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The workforce of a certain company comprised exactly 10,500

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The workforce of a certain company comprised exactly 10,500 [#permalink]

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09 Sep 2005, 07:32
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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
[Reveal] Spoiler: OA

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09 Sep 2005, 08:43
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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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09 Sep 2005, 10:21
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
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09 Sep 2005, 10:35
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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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09 Sep 2005, 11:21
Good...but hint is you should know your number properties and prime factors./...will post OA later today...

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Re: The workforce of a certain company comprised exactly 10,500 [#permalink]

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19 Apr 2014, 09:17
Any alternative approach to solve this backsolving?

Thanks much!
Cheers!
J

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Re: The workforce of a certain company comprised exactly 10,500 [#permalink]

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24 Sep 2014, 04:37
jlgdr wrote:
Any alternative approach to solve this backsolving?

Thanks much!
Cheers!
J

make a table according to ratios

x 5x 1750 70x 10500

now put options in place of 5x (i.e. multiply each option by 14)
you can eliminate A & B just by looking
C gives 3500
D gives 490
E gives 10500 (Eliminated)
D will not give an integer ratio
It will take less than 90 seconds

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Re: The workforce of a certain company comprised exactly 10,500 [#permalink]

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08 Nov 2014, 02:34
sanket1991 wrote:
jlgdr wrote:
Any alternative approach to solve this backsolving?

Thanks much!
Cheers!
J

make a table according to ratios

x 5x 1750 70x 10500

now put options in place of 5x (i.e. multiply each option by 14)
you can eliminate A & B just by looking
C gives 3500
D gives 490
E gives 10500 (Eliminated)
D will not give an integer ratio
It will take less than 90 seconds

I didn't get it ?
checking each option w.r.t to 5x, will give x=10 for option A. This will satisfy all the ratios. May be I am not getting it?
Can any one explain this.
Thanks

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Re: The workforce of a certain company comprised exactly 10,500 [#permalink]

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09 Nov 2014, 18:20
solitaryreaper wrote:
sanket1991 wrote:
jlgdr wrote:
Any alternative approach to solve this backsolving?

Thanks much!
Cheers!
J

make a table according to ratios

x 5x 1750 70x 10500

now put options in place of 5x (i.e. multiply each option by 14)
you can eliminate A & B just by looking
C gives 3500
D gives 490
E gives 10500 (Eliminated)
D will not give an integer ratio
It will take less than 90 seconds

I didn't get it ?
checking each option w.r.t to 5x, will give x=10 for option A. This will satisfy all the ratios. May be I am not getting it?
Can any one explain this.
Thanks

Well 5x is the number of first year students (as x is number before first year)
What I meant was ,
Options are given for first year, and I have take first year as 5x
multiply by 14 will give you number of third year students
A and B are so low that they wont reach to 10500 when multiplied by these numbers.
I hope you understand...

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Re: The workforce of a certain company comprised exactly 10,500 [#permalink]

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24 Jan 2016, 10:23
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The workforce of a certain company comprised exactly 10,500 [#permalink]

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15 Jul 2016, 00:15
Is this Question Correct?
Question says " the ratio of the number of workers from one year to the next was always an integer."
i.e. (After 2nd Year): (After 3rd year):: (integer):1

This statement should be " the ratio of the number of workers from one year to the PREVIOUS was always an integer."

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The workforce of a certain company comprised exactly 10,500 [#permalink]

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04 Aug 2016, 00:21
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FN wrote:
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 very trickily-worded question.
I'll use the notation $$Y_n \to Y_k = R$$ to signify the ratio increase from year $$n$$ to year $$k$$, where $$R$$ is an integer $$> 1$$ (the number of workers increased every year and is an integral ratio).

Determine the number of workers on year 2:
$$Y_2 = \frac{Y_4}{Y_2 \to Y_4} = \frac{10500}{6} = 1750$$

Since $$Y_2 \to Y_4 = 6$$ is a 2-year gap, there must be two intermediate ratios, the product of which is 6.
Therefore each ratio is a combination of the prime factors of 6. We do not currently know this combination.
$$Y_2 \to Y_3 \,\bigg\vert\, Y_3 \to Y_4 = \{2,3\}$$.

Since $$Y_1 \to Y_3 = 14$$ is a 2-year gap, there must be two intermediate ratios, the product of which is 14.
$$Y_1 \to Y_2 \,\bigg\vert\, Y_2 \to Y_3 = \{2,7\}$$.

From the intersection of the above two statements, we can see that: $$Y_2 \to Y_3 = 2$$. From this, we can conclude:
$$Y_1 \to Y_2 = 7\\ Y_3 \to Y_4 = 3$$

Since we know $$Y_2$$ from above, we can obtain $$Y_1$$ as follows:
$$Y1 = \frac{Y_2}{Y_1 \to Y_2} = \frac{1750}{7} = 250$$

(C) 250

[Reveal] Spoiler: Footnote
$$Y_0 \to Y_3 = 70\\ Y_1 \to Y_3 = 14\\ \therefore Y_0 \to Y_1 = 5$$

$$Y_0 = 50$$
$$Y_0 \to Y_1 = 5 \,\vert\, Y_1 = 250\\ Y_1 \to Y_2 = 7 \,\vert\, Y_2 = 1750\\ Y_2 \to Y_3 = 2 \,\vert\, Y_3 = 3500\\ Y_3 \to Y_4 = 3 \,\vert\, Y_4 = 10500$$

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Re: The workforce of a certain company comprised exactly 10,500 [#permalink]

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11 Jan 2017, 07:38
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

since the question states that the ratio between any two years is always an integer, we know that 1,750 must be a multiple of the number of workers after the first year. Since only 70 and 250 are factors of 1750, we know the answer must be either choice B or choice C. If we assume that the number of workers after the first year is 70, however, we can see that this must cannot work. The number of workers always increases from year to year, but if 70 is the number of workers after the first year and if the number of workers after the third year is 14 times greater than that, the number of workers after the third year would be 14 x 70 = 980, which is less than the number of workers after the second year. So choice B is eliminated and the answer must be choice C.
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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Re: The workforce of a certain company comprised exactly 10,500   [#permalink] 11 Jan 2017, 07:38
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