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

In Aptitude by Support Staff

  1. Case: A & B work start together but B leaves
  2. Case: A started work, B joined after 4 days, how days work lasted?
  3. Case: A starts, B joins, How long did B work?
  4. Case: A and B start together but A leaves
  5. 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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