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If Dave works alone he will take 20 more hours to complete a

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If Dave works alone he will take 20 more hours to complete a [#permalink] New post 29 Apr 2009, 00:08
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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
[Reveal] Spoiler: OA

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Re: Working Together [#permalink] New post 02 May 2009, 22:06
B. 2: 1

Wd = Time to complete when Dave works alone
Wn = Time to complete when Donna works alone
Wt = Time to complete when work together

Wd-Wt = 20 ---(1)
Wn-Wt = 5 ----(2)
1/Wd + 1/Wn = 1/Wt ---(3)

solve these 3 equations , you'll get (Wd)^2 = 20 (Wn)^2 = 5
so Wd/Wn = 2:1
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Re: Working Together [#permalink] New post 03 May 2009, 00:15
Lets say Dave takes 50 hrs.
then Dave + Diane takes 30 hrs.
Since Diane take 5 hrs less so she takes 25 hrs.

Ratio of Dave/Diane = 50/25 = 2/1.
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Re: Working Together [#permalink] New post 03 May 2009, 12:54
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meaningful wrote:
B. 2: 1

Wd = Time to complete when Dave works alone
Wn = Time to complete when Donna works alone
Wt = Time to complete when work together

Wd-Wt = 20 ---(1)
Wn-Wt = 5 ----(2)
1/Wd + 1/Wn = 1/Wt ---(3)

solve these 3 equations , you'll get (Wd)^2 = 20 (Wn)^2 = 5
so Wd/Wn = 2:1


Hi meaningful, can you explain a bit on how you get equation (3) 1/Wd + 1/Wn = 1/Wt. I'm not completely following the logic.

mdfrahim wrote:
Lets say Dave takes 50 hrs.
then Dave + Diane takes 30 hrs.
Since Diane take 5 hrs less so she takes 25 hrs.

Ratio of Dave/Diane = 50/25 = 2/1.

Hi mdfrahim, I don't think this is quite right. Diane would have taken 5 more hours to complete the task, if she worked alone, so it would be 35 hours. I don't think Plug and chug method would work on this problem unless you are able to pick out the correct starting time.
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Re: Working Together [#permalink] New post 03 May 2009, 16:08
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meaningful wrote:
B. 2: 1

Wd = Time to complete when Dave works alone
Wn = Time to complete when Donna works alone
Wt = Time to complete when work together

Wd-Wt = 20 ---(1)
Wn-Wt = 5 ----(2)
1/Wd + 1/Wn = 1/Wt ---(3)

solve these 3 equations , you'll get (Wd)^2 = 20 (Wn)^2 = 5
so Wd/Wn = 2:1


I agree with your setup, but you should not at the end find that (Wd)^2 = 20 or that (Wn)^2 = 5 -- clearly in that case Diana doesn't take 15 hours less time than Dave. You should find that together they take 10 hours, that Diana alone takes 15, and that Dave alone takes 30.

meaningful is just using the standard combined rates formula here: if Dave finishes a job in x hours, and Diana finishes a job in y hours, and together they finish the job in t hours, then

1/x + 1/y = 1/t

Here, x = t + 20, and y = t + 5. Solve the following:

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

and you'll find that t = 10.

______

There's another way one could look at the problem. If together they take t hours, then Diana takes t+5, and Dave takes t+20. If we had two workers like Diana, they would take (t+5)/2 hours. We don't - Dave is slower than Diana - so t must be greater than (t+5)/2. Similarly, two workers like Dave would take (t + 20)/2 hours, but we don't have two workers like Dave - Diana is faster - so t must be less than (t + 20)/2. So,

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

Notice then that Diana's time (t+5) is between 10 and 25, Dave's time (t+20) is between 25 and 40, and the ratio of their times must be between 25/10 and 40/25. Only one answer is in that range.
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Re: Working Together [#permalink] New post 03 May 2009, 21:27
Thx. Ian for such a nice explanation.
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Re: Working Together [#permalink] New post 07 Oct 2009, 10:57
I was doing this via algebra and it was taking me forever, thanks for such a fast shortcut...Picking #'s is the way to go!

