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Work rate

by JeetGulia » Sat Jul 31, 2010 12:50 pm
If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?


A. 4 : 1


B. 2 : 1


C. 10 : 1


D. 3 : 1


E. 1 : 2
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by dinesh19aug » Sat Jul 31, 2010 1:48 pm
Is the answer 1:2??

There are two ways to solve this problem.

Approach 1: This question is a good candidate for a guess question.
If you see that Dave take 20 hours more
Diana takes 5 hours more to complete the same task.
This means that Diana rate is faster than Dave's rate
So Dave rate : Dian rate = x : y, which means that x < y
The only choice that reflects this is Answer choice E, (1:2) All other choice indicate that Dave is faster than Diana which is not true.

Approach 2:
Draw the matrix
==============================
| R | T | W
Dave | | |
Diana | | |
Together | | |

Now pick a number say 10 and assumes that this is the amount of time taken when both work together. Now fill in the values in matrix above


==============================
| R | T | W
Dave | 1/30| 10+20 = 30 | 1
Diana | 1/15| 10+5 = 15 | 1
Together | 1/10 | 10 | 1


So you can see that Dave rate : Diana Rate = 1:2 Answer

Note: This answer can also be solved by basic algebra by taking values as t1, t2 t etc and solving it, but that is too cumbersome and lengthy way. Don't waste your time on it.

- Dinesh
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by clock60 » Sat Jul 31, 2010 2:45 pm
hard to me, spend plenty of time and not sure that i am right
my answer B, reasoning
let it be
time dave=x
time diana=y
time when they work together=a
according to the problem:
x=a+20
y=a+5
substract
x-y=15
and now to the answer choices
a) x/y=4/1
x=4k, y=1k, where k +ve integer

4k-k=3k. 3k=15. k=5. it means that x=4*5=20, and y=5. now let us find time when both work together
(1/20+1/5)*a=1.a=4
but 20-4=16,and we need 20 (as dave take 20 more hours to complete alone)
so option 1 is wrong

b)x=2k,y=k. 2k-k=15, k=15, x=30.y=15
time together
1/30+1/15=3/30, a=10
30-10=20
15-10=5
all answers coincide with the problem
answers c,d,e,can be neglected
as c and d give fraction as a result . and e gives -ve number
so my answer is B
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by Asif » Sat Jul 31, 2010 8:35 pm
Lets say together their rate is x hours.

1/(x+20) + 1/(x+5) = 1/x

Solving for x we get x= +10 or -10 (we shall take +10)

now the ratio is (10+20)/(10+5) = 2/1

thus its B.
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by kmittal82 » Sun Aug 01, 2010 12:51 pm
I'm trying to do this using pure algebra, but getting a diff. answer. Can someone find out the flaw?

Time taken by Dave = x
Diana = y

Time taken when working together = xy/(x+y)
[ Because 1/x + 1/y = 1/T where T is the total time taken to finish the task]

So, we have 2 equations i.e.

x = xy/(x+y) + 20
y = xy/(x+y) + 5

Also, using these, we can see that x - y = 15

Now, lets take some simple numbers and satisfy the above. Let x = 16, y = 1

xy/(x+y) = 16/17

x/y = a + 20 / a + 5 = 16+340/16+85 = 356/101 = 3.5:1 = 7:2

Plugging in different values of x and y offer different results

Any help appreciated.
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by zareentaj » Thu Aug 12, 2010 12:29 am
Asif wrote:Lets say together their rate is x hours.

1/(x+20) + 1/(x+5) = 1/x

Solving for x we get x= +10 or -10 (we shall take +10)

now the ratio is (10+20)/(10+5) = 2/1

thus its B.
Thanks Asif.Can you explain step by step.
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by chris@veritasprep » Thu Aug 12, 2010 5:11 am
This is a great (and more difficult) work/rate question that is typical of what you might see on the test. When I teach work/rate problems I try to avoid any gimmicks and have students use the following approach, which will work for any work/rate problem:

1. Always convert times to rates. If times are given for a complete job then the rate is always equal to the inverse of the time: r = 1/t While many of these problems can be solved logically using times, it is often problematic.

