Mary and Nancy can each perform a certain task in m and n

D..
A results in n>m
B results in m<n

Can you let me know how do you get the expression 2mn/(m+n)??
Thanks

Posted from GMAT ToolKit

D is the correct answer.

1- Leads for n>m
2- Leads for m<n

Using either of them the question can be answered

Mary takes m hrs to complete a job and Nancy takes n hrs. Together, how long will they take?

Time taken by Mary = m hrs
Time taken by Nancy = n hrs

Time taken is inverse of Rate of Work which we get from the expression Work = Rate * Time. If work done is 1, Rate = 1/Time. Hence,

Rate of work of Mary = 1/m
Rate of work of Nancy = 1/n
When they are working together, their rate of work = 1/m + 1/n = (m+n)/mn
Time taken by them together = mn/(m+n)

2mn/(m+n) is 'Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates'

D it is.

The basic equation for RTD problems is work = rate X time
Let work be 1, so the rate by which Mary works per hour is 1/m
Similarly, the rate by which Nancy works per hour is 1/n

Statement 1:
With the information given in the question and using the RTD formula, the combined time to complete the same work (say 1) can be calculated as:

1= (1/m+1/n)t (t is the time)
Therefore, t = mn/(m+n)

Now, according to the statement, 2mn/(m+n) > m
Here we need to understand that m can not be negative. So we can divide both sides of the equation by m. By solving the equation we get,
n>m
Hence Sufficient.

Statement 2:
We can prove the statement to be true as Statement 1.
Hence Sufficient

Both statements are sufficient, so the answer is D.

Since the question asked " Is m<n?", I see we can get both answers YES and NO for the question . Because we do not have enough information, using all given info in 1 and 2 may result in both m>n and n>m.... think about it
confusing ?
All posted solutions assume that Mary's rate is 1/m (1 over m) and Nancy's rate is 1/n, but what is Mary's rate would be m or 2/m or ... so we may have different answers in these diff. cases, do not we ?

The question tells you that Mary takes m hours to complete 1 work.

We know

Work = Rate * Time

1 = Rate * m

Rate = 1/m

So the question tells you that Mary's rate is 1/m only.

My 2 cents:

We could probably use smart numbers here and get D as well. If we just use 1 and 2 hours -> rates are 1/2 and 2/2 -> combined rate is 3/2 -> combined time 2/3 hours.

1. twice the time is greater than m, so 2/3*2 = 4/3 or 1.33, so 1< 1.33 < 2. We can conclude that 1 = m and 2 = n, so n>m
2. twice the time is less than n, so the same, 2/3*2 = 4/3 or 1.33, so 1< 1.33 < 2 We can conclude that 1 = m and 2 = n, so n>m

Twice the time means "half the rate"
We can't add times directly, so we convert them into rates and then add them.
So we get (1/m+1/n) < 2/m and (1/m+1/n) > 2/n
Answer is D
Hope this helps!

Solution:

Mary’s rate is 1/m, and Nancy’s rate is 1/n. Their combined rate is 1/m + 1/n = (n + m)/mn. Since time is the inverse of rate, their combined time to perform the task is mn/(n + m). We need to determine if m < n.

Statement One Alone:

Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is greater than m.

We are told that twice the combined time is greater than Mary’s time alone. We express this as:

2mn/(n + m) > m

2mn > nm + m^2

mn > m^2

n > m

The answer to the question is yes. Statement one is sufficient.

Statement Two Alone:

Twice the time it would take both Mary and Nancy to perform the task together, each working at their respective constant rates, is less than n.

We are told that twice the combined time is less than Nancy’s time alone. We express this as:

2mn/(n + m) < n

2mn < n^2 + mn

mn < n^2

m < n

The answer to the question is yes. Statement two is sufficient.

Answer: D

VeritasKarishma In considering time = 1/((1/m)+(1/n)) , aren't we assuming that both Mary and Nancy did equal proportion of work?

Note how we come to this expression of time taken.

Individual rates of Mary and Nancy are 1/m and 1/n. This means that Mary does (1/m)th of the work every hour and Nancy does (1/n)th of the work every hour.
When they work together, Mary does (1/m)th of the work and Nancy does (1/n)th in one hour i.e. they together complete (1/m)+(1/n) of the work every hour.

So time taken together to complete 1 full work = 1/(1/m + 1/n) = mn/(m+n)

[/quote]

VeritasKarishma While I was reverting, I realised where I was going wrong! Thanks!

Mary's rate=a
Nancy's rate=b
W/a=m; W/b=n;
a/W=1/m, b/W=1/n
S1)2W/(a+b)>m
=>2/(1/m+1/n)>m
=>2mn/(m+n)>m
=>m(2n/(m+n)-1)>0
=>(n-m)/(m+n)>0
=>n>m

S2) 2W/(a+b)<n
=> 2mn/(m+n)<n
=> 2m/(m+n)-1<0
=> (m-n)/(m+n)<0
=>m<n