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805+ (Hard)|   Work and Rate Problems|                                 
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Running at their respective constant rates, Machine X takes 2 days 26 Apr 2022, 13:12
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Rates are ADDITIVE:
X's rate + Y's rate = combined rate for X and Y
Rate formula:
\(rate = \frac{work}{time}\)

TalonShade
Maybe its a super ignorant doubt but what am I doing wrong if I do the following -

If Y does W work - its rate is 1/W so the other one is 1/(W+2)

Giving us

1/W + 1/(W+2) = 5W/12

Not sure why this is throwing me off
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Here, \(w\) represents the amount of work, so it does not belong in the denominator.

Let the job = w = 1 widget
Let X's time = x

Since X takes x days to produce 1 widget, we get:
X's rate \(= \frac{work}{time} = \frac{1}{x}\)
Since Y takes 2 fewer days to produce 1 widget, we get:
Y's rate = \(\frac{work}{time} = \frac{1}{x-2}\)
Since X and Y together take 3 days to produce 5/4 of 1 widget, we get:
Combined rate for X and Y \(= \frac{work}{time} = \frac{5}{4} ÷ 3 = \frac{5}{12}\)
Since X's rate + Y's rate = combined rate for X and Y, we get:
\(\frac{1}{x} + \frac{1}{x-2} = \frac{5}{12}\)
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Running at their respective constant rates, Machine X takes 2 days 03 Dec 2022, 17:08
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avigutman

Loved your approach to this question. 😀 A lot of critical thinking involved rather than going a quadratic route! Would it please be possible for you to explain how you came up with your statement "Product of individual times divided by sum of times gives us the total time of the combined workers"?

P.S: I tested your statement and it works but curious to know how it is derived.
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Running at their respective constant rates, Machine X takes 2 days 03 Dec 2022, 17:24
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Vegita
avigutman

Loved your approach to this question. 😀 A lot of critical thinking involved rather than going a quadratic route! Would it please be possible for you to explain how you came up with your statement "Product of individual times divided by sum of times gives us the total time of the combined workers"?

P.S: I tested your statement and it works but curious to know how it is derived.
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Sure thing, Vegita. Here's an excerpt from my book.
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Running at their respective constant rates, Machine X takes 2 days 28 Dec 2022, 12:58
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Bunuel
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Can somebody please explain the following:

How does this algebra work: 1/t + 1/t+2 = 5/12 is translated to 5t^2 + 34t - 24.

Thanks in advance ! Chris

\(\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}\);

\(\frac{(t-2)+t}{t(t-2)}=\frac{5}{12}\);

\(\frac{2t-2}{t^2-2t}=\frac{5}{12}\);

\(24t-24=5t^2-10t\);

\(5t^2-34t+24=0\).

Hope it's clear.
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Bunuel

Why can't you do

(w/x)+ w/(x+2)=(5/12)*w

Essentially, why do you need to subtract the 2 versus adding the two? I let x = machine y's time and x+2 = machine X's time because it takes machine X two more days than machine Y. When I went ahead and tried to solve though, I could not get the answer into the same quadratic equation because there were w's and x's whereas the w's canceled out when you do w/(x-2) compared to w/(x+2). Thank you for your help.
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Running at their respective constant rates, Machine X takes 2 days 31 Dec 2022, 16:27
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woohoo921


Your approach is fine. The w's should all cancel out of your equation. In fact, you can cancel them directly from the form you have right now!

(w/x)+ w/(x+2)=(5/12)*w
(1/x) + 1/(x+1)= 5/12

From there the math is much the same as for Bunuel's solution. You should end up here:

5x^2 -14x - 24 = 0
(5x+6)(x-4) = 0

Positive solution for x = 4, so X's actual time is x+2 = 6.

However, notice that you are setting yourself up for trouble here. If you represent Machine Y as "x" and Machine X as "x+2," you are far more likely to end up providing the wrong answer after all that calculation! Two valuable principles here:
1) Choose variables that match what they represent. You were really solving for Y here, so call it y.
2) When possible, solve directly for what you want. The question is about X's time, so set up and solve for that.
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Running at their respective constant rates, Machine X takes 2 days 31 Dec 2022, 16:34
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Quick approximation:

First, get a value for how long the 2 machines together would take to produce 2w widgets. We're told that it takes 3 days to produce 5w/4. That means they will take 3(4/5) = 12/5 days to produce w widgets and 2(12/5) = 24/5 days to produce w widgets. So basically, together the 2 machines take almost 5 days to produce 2w widgets.

