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VP  B
Joined: 07 Apr 2009
Posts: 1198
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
If Dave works alone he will take 20 more hours to complete a  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 50% (02:44) correct 50% (02:47) wrong based on 376 sessions

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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
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
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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

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
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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Karishma
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##### General Discussion
Intern  Joined: 05 Dec 2008
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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
Manager  Joined: 28 Jan 2004
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4
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.
VP  B
Joined: 07 Apr 2009
Posts: 1198
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012

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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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Joined: 24 Jun 2008
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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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Manager  Joined: 28 Jan 2004
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Thx. Ian for such a nice explanation.
Intern  Joined: 07 Oct 2009
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Location: New York, NY
Schools: NYU Bound!

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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! Director  Joined: 01 Apr 2008
Posts: 584
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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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
Manager  Joined: 18 Jul 2009
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Location: India
Schools: South Asian B-schools

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Nice one economist
Manager  Joined: 25 May 2011
Posts: 93

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1
2
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.
Intern  Joined: 07 Dec 2013
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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

Intern  Joined: 06 Jan 2014
Posts: 35
Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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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    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
Veritas Prep GMAT Instructor V
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Location: Pune, India
Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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TroyfontaineMacon wrote:
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    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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Manager  B
Joined: 26 May 2013
Posts: 91
Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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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 Originally posted by ak1802 on 06 Feb 2014, 11:40.
Last edited by ak1802 on 06 Feb 2014, 18:58, edited 1 time in total.
Intern  Joined: 15 Nov 2013
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Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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For 1/(t+20)+1(t+5) = 1/t .... how does t = 10? I don't know how to calculate it.
Manager  B
Joined: 26 May 2013
Posts: 91
Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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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.
Intern  Joined: 01 Aug 2006
Posts: 31
Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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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
Intern  Joined: 31 Jul 2014
Posts: 1
Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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ak1802 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?
Math Expert V
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Posts: 62379
Re: If Dave works alone he will take 20 more hours to complete a  [#permalink]

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KingMike782 wrote:
ak1802 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: $$t*(t+20)+t*(t+5)=(t+5)*(t+20)$$.
_________________ Re: If Dave works alone he will take 20 more hours to complete a   [#permalink] 31 Jul 2014, 02:54

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