
poda
October 15th, 2010 - 21:25
I have an entire book to add
Now suggest how it can be distributed.
October 16th, 2010 - 22:30
sorry i didnt catch u…u an just name one r two
October 15th, 2010 - 20:06
AND PLZ IF U CAN SUGGEST ME SOME GOOD COACHING FOR CAT IN DELHI ..?
October 14th, 2010 - 20:54
Thank you for give very nice info. I am impressed by the information that you have on this blog. It shows how well you understand this subject. Bookmarked this page, will come back for more. You, my friend, ROCK! I found just the information I already searched everywhere and just couldn’t find. What a perfect site. Like this website your website is one of my new favs.I like this website shown and it has given me some sort of desire to succeed for some reason, so keep up the good work.
October 12th, 2010 - 16:42
hi Sir
im Planning to give my Cat exam in 11…nd would b compleating my b.com this year..I have to understand should i give my whole year to the prepration of cat or an ny thing ould b done side by side with it plz do advie …?nd i left my maths in lass 8th nd now i feel i just forgot maths nd all those calculations…sir plz advice me how should i make a good nd effcetive start in maths coz i have nt been in touch with itplz suggest me some books or tell me ur address ny where in India i will b there…plz help
October 12th, 2010 - 17:54
@Abhinav – The only key to success is hard work. So let the process begin.
Lot of people manage it with work. There work ex. helps them a lot in B-Schools and there after. But it depends on you, if you can cope up with both.
I am suggesting you few books which can help you in preparation –
Suggested Books for CAT –
http://www.quickermaths.com/suggested-books-for-cat/
Logical Reasoning Books for CAT –
http://www.quickermaths.com/logical-reasoning-books/
Books to improve your calculation speed –
http://www.quickermaths.com/list-of-best-quicker-maths-books/
Join this free SMS service to improve your vocab –
http://labs.google.co.in/smschannels/subscribe/FIREUPCAT
October 15th, 2010 - 15:18
great advice ..im highly thankful dear sir..
October 12th, 2010 - 10:36
Similar problems relate to filling or emptying of water tank by a set of inlets and outlets.
Suppose a tap fills a tank in 9 hrs, another tap fills the same tank in 6 hrs, a third tap empties the tank in 8 hrs. If all taps are open what is the water volume in the tank after 5 hrs?
Take the LCM of 9, 6 and 8. It is 72.
In 72 hrs, the first tap fills 72/9 = 8 tanks
the second tap fills 72/6 = 12 tanks
the third tap empties 72/8 = 9 tanks.
Net effect after 72 hrs = 8+12-9 = 11 tanks.
In 5 hrs the volume filled = 11/72*5 = 55/72 tank.
My question is whether the time and work equation can be modified to solve problems of this type.