- 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

**visit ****Mrunal.org/aptitude****, For more aptitude articles **

## 88 Comments on “[Speed Time Work] Two Men can finish a work, A is 3x more efficient than B, B leaves before completion & variety of cases”

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