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very tricky thing

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very tricky thing

by tanviet » Sat Sep 12, 2009 8:02 pm
Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 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

remember you have only 5 minites to solve this one. Pls, realize the problem inhere
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by gmat740 » Sat Sep 12, 2009 8:31 pm
First of all, your target time must be less than 2 mints!

X=> d+2
Y => d

Let time to produce together w widgets = t
=> 1/t = 1/d+2 + 1/d
=> t = d(d+2)/2d+2

So, time to produce 5/4w widgets = 5/4t
= 5/4[d(d+2)/2d+2]
Its given that this 5/4t = 3 days

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

5d^2 + 10d -24 = 0
(5d+12)(d-2)

d=2

X = d+2 = 4 days to produce w widgets
thus, 8 days to produce 2w widgets.

The solving is very simple if you do on paper

Hope this Helps
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Re: very tricky thing

by umaa » Sat Sep 12, 2009 9:30 pm
duongthang wrote:Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 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

remember you have only 5 minites to solve this one. Pls, realize the problem inhere
IMO E.

The number of days x takes to produce w widgets is, A
The number of days y takes to produce w widgets is, B

A = B+2

For 3 days,

3*W/A + 3*W/B = 5W/4

Substitute, B = A-2,

3*W/A + 3*W/(A-2) = 5W/4

24WA - 24W = 5WA^2 - 10WA

5WA^2 - 34WA + 24W = 0

Divide W,

5A^2 - 34A + 24 = 0

Approximately, A = 6 fpr W

For 2W, its 12.

OA pls. Also, pls post the source of the question.
What we think, we become
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by tanviet » Tue Sep 15, 2009 8:40 pm
OA is E

this question from 15 test set. a good sourse.

the matter is that

how can you factor the operation with x^2 to find the solution

this can be done easily if the operation is simple as in almost OG questions. however. is this case, factoring is not easy and take a lot of time to do. we do not have to solve by counting M=B^2-4ac

pls comment.
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by blr_gmat_prep » Tue Sep 15, 2009 10:09 pm
Production of w widgets:
Machine X (from the options): 2,3,4,5,6
Machine Y:0,1,2,3,4

Since answer is given in terms of 2w , we can half the values to get for w widgets.

Rule out option A since Y cant be 0.

we know that together X,Y produce W widgets in XY/(X+Y) days.
So they will produce 5w/4 in 5w/4*(XY)/((X+Y)) => 3 (As given in the question)

Putting C. we get 10/6 days so we have to go higher.
Putting E we get the value as 3 days.
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by vivekjaiswal » Tue Sep 15, 2009 10:28 pm
blr_gmat_prep wrote:Production of w widgets:
Machine X (from the options): 2,3,4,5,6
Machine Y:0,1,2,3,4

Since answer is given in terms of 2w , we can half the values to get for w widgets.

Rule out option A since Y cant be 0.

we know that together X,Y produce W widgets in XY/(X+Y) days.
So they will produce 5w/4 in 5w/4*(XY)/((X+Y)) => 3 (As given in the question)

Putting C. we get 10/6 days so we have to go higher.
Putting E we get the value as 3 days.
This is by far the best way to solve the problem and would be the only one to give an answer within 2 mins.

Thanks blr_gmat_prep

Cheers,
Vivek
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