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If Dave works alone he will take 20 more hours to complete a [#permalink]

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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?

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.

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.

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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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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]

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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

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

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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