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Re: Working Together [#permalink] New post 07 Oct 2009, 11:12
I used backsolving:)
dave = combined+20
diana = combined+5

hence, dave-diana = 15.
Then I checked the ratios to see if I can get some multiples where the difference is 15.
2*15 - 1*15 = 15...so 2:1

However, this is obviously not the best way; just in case you are left with no weapon
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Re: Working Together [#permalink] New post 07 Oct 2009, 12:27
Nice one economist
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Re: Working Together [#permalink] New post 30 Oct 2011, 08:49
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we know that:

\frac{1}{T (diana)}+\frac{1}{T (dave)}=\frac{1}{T (together)}

\frac{1}{T+5}+\frac{1}{T+20}=\frac{1}{T}

==> T=10

So, \frac{T+20}{T+5}=\frac{30}{15}=2

Picking numbers works better and faster though.
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Re: Working Together [#permalink] New post 20 Jan 2014, 20:56
i dont understand why we are taking like this
Wd-Wt = 20 ---(1)
Wn-Wt = 5 ----(2)

Coz I wrote this step like (1/Wd)- (1/Wt)= 1/20
similarly 1/Wn - 1/Wt = 1/5

i dont understand when to use 1/x + 1/y and when to take it as x+y

Please explain
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Re: Working Together [#permalink] New post 20 Jan 2014, 22:26
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Nivigmat wrote:
i dont understand why we are taking like this
Wd-Wt = 20 ---(1)
Wn-Wt = 5 ----(2)

Coz I wrote this step like (1/Wd)- (1/Wt)= 1/20
similarly 1/Wn - 1/Wt = 1/5

i dont understand when to use 1/x + 1/y and when to take it as x+y

Please explain


Note that Wd is the "time taken by Dave alone".
Also Wt is "time taken by both together"
We are given that Dave takes 20 more hours when he works alone. So Wd = Wt + 20

You use 1/Wd and 1/Wn when working with rates.
The rate of work of Dave = 1/Wd
The rate of work of Diana = 1/Wn
Combined rate of work = 1/Wd + 1/Wn = 1/Wt
The combined rate of work will be more than the rate of work of each person alone.
This is not correct: (1/Wd)- (1/Wt)= 1/20
1/Wt - 1/Wd = 1/Wn ----> We don't know what 1/Wn is. When Dave finishes the rest of the work, he takes 20 hrs. We can't say that Diana takes 20 hrs to finish the work alone.

Another Method:
Instead, say time taken together is T hrs. Dave working alone takes T+20 hrs and Diana working alone takes T+5 hrs.
So 1/(T+20) + 1/(T+5) = 1/T
(Rates are additive)
Here you get T = 10 so time taken by Dave/time taken by Diana = 30/15 = 2/1

Yet another Method:
Or think logically
They take T hrs together to complete the work. Dave takes 20 extra hrs because Diana is not working with him. If she were working with him, in T hrs, she would have finished the work that Dave does in 20 hrs.
Time taken by Dave:Time taken by Diana = 20/T
Also, in 5 hrs, Diana completes the work that Dave was doing in T hrs.
So Time taken by Dave:Time taken by Diana = T/5

20/T = T/5
T = 10

time taken by Dave/time taken by Diana = 20/10 = 2/1
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 21 Jan 2014, 12:58
asimov 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


I am so FRUSTRATED :evil: :evil: :evil: :cry:

I keep getting \frac{1}{t+20} + \frac{1}{t+5}= \frac{1}{4}

because \frac{1}{20}+ \frac{1}{5}=\frac{1}{4}

none of these explanations are clicking

furthermore, Dave has the slowest rate...and if it took them both 10 hours to complete the task, it took Dave alone 30 hours....Diana 15 hours

that's why its 2:1

none of this crap you guys are saying...its logic based
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 21 Jan 2014, 20:15
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asimov 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


