This is merely a subtype of Time n Work problems. and can be solved using our good ol’ STD table Method.
Case 1: Women finish entire job
4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?
Suppose 1 man can pour 1 bucketful of water in the tank in 1 minute.
If 6 men work together, they pour 6 buckets in a tank in 1 minute.
In short, you multiply the speed with number of person.
Our usual table
1 Man | 1 Woman | 4 men & 6 women | 3 men & 7 women | |
Speed | m | w | 4m+6w | 3m+7w |
Time | 8 | 10 | ||
distance |
Distance covered in each column is same.
So compare last two columns
Distance = distance
Speed x time = speed x time
(4m+6w)*8=(3m+7w)*10
Solve this equation and you get m=11w
Replace this value of w in last column
1 Man | 1 Woman | 4 men & 6 women | 3 men & 7 women | |
Speed | m | w | 4m+6w | 3(11w)+7w |
Time | 8 | 10 | ||
distance |
3(11w)+7w
=33w+7w
=40w
1 Man | 1 Woman | 4 men & 6 women | 3 men & 7 women | |
Speed | m | w | 4m+6w | 40w |
Time | 8 | 10 | ||
distance | 40w x 10= 400w |
So the total work is
Speed x time = distance
40w x 10= 400w
If 10 women work together, they’ve to cover 400w kms. Make a new column, run STD formula
1 Man | 1 Woman | 4 men & 6 women | 3 men & 7 women | 10 women | |
Speed | m | w | 4m+6w | 40w | 10w |
Time | 8 | 10 | ?? | ||
distance | 40w x 10= 400w | 400w |
10w x time =400w
Time =400w/ 10w=40days.
Answer: If 10 women work together, it’ll take 40 days to finish the job.
Case 2: Women to finish remaining job
12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days, all of them stopped working. How many women should be put on the job to complete the remaining work, if it is to be completed in 3 days?
12 Man | 15 Woman | |
Speed | 12m | 15w |
Time | 4 | 4 |
distance |
Distance covered in both column is same so compare them
Speed x time = speed x time
12m x 4= 15w x 4
4m=5w
M=(5/4)w
The question is asking about number of women in the end, so better convert everything in terms of women.
Given: 6 men start working on the job and after working for 2 days, all of them stopped working
Make a new column.
12 Man | 15 Woman | 6men | |
Speed | 12m | 15w | 6x (5/4)w |
Time | 4 | 4 | 2 |
distance | 60w | 6x (5/4)w x2=15w |
As you can see, I’ve applied STD formula in last two columns to find distance simultaneously.
Remaining work=60w minus 15w=45w
Required: we want to finish this work in 3 days.
Asked: How many women should be put on the job to complete the remaining work, if it is to be completed in 3 days?
Make a new column. Suppose we need “F” number of women so speed = f x w = Fw.
12 Man | 15 Woman | 6men | Find women | |
Speed | 12m | 15w | 6x (5/4)w | Fw |
Time | 4 | 4 | 2 | 3 |
Distance | 60w | 15w | 45w |
Time limit is given to us: complete remaining work in three days. So t=3 for last column.
Run STD on last column
Fw x 3 =45w
F=45w /3w
F=15
Answer : 15 women required.
Important: Remaining work is to be completed in 3 days. So t=3
If they had asked “total work is to be completed in 3 days”, we’ll need to take t=1, because those men already worked for 2 days to only 1 day left to complete the job (2+1=3 days)
Case: Child labour
Twelve children take sixteen days to complete a work which can be completed by eight adults in twelve days. Sixteen adults started working and after three days ten adults left and four children joined them. How many days will they take to complete the remaining work?
12 kids | 8 men | |
Speed | 12k | 8m |
Time | 16 | 12 |
Distance |
Since work done in both columns is same
Speed x time = speed x time
12k x 16= 8m x 12
K=(1/2)m
Concentrate on middle column (8 men) run STD table on it. You get distance = 8m x 12 =96m
So total distance to be covered is 48m
Given: Sixteen adults started working and after three days ….
Means 16 men worked for 3 days.
12 kids | 8 men | 16 men | |
Speed | 12k | 8m | 16m |
Time | 16 | 12 | 3 |
Distance | 8m x 12=96m | 16m x 3=48m |
We’ve run STD on last column and got that 16 men covered 48m kms.
