How long would each take, working alone?

[Chikis]

A and B can do a piece of work together in 15 days. They both started, but after 6 days B gets ill and then A takes 30 more days to finish it by himself. How long would each take, working alone? How do I tackle this?

 

Am havina a hard time dealing with this problem.

 

They can do the job together in 15 days.

When they both started, they did the job for 6 days. But B could not complete the job due to ill health. This means that it remaining more 9 days to complete the job if both of them are working together. But in this case A spent 30 days doing the 9 days work.

Ijust have had time dealing with this.

[studiot]

This problem is about ratio, proportion or fractions - whatever you like to call them.

 

You have nearly done the first bit, I will start you off with some notation

 

Let 'a' be the number of days A takes to complete alone

 

Let 'b' be the number of days B takes to complete alone

 

So taking the information you have deduced from the problem can you think of any equation involving a fraction and 'a'?

 

Also can you tell me what is the ratio of speeds at which A and B work?

Edited October 16, 2014 by studiot

[Chikis]

If this is a problem on ratio or propotion, it would have been more easier for me to handle. Am thinking this is a harder problem on rate.

 

For example, I would find it more easier to handle this problem:

12 men complete a job in 9 days. How many men working at the same rate, would be required to complete the job in 6 days?

 

Could you ask me to show you my work on this problem? Maybe you now show me how can apply the same method to solve the problem in op.

Edited October 16, 2014 by Chikis

[studiot]

 

Could you ask me to show you my work on this problem? Maybe you now show me how can apply the same method to solve the problem in op.

 

 

Better would be if you told me what fraction of the work remained when B fell ill.

[Chikis]

Let's the work is 1 whole. 1/9 is remaining after B fell ill because they had already done 1/6 of the work together. Am I right?

[studiot]

 

Let's the work is 1 whole. 1/9 is remaining after B fell ill because they had already done 1/6 of the work together. Am I right?

 

 

Unfortunately not. Neither the 1/9 nor the 1/6 is correct (do 1/9 + 1/6 make 1 ?)

 

Edit divide the '1 whole' into days.

 

How many days were there when the job was completed when they were both working together?

Edited October 16, 2014 by studiot

[Chikis]

You mean I should divide the 1 whole by the days?

[studiot]

 

You mean I should divide the 1 whole by the days?

 

 

Well yes that is what it comes down to, but you need to understand the answer to

 

If it takes 15 days to complete the job what fraction of the job is complete after 6 days?

 

and also

 

What fraction of the job remains to be completed?

Edited October 16, 2014 by studiot

[Chikis]

We were told that they could finish the job in 15 days. They did 6 days work remaining 9 days work, which A complete in 30 days?

 

9 days work took A 30 days.

6 days work will take A [math]\frac{6}{9}\times30[/math] = 20 days

It will take A 20 + 30 = 50 days. I believe it will take B 50 days as well.

[studiot]

 

It will take A 20 + 30 = 50 days

 

Yes correct. You are getting there, although I wanted you to say that A working alone takes 30 days to complete 9/15 of the total

 

[math]\left( {\frac{9}{{15}}} \right)a = 30[/math]

 

so

[math]a = 50[/math]

 

This was the point of my first question in post#2

 

 

 

I believe it will take B 50 days as well.

 

 

We need to do some more work on this part to find b.

 

Please use my 'a' and 'b' notation from post#2 as it will make the next bit easier.

 

 

Now that we have found 'a', can you answer my second question in post#2?

 

Edited October 16, 2014 by studiot

[Chikis]

Am assuming that since A can do 6 days work in 20 day, B could as well do 6 days work in 20 days.

[studiot]

But A and B work at different rates so each accomplishes a different amount in 6 days or 50 days or 15 days.

 

We have calculated that Mr A accomplishes the job on his own in 50 days.

 

If Mr B worked at the same rate he would also finish in 50 days on his own so together they would finish in half that time ie 25 days.

But we are told they finish faster in only 15 days so Mr B must work faster than Mr A.

 

So I ask again how much faster does Mr B work than Mr A?

[Chikis]

How do we know that they working at diffrent rates?

[studiot]

Let us do the second part of the question in two stages and it should become clear why they are working at different rates.

 

 

Let us first consider the situation where m workers are working and all m workers work at the same rate.

 

So if you like they are all clones or copies of each other.

 

So if one worker is working then the task takes T days.

 

If two workers are working the task takes T/2 days

 

If three workers are working the task takes T/3 days

 

If m workers are working the task takes T/m days.

 

We can consider this as one worker worker working m times as fast.

 

So one worker working m times as fast will complete in T/m days

 

So if one worker works 1.5 times as fast m = 1.5 and will complete in T/1.5 days

 

So days = T/m or m = T/days

 

In our situation, T =50, days =15

so m =50/ 15 = 10/3 times as fast (note this is not a whole number don't worry it will become clear)

Edited October 16, 2014 by studiot

[Chikis]

Since they are working at diffrent rates. How do I know, how long it would take B doing 6 days work all alone. If I can compute that, then I would be able to compute how long it would take B to do 15 days work all alone?

[studiot]

 

If I can compute that, then I would be able to compute how long it would take B to do 15 days work all alone?

 

 

Surely it would take B 15 days to do 15 days work?

 

This is not much better than your last comment, which I pretended didn't exist.

 

 

B could as well do 6 days work in 20 days.

 

 

 

Just bear with me and follow through what I have already said and we will get to the end.

 

Edited October 16, 2014 by studiot

[Chikis]

Am almost giving up on this.

[studiot]

 

Am almost giving up on this

 

Can we go we the idea that if C works twice as fast as A he takes half as long?

[davidivad]

two men can do the work of three. this should be accounted for...

lol

[studiot]

 

two men can do the work of three. this should be accounted for...

lol

 

 

If you can explain it better, please do, you are welcome.

 

But please don't disrupt the thread.

[Chikis]

Surely it would take B 15 days to do 15 days work?

 

This is not much better than your last comment, which I pretended didn't exist.

 

 

 

 

Just bear with me and follow through what I have already said and we will get to the end.

 

Could this be the fraction you want me to show?

We were told that they did six days work out of 15 days ways. This means it remains 9/15 days work.

Thus

9/15 + 6/15 = 1

[davidivad]

 

If you can explain it better, please do, you are welcome.

 

But please don't disrupt the thread.

my appologies.

i just felt a bit of humor. please carry on.

[studiot]

 

9/15 + 6/15 = 1

 

 

Yes this is perfectly true and you have already used it (indirectly) in solving the first part of the question to get to the 50 days.

 

The original fraction I wanted and why I said

 

 

Let 'a' be the number of days A takes to complete alone

 

Let 'b' be the number of days B takes to complete alone

 

Now you found a and have to find b.

 

Since A completes the task in 50 days

 

If B completed the task in 1 day he would be 50 times as fast.

 

If he completed it in 10 days he would be 5 times as fast.

 

So B is a/b = 50/b times as fast as A

 

Which of course means that if b =50 (it does not) the B is exactly as fast as a since 50/b = 50/50 = 1

 

This ratio or fraction is important because it connects B's contribution to A's contribution so that we can add them together as the next step.

[Chikis]

6/15 + 9/15 = 1

6/15 = 6/b + 6/a

9/15 = 30/a

6/b + 6/a + 30/a = 1

where a and b are the work are the number days A and B did the job alone? I believe this equation are okay enough to get whatever number of days B did the job alone?

[studiot]

Yes it does.

 

Well done you have produced a different method than mine for both parts of this question, although I hope I have helped.

 

+1

Edited October 17, 2014 by studiot