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OG 12 problem

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OG 12 problem

by independent » Fri Apr 05, 2013 10:43 am
OA: E

I understand how to solve this problem, but I'm stuck in one of the final steps. I'll post the SS in the next post because I don't want to confuse someone with a spoiler.



EDIT: the problem is from the quantitative guide 2nd edition. Not OG12.
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Last edited by independent on Fri Apr 05, 2013 10:48 am, edited 1 time in total.
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by independent » Fri Apr 05, 2013 10:47 am
Now, normally I know how to solve this step when there is no number in front of x^2. I'm confused with the 5x^2 and how the following step is derived. There is probably a rule I don't know so I'm asking for your help.
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by srcc25anu » Fri Apr 05, 2013 11:04 am
Mach X takes 'X+2' days to produce W widgets
Mach Y should takes 'X' days to produce W widgets

In 1 Day:
Mach X produces w/(x+2) widgets --- Eq 1
Mach Y produces w/x widgets
Mach X + Mach Y produce [w/(x+2) + w/x] widgets in a day
In 3 days they produce 3 * [w/(x+2) + w/x] widgets which is eual to 5/4w

Solving we get: w ( 5x^2 - 14x - 24) = 0
or w (x-4)(5x+6) = 0

since x is days it cannot be negative (or -6/5)
hence x has to be 4 days

Now we know X produces w/6 widgets in 1 day (from Eq 1)
so W widgets must be produced in 6 days
and 2w widgets will get produced in 12 days
hence E
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by srcc25anu » Fri Apr 05, 2013 11:09 am
independent wrote:Now, normally I know how to solve this step when there is no number in front of x^2. I'm confused with the 5x^2 and how the following step is derived. There is probably a rule I don't know so I'm asking for your help.
the equation that you have arrived at is incorrect.
instead of 5x^2 - 34x + 24 = 0 it should instead be 5x^2 - 14x - 24 = 0 (so please check your calculations one more time)

Now that you have the correct eqn to work with, the rule is sum of roots must equal -14 and product of roots must equal co-effecients of x^2 and c that is 5*(-24) = -120.
Only combination to satify this is -20 and 6 which will give sum of roots as -14 and product of roots as -120.

this is no different rule than when you see a 1x^2 or simply x^2 + x + c = 0 equation. its just that since multiplication by 1 doesnt change the product, we easily ignore the co-effecient of x^2 while showing our calcs.
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by GMATGuruNY » Fri Apr 05, 2013 5:52 pm
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
An efficient way to solve this problem is to plug in the answer choices.

Let w=12.
In 3 days, X and Y need to produce 5/4*w = 5/4*12 = 15 widgets.
The answer choices represent the time for X to produce 2w=24 widgets.

Answer choice C: 8 days for X to produce 24 widgets
Thus, X produces 12 widgets in 4 days.
Since X takes 2 days longer, Y produces 12 widgets in 4-2=2 days.
Rate for X = w/t = 12/4 = 3 widgets per day.
Rate for Y = w/t = 12/2 = 6 widgets per day.
Combined rate for X+Y = 3+6 = 9 widgets per day .
Work completed in 3 days = r*t = 9*3 = 27 widgets.
Incorrect: 27 is almost twice the number of widgets that need to be produced (15).
Thus, X and Y need to work MUCH MORE SLOWLY, implying that the time for X to produce 24 widgets must be MUCH LONGER.

Answer choice E: 12 days for X to produce 24 widgets
Thus, X produces 12 widgets in 6 days.
Since X takes 2 days longer, Y produces 12 widgets in 6-2=4 days.
Rate for X = w/t = 12/6 = 2 widgets per day.
Rate for Y = w/t = 12/4 = 3 widgets per day.
Combined rate for X+Y = 2+3 = 5 widgets per day.
Work completed in 3 days = r*t = 5*3 = 15 widgets. Success!

The correct answer is E.
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