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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?

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.

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

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

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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