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# Working together, Jose and Jane can complete an assigned task in 20

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Working together, Jose and Jane can complete an assigned task in 20 [#permalink]  09 Apr 2007, 19:23
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Question Stats:

54% (03:09) correct 46% (03:24) wrong based on 59 sessions
Working together, Jose and Jane can complete an assigned task in 20 days. However, if Jose worked alone and complete half the work and then Jane takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Jose take to complete the task if he worked alone? Assume that Jane is more efficient than Jose

A. 25 days
B. 30 days
C. 60 days
D. 65 days
E. 36 days
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Jul 2015, 03:41, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Working together, Jose and Jane can complete an assigned task in 20 [#permalink]  09 Apr 2007, 19:46
ggarr wrote:
Working together, Jose and Jane can complete an assigned task in 20 days. However, if Jose worked alone and completes half the work and then Jane takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Jose take to complete the task if if he worked alone? Assume that Jane is more efficient than Jose?

Let Jose finish work in x days and Jane in y days. So to finish half task jose will take x/2 days and Jane will take y/2 days.

By given info we have x/2 + y/2 = 45 or (x + y) = 90 ---(1)---

Also when both work together we have 1/x + 1/y = 1/20
or (x+y)/xy = 1/20
0r xy = 20 * (x + y) = 1800

Since jane is efficient this means x > y

we know (x - y)^2 = (x + y)^2 - 4xy
or (x - y)^2 = 8100 - 7200 = 900
or x - y = 30
Since x + y = 45 on solving we get x = 75/2 = 37.5 and y is 7.5
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Re: Working together, Jose and Jane can complete an assigned task in 20 [#permalink]  09 Apr 2007, 20:50
Assume:

Jose does 1 job in x days, so Jose does 1/x job in a day
Jane does 1 job in y days, so Jane does 1/y job in a day

Together, they does (x+y)/xy job in a day. This is equals to 1/20. So

(x+y)/xy = 1/20
20(x+y) = xy

Next, we're told 1 job takes 45 days to complete if Jose and Jane each does half the work. So since Jose does 1 job in x days, he wil need x/2 days to do half the job. Jane similarly will need y/2 days to do the other half.

x/2 + y/2 = 45
x+y = 90

So
xy = 1800

25 days
30 days
60 days
65 days
36 days

From the answer choices, 25,30 and 36 days are out since they result in Jane being less efficient. 65 days gives a fractional days (technically allowed, but I think we're thinking of whole numbers). So I'll go for 60 days for Jose and 30 days for Jane.
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Re: Working together, Jose and Jane can complete an assigned task in 20 [#permalink]  11 Apr 2007, 12:37
jose = x days
jane = y days

Jose rate = Rx = 1/x
Jane rate = Ry = 1/y

Rx+y = 1/20

Let Tx be the time for Jose to finish the work by himself. When half the work is done,
Rx = (1/2)/Tx
Ry = (1/2)/(45-Tx)

Rx+y = Rx + Ry
1/20 = 1/2*Tx + 1/2*(45-Tx)

Solve for Tx,
Tx = 30 or 15 days

If Tx = 15 days, then Ty = 30 days. Not possible since Jane is more efficient than Jose.

Tx = 30 days.

That is for half the work. For the entire amount, Tx = 60days.
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Jose -> 1 job in x days -> 1/x job in 1 day
Jane -> 1 job in y days -> 1/y job in 1 day

Together, 1/x + 1/y = 1/20
x+y/xy = 1/20
20x+20y = xy

Jose takes x/2 days to finish 1/2 a job. Jane takes y/2 days to finish 1/2 a job. Together, x/2 + y/2 = 45 --> x+y = 90

So

20(x+y) = xy
20(90) = x(90-x)
1800 = 90x - x^2
x^2 - 90x + 1800 = 0
x = 60 days or 30 days. But we are told Jane is more efficient, so x = 60 days.
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Expert's post
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C

1. 1/2Jose+1/2Jane=45 => Jose+Jane=90 => Jose>45 (C or D)

for C:

Jose=60 => Jane=30 => 20*(1/60+1/30)=1 therefore C correct.
VP
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walker wrote:
C

