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Current Student
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lets do some real problems...shall we?...math section of [#permalink]
 09 Sep 2005, 07:32
lets do some real problems...shall we?...math section of late has become relatively dormant...
The workforce of a certain company comprised exactly 10,500 employees after a four-year period during which it increased every year. During this four-year period, the ratio of the number of workers from one year to the next was always an integer. The ratio of the number of workers after the fourth year to the number of workers after the the second year is 6 to 1. The ratio of the number of workers after the third year to the number of workers after the first year is 14 to 1. The ratio of the number of workers after the third year to the number of workers before the four-year period began is 70 to 1. How many employees did the company have after the first year?
(A) 50
(B) 70
(C) 250
(D) 350
(E) 750
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Senior Manager
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10,500/6 = 1,750 = number of workers after the second year
Now we need a number of workers after the third year. Let it = X. We have some clues to get to X:
- 10,500/X is an integer greater than 1.
-X/1,750 is an integer greater than 1.
We note that 10,500/1750 = 6. So we need two integer factors of 6, and neither factor can be 1. This means that the factors are 2 and 3.
Therefore we have two possibilities: either X is 3,500 or X is 5,250.
But we know that the ratio of the number of workers after the third year to the number of workers after the first year is 14 to 1.
3,500/14 = 250
5,250/14 = 375
So the number of workers after the first year is either 250 or 375. But this has to be an integer ratio with the number of workers after the second year, which is 1,750. Of the two possibilities, only 250 satisfies the condition, because 1,750/250 = 7 and 1,750/375 = 4 2/3.
Therefore the number of workers after the first year is 250. Answer C.
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Senior Manager
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A and B are too low it gives less employess in 3rd year than second year so chuck it
250 gives
50 - 250 - 1750 - 3500 - 10500
(all ratios in integer)
D, E does not give integer ratios...so C
_________________
Fear Mediocrity, Respect Ignorance
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Intern
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My answer is also c.
x --> x1--> x2 --> x3 --> x4
x4 is 10,500 & x4:x2=6:1 i.e. x2=1750
x3:x1=14:1
x3:x=70:1
I started with choice c. x1=250
50--> 250 --> 1750 --> 3500 --> 10,500
Chocie A returns x3=700 which is less than x2=1750, so it has to be ruled out and Choice B can also be ruled out.
No other answer choice returns an integer ratio.
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Current Student
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Good...but hint is you should know your number properties and prime factors./...will post OA later today...
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