2. Rates are additive. The combined rate is always equal to the sum of all the rates of the machines/people/etc. working in the problem. Rate Combined = r1 +r2 +r3.....

3. Use the Work = Rate x Time (W=RT) formula to solve for the necessary unknown in the problem at hand.

4. On more complicated problems where conditions are changing throughout the problem (machines/workers are being added or taken away) do not try to solve the problem with one equation. Treat it as multiple problems and account for changes with each new equation.

For this problem you only need the first two steps. While there are many different ways you could approach this, I prefer Asif's approach, which relies on an understanding that rates are additive. As there were some questions about the details of his approach I will elaborate.

As is the case with most work/rate problems, the information in the problem is given in times, which we must convert to rates.

Let t = the time together to complete the job. That means that time dave = t +20 and time diana = t + 5.

If we invert each of these to get the rates of dave and diana we can see that rate dave = 1/(t +20) and rate diana = 1/(t+5) and rate together = 1/t Using our understanding that rates are additive, we know that the rate dave + rate diana = rate combined. This gives us the equation that asif used in his post:

1/(t+20) + 1/(t+5) = 1/t

Now there is a little tedious algebra to isolate t and complete the problem, but it is not too bad.

As with any algebraic equation involving fractions and variables, multiply through the equation with the LCM of the three denominators to get rid of the fractions. Here that involves multiplying through by the product of the three denominators which is (t+20)(t+5)t

After this manipulation, you are left with (t)(t+5) + (t)(t+20) = (t+5)(t+20)

Simplify to get t^2 + 5t +t^2 +20t = t^2 +25t +100

Combine/Subtract like terms to get 2t^2 + 25t = t^2 + 25t + 100

Simplify again to see that t^2 = 100 and t = 10 (t cannot be -10 as it must be positive here).

If t = 10 then time dave = 30 and time diana = 15. Ratio of time dave to time diana = 30: 15 or 2:1 and the answer is B. Hope that helps!
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by missrochelle » Wed Aug 25, 2010 7:51 pm
Im looking for a few rate/work additive problems to practice as that is one of my weaknesses. Has anyone seen a good comprehensive post/article on this ? Or have any good problems to share that test the concepts mentioned above?
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by Ian Stewart » Wed Aug 25, 2010 10:17 pm
If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?
If we let t represent their time together, then Dave alone takes t+20 hours, and Diana alone takes t+5 hours.

There's a trick that can be used in a surprising number of rates problems: if we had two workers like Diana, they would do the job in half the time, so in (t+5)/2 hours. If we had two workers like Dave, they'd do the job in (t+20)/2 hours. We have one Dave and one Diana, so their time together is somewhere in the middle:

(t+5)/2 < t < (t+20)/2
t+5 < 2t < t + 20
5 < t < 20

So Diana's time, t+5, is somewhere between 10 and 25, and Dave's time, t+20, is somewhere between 25 and 40. The ratio of Dave's time to Diana's will be largest when their times are as small as possible, so must be less than 25/10 = 2.5 to 1. Similarly, the ratio must be greater than 40/25 = 1.6 to 1. So B is the only reasonable answer.
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by roh00kan » Sat Jan 29, 2011 12:15 pm
Chris (or anyone),
I need some clarifications.
Could you explain what is meant by "rates are additive"? Did you mean "rates are additive" in "work done" type problems?

In normal rate-distance type problems, rates cannot be added, right?

For example: A bus goes from Station A to Station B in 40 miles/hr and returns to the Station A taking the same route, at a speed of 60 miles/hour. What is the overall rate of the entire trip?
The overall rate will be 48 miles/hour.

In these kind of problems, we can add the times and then calculate the over all rate of the entire trip.

So can we conclude in this way,
times are additive in rate-time-distance type problems and rates are additive in rate-time-work_done type problems?

