- Case: A & B work start together but B leaves
- Case: A started work, B joined after 4 days, how days work lasted?
- Case: A starts, B joins, How long did B work?
- Case: A and B start together but A leaves
- Case: Efficiency given: A is thrice good than B
Case: A & B work start together but B leaves
Q.Kalmadi can build a stadium in 14 days, while Raja can do the same in 21 days. They started working together.
- Case #1: 3 days before the completion of work, Kalmadi was sent to Tihad jail. So find out Total Number of days to complete the work?
- Case #2: Same question but Raja goes to jail.
Step 1: Draw the STD Table
Whether it is Pipes and Cistern problem or it is Time and Work problem, draw the “STD” table, fill up the values
| Kalmadi | Raja | Kalmadi + Raja | |
| Speed | |||
| Time | 14 | 21 | |
| Distance |
Step 2: Take LCM, update Table
Now we’ll take the LCM of 14, 21
14=2×7
21=3×7
LCM (14,21)= 2x3x7=42
Let’s just visualize this stadium has just 42 seats. And these fine gentlemen have to place the seats, and drill the bolts.
So 42 seats is the “Total work to be done” or Total Distance to be covered.
Fill up the table
| Kalmadi | Raja | Kalmadi + Raja | |
| Speed | |||
| Time | 14 | 21 | |
| Distance | 42 | 42 |
Now the speed of Kalmadi
Sx T= D
S x 14=42 ; because we know Kalmadi takes 14 days to finish this work
Speed of Kalmadi= 42/14= 3 seats a day
Same way Raja can fit 2 seats a day (because 21 x 2 = 42)
Update the Table
| Kalmadi | Raja | Kalmadi + Raja | |
| Speed | 3 | 2 | |
| Time | 14 | 21 | |
| Distance | 42 | 42 |
Step 3: Kalmadi + Raja Together?
What if Kalmadi + Raja work together? Obviously their speed will increase.
Now they can fit 3+2 = 5 seats per day.
| Kalmadi | Raja | Kalmadi + Raja | |
| Speed | 3 | 2 | 3+2=5 |
| Time | 14 | 21 | ?? |
| Distance | 42 | 42 | ?? |
BUT we don’t know for how many days did they work together?
So we cannot find how many seats they fixed. (STD formula requires two known values!)
So leave the (Kalmadi+Raja) column and move ahead.
Step 4: Kamadi goes to Jail, Raja Works Alone
We are given that “3 days before the completion of work, Kalmadi was sent to Tihad jail.”
It means for the last three days, Raja worked alone, right ?
So, Expand the table, create one more column for Raja Alone
| Kalmadi | Raja | Kalmadi + Raja | Raja alone | |
| Speed | 3 | 2 | 3+2=5 | 2 |
| Time | 14 | 21 | ?? | 3 |
| Distance | 42 | 42 | ?? | 6 |
So, how many seats did Raja fix in last three days?
Speed x time = distance
2 x 3 = 6. ; Because Raja’s speed Is 2 and he works for 3 days.
Step 5: Remaining seats
We know that total seats= 42, and Raja fixed 6.
Meaning 42 minus 6 = 36 seats were fixed by Kalmadi +Raja together.
Plug the value back in the Table
| Kalmadi | Raja | Kalmadi + Raja | Raja alone | |
| Speed | 3 | 2 | 3+2=5 | 2 |
| Time | 14 | 21 | ?? | 3 |
| Distance | 42 | 42 | 36 | 6 |
Again STD formula to find how many days did they work together ?
Speed (Kalmadi + Raja) x Time = Distance
5 x Time = 36
Time = 36/5=7.2
Thus, our final table looks like this
| Kalmadi | Raja | Kalmadi + Raja | Raja alone | |
| Speed | 3 | 2 | 3+2=5 | 2 |
| Time | 14 | 21 | 7.2 | 3 |
| Distance | 42 | 42 | 36 | 6 |
We’re about to solve the problem now.
What is the Total time taken by these two fine gentlemen of impeccable integrity, to build this stadium?
Look at the finished table.
7.2 days they worked together
Plus 3 days Raja worked alone.
=7.2+3=10.2 days.
Final answer= 10.2 days
Similarly you do for case#2 when Raja is sent to Jail, 3 days before the completion of work.
Case: A started work, B joined after 4 days, how days work lasted?
Abdul and Bhide can do a piece of work in 20 days and 12 days respectively. Abdul started the work alone and then after 4 days Bhide joined him till the completion of the work. How long did the work last?
Step1: Create STD table
| Abdul | Bhide | |
| Speed | ||
| x Time | 20 | 12 |
| = distance |
Step2: Take LCM of time
LCM (20,12)
in other words, which is the first number that comes in multiplication table of both 12 and 20?