I am so FRUSTRATED :evil: :evil: :evil: :cry:

I keep getting \frac{1}{t+20} + \frac{1}{t+5}= \frac{1}{4}

because \frac{1}{20}+ \frac{1}{5}=\frac{1}{4}

none of these explanations are clicking

furthermore, Dave has the slowest rate...and if it took them both 10 hours to complete the task, it took Dave alone 30 hours....Diana 15 hours

that's why its 2:1

none of this crap you guys are saying...its logic based


There is one thing wrong here: \frac{1}{t+20} + \frac{1}{t+5}= \frac{1}{4}

You are adding the rates of Dave and Diana which is perfectly fine but how do they add up to 1/4? They add up to 1/t because we are assuming that it takes them t hrs to complete the work together. You can write this:
\frac{1}{t+20} + \frac{1}{t+5}= \frac{1}{t}
and then you will get t = 10

(This is 'Another Method' discussed in my post above)
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 06 Feb 2014, 11:40
Can't grasp what anyone is explaining here so here's my explanation:

To complete a job Dave must work: 1/(t+20)

To complete the same job, Diana must work: 1/(t+5)

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

Solve the equation and you get that time = 10

1/(10+20) + 1/(10+5) = 1/10 job

Therefore it takes Dave time of 30 and Diana time of 15.

This works nicely and follows the standard Work/Rate formulas.

Kudos if this helped, quote & yell at me if this doesn't make sense

:lol:

Last edited by ak1802 on 06 Feb 2014, 18:58, edited 1 time in total.
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 06 Feb 2014, 13:28
For 1/(t+20)+1(t+5) = 1/t .... how does t = 10? I don't know how to calculate it.
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 06 Feb 2014, 19:02
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bradburyj wrote:
For 1/(t+20)+1(t+5) = 1/t .... how does t = 10? I don't know how to calculate it.


Hey, I hope this helps:

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

(t+20+t+5) / [(t+5)*(t+20)] = 1/t

(t+5)*(t+20) = t(t+20) + t(t+5)

t^2 + 25t + 100 = t^2 + 20t + t^2 + 5t

100 = t^2

10 = t , plug 10 = t back into the original set-up.
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 07 Feb 2014, 17:46
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short cut for such problems:

time taken to complete the task working together = sqrt (20*5) = 10.
Dave takes 20 more hours = 10 + 20 = 30; Diane takes 5 more hours = 5 + 10 = 15.

Ratio = 30/15 = 2:1
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 31 Jul 2014, 02:33
ak1802 wrote:
bradburyj wrote:
For 1/(t+20)+1(t+5) = 1/t .... how does t = 10? I don't know how to calculate it.



(t+20+t+5) / [(t+5)*(t+20)] = 1/t

(t+5)*(t+20) = t(t+20) + t(t+5)




Could you please explain this step in detail?
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Re: If Dave works alone he will take 20 more hours to complete a [#permalink] New post 31 Jul 2014, 02:54
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KingMike782 wrote:
ak1802 wrote:
bradburyj wrote:
For 1/(t+20)+1(t+5) = 1/t .... how does t = 10? I don't know how to calculate it.



(t+20+t+5) / [(t+5)*(t+20)] = 1/t

(t+5)*(t+20) = t(t+20) + t(t+5)




Could you please explain this step in detail?


\frac{t+20+t+5}{(t+5)*(t+20)} = \frac{1}{t};

Cross-multiply: ((t+20)+(t+5))t=(t+5)*(t+20);

Expand the left-hand side: t*(t+20)+t*(t+5)=(t+5)*(t+20).
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Re: If Dave works alone he will take 20 more hours to complete a   [#permalink] 31 Jul 2014, 02:54
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