- 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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Mrunal sir.. I’m deaf man. I understand your explanation clearly. Now I’ve some doubts. Please clear my doubts before having bank exam next week.
Here some questions :
1) A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
2) 4 men can complete a piece of work in 2 days. 4 women can complete the same piece of work in 4 days
whereas 5 children can complete the same piece of work in 4 days. If 2 men, 4 women and 10 children
work together, in how many days can the work be completed?
10 men and 15 women can finish a work in 6 days. One man can alone finish a work in 100 days. In how many days will a woman finish the work ?
sir how will it be solved by Std table?
Can someone solve this question with the above mentioned technique:
Two pipes A and B can fill a tank in 24mins and 32 mins. If both pipes are opened together,after how much time B should be closed so that tank is full in 18 mins. [Ans. 8mins]
(18/24)+(x/32)=1
its very easy. speed of a comes out to be 4 . speed of b= 3 . total distance or time = 96. a+b = 7 . now take time of a+b = x and a alone to be 18 – x . now the equation will be 7x + 4 ( 18 – x ) = 96 . solve it and you will get the value of x = 8 min .
total distance =96 and not time
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Sir, how to solve using STD table A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is?
please tell me ho to solve by std approach
a and b can do a job in 16 and 12 days respectively 4 days before completion of job a joins b find out how many days b has worked alone if b has started work alone ???????
a 6
b 4
c5
d7
answer is c ie 5 days???
A can do a piece of work in 80 days.he works at it for 10days and then B alone finishes the remaining work in 42 days .in how much time will A and B working together , finish the work????
pls solve this using STD method …..
ANS: 30 DAYS
Tejaswi yu got the answer????,
yes answer is 30 days
i got it
Hello Mrunal Sir,
Thanks for these great tricks. These are very helpful and I could understand the topic in very small time.
4men and 6women get RS.1600 by doing a piece of work in 5days.3men and 7women get RS.1740 by doing the same work in 6 days.In how many days,7 men and 6 women can
complete the same work getting RS. 3760
10M+ 15W=6days, 1M=100days hence 10M=10 days
1/10 days + 15W= 1/6—-> 15W=15 days , 1W= 225 days
A can do a job in 24 day,B is 60% efficent than A. In how many days B will complete the work please help but through ISD table
A B
S a 1.6a
T 24
D 24a 24a
So time taken by B= 24a/1.6a =15 days.
Let me know if the answer is wrong. :)
my name is manisha ,i applied for ibps po . please guiide me how to prepare for exam .i am 27 years old so i am so worried about job as i am in general category
Sir, how to solve A&B can do work in 12 days,B&C IN 15 DAYS,C&A IN 20 DAYS.ALLCOMBINED complete IN HOW MANY DAYS.SOLVE BY STD TABLE
A+B B+C C+A A+B+C
S 5 4 3 6 (12/2)
T 12 15 20
D 60 60 60 (LCM OF 12,15,20) 60
Time for A+B+C = 60/6= 10 Days
Sir in the last question if we take speed of A as “a” and speed of B as ” a/3″ then final answer will be different. How to know the right method for this problem? Thanks
Sir, unable to find pipes and cistern article…
A can do a piece of work in 80 days.he works at it for 10days and then B alone finishes the remaining work in 42 days .in how much time will A and B working together , finish the work????
Please anyone can help to solve this by STD method?
answer is 30 days???
We consider speed of A as ‘a’
A After 10 days(A) Remaining work(done by b)
S a a
T 80 10 42
D 80a 10a 80a-10a= 70a
so speed of b= 70a/42= 5a/3
Now
A+B
S a+5a/3 = 8a/3 so time= D/S = 30 days
T
D 70a
Hope this helps.. :)
Krishna I need STD method for that question can you help me???
What the hell there is no other person Active hereto solve us problems. Please Atleast thoes people are interested in cracking Apti please be here or tell any best blog where we can practice the questions by discussing..
Mrunal Sir …you are really smart teacher…what the best method you gave to us..thanks a lot
Mrunal sir …pleases
suggest me list of books for reasoning and aptitude
A is twice as efficient as B. B started the work and after 4 days A joins and the total work completed in 9 days. In how many days B can complete the work alone.
Can you let me know the answer first?
Let, B’s rate of doing work be r
Then A’s – 2r
4r + 5(3r) =1
r= 1/19
B takes 19 days
Mrunal Sir.. I m stuck in between when trying to solve the question by ur approach.
Two pipes can fill a small tank in 7 hrs and 8hrs respectively. A leakage was found when two pipes are opened together and it took 16min more to fill it up. How much time the leakage will take to empty the full tank?
Answer – 56 hrs.
let work be 56 units.
Together they complete in 4 hrs – Pipe A & B plus the leak.
But had only Pipe A & B open they would have taken 56/15 hrs.
Implies they had to work more for 4/15 hrs.
In that duration they work 4/15*15 = 4 units more.
This is the negative work done by the leak in 4 hrs.
Thus to complete 56 work it takes 56 hrs
A and B together can complete a task in 18 hours. A is 20% more efficient than B. A total of 66000 words have to be typed. How many words can A type in one hour?
The answer should be 2000 words
Solution :
The total effort(S) is A+B = 1.2 B+ B= 2.2 B
Total Time(T) is 18 h
Total Work (D) is 66000
2.2 B= 66000/18
B=1666.66
A= B* 1.2
A=2000
23 days