Well, we know that
12 x 5 =60 and
20 x 3 = 60
Therefore, 60 is the least common multiple of (12, 20)
Assume total distance = 60
Update table
| Abdul | Bhide | |
| Speed | ||
| Time | 20 | 12 |
| distance | 60 | 60 |
Step3: Take ratio of speed
now we know that ‘total distance’ is 60. and if Abdul took 20 minutes to finish, what is his speed?
Speed x time = distance
Speed x 20 = 60
Therefore, Abdul’s speed is 3; similarly Bhide’s speed is 5.
Update the table.
| Abdul | Bhide | |
| Speed | 3 | 5 |
| x Time | 20 | 12 |
| = distance | 60 | 60 |
in the exam, just do it in your head. you took LCM so you’d already know that 20×3=60 and 12×5=60 so automatically 3:5 is the ratio
Step 4: start the work
Given: Abdul started the work alone and then after 4 days Bhide joined him till the completion of the work
Means, for the first four days, Abdul worked alone.
| Abdul | Bhide | Abdul started | |
| Speed | 3 | 5 | 3 |
| Time | 20 | 12 | 4 |
| distance | 60 | 60 | 3×4=12 |
In 4 days, he covered 12 kms.
So remaining work = 60 minus 12 =48.
Which was completed by X+Y
| Abdul | Bhide | Abdul started | Abdul + Bhide | |
| Speed | 3 | 5 | 3 | 3+5=8 |
| Time | 20 | 12 | 4 | ?? |
| distance | 60 | 60 | 3×4=12 | 48 |
Run std on last column
Speed x time = distance
8 x time = 48
Time =48/ 8 =6 days
Final table looks like this
| Abdul | Bhide | Abdul started | Abdul + Bhide | |
| Speed | 3 | 5 | 3 | 3+5=8 |
| Time | 20 | 12 | 4 | 6 |
| distance | 60 | 60 | 3×4=12 | 48 |
Question: How long did the work last?
X started and X+Y finished.
Look at the “time” cells of their respective columns. (i.e. last column and second last column)
Total time= 4 + 6=10 days.
Final Answer: work lasted for 10 days.
Case: A starts, B joins, How long did B work?
Q. Abdul & Bhide can do a job alone in 10 days and 12 days respectively. Abdul starts the work & after 6 days Bhide also joins to finish the work together. For how many days Bhide actually worked on the job ?
Take LCM of (10,12)=60
After 6 days, Bhide also joins. That means for the first 6 days, Abdul worked alone.
Fill up the table.
| Abdul | Bhide | Abdul starts | Abdul+Bhide | |
| Speed | 6 | 5 | 6 | (6+5)=11 |
| Time | 10 | 12 | 6 days | ?? |
| Distance | 60 | 36 | (60-36)=24 | |
In the first 6 days, Abdul finished 6 x 6 = 36 kilometers. Hence remaining work= 60 minus 36= 24
This 24 kilometers were done by Abdul+Bhide together. (last column)
Run the STD on last column
Speed x time = distance
11 x Time =24
Time =24/11= approx. 2.18
Final answer: Bhide worked for 2.18 days.
Case: A and B start together but A leaves
Abdul and Bhide can do a job in 15 days & 10 days respectively. They began the work together but Abdul leaves after some days and Bhide finishes the remaining job in 5 days. After how many days did p leave?
Take LCM of (15, 10)=30 and fill up the table
| Abdul | Bhide | A+B work together | only Bhide | |
| Speed | 2 | 3 | 3+2=5 | 3 |
| xTime | 15 | 10 | ?? | 5 |
| Distance | 30 | ?? | 15 | |
According to last column, “Bhide” covered 3x 5= 15 kilmeteres out of total 30 km.
Hence the remaining 30 minus 15 = 15 kilometers were covered by A+B together
their combined speed is (3+2)= 5.
Run the STD formula on (A+B) column
5 x time = 15
Time = 15/5=3
Means P+Q worked together for 3 days. And then P left.
Final answer: P left after 3 days.
Case: Efficiency given: A is thrice good than B
Q. Abdul is thrice as good as workman as Bhide and therefore is able to finish a job in 60 days less than Bhide. Working together, they can do it in how many days?
Abdul is thrice as good means if speed of Bhide is “b”, then
Speed of Abdul = 3b
Given: Abdul can finish job in 60 days less than Bhide
Means if B can do work in t days, then A can do it in t-60 days
Fill up the table
| A | B | |
| Speed | 3b | b |
| Time | (t-60) | t |
| distance |
The distance covered in each column is same, therefore
(Speed x time)A’s column = (speed x time) B’s column
3b x (t-60)=b x t
3(t-60)=t
t=90
Update the table with this value of “t”
| A | B | |
| Speed | 3b | b |
| Time | 90-60=30 | 90 |
| distance |
In B’s column apply STD formula so you get
b x 90 =90b that is our total distance.