1. 1/2Jose+1/2Jane=45 => Jose+Jane=90 => Jose>45 (C or D)

for C:

Jose=60 => Jane=30 => 20*(1/60+1/30)=1 therefore C correct.

this is based on assumptions,

any other inputs
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Expert's post
OK

Jone- x (days), Jane -y (days)

1. xy/(x+y)=20
2. 1/2*(x+y)=45 => x+y=90

for first: xy=20(x+y)= 1800

xy=1800 and x+y=90 => x,y e {30,60} => x=60, y=30 (x>y)

C.

first version seems is faster
VP
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walker wrote:
OK

Jone- x (days), Jane -y (days)

1. xy/(x+y)=20
2. 1/2*(x+y)=45 => x+y=90

for first: xy=20(x+y)= 1800

xy=1800 and x+y=90 => x,y e {30,60} => x=60, y=30 (x>y)

C.

first version seems is faster

i guess this is the only sound way to come to the conclusion on this problem. as it says one of the guys is more efficient than the other. so 30 and 60 is logical conclusion.

thanks Walker!
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Posts: 3573
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Kudos [?]: 2503 [0], given: 359

Expert's post
pmenon wrote:
isnt it 1/x + 1/y = 1/20 ?

sp x+y/xy = 1/20 ?

1/x+1/y=1/20 <==> xy/(x+y)=20. they are equivalent.
VP
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yeh, it might nt be from gmatleague but it is included as gmatquant practice PS
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Re: Work/Rate [#permalink]  16 Nov 2011, 01:36
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Let A represent the number of days it would take Jose to complete the task if he worked alone (i.e. 1/A is Jose's work rate).
Let B represent the number of days it would take Jane to complete the task if he worked alone (i.e. 1/B is Jane's work rate).

Working together, Jose and Jane can complete an assigned task in 20 days.
20 (1/A + 1/B) = 1 (1)

However, if Jose worked alone and complete half the work and then Jane takes over the task and completes the second half of the task, the task will be completed in 45 days.

A/2 + B/2 = 45
A + B = 90
B = 90 - A (2)

Sub in (2) into (1):

$$20 (\frac{1}{A} + \frac{1}{{90-A}}) = 1$$

$$\frac{{(90-A)+A}}{{A(90-A)}} = \frac{1}{20}$$

$$\frac{{90}}{{90A-A^2}} = \frac{1}{20}$$

$$90A - A^2 = 1800$$

$$A^2 - 90A + 1800 = 0$$

$$(A - 60)(A - 30) = 0$$

So Jose's completion time on his own can be either 30 or 60 days. However, since the problem states that he is more inefficient than Jane, we can eliminate 30. Therefore, the answer is C: 60.
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Re: Work/Rate [#permalink]  16 Nov 2011, 01:41
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Hi

As you've got numbers in the answer choices, I'd prefer using the answer choices and working backwards to arrive at the required answer. It's always good to start with the middle/third option.
Also, if you can see numbers that might be related in some way (multiples, divisors, etc), try to use them.

In the question stem, we've got the numbers 20 and 45 - both are multiples of 5. So, I'd personally use a multiple of 5 present in the answer choices to begin with. Here, 4 of the 5 choices are multiples of 5, making it a bit difficult to choose a number - so I'd go with option 3.

Let the days taken by Jose be 'A' days and that taken by Jane be 'B' days. We're told that Jane is more efficient than Jose which means that Jane will take fewer number of days to complete the work as compared to Jose. Therefore, A > B

Now, using option C i.e. 60 days to be the time taken by Jose to complete the work,

Working together, A and B take 20 days to complete the work.
Therefore, 1/A + 1/B = 1/20
Substituting A = 60 and solving for B, we get B = 30 days

We know that Jose works and completes half of the work. Since Jose takes 60 days to complete the work, he would take 30 days to complete half the work.
Similarly, since Jane takes 30 days to complete the work, she would take 15 days to complete half the work.
So, in total, Jose and Jane take 45 days to complete the work - which is what is given in the question stem.

Hence Option C is the correct answer.

Hope this helps

Cheers!
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Re: Work/Rate [#permalink]  07 Jul 2015, 03:22
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Re: Work/Rate   [#permalink] 07 Jul 2015, 03:22
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