So remaining work is 96m minus 48m=48m
Given: after three days ten adults left and four children joined them
So now men left =16 minus 10 =6 men and 4 kids joins them so speed is 6m+4k
Make a new column
12 kids | 8 men | 16 men | Men & kids | |
Speed | 12k | 8m | 16m | 6m+4k |
Time | 16 | 12 | 3 | ?? |
Distance | 96m | 48m | 48m |
Run STD on last column
(6m+4k) x time =48m
But We already calculated that K=(1/2)m
Apply it in above equation{6m+4(1/2)m} x time =48m
{6m+2m} x time = 48m
8m x time = 48m
Time =48m /8m
Time=6 days.
beautifully explained, aptitude seems to be very easy by your explainations sir
concepts made very easily, aptitude seems to very easy by your explaination sir
i dnt knw wats ur future plans , but if u stand for election .. whole youth of india will vote for u !!! mrunal long live …
i dnt knw wats ur future plan , but if u stand for election whole youth of this country will vote for u . mrunal long live
Plz solve dis by ur method: Q) A,B,C can do a piece of work in 12,15,20 days respectively..If A is working all the time, B and C are working alternatively,then in how many days work wil be done?
ans- 120/17
as per the technique,
A B C A+(B/2)+(C/2)
speed A B C A+(B/2)+(C/2)
Time 12 15 20 ??
Distance 12A 15B 20C 12A
12A=15B=20C
Now, Time= 12A/[A+(B/2)+(C/2)] _____since time= dist./speed
By solving above equation u’ll get 120/17 ans.
Whats the answer??
I am getting 7 days..I may be wrong..I am known to do silly mistakes!!! :)
u r right …………ans is 7 days
Answer is 7 days
Ans is 7
And pls don’t makes it complicated..
Find,
1 day work of A+B and
1 day work of A+C
A+B—9/60 work in one day
A+C—8/60 work in one day
Working luklike..
(A+B), (A+C)……
9/60 + 8/60 +………..
After 7th day work done (u get 60/60 i.e. 1 work)
Plz also solve other models in work,time,wages like A can do 25 percent work,b does 1/6th work……so on
THANK U!!!! so much for giving a such wonderful tricks to solve the problem…………………..THANK u sooo much……….
Mr. A finish a worked on 12 days. when Mr. B finished this work on 15 days and Mr. C on 20 days separately. but after 2days A left the work and B left the work before 2days finished the work. How many day worked will be finished?
Plz solve this question ?
hello Megha!
A – 12
B – 15
C – 20
in question they say that after 2 days A left
now we will find how much work they have done in 2 days
2*( 1/12 + 1/15 + 1/20 ) = 2/5 it means they have completed 2/5 work .now remaining work
is 1-2/5 = 3/5
let form here own-word that they (B and C) finish the work in D days .
D* 1/20 + (D-2)* 1/15 = 3/5
form here D comes ——————– 44/7 days.
final answer is 2 + 44/7 = 58/7 days .
please replay if i am wrong .
Ur right mohit kumar….I solve it in std technique.I got same answer…:-)
Please solve
5 women and 3 men done any work in 10 days. 4 women and 4 men done this work in 3 days. How many days that 10 women take to done this work?
Mr. A finish a worked on 12 days. when Mr. B finished this work on 15 days and Mr. C on 20 days separately. but after 2days A left the work and B left the work before 2days finished the work. How many day worked will be finished?
sol : (more simple)
let they complete the work in x days then,
x/20+(x-2)/15+2/12=1;
=> 58/7 days (ans)
x/20= C’s total work
(x-2)/15= B’s total work
2/12 = A’s total work
(you can do such questions with approach)
Mr. A finish a worked on 12 days. when Mr. B finished this work on 15 days and Mr. C on 20 days separately. but after 2days A left the work and B left the work before 2days finished the work. How many day worked will be finished?
sol : (more simple)
let they complete the work in x days then,
x/20+(x-2)/15+2/12=1;
=> 58/7 days (ans)
x/20= C’s total work
(x-2)/15= B’s total work
2/12 = A’s total work
(you can do such questions with approach)
Hello friends I am from vocational Background but working really hard from SSC preparation. Please hint me on the solving following time and work problems.
1. If 5 men or 8 women can do a piece of work in 12 days, how many days will be taken by 2 men and 4 women to do the same work
2. 15 women can do 1/2 work in 20 days. In how many days will 20 women do the full work?
3. A and B can be a piece of work in 10 days and C and A in 20 days. C alone can do the work in `
4.Two taps can separately fill a cistern in 6 and 7 minutes respectively. If the taps are opened turn by turn each for a minute, find the time taken by them to fill the cistern.
1st sol: since work is same apply speed*time-distance , so 5m*12w=8w*12,solve for 1 w=5/8m asks:2m+4w=days? Put 1w=5/8m n sol ve tht is 9/2m nw 5m do in 12 days n 9/2 m will do in 40/3 d ..its min whts urs
i have solved 2nd
ans is 120?
nice stuff admin .thank u..
