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by Gurpinder » Wed Jul 07, 2010 5:38 pm
One work crew can do a job in 8 days. After the first crew worked 3 days,
a second crew joins them, and together, the two crews finish the job in 3
more days. How long would it take the second crew to do the job alone?



i am confused how you would write the part where the second crew comes in after 3 days.?!?!?!?!?

thanks for all your help!
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by Rich@VeritasPrep » Wed Jul 07, 2010 5:57 pm
Work = Rate * Time

w_1 = r_1 * t_1

w_1 = (1 job / 8 days) * (3 days)

w_1 = 3/8 job

There is 5/8 of the job left:

5/8 job = (r_1 + r_2) * 3

5/8 job = (1/8 + r_2) * 3

5/8 job = 3/8 + 3r_2

3r_2 = 1/4

r_2 = 1 job / 12 days

Answer: 12 days

Make sense?
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by Gurpinder » Thu Jul 08, 2010 6:28 am
raz1024 wrote:Work = Rate * Time

w_1 = r_1 * t_1

w_1 = (1 job / 8 days) * (3 days)

w_1 = 3/8 job

There is 5/8 of the job left:

5/8 job = (r_1 + r_2) * 3

5/8 job = (1/8 + r_2) * 3

5/8 job = 3/8 + 3r_2

3r_2 = 1/4

r_2 = 1 job / 12 days

Answer: 12 days

Make sense?

Hey raz,

I am sorry but I am still confused. I use the table method to solve these problems. can you please explain this one using the table method. (alone, rate,time,work).

your help is greatly appreciated.
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by silencz » Thu Jul 08, 2010 7:29 am
yes, can this answer please be explained in another way. I am not familiar with these "_" symbols and there seems to be alot of these type of questions on the gmat.
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by GMATGuruNY » Thu Jul 08, 2010 7:54 am
Gurpinder wrote:One work crew can do a job in 8 days. After the first crew worked 3 days,
a second crew joins them, and together, the two crews finish the job in 3
more days. How long would it take the second crew to do the job alone?



i am confused how you would write the part where the second crew comes in after 3 days.?!?!?!?!?

thanks for all your help!
First, the rate formula. If r = rate, t = time, and w = work:

r * t = w.

Now a tip for work problems:

When the job is undefined, plug in your own number for the job.

Since the problem discusses a time of 8 days and a time of 3 days, the value of the job should be divisible by 8 and 3 in order to make the math easier. Let's say job = 24 bricks.

The first crew can complete the job of 24 bricks in 8 days. r = w/t = 24/8 = 3 bricks/day for the first crew.

The first crew works for 3 days. So it lays 3 bricks/day for 3 days. w = r * t = 3 * 3 = 9 bricks. So the first crew lays 9 of the 24 bricks. 24 - 9 = 15 bricks left to lay.

The two crews together to lay the remaining 15 bricks in 3 days. r = w/t = 15/3 = 5 bricks/day.

So when the 2 crews work together, they lay 5 bricks/day. We know that the first crew lays 3 bricks/day. So the second crew must lay 5 - 3 = 2 bricks/day.

Since we have 24 bricks, and the 2nd crew lays 2/day, to complete the whole job by itself the second crew will need t = 24/2 = 12 days.

Clear?
Last edited by GMATGuruNY on Thu Jul 08, 2010 8:05 am, edited 2 times in total.
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by kvcpk » Thu Jul 08, 2010 8:02 am
Gurpinder wrote:One work crew can do a job in 8 days. After the first crew worked 3 days,
a second crew joins them, and together, the two crews finish the job in 3
more days. How long would it take the second crew to do the job alone?



i am confused how you would write the part where the second crew comes in after 3 days.?!?!?!?!?

thanks for all your help!
Let me explain a simple strategy for such problems.. The fractions are more confusing.
So let us forget fractions for sometime.

Take LCM of all the numbers you see in the question for days.
8 and 3 .. So LCM is 24.