Now we know that X is slower than Y. If they equally fast, one machine alone would take twice as long, or almost 10 days, to produce 2w widgets. However, we know that X is considerably slower than Y. So it will take more than 10 days. E.

(This would definitely be a weird approach to invent on the fly to solve this one problem, but you might be surprised how often this kind of approximation will get you to--or near--the right answer.)
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Running at their respective constant rates, Machine X takes 2 days 15 Jan 2024, 07:17
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KarishmaB
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Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A. 4
B. 6
C. 8
D. 10
E. 12

Please show the whole calculation.

Together they make 5/4 w widgets in 3 days. So they make w widgets in 3*4/5 = 12/5 days.

1/(t+2) + 1/t = 5/12

Now, calculating this is really cumbersome so try to plug in options to get to the answer.
If machine X takes 12 days to produce 2w widgets, it would take 6 days to make w widgets, t would be 4.

1/6 + 1/4 = 5/12
It works so t = 4.

Answer is 12.
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Why do you put 12/5 in the first lane adn change it to 5/12 in the second?

Thanks!
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Running at their respective constant rates, Machine X takes 2 days 30 Jan 2024, 23:17
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This solution is relying on the idea that rate and time are reciprocal. It's the same reason that if they make 5/4 w widgets per day, they take 4/5 of a day to make w widgets.

In the part you're asking about, 12/5 days is the time to make w widgets. If they take 12/5 days to make w, then they do 5/12 of w per day. The rates 1/t and 1/(t+2) must add to this combined rate.

Thib33600
KarishmaB
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Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A. 4
B. 6
C. 8
D. 10
E. 12

Please show the whole calculation.

Together they make 5/4 w widgets in 3 days. So they make w widgets in 3*4/5 = 12/5 days.

1/(t+2) + 1/t = 5/12

Now, calculating this is really cumbersome so try to plug in options to get to the answer.
If machine X takes 12 days to produce 2w widgets, it would take 6 days to make w widgets, t would be 4.

1/6 + 1/4 = 5/12
It works so t = 4.

Answer is 12.

Why do you put 12/5 in the first lane adn change it to 5/12 in the second?

Thanks!
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Running at their respective constant rates, Machine X takes 2 days 04 Apr 2024, 09:09
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Bunuel
Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?

(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
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­Let us assume, Y takes n days to produce w widgets. Then X will take (n+2) days to produce w widgets.
Efficiency of Y=1/n, efficiency of X=1/(n+2). So total efficiency to produce w widgets is 1/n + 1/(n+2)=(2n+2)/(n(n+1)).

Take the reciprocal of the total efficiency, we will get total days i.e. (n(n+1))/(2n+2). This is the time to produce w widgets ... (i)

Given 5w/4 widgets are produced in 3 days, then w widgets are produced in 3*w*4/(5*w)=12/5 days ... (ii)

(i) and (ii) represent the same, let's equate them. \((n(n+1))/(2n+2)\)=12/5
We get 5\(n^2\)-14n-24=0. Solve it for the positive value, n will be 4.

Therefore, X takes 4+2=6 days to produce w widgets. Then for 2w widgets, X will take 6*2=12 days. Option (E) is correct.
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Running at their respective constant rates, Machine X takes 2 days 16 Feb 2025, 23:39
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I was running out of time. Made the following observation o solve this in under 30sec. Sharing it here in case it helps anybody.
given, 5/4w in 3 days.
hence 2w in 24/5 days which is roughly 5 days.
This means if both the machines work together, and if they have similar rates, then it takes about 5 days to make 2w.
Meaning both machines would take 10 days each for the job. But Machine X is 2 days slower. Hence it would take more than 10 days for machine X to complete the job. Among the answer choices, only 12 is greater than 10.
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Running at their respective constant rates, Machine X takes 2 days 21 Aug 2025, 11:03
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I did something, i assumed a value for w. I took w=12.
So, A can do the work in 6 days and B in 4 days,
hence A produces 12/6=2 per day, B=12/4=3 per day.
In 1 Day A+B together produce 2=3=5, So checks out for 5/4.w days as 5/4 x 3 = 15 =(A+B)(3)
Now, we know A does 2 units a day and we have to find for 2W units i.e. 2x12(remember i assumed w as 12) so 24 units
Time taken by A = Total units/ Units A does per Day
= 24/2
=12
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Running at their respective constant rates, Machine X takes 2 days 20 Sep 2025, 03:20
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If you’re not a fan of working with variables or if the variable approach doesn’t come to mind right away, you can still solve this problem using trial and error. The key here is that we're dealing with natural numbers in the options and we don’t need many trials to narrow down the correct answer, especially since the total work is done in just 3 days.

