Running at their respective constant rates, Machine X takes 2 days

QE: 5t^2 - 14t - 24 = 0, How do we get solution = 4?

You can also give the quadratic route a miss.
Once you get (1/t) + (1/(t-2)) = 5/12, you can try out some values for t. t is an integer since it is one of the given 5 options.

To get 12 in denominator on the right hand side, t*(t-2) should give you 12 or a multiple.
12 = 4*3 or 6*2 (not of the form t*(t-2))
24 = 6*4 (this is possible)

Check 1/6 + 1/4 = 10/24 = 5/12
You get the value of t.

Or you can also try substituting the value of t from each option.

Tough one! My recommended solution is plugging in 1 for w. Attached is a visual that should help.

Strictly speaking, one doesn't need a formula for rates problems. The well-known formula works by expressing the amount of work each machine does in the *same* amount of time- one hour. If you know this, you can do any rates problem without a formula; just get the same time (rather than the same number of widgets) for each machine. Algebraically, this produces the same equation as the formula (though without the fractions), but is more flexible in case you encounter some kind of twist. Applied to this problem, it's still quite abstract, but it works:

Y produces w widgets in d days
X produces w widgets in d+2 days
X+Y produce w widgets in 12/5 days

Getting the same time for each:

Y produces dw + 2w widgets in d(d+2) days
X produces dw widgets in d(d+2) days
X+Y produce (5/12)*d*(d+2)*w widgets in d(d+2) days

So:
dw + 2w + dw = (5/12)*d*(d+2)*w
(24/5)w(d+1) = dw(d+2)
24d + 24 = 5d^2 + 10d
5d^2 - 14d - 24 = 0
(5d + 6)(d - 4) = 0

And since d > 0, we find d = 4.

The question asks us to find 2d+4, which is 12.

Because the data given says that one machine takes t days and another t + 2 days, there will be a variable and equation involved. When we have rates in multiples (say time taken is t days and 2t days), we can often make-do without any variable and equation.

Thanks a lot, this was very helpful! I know how to factorize but as soon as it looks like this I can"t handle it anymore 5t^2-34t+24=0 After I found the factors for 24, there is no way getting around trial and error, is there?

And is this a formula to compute the times 2 machines need to finish the same job? I always thought the times of the machines do NOT add up in work problems..
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)

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

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

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

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

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

Hope it's clear.

Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Hope it helps.

Sure thing, Vegita. Here's an excerpt from my book.

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.

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.)

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.

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

Hi,

I am new here and I have solved this math correctly. However, here's my problem:

If I consider the number of days required to produce 'W' widgets as 'X' and 'X-2' for X and Y respectively, I end up with the equation 5t^2 - 34t + 24 which yields me 2 solutions, one of which is correct (x=6)

However, when I take 'X' and 'Y' as 'X+2' and 'X', which are the same as above, because either ways, X is taking 2 days longer, I end up with the equation 5x^2 - 14x - 24 which has no solutions (I hope I'm not making some stupid mistake here)

Can someone please clarify this and correct me where I'm wrong.

can someone pelase explain how they got from (1/t) + (1/((t-2)) = 5/12
TO
5t^2-34t+24=0

1. Do all the calculations for 2w widgets
2. It takes x 4 days more to produce 2w widgets'
3. Together they take 4.8 days to produce 2w widgets
4. 1/y + 1/x=1/4.8 or 1/y + 1/(y+4) =5/24, y=8 . Therefore y+4=12.
X takes 12 days to produce 2w widgets

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

So far I have
X's rate = w/(d+2)
Y's rate = w/d

[w/d + w/(d+2)] * 3 = 5w/4

Not sure what to do with those fractions - how do you get a common denominator??