Thank You!
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by GMATGuruNY » Sun Jan 30, 2011 5:18 am
JeetGulia wrote:If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?


A. 4 : 1


B. 2 : 1


C. 10 : 1


D. 3 : 1


E. 1 : 2
The algebra in this problem becomes a bit messy. The good news is that we can turn this into a straight arithmetic problem by plugging in the answer choices and a value for the job.

Since Dave takes longer to complete the job alone than does Diana, eliminate E.
Given that Dave's time alone is only 20-5 = 15 more hours than Diana's time alone, it is impossible that (Dave's time alone):(Diana's time alone) = 10:1. Eliminate C.

Answer choice D:
(Dave's time alone):(Diana's time alone) = 3:1

To determine everyone's respective rates, let's plug in a temporary value for the job. Let job = 3.
Rate for Dave alone = w/t = 3/3 = 1 per hour.
Rate for Diana alone = w/t = 3/1 = 3 per hour.
When people work together, we add their rates. Combined rate for Dave and Diana together = 1+3 = 4 per hour.

Now that we've determined everyone's respective rates, we need to plug in an actual value for the job. In order for Diana to take 5 hours longer to complete the job alone, the actual value of the job should be a multiple of all the values above (1, 3, 4, 5).

Plug in job = 60.
Time for Diana alone = w/r = 60/3 = 20 hours.
Time for Dave and Diana together = w/r = 60/4 = 15 hours.
Diana alone - Dave and Diana together = 20-15 = 5 more hours for Diana alone. So far so good.

Time for Dave alone = w/r = 60/1 = 60 hours.
Dave alone - Dave and Diana together = 60-15 = 45 more hours for Dave alone.
Since Dave should take only 20 more hours when he works alone, (Dave's time alone):(Diana's time alone) must be smaller.

The correct answer is B.
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by Zerks87 » Sun Jan 30, 2011 9:20 am
GMATGuruNY wrote:
JeetGulia wrote:If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?


A. 4 : 1


B. 2 : 1


C. 10 : 1


D. 3 : 1


E. 1 : 2
The algebra in this problem becomes a bit messy. The good news is that we can turn this into a straight arithmetic problem by plugging in the answer choices and a value for the job.

Since Dave takes longer to complete the job alone than does Diana, eliminate E.
Given that Dave's time alone is only 20-5 = 15 more hours than Diana's time alone, it is impossible that (Dave's time alone):(Diana's time alone) = 10:1. Eliminate C.

Answer choice D:
(Dave's time alone):(Diana's time alone) = 3:1

To determine everyone's respective rates, let's plug in a temporary value for the job. Let job = 3.
Rate for Dave alone = w/t = 3/3 = 1 per hour.
Rate for Diana alone = w/t = 3/1 = 3 per hour.
When people work together, we add their rates. Combined rate for Dave and Diana together = 1+3 = 4 per hour.

Now that we've determined everyone's respective rates, we need to plug in an actual value for the job. In order for Diana to take 5 hours longer to complete the job alone, the actual value of the job should be a multiple of all the values above (1, 3, 4, 5).

Plug in job = 60.
Time for Diana alone = w/r = 60/3 = 20 hours.
Time for Dave and Diana together = w/r = 60/4 = 15 hours.
Diana alone - Dave and Diana together = 20-15 = 5 more hours for Diana alone. So far so good.

Time for Dave alone = w/r = 60/1 = 60 hours.
Dave alone - Dave and Diana together = 60-15 = 45 more hours for Dave alone.
Since Dave should take only 20 more hours when he works alone, (Dave's time alone):(Diana's time alone) must be smaller.

The correct answer is B.

I got the same answer but I did it a little differently and I was wondering whether my method was sound because I feel like for a problem like this it would save me alot of time on the test.

I picked a good number for the combined time, just set combined time = to 10, then Dave's time alone must be 30 and Diana's time alone would be 15. Therefore the ratio of time Dave took to the time Diana took is 2:1, picked B.

