If 12 men and 16 women can do a piece of work in 5 days and 13 men and

Here is how it is solved:

5(12 M + 16 W) = 4(13 M + 24 W)
60M + 80W = 52M + 96W

That gives, 8M = 16W, which is , 1M=2W

Now, 12 M + 16W can do work in 5 days.
So, 12 M + 8 M can do work in 5 days.
So, 20M can do work in 5 days.

Now, 7M + 10W = 7M + 5M = 12 M

Hence, if 20M can do work in 5 days, 12M can do work in (20*5)/12 = 8.33 days.

You will often be able to ballpark or find the exact solution by just working on the multiples of the given equations.

Say rate of work of each man is M and that of each woman is W.

Given: 12M + 16W = 1/5 (Combined rate done per day)
______3M + 4W = 1/20 (in lowest terms) ...........(I)
Given: 13M + 24W = 1/4 ......................................(II)

Reqd: 7M + 10W = ?

Using equations (I) and (II) we need to find the sum of 7M and 10W. We can get a multiple of 7M in various ways:
(3M + 4W = 1/20) * 5 gives 15M + 20W = 1/4
Adding this to equation II, we get 28M + 44W = 1/2
7M + 11W = 1/8

7 men and 11 women complete the work in 8 days. 7 men and 10 women will take a little more than 8 days to complete the work. If you do get such a question in GMAT, you will easily be able to manipulate the equations to get the desired equation. Finding the values of M and W is way too painful to interest GMAT.

Let a man do \(\frac{1}{x}\) work per day and

each woman do \(\frac{1}{y}\) work per day

\(\frac{12}{x} + \frac{16}{y} = \frac{1}{5}............\)(1)
\(\frac{13}{x} +\frac{24}{y} = \frac{1}{4}\) ............ (2)

Multiply (1) by 3 & (2) by 2

\(\frac{36}{x} + \frac{48}{y} = \frac{3}{5}\) ....... (3)

\(\frac{26}{x} + \frac{48}{y} = \frac{1}{2}\) ........ (4)

Equation (3) - (4)

\(\frac{10}{x} = \frac{1}{10}\)

x = 100

y = 200

We require to find z; substituting the values

\(\frac{7}{x}+\frac{10}{y}= \frac{1}{z}\)

\(\frac{7}{100}+ \frac{10}{200} = \frac{1}{z}\)

\(\frac{7}{100}+ \frac{5}{100} = \frac{1}{z}\)

\(\frac{12}{100} = \frac{1}{z}\)

\(z = \frac{100}{12} = 8.33 = Answer\)

If you solve the two equations:

12M + 16W = 5
13M + 24W = 4

and then plug M and W into

7M + 10W = X,

X should be the amount of days.

Thats too much math for me, and the numbers don't look too clean either.

Here is how I solved it:

Create two equations based on the information given. Start from the basics. Each man can do 1/x work per day and each woman can to 1/y work per day. If we have 12 men and 16 women, then in total they can do 12/x work per day and 16/y work per day together. We also know that they can do the work together in 5 days. So in 1 day, how much of the job did they finish? 1/5 or 20%. Repeat this logic to create the second formula.
12/x + 16/y = 1/5
13/x + 24/y = 1/4

Solve for common denominators in each formula:
12y+16x=((xy)^2)/5
13y+24x=((xy)^2)/4

Multiply by denominator of fraction in each formula to get nice numbers. Since both would equal the same variable, make them equal to each other:
60y+80x=52y+96x
8y=16x
y=2x

Plug y=2x into the first equation to get:
12/x+16/2x=1/5
12/x+8/x=1/5
20/x=1/5
x=100

If x=100 and y=2x, y=200

Now in the original equation, it asks us how long 7 men and 10 women can do the work. We can create the following formula, similar to what we did in the first step. This formula tells us that in one day 7 men and 10 women can do X% of the job. We can simply take the reciprocal of the fraction to get the number of days the job will be completed in:
7/x+10/y=1/z (we want to solve for z)
7/100+10/200=1/z
7/100+5/100=1/z
12/100=1/z
z=100/12 or 8.33

Similar questions to practice:

it-takes-6-days-for-3-women-and-2-men-working-together-to-82718.html
if-w-women-can-do-a-job-in-d-days-then-how-many-days-will-83771.html
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Answer: Option C

Please find solution as attached.

