How Cheenta works to ensure student success?

Explore the Back-StoryProblems and Solutions from CMI Entrance 2022. Learn More

Contents

[hide]

Try this beautiful problem from PRMO, 2017 based on Time and work.

A contractor has two teams of workers : team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins team A after four days. The team A withdraws after two more days. For how many more days should team B work to complete the job ?

- $20$
- $16$
- $13$

Arithmetic

multiplication

unitary method

But try the problem first...

Answer:$16$

Source

Suggested Reading

PRMO-2017, Problem 3

Pre College Mathematics

First hint

In the problem,we notice that first 4 days only A did the work.so we have to find out A's first 4 days work done.next 2 days (A+B) did the work together,so we have to find out (A+B)'s 2 days work.

so we may take the total work =1

A's 1 day's work= \(\frac{1}{12}\) and B's 1 day's work=\(\frac{1}{36}\)

Can you now finish the problem ..........

Second Hint

Now B did complete the remaining work.so you have to find out the remaining work and find out how many more days taken....

so to find the remaining work subtract (A's 4 day;s work + (A+B)'S 2 days work)) from the total work

Can you finish the problem........

Final Step

Let the total work be 1

A can complete the total work in 12 days,so A'S 1 day's work=\(\frac{1}{12}\)

B can complete the total work in 36 days, so B's 1 day's work=\(\frac{1}{36}\)

First 4 days A's workdone=\(\frac{4}{12}=\frac{1}{3}\)

After 4 days B joined and do the work with A 2 days

So \((A+B)\)'s 2 day's workdone=\(2 \times( \frac{1}{12}+\frac{1}{36})\)=\(\frac{2}{9}\)

Remaining workdone=\((1-\frac{1}{3}-\frac{2}{9}\))=\(\frac{4}{9}\)

B will take the time to complete the Remaining work=\(36 \times \frac{4}{9}\)=16

Hence more time taken=16

- https://www.youtube.com/watch?v=i_pmSwUO4LA
- https://www.cheenta.com/probability-problem-amc-8-2016-problem-no-21/

Contents

[hide]

Try this beautiful problem from PRMO, 2017 based on Time and work.

A contractor has two teams of workers : team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins team A after four days. The team A withdraws after two more days. For how many more days should team B work to complete the job ?

- $20$
- $16$
- $13$

Arithmetic

multiplication

unitary method

But try the problem first...

Answer:$16$

Source

Suggested Reading

PRMO-2017, Problem 3

Pre College Mathematics

First hint

In the problem,we notice that first 4 days only A did the work.so we have to find out A's first 4 days work done.next 2 days (A+B) did the work together,so we have to find out (A+B)'s 2 days work.

so we may take the total work =1

A's 1 day's work= \(\frac{1}{12}\) and B's 1 day's work=\(\frac{1}{36}\)

Can you now finish the problem ..........

Second Hint

Now B did complete the remaining work.so you have to find out the remaining work and find out how many more days taken....

so to find the remaining work subtract (A's 4 day;s work + (A+B)'S 2 days work)) from the total work

Can you finish the problem........

Final Step

Let the total work be 1

A can complete the total work in 12 days,so A'S 1 day's work=\(\frac{1}{12}\)

B can complete the total work in 36 days, so B's 1 day's work=\(\frac{1}{36}\)

First 4 days A's workdone=\(\frac{4}{12}=\frac{1}{3}\)

After 4 days B joined and do the work with A 2 days

So \((A+B)\)'s 2 day's workdone=\(2 \times( \frac{1}{12}+\frac{1}{36})\)=\(\frac{2}{9}\)

Remaining workdone=\((1-\frac{1}{3}-\frac{2}{9}\))=\(\frac{4}{9}\)

B will take the time to complete the Remaining work=\(36 \times \frac{4}{9}\)=16

Hence more time taken=16

- https://www.youtube.com/watch?v=i_pmSwUO4LA
- https://www.cheenta.com/probability-problem-amc-8-2016-problem-no-21/

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
Google