Any reply would be greatly appreciated
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by GMATGuruNY » Sun Jan 30, 2011 1:10 pm
Zerks87 wrote:
GMATGuruNY wrote:
JeetGulia wrote:If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?


A. 4 : 1


B. 2 : 1


C. 10 : 1


D. 3 : 1


E. 1 : 2
The algebra in this problem becomes a bit messy. The good news is that we can turn this into a straight arithmetic problem by plugging in the answer choices and a value for the job.

Since Dave takes longer to complete the job alone than does Diana, eliminate E.
Given that Dave's time alone is only 20-5 = 15 more hours than Diana's time alone, it is impossible that (Dave's time alone):(Diana's time alone) = 10:1. Eliminate C.

Answer choice D:
(Dave's time alone):(Diana's time alone) = 3:1

To determine everyone's respective rates, let's plug in a temporary value for the job. Let job = 3.
Rate for Dave alone = w/t = 3/3 = 1 per hour.
Rate for Diana alone = w/t = 3/1 = 3 per hour.
When people work together, we add their rates. Combined rate for Dave and Diana together = 1+3 = 4 per hour.

Now that we've determined everyone's respective rates, we need to plug in an actual value for the job. In order for Diana to take 5 hours longer to complete the job alone, the actual value of the job should be a multiple of all the values above (1, 3, 4, 5).

Plug in job = 60.
Time for Diana alone = w/r = 60/3 = 20 hours.
Time for Dave and Diana together = w/r = 60/4 = 15 hours.
Diana alone - Dave and Diana together = 20-15 = 5 more hours for Diana alone. So far so good.

Time for Dave alone = w/r = 60/1 = 60 hours.
Dave alone - Dave and Diana together = 60-15 = 45 more hours for Dave alone.
Since Dave should take only 20 more hours when he works alone, (Dave's time alone):(Diana's time alone) must be smaller.

The correct answer is B.

I got the same answer but I did it a little differently and I was wondering whether my method was sound because I feel like for a problem like this it would save me alot of time on the test.

I picked a good number for the combined time, just set combined time = to 10, then Dave's time alone must be 30 and Diana's time alone would be 15. Therefore the ratio of time Dave took to the time Diana took is 2:1, picked B.

Any reply would be greatly appreciated
If you reasoned that the combined time had to be 10 hours, then your method is sound. If you just randomly plugged in 10 for the combined time, then you were lucky. Had you plugged in a different number, you would not have determined the correct ratio. For example, if you had said that the combined time = 15, then Dave's time alone would be 15+20 = 35 hours, Diana's time alone would be 15+5 = 20 hours, and the ratio would be 35:20 = 7:4.
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by tgou008 » Mon May 02, 2011 7:27 am
If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?

A. 4 : 1

B. 2 : 1

C. 10 : 1

D. 3 : 1

E. 1 : 2

MY APPROACH:

DA = Time taken by Dave working alone
DI = Time taken by Diana working alone
T= Total time taken when DA and DI work together

DA = T + 20
DI = T + 5

What is the ratio of DA / DI?
I took a little bit of a different approach for this; I plugged in numbers and solved that way. I also decided to use 10 because it is either a factor or multiple of 5 and 20.

Therefore
DA = 10 + 20 = 30
DI = 10 + 5 = 15

Therefore, DA / DI = 30/15 = 2/1 = OPTION B

I see from reading this thread that if I had picked another number, say 100 I would have arrived at quite a different answer. What then should have been the correct justification for picking the number 10? As I may not be soo lucky going forward.

Thanks
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by mannuunnam » Fri Jun 03, 2011 8:04 am
If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?

A. 4 : 1

B. 2 : 1

C. 10 : 1

D. 3 : 1

E. 1 : 2

My approach is

let D = dave; Di diana

Given that,

D = (D+Di) + 20 and

Di = (D+Di) + 5

Pick a number for (D+Di), let say 10 which is easy to calculate.

D = 10 + 20 => 30

Di = 10 + 5 => 15

D/Di = 30/15 => 2:1

I hope it can be done this way...
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