If 12 men and 16 women can do a piece of work in 5 days and 13 men and 24 women can do it in 4 days, how long will 7 men and 10 women take to do it?

(A) 4.2 days
(B) 6.8 days
(C) 8.3 days
(D) 9.8 days
(E) 10.2 days

We can solve this using algebra but as others have noted doing this on GMAT may not be feasible.
So I thought about an intuitive explanation and so forum members let me know if you concur.
So we have 13 Men and 24 Women who together will finish the task in 4 days (per the prompt). Now if we reduce women and men by half. The task is going to take twice the time.
We see that we reduce Men to 7 (reduce by slightly less than half 6/13) and Women to 10 (reduce by slightly more than half 14/24 = 7/12). So my time would have nearly doubled. 9.8 is next nearest to 8 , but that would imply that time required has become 2.5 times. 8.8 is a good guess using this reasoning and turns out to be the correct answer.

Nothing wrong with the solution but I would be a little uncomfortable making this approximation. If I know the approximate relation between that the rate of work of men and women, then perhaps I will have an easier time making these assumptions.

1. Let us take the same number of days in both the cases, say 20 days
2. In the first case 3 men and 4 women can do the work in 20 days and in the second case 2.6 men and 4.8 women can do the work in 20 days.
3. 0.8 more women does the work of 0.4 less men or 2 women does the work of 1 man.
4. We can find the number of days taken by 1 man and 1 woman as 100 and 200 days
5. So the time taken by 7 men and 10 women is 1/ (7/100+10/200)=8.3 days

Do not mind the decimals if you can reach the solution faster.
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We can let the time it takes 1 man to finish the work = m, and thus the rate of 1 man = 1/m. Likewise, we can let the time it takes 1 woman to finish the work = w, and thus the rate of 1 woman = 1/w.

Thus, the combined rate of 12 men and 16 women is 12/m + 16/w. Since they can finish the work in 5 days, their combined rate is also equal to 1/5. Thus, we have:

12/m + 16/w = 1/5

Multiplying both sides of the equation by 5mw, we have:

60w + 80m = mw

Similarly, the combined rate of 13 men and 24 women is 13/m + 24/w. Since they can finish the work in 4 days, their combined rate is also equal to 1/4. Thus, we have:

13/m + 24/w = 1/4

Multiplying both sides of the equation by 4mw, we have:

52w + 96m = mw

So, we have 60w + 80m = 52w + 96m (since they both equal mw).

60w + 80m = 52w + 96m

8w = 16m

w = 2m

We can now substitute w = 2m into the first equation, 12/m + 16/w = 1/5, to solve for m:

12/m + 16/(2m) = 1/5

12/m + 8/m = 1/5

20/m = 1/5

m = 100

Since m = 100 days, w = 200 days. The rate of 1 man is 1/100 and the rate of 1 woman is 1/200. Thus, the rate of 7 men and 10 women is 7/100 + 10/200 = 7/100 + 5/100 = 12/100, and the time for them to finish the same work is 1/(12/100) = 100/12 = 8.3 days.

Answer: C
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Dear GMATinsight

I liked you solution. But I do not understand the abbreviation used in the equation compared to other problems in which you used this equation.

In the equation (M*T)/w, I understand 'M' as pure manpower, but it seems you used (12M+16w) as rate multiplied by # of days (5) which work equation and at end Work/Work. I have real confusion about the abbreviations used in the equation and the solution.

Can you elaborate please ?
Thanks in advance

Mo2men

M = manpower/Machine power here total manpower is 12 Men and 16 women hence wrote it as 12M+16W
T = Time that Man/machine power will take to finish the work
W = Amount of work to be finished

M is directly proportional to W and
M is inversely Proportional to T
hence, M = k(W/T)
CONCEPT: \(\frac{(Machine_Power * Time)}{Work} = Constant\)
i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)

Hope this explains your doubt

Dear GMATinsight

Thanks for your care to reply. I still have confusion. Let me try to explain.

The highlight past is comprise of 12M so it is 12 multiplied by variable M and 16 multiplied by variable W and hence you concluded a relation between them. I do not know why you multiplied by those variables whereas the original formula (as stated by you and your solution in other problems like below) contains only machine/man power without multiplication in any variables). Please look below

https://gmatclub.com/forum/if-it-takes- ... s#p1729381

While I understood your solutions in the link, I could not comprehend what you provided for the question at hand.

