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

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

The answer choices are:
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.

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.

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.

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

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!

1/x+1/y=1/20 <==> xy/(x+y)=20. they are equivalent.

yeh, it might nt be from gmatleague but it is included as gmatquant practice PS

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!

let x=jose's rate
let y=jane's rate
let d=jose's days
x+y=1/20
x+y=(1/2)/d+(1/2)/(45-d)
1/20=(1/2)/d+(1/2)/(45-d)
d^2-45d+450=0
d=15 and 30
because we know that jose's days will be greater,
30x=1/2
x=1/60
60 days

a and b together do it in 45 days
first x days a works and then 45-x b works .....(remember this time for 1/2 work) for full
1/2x + 1/2(45-x) = 1/20
basically
1/x + 1/45-x = 1/10 ...wait dont need to solve
just try by checking answers .... by making each value half because answer is for full work

When both work together and complete the task, the equation is:
time together/ time taken by Jane + time together/ time taken by Jose = 1
This is : 20/Jane +20/Jose = 1 -- (1)

When both complete half the work and the job is completed in a certain number of days, the equation is
Time taken by Jane/2 + time taken by Jose/2 = total number of days
This is Jane/2 + jose/ 2=45 -- (2)

Solving (1) and (2) we get time taken by Jose=60 days

because their combined rate=1/20 and she works faster than he does, their individual rates can't be 1/40 each
if we assume integers and look at 1/20 as 3/60, his rate has to be half as fast as hers--1/60:1/30
confirming, 1/2 task*60 days+1/2 task*30 days=45 days, per problem stem
1/60 rate➡60 days time
C

Can anyone do this using the LCM approach to find efficiency? That'd be really helpful

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