Hai, i have doubt on following question, please help me to solve quickly,
A alone can complete a work in 16 days and B alone can do in 12 days. Starting with A, they work on alternate days. The total work will be completed in how many days?
I think the answer will be 94/7.
LCM(16,12)=48.
A’s 1 day work = 48/16 =3 ; B’s 1 day work = 48/12 =4.
In 2 days their combined work = 3+4=7.
Since 7*6=42 and 7*7=49, hence our answer must be between 6*2=12 days and 7*2=14 days.
In 12 days the work finished = 42 ; remaining work = 48-4=6.
Now its A’s turn to work for 1 day . It will do ‘3’ units of work. Remaining work =3. Total days taken till now =13.
Now its B’s turn . Since it does 7 units of work in 1 day , it will do 3 units of work in 3/7 day.
hence final answer = 13+3/7 OR 94/7 days.
#HAPPY LEARNING:)
Pls help me solve this question using above STD method
A alone can complete work in 16 days and B alone in 12 days.Starting with A ,they work on alternate days.The total work will be complted in how many days?
20 women can do a job in 20 days. after each day, a woman is replaced by a man is twice as efficient as a women on which day does the job gets completed??? anyone
well ans-10 days..becz if each woman replaced by a man-tht mean 20 men,but efficiency is high time will be less.so half time in my opinion
Let efficiency of 1 women = w then effic. of 1 men =2*w.
Total work =20*w*20=400*w.
Since after each day 1 w is replaced by 1 m i.e. 2*w.
After 1st day 20*w – w +2*w =21*w.
Similarly after 2nd day 21*w-w+2*w=22*w.
Clearly its an A.P. with a = 20 , d = 1 and sum = 400. The value of n will be the answer , which can be find out as 400 = n*(2*a + (n-1)*d)/2.
#HAPPY LEARNING:)
Respected sir, I was unable to solve following 2 problems from Time and work. It looks complicated to me. Plz,help.
problem 1- B can do a job in 6 hrs.B and C can do it in 4 hrs, and A,B,C in 2*(2/3)hrs.In how many hrs can A and B do it? option (a)3*(3/4) (b)3*(3/7) (c)2*(3/7) (d)4*(3/7)
problem 2- A and B can do a piece of work in 12 days.A and C together work twice as much as B, A and B together work thrice as much as C. In what time could each do it separately?
can any one solve the question by using STD table method
urgent coz i got some confusion-”
A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in:
Question : 6 men can complete the electric fitting in a building in 7 days .how many days will it take 21 men to do the job?
mob : 9799422423
plz reply……………………..fast
how we calculate
Remaining work=60w minus 15w=45w
in 2nd Que explain above by mrunal sir.
easy method is take out efficiency
let men no be =x
and womwnen = y
efficiency of 4m and 6w together = 100/8 = 12.5%
efficiency of 3m and 7w together = 100/10 = 10 %
so,
4x+6y=12.5 – 1
3x+7y= 10 – 2
solving one and two u get y = 1/4% (efficiency of women per day)
for 10 women multiply by 10 = 10/4%
so number of days would be 100/10/4 = 40 days
Thanks sir becoz of only u now m able to solve questions VERY easily…and concepts also get cleared. Thanku…
Mrunal sir,how to solve this problem?
Two persons A and B together can do a work in 10 days.The same work can be finished in 18 days if A works alone for 12 days and B works alone for 6 days.find the time required for each of them to finish the work if they work alone?
really u r einstein’s incarnation
Thank you, you teach and explain in a way that is so simple yet very effective. I wish you all the best. Thanks again.
mrunal sir or any1 plz solv dis from rajesh verma..a certain persons cn dig a trench 200 m long ,20 m deep ,100 m broad.thn the same no of persons dig another trench 40 m broad 30 m deep in 60 days..find length of 2nd trench..explain step by step ..if possible unitary method..
ji i tried….but i didnt get…. if u got answer…pls explain me
Thanks a lot Mrunal.ORG :)
Twelve men can complete a work in 8 days. Three days after they stated the work, 3 more men joined them. In how many days will all of them together complete the remaining work?
plz here me to solve the above problem in a simple way Twelve men can complete a work in 8 days. Three days after they stated the work, 3 more men joined them. In how many days will all of them together complete the remaining work?
SOLUTION :-. APPLY STD TABLE
MEN MEN MEN
S 12 12 12 + 3 = 15
T 8 3 60/15 = 4 DAYS
D 12X8 = 96 36 96 – 36 = 60 ANS. 4 DAYS
SOLUTION IS NOT SHOWING PROPERLY
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