When A+B work together, they’ve to cover the same distance (90b)
| A | B | A+B | |
| Speed | 3b | b | 3b+b=4b |
| Time | 90-60=30 | 90 | ?? |
| distance | b x 90=90b | 90b |
Apply STD in last column
4b x time =90b
Time =22.5 days
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Thank you so much for ur constructive support and all ur efforts
I think answer to 1st case,1 subquestion I.e Kalamadi left the work 3days before should be..12.9 days.
Check and correct me , if I m wrong
@Ashish, you have mis-read the question, like I did in first attempt and ended up with the same answer. You must have calculated total time as 42/5 and substracted 3 from that. There is the problem, the time is given considering he already left 3 days before. So equation would be like this,
[(t-3)/14]+[t/21] = 1
t = 10.2
the answer (10.2) is correct.
If raja goes to jail, the answer would be 9.6 days
Question 1 can also be interpreted in this manner that they were supposed to finish the work together in three days, but one was required to go to jail three days before project completion?
A does half as much work as B in three fourth of the time. IF together they take 18 days to complete the work, how much time shall B take to do it?
may i know whether you found how to get the answer
time 2.18 of q 3 is of both not only of bhind
kindly help with below question-
my answer- 48.
actual answer-24
twenty four men can complete a work in 16 days.
thirty two women can complete the same work in 24 days.
sixteen men & sixteen women started working and worked for twelve days
how many more men are to be added to complete the remaining work in 2 days?
thanks.
it will be better if u could also write ur procedure of solving this Q. Nevertheless,I think u made mistake by not subtracting 48(men req to complete work in 4 days)-24(men already present)=24(men req)………………….hope this will help
24M*16=32W*24
M=2W
So 16*2W+16W = 48(Total work)
w= 1
m=2
now no work left to be done as m again getting 48.
please help
M=2W
16M+16W=24M
we already know 24M can complete work in 16 days
24M work for only for 12days and remaining needs to be completed by additional men + existing men to complete in 2 days
24Mx4=nx2
n=48(this is existing men of 24 + additional men)
24 M*16=32 W* 24
therefore 1 M = 2W
( 16 M +16 W )12 + ( n M+ 16 W ) 2=24 * 16 M
substituting M = 2 W, and solving you’ll get n=40
now this 40= existing 16 Men + extra men
extra men= 40 -16 = 24
A B C together do work in 10 days. B C together work thrice as much as A and A and B together work four time as much as C in what time thay can do each other ?
sir i follow your method to solve time work quesiton but in this method how to put your method plz guide me
How to solve this one with STD approach:
A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
How to solve this one also with STD approach:
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
A. 6 hours B. 10 hours
C. 15 hours D. 30 hours
14 hours
fill up the std table with values
what will be the distance ??
sorry :P i got it..silly me
How to solve this one with STD approach:
A pump can fill a tank with water in 2 hours. Because of a leak, it took (7/3) hours to fill the tank. The leak can drain all the water of the tank in:
How to solve this one also with STD approach:
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
A. 6 hours B. 10 hours
C. 15 hours D. 30 hours
I would like to ask one question, for which i am not able to get the solution from above cases. Please help me in solving the question.
Ques: Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked 1/3 as efficiently as he actually did, the work would have been completed in 3 days. To complete the job alone A would require how many days?
Reply ASAP.
Suppose A alone can do the work in a days.
Suppose B alone can do the work in b days.
By First statement —- 1/a + 1/b = 1/5
Second statement — A works twice as efficiently as he did means if earlier he was taking a days now he will take only a/2 days. Similarly B will take 3b days .
Thus,
2/a + 1/(3b) = 1/3
Solve above two to get value of a.
Suppose A alone can do the work in a days.
Suppose B alone can do the work in b days.
By First statement —- 1/a + 1/b = 1/5
Second statement — A works twice as efficiently as he did means if earlier he was taking a days now he will take only a/2 days. Similarly B will take 3b days .
Thus,
2/a + 1/(3b) = 1/3
Solve above two to get value of a.