Now, assume that there is 24 units of work to be done.

Let the workers be A and B.
It is given that A can complete the work in 8 days.

So 24 units can be done in 8 days implies that per day A does 3 units of work.

A worked for 3 days.
So how many units of work is over? 3*3units = 9 units.

How many units is left? 24-9 =15 units.

Now B joins A. Let us assume that B does x units per day.
then together they will do (x+3) units per day.

They completed the work together in 3 days.
So per day how many units they have done?

15/3 = 5 units per day.
so (x+3)=5 -> x=2

hence B can do 2 units per day.
Now we have 24 units. So B can complete the work in 24/2 = 12 days.

Hope this helps!! Let me know if you have any difficulty. Solving work-rate problems this way saves time and calculaltions are very easy.
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by Gurpinder » Thu Jul 08, 2010 8:27 am
Thanks for the awsomeeee tip GmatGURUNY. "When the work is undefined, plugin a value." I will remember that.


@ kvcpk

thxx for the solution. your way of solving the problem or explaning how to solve it was reallyyyyyyy simple. I actually get it now.

The thing is, i have been doing math refreshing for some time now and every guide taught me to take the table approach:

Image


i am in fact very very confortable using the table. would it be possible to tackle this problem using the table?

if so, can you please explain

thank you soo much
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by silencz » Thu Jul 08, 2010 9:09 am
Thank you, that is very clear. To make sure I'm getting this, let me ask a simpler question.

If Susie takes 4 hours to paint an entire house and Bob takes 3 hours to paint a house, how long would it take if they were to paint the house together?

Use the number 12 for work to simplify the math

r=12/4, so Susie paints portions of the house at a rate of 3 per hour

r=12/3, Bob paints at 4 per hour

together they paint at a rate of 4+3=7 per hour.

w=r*t
12=7*t
so it would take them 1.71 hours, so a little under 2 hours together?
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by Gurpinder » Thu Jul 08, 2010 9:12 am
silencz wrote:Thank you, that is very clear. To make sure I'm getting this, let me ask a simpler question.

If Susie takes 4 hours to paint an entire house and Bob takes 3 hours to paint a house, how long would it take if they were to paint the house together?

Use the number 12 for work to simplify the math

r=12/4, so Susie paints portions of the house at a rate of 3 per hour

r=12/3, Bob paints at 4 per hour

together they paint at a rate of 4+3=7 per hour.

w=r*t
12=7*t
so it would take them 1.71 hours, so a little under 2 hours together?


the problem you outline is a simple work problem silencz. this can be solved easily. the tricky thing with the one i posted was that one of the crew HAD already been working on a project for 3 days and the next crew joined them after 3 days.
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by GMATGuruNY » Thu Jul 08, 2010 9:16 am
silencz wrote:Thank you, that is very clear. To make sure I'm getting this, let me ask a simpler question.

If Susie takes 4 hours to paint an entire house and Bob takes 3 hours to paint a house, how long would it take if they were to paint the house together?

Use the number 12 for work to simplify the math

r=12/4, so Susie paints portions of the house at a rate of 3 per hour

r=12/3, Bob paints at 4 per hour

together they paint at a rate of 4+3=7 per hour.

w=r*t
12=7*t
so it would take them 1.71 hours, so a little under 2 hours together?
Perfect!
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by Balla » Wed Aug 09, 2017 8:21 pm
One work crew can do a job in 8 days. After the first crew worked 3 days,
a second crew joins them, and together, the two crews finish the job in 3
more days. How long would it take the second crew to do the job alone?


after 3 days they do 3/8 of the job so have 5/8 of the job left

together they do 5/8 of job in 3 days

(5/8)/3 gives how much of the job they do in one day together.

they do 5/24 of the job per day together and one work crew does 1/8 or 3/24 of the job per day.

The second crew must do 5/24 - 3/24 of the job per day so 2/24 or 1/12 of the job per day.

Hence, they can do the entire job in 12 days.
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