I hope you understand where my confusion is.
Thanks in advance

Mo2men

Let me attempt to answer this again

The relationship that I explained MT/W = Constant is applicable only if the work is of same nature in both scenarios and the Manpower also is of same nature in both scenario only their quantities differ.

When I say 12M+16W then I am considering the manpower of different nature hence it's important for me to use this relationship to determine the ratio of efficiencies of different manpowers.

Similarly if two machines of different types (e.g. old and new) are used together then it's paramount that we convert the machines into equivalent machines of one nature only.

12M refers to efficiency of 12 men and 16W refers to efficiency of 16 women here.

\(7\,\,{\text{men}}\,\, \cup \,\,\,{\text{10}}\,\,{\text{women}}\,\,\, - \,\,\,1\,\,{\text{work}}\,\,\,\, - \,\,\,?\,\,{\text{days}}\)

Is there a systematic way of dealing with this kind of problem, to be able to do it in a few minutes "naturally"?

Certainly! Let´s do it:

Let "task" be the fraction of this (piece of) work that one man can do in 1 day, hence:

\(1\,\,{\text{man}}\,\,\, - \,\,\,1\,\,{\text{day}}\,\,\,\, - \,\,\,1\,\,{\text{task}}\)

Let k (k>0) be the fraction of the "task" defined above that one woman can do in 1 day (where k may be between 0 and 1, or equal to 1, or greater), hence:

\(1\,\,{\text{woman}}\,\,\, - \,\,\,1\,\,{\text{day}}\,\,\, - k\,\,{\text{tasks}}\,\)

Now the long-lasting benefit of this "structure": everything else becomes easy and "automatic":

\(\left. \begin{gathered}\\
{\text{12}}\,\,{\text{men}} - \,\,\,5\,\,{\text{days}}\,\,\, - \,\,\,\,12 \cdot \,5 \cdot 1\,\,\,{\text{tasks}}\, \hfill \\\\
{\text{16}}\,\,{\text{women}} - \,\,\,5\,\,{\text{days}}\,\,\, - \,\,\,\,16 \cdot \,5 \cdot k\,\,\,{\text{tasks}}\,\,\, \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{question}}\,\,{\text{stem}}} \,\,\,\,5\left( {12 + 16k} \right)\,\,{\text{tasks}}\,\,\,\, = \,\,\,\,1\,\,{\text{work}}\,\,\,\,\left( * \right)\)

\(\left. \begin{gathered}\\
{\text{13}}\,\,{\text{men}} - \,\,\,4\,\,{\text{days}}\,\,\, - \,\,\,\,13 \cdot \,4 \cdot 1\,\,\,{\text{tasks}}\, \hfill \\\\
{\text{24}}\,\,{\text{women}} - \,\,\,4\,\,{\text{days}}\,\,\, - \,\,\,\,24 \cdot \,4 \cdot k\,\,\,{\text{tasks}}\,\,\, \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{question}}\,\,{\text{stem}}} \,\,\,\,4\left( {13 + 24k} \right)\,\,{\text{tasks}}\,\,\,\, = \,\,\,\,1\,\,{\text{work}}\,\,\,\,\left( {**} \right)\)

\(\left( * \right) = \left( {**} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,5\left( {12 + 16k} \right) = 4\left( {13 + 24k} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,k = \frac{1}{2}\)

\(?\,\,\,\,:\,\,\,\,\,\left. \begin{gathered}\\
\boxed{{\text{7}}\,\,{\text{men}}} - \,\,\,1\,\,{\text{day}}\,\,\, - \,\,\,\,7\,\,\,{\text{tasks}}\, \hfill \\\\
\boxed{{\text{10}}\,\,{\text{women}}} - \,\,\,1\,\,{\text{day}}\,\,\, - \,\,\,\,10 \cdot k = 5\,\,\,{\text{tasks}}\,\,\, \hfill \\ \\
\end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\boxed{{\text{FOCUSED - GROUP}}}} \,\,\,\,\frac{{12\,\,{\text{tasks}}}}{{1\,\,\,{\text{day}}}}\,\,\,\,\left( {***} \right)\)

\(\left( * \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,5\left( {12 + 16 \cdot \frac{1}{2}} \right) = 100\,\,{\text{tasks}}\,\,{\text{ = }}\,\,{\text{1}}\,\,{\text{work}}\,\)

And we finish in "high style", using UNITS CONTROL, one of the most powerful tools of our method!

