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The workforce of a certain company comprised exactly 10,500
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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 fouryear period during which it increased every year. During this fouryear 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 fouryear 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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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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Re: The workforce of a certain company comprised exactly 10,500
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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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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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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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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
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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
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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
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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
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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
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24 Jan 2016, 10:23
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The workforce of a certain company comprised exactly 10,500
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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."
Please clarify.



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The workforce of a certain company comprised exactly 10,500
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04 Aug 2016, 00:21
FN wrote: The workforce of a certain company comprised exactly 10,500 employees after a fouryear period during which it increased every year. During this fouryear 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 fouryear period began is 70 to 1. How many employees did the company have after the first year? A very trickilyworded 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 2year 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 2year 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: \(Y1 = \frac{Y_2}{Y_1 \to Y_2} = \frac{1750}{7} = 250\) \(Y_0 \to Y_3 = 70\\ Y_1 \to Y_3 = 14\\ \therefore Y_0 \to Y_1 = 5\)
\(Y_0 = 50\) \(Y_0 \to Y_1 = 5 \,\vert\, Y_1 = 250\\ Y_1 \to Y_2 = 7 \,\vert\, Y_2 = 1750\\ Y_2 \to Y_3 = 2 \,\vert\, Y_3 = 3500\\ Y_3 \to Y_4 = 3 \,\vert\, Y_4 = 10500\)
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Re: The workforce of a certain company comprised exactly 10,500
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24 May 2018, 19:28
After all said, you have:
x  5x  1750  70x  10500
You know that: 5x * 2 * 7 = 70x 1750 * 2 * 3 = 10500
So the only possible option is that 1750 * 2 = 70 x, and thus 5 x * 7 = 1750 and 70 x * 3 = 10500 Anyway, you get x = 50 and then 5x = 250.
Hope it is clear.



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Re: The workforce of a certain company comprised exactly 10,500
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20 Aug 2018, 00:36
FN wrote: The workforce of a certain company comprised exactly 10,500 employees after a fouryear period during which it increased every year. During this fouryear 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 fouryear 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 Given \(N_4 = 10500\), \(\frac{N_4}{N_2} = 6\), hence \(N_2 = 1750\) \(\frac{N_3}{N_1} = 14\) & \(\frac{N_3}{N_0} = 70\) We have \(N_0\) > \(N_1\) > \(1750\) > \(14\)\(N_1\) >\(10500\) \(N_1\) is a factor of 1750, hence can be answer choices 50, 70 & 250 We also have \(14\)\(N_1\) \(> 1750\) > \(N_1\)\(> 125\) Hence \(N_1 = 250\) Answer C. Thanks, GyM
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Re: The workforce of a certain company comprised exactly 10,500 &nbs
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