Here’s my thought process:

I started by considering the time it would take Machine Y and Machine X to complete the task individually. My first instinct was to test the idea that:
  • Machine Y takes 4 days to produce w widgets, and
  • Machine X takes 6 days to produce w widgets (since Machine X is 2 days slower than Machine Y).

I ruled out Machine Y taking 3 days, because if that were the case, Machine X would take 5 days, and together they would produce more than 5/4w in 3 days.

Now, with Y = 4 days and X = 6 days:
  • In 3 days, Machine Y would produce 3/4 of the total task (because it takes 4 days for Y to complete the full task).
  • In 3 days, Machine X would produce 1/2 of the total task (since it takes 6 days for X to complete the full task).

When we add both parts together:
3/4 + 1/2 = 5/4

This matches the total amount of work done in 3 days, which is 5/4w.

Therefore, with this info, we can now easily calculate how long it would take Machine X to produce 2w widgets. Since Machine X takes 6 days to produce w widgets, it would take: 6×2=12 days to produce 2w widgets.

Thus, the correct answer is (E) 12.
This trial and error approach was quick and effective for me, especially cuz the problem deals with manageable numbers like these!
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Running at their respective constant rates, Machine X takes 2 days 20 Sep 2025, 04:08
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Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y.

At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets?

The two machines together produce in 1 day = (5/12)w widgets
The two machines together produce (5/12)w widgets in = 1 day
The two machines together produce w widgets in = 12/5 days

Let Machine X alone take x days to produce w widgets.
Machine Y alone takes (x-2) days to produce w widgets.

They together take x(x-2)/(2x-2) days to produce w widgets

x(x-2)/2(x-1) = 12/5

x = 6 days

Machine X alone will take 6*2 = 12 days to produce 2w widgets

IMO E
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Running at their respective constant rates, Machine X takes 2 days 25 Dec 2025, 13:55
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x+y took 3 days to do 5w/4 widget
x+y took 1 day to do 5w/4 divided by 3 = 5w/12 widget (1)

if x took d days to do w widget
then y took d-2 days to do w widget

In 1 day, x would finish w/d widget
and in 1 day, y would finish w/(d-2) widget (2)

From (1) and (2), we have:
w/d + w/(d-2) = 5w/12
=> 1/d + 1/(d-2) = 5/12 (divided both sides by w)
=> 12(d-2) + 12(d) = 5(d)(d-2)
=> 5d^2 - 34d + 24 = 0
=> 5d^2 - 30d - 4d + 24 = 0
=> 5d(d-6) - 4(d-6) = 0
=> (d-6)(5d-4)=0
=> d can be 6 or 4/5

so it took 6 days for machine x to do w widget
so it took 6*2=12 days for machine x to do 2w widget

if we use d=4/5, there's no answer for that, but I'm not sure why else to not use this solution. Hope this helps :)
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Running at their respective constant rates, Machine X takes 2 days 28 Dec 2025, 22:31
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Considering machine Y takes x days to complete w widgets, it's safe to say that X takes (x+2) days.
Therefore, rate of X = 1/(x+2) and rate of Y=1/x

Together, they take 3 days to complete (5/4)w widgets

That means \((\frac{1}{{x+2}} + \frac{1}{x}).3 = \frac{5w}{4}\)

Simplifying (skipping a few steps here because typing it out in mobile is hard T_T), we get:

\(\frac{x+1}{x(x+2)} = \frac{5w}{24}\)

This can be expanded to look like:

\(\frac{1}{x+2}.\frac{x+1}{x} = \frac{5w}{24}\)

Hey, that's the work equation for just machine X (RateX . Time taken = 5w/24)

We can manipulate this equation to establish one that requires machine X to produce 2w widgets. That means multiplying the RHS by a value that turns it into 2w

And that value would be \(\frac{48}{5}\)

Now that needs to go in the LHS as well. Neatly arranging our findings, we arrive with:

\(\frac{1}{x+2}.(\frac{48}{5}.\frac{x+1}{x}) = 2w\)

The bracketed part is the time taken by X alone to complete 2w widgets. Now it's just a matter of equating that time taken bit to each of the options and finding which works. Mind you, since there's a 48 in the numerator, the right answer will need to be a factor of 48, which leaves us with options 4, 8 and 12.

On checking, 12 is the only one that gives us a positive value for x. Therefore, that's our answer.
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