\(\left( {***} \right)\,\,\,\,?\,\,\, = \,\,\,100\,\,{\text{tasks}}\,\,\,\,\left( {\frac{{1\,\,\,\,{\text{day}}}}{{12\,\,{\text{tasks}}}}\begin{array}{*{20}{c}}\\
\nearrow \\ \\
\nearrow \\
\end{array}} \right)\,\,\,\, = \,\,\,\,\frac{{100}}{{12}} = \frac{{25}}{3} = \frac{{24 + 1}}{3} = 8\frac{1}{3}\,\,\,\,\left[ {{\text{days}}} \right]\)
Obs.: arrows indicate licit converter.



This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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This is a problem that the testmaker is trying to fool you into completing. You can definitely find the correct answer with math, however for most students this will take way too much time and is fraught with the opportunity for math errors. However, with 45 seconds or so of logic and estimation you can get this down to two possible answers. Strategically, this is a better choice here.

Here is the logic that you need to know:

Twice as many people complete the job in half of the time
Half of the people complete the job in twice of the time.

Pretty simple, right? Let's apply it to this question.

You know that 12 men and 16 women do the job in 5 days. So, 6 men and 8 women do the job in 10 days. Are 7 men and 10 women faster or slower? Well, faster! Eliminate E

You know that 13 men and 24 women do the job in 4 days. So, 6.5 men and 12 women do the job in 8 days. Are 7 men and 10 women faster or slower. While not as exact as the first estimation, probably a bit faster. A and B look pretty unlikely at this point.

So you are down to C and D. Is the answer closer to 8.3 or 9.8? From the 6.5 men and 12 women scenario, probably 8.3, which turns out to be the right answer.

You are likely to get only one or two work equation questions on test day. These questions typically involve a lot of math and are time consuming. Strongly consider opting for a logic approach on these questions. This approach will save you time that you may allocate to other questions.

Take a look at OG2019 PS 88. A very similar logic can be used to quickly eliminate c, d and e.

Hi All,

We're told that 12 men and 16 women can do a piece of work in 5 days and 13 men and 24 women can do it in 4 days. We're asked how long it will take 7 men and 10 women to complete that same task. This question can be approached in a couple of different ways (some of which involve a lot of calculations). The answer choices are sufficiently 'spread out' that you can use a bit of 'ratio math' and a little logic to get to the correct answer.

We're going to focus on just the first piece of information: it takes 12 men and 16 women a total of 5 days to complete a task. If you were to DOUBLE the number of workers, then you would HALVE the amount of time (re: 24 men and 32 women would take 2.5 days to complete the task). If you were to HALVE the number of workers, then you would DOUBLE the amount of time that it takes to complete the task:

6 men and 8 women would take a total of 10 days to complete a task.

If we HALVE the number of workers again, we would again DOUBLE the amount of time needed to complete the task:

3 men and 4 women would take a total of 20 days to complete a task.

We're asked how long it would take 7 men and 10 women to complete that task. If we multiply the above 'work information' by 2.5, we get...

(2.5)(3) men and (2.5)(4 women) would take a total of 20/2.5 days to complete a task...
7.5 men and 10 women would take a total of 8 days to complete a task.
Notice how this is almost the exact question we were asked to solve for. The difference is that we're including an extra "1/2 of a man" in this calculation. With just 7 men (instead of 7.5 men), we would need slightly more than 8 days to complete the task. There's only one answer that matches....

Final Answer:

GMAT assassins aren't born, they're made,
Rich

This is a standard, no-complication work-rate problem.

If you practice enough work-rate problems, you will be able to do this question in less than two minutes: because you will set up the equations quickly, without having to think too much.

Let us say,
A man can do m amount of work in one day,
a woman can do n amount of work in one day

So 12 men and 16 women do (12m + 16n) work in one day.
In 5 days they do 5(12m + 16n) work

We know: 13 men and 24 women do the same work in 4 days
So 5(12m + 16n) = 4(13m + 24n)

We want to know how long 7 men and 10 women would take.
Let that number of days be d

So 5(12m + 16n) = 4(13m + 24n) = d(7m + 10n)

All simple now, start solving
60m + 80n = 52m + 96n
8m=16n
m=2n

5( 40n ) = 4( 50n ) = d( 24n )
200n = 24nd
25 = 3d
d = 25/3 = 8.33

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