Thanks Mrunal
person A can do a job in x days
person B can do a job in y days
part of job done by person A alone in one day is 1/x th part
part of job done by person B alone in one day is 1/y th part
part of job done by A and B together in one day is 1/x + 1/y th part
So total days required to do the job is 1/(1/x +1/y) { Suppose I can do 1/5th part of job in one day so I will need 1/(1/5) days i.e 5 days to complete the job}
This is the only theory …there are no other rules or tricks
The origin to this theory is that you cannot add inversely proportional relations
mukul and vipin together do a work in 6 days. if mukul does 1/5 part of the work in 2 days, in how many days will vipin alone do the remaining work ?
Mukul does 1/5 work in 2days therefore in one day he can do 1/10 work. So he alone can complete work in 10 days
Vipin and mukul together= 1/6
Vipin alone= 1/6 – 1/10 = 1/15
So vipin alone can do in 15 days.
But 1/5 work has already been done by Mukul so the remaining work is 4/5
Vipin does 1 work in 15 days.
He can do 4/5 work in 15X4/5 days = 12 days
12??
Total combined efficiency= 100/6= 16.67
Mukul can do 100 percent wrk in = 5*2=10
Mukul efficiency= 100/10=10
therefore Vipin efficiency= 16.67-10= 6.67
Work left=4/5=80%
therefore numberbof days for vipin to complete= 80/6.67= 12 days??
m’s 1day work= 1/10
(m+v)’s 1day work=1/6
v’s 1day work= 1/6-1/10 = 4/60
v takes to complete the work alone= 60/4=15 days.
hope im clear
i think 7 days
12 is the right ans
The shape of a garden is rectangular in the middle and semi circular
at the both ends . Find the area and the perimeter
of this garden.
total length of garden- 20 m
bredth of garden- 7 m
i got the area right( 91+38.5)
but not getting perimeter.
my answer- perimeter of rectangl portion=40 & perimtr of circle=22, so in total =62
but answer is 48
kindly help with this…
thanks…
You are getting the area right so you are clear that garden is a rectangle of 7*13 and 2 semi circles of radius 3.5 . These semi circles are adjacent to the sides of length 7.
Now perimeter 13*2 (free sides) + 2*22/7*3.5 (2*22/7*r) = 48
24M*16=32W*24
M=2W
So 16*2W+16W = 48(Total work)
w= 1
m=2
now no work left to be done as m again getting 48.
please help
sunny complete que liko plz y tnw ka
got it.i was double counting the perimeter.
thanks!
permiter of the garden is 76m
1 more for you-
kindly help with below question-
my answer- 48.
actual answer-24
twenty four men can complete a work in 16 days.
thirty two women can complete the same work in 24 days.
sixteen men & sixteen women started working and worked for twelve days
how many more men are to be added to complete the remaining work in 2 days?
thanks.
16x24M=32x24W => M=2W ….Now the total work is 16x24M …. 16M+16W = 24M …work completed 24x12M …remaining 24x4M means 96M work still remaining , 24M can do 24M work per day,means they will complete 48M work,remaining is 48M it needs 24 more Men or 48 more women.
can any1 plzz help me with this ques?
20 women can finish a job in 20 days.after each day, a woman is replaced by a man and a man is twice as efficient as a woman.on which day does the job get completed?
ans is 18th day.
In question, 20 women can finish a job in 20 days. assume every women has 2 units then total unit of work is 20*2*20=800 unit(20 women has 2 unit each and work for 20 days).
Second term of question, After each day a women is replaced by a man and a man is twice as efficient as a women,
hence, man unit is 4 (because it is twice as efficient as a women)
so a woman does 2 units of work in a day and man does 4 units of work in a day
starting with 20 women initially
day 1…40 units completed
day 2…38 + 4 = 42 units (1 women replace by 1 men)
day 3 ..36 + 8 = 44 units (2 women replace by 2 men)
day 4 ..34 +12 = 46 units(3 women replace by 3) men) etc.. etc…
now we have to find in how many days this will add to 800 unit
easy way to do this,
adding each day
day1+day2+day3+day4…..=800 unit.
in 15 days … total work done = 810 unit.
so work gets over on the 15th day
Answer is 15th days not 14th day
sorry typeing errror…answer to above question is 14th day…..
A can do a piece of work in 36 days, B in 54 days and C in 72 days. All the three began the work together but A left 8 days and B 12 days before the completion of the work. How many days in all did C put in till the entire work was finished?
pls solve it in STD method
in last question why a and b work together for 90 days,in rest of the questions its not like that
Two pipes L and M can fill a tank in 15 and 12 h respectively and a 3rd pipe N can empty it in 4 h. If the pipes are opened at 8 am 10 am and 11 am respectively find the time the tank will be emptied.
the answer is 16.50.
thanku mrunal sir
Bhai thanks
Itna bhadiya concept samjhane k liye
Mrunal Bhai, I am expert in Time and Work type question. all because of you.
Thank you so much.