The workforce of a certain company comprised exactly 10,500

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.

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.

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

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.

Good...but hint is you should know your number properties and prime factors./...will post OA later today...

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...

Please see attachment. Thanks

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.

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:
\(Y1 = \frac{Y_2}{Y_1 \to Y_2} = \frac{1750}{7} = 250\)

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.

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
_________________

Key Facts:

Ratio of: (Each End of Year) / (End of PRIOR Year) = Integer

and

the No. of Employees INCREASED Each Year:


(1st) End of 4th : End of 2nd = 6a : 1a

since End of 4th = 10, 500 ppl

End of 2nd = 1 * (10,500 / 6) = 1,750 ppl


(2nd) Ratio of:

End 3rd: End 1st = 14b : 1b

End 3rd : Beg of 1st = 70c : 1c
________________________
Combining the 2 Ratios we have:

End 3rd: End 1st : Beg. of 1st = 70x : 5x : 1x


(3rd) Putting All the Information together:

Actual ppl End of 4th: 10,500

Ratio Units of ppl End of 3rd = 70x

Actual ppl End of 2nd = 1,750

Ratio Units of ppl End of 1st = 5x

Ratio Units of ppl at Beg. of 1st = 1x



"the Ratio of Workers from 1 year to the Next was always an INTEGER"

THUS------>

I.
(End of 4th) / (End of 3rd) = (10,500) / (70x) = Integer

150/x = Integer ---------> Unknown Ratio Multiplier must be a Factor of 150



II.
(End of 3rd) / (End of 2nd) = (70x) / (1,750) = Integer

x/25 = Integer -------> Unknown Ratio Multiplier must be a Multiple of 25


Possible Ratio Multipliers thus far: 25 ---- 50 ---- 75 ----- 150


III.
(End of 2nd) / (End of 1st) = (1,750) / (5x) = Integer

350/x = Integer -----> Unknown Ratio Multiplier must be a Factor of 350

This Fact Eliminates 75 and 150 as Possible Unknown Ratio Multiplier = X


only 2 Possibilities Remain -----> X = 25 ----- or ----- X = 50



IV. if X = 25, then from the Ratio we found Above, the employees at the End of the 3rd Year would be:

70x = End of 3rd year employees = 70 * 25 = 1,750

But this is the Number of employees at the End of 2nd year. And we are told that EACH Year the Number of Employees INCREASES.....Therefore:



*SUMMARY*
the Unknown Ratio Multiplier = X= must be 50


The No. of employees at the End of the 1st Year in Ratio Units = 5x


5(50) = 250 employees at the End of the 1st Year


-Correct Answer C-


geesh....who came up with this question??

i dont think the line which says "only 70 and 250 are factors of 1750" is correct !!, this is so misleading, all the answer choices are factors of 1750 right, this has made this question all the more confusing, can anyone help?