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From 2000 to 2003, the number of employees at a certain company increa

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From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 22 Jun 2018, 01:21
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From 2000 to 2003, the number of employees at a certain company increased by a factor of 1/4. From 2003 to 2006, the number of employees at this company decreased by a factor of 1/3. If there were 100 employees at the company in 2006, how many employees were there at the company in 2000 ?


A. 200

B. 120

C. 100

D. 75

E. 60



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(PS15957)

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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 22 Jun 2018, 01:51
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Bunuel wrote:
From 2000 to 2003, the number of employees at a certain company increased by a factor of 1/4. From 2003 to 2006, the number of employees at this company decreased by a factor of 1/3. If there were 100 employees at the company in 2006, how many employees were there at the company in 2000 ?


A. 200

B. 120

C. 100

D. 75

E. 60



NEW question from GMAT® Official Guide 2019


(PS15957)


5/4 x 4/3 (Population in 2000) = 100, solving we get X = 60
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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 22 Jun 2018, 01:59
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Answer is B.

Let population in 2000 be x.
Let population in 2003 be y.
Population in 2006 is 100
2y/3 = 100
y = 150

Also, population in 2003 is 5x/4
So, 5x/4 = 150
x = 120
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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 22 Jun 2018, 02:51
1
In 2000 Let employees x
In 2003 employees then 5x/4
In 2006 employees then (5x/4-5x/12)=10x/12
10x/12=100
X=120


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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 22 Jun 2018, 05:01
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Solution



Given:
    • From 2000 to 2003, the number of employees at a certain company increased by a factor of \(\frac{1}{4}\)
    • From 2003 to 2006, the number of employees at this company decreased by a factor of \(\frac{1}{3}\)
    • There were 100 employees at the company in 2006

To find:
    • The number of employees in the company in 2000

Approach and Working:
Let us assume the number of employees in the company in 2000 is 12x [12 = LCM (4,3)]

Given that, from 2000 to 2003, the number of employees at a certain company increased by a factor of \(\frac{1}{4}\)
    • Therefore, the number of employees in 2003 = \(12x + \frac{1}{4} * 12x = 12x + 3x = 15x\)

It is also given that, from 2003 to 2006, the number of employees at this company decreased by a factor of \(\frac{1}{3}\)
    • Therefore, the number of employees in 2006 = \(15x – \frac{1}{3} * 15x = 15x – 5x = 10x\)

As there were 100 employees in 2006, we can write
    • 10x = 100
    Or, x = 10
    Or, 12x = 10 * 12 = 120

Hence, the correct answer is option B.

Answer: B
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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 22 Jun 2018, 05:31
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Bunuel wrote:
From 2000 to 2003, the number of employees at a certain company increased by a factor of 1/4. From 2003 to 2006, the number of employees at this company decreased by a factor of 1/3. If there were 100 employees at the company in 2006, how many employees were there at the company in 2000 ?


A. 200

B. 120

C. 100

D. 75

E. 60


We can PLUG IN THE ANSWERS, which represent the number of employees in 2000.

Since the number of employees increases by 1/4 and then decreases by 1/3, the correct answer must be divisible by 4 and 3.
In other words, the correct answer must be a MULTIPLE OF 12.
Eliminate A, C and D.

Since the decrease (1/3) is greater than the increase (1/4), the number of employees must DECREASE from 2000 to 2006.
Implication:
The number of employees in 2000 must be GREATER than the number of employees in 2006 (100).
Eliminate E.


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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 23 Jun 2018, 03:16
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2000 -> 2003: x -> x + x/4
2003 -> 2006: (x + x/4) -> (x + x/4) - (x +x/4)/3
2006: 100

100 = (x + x/4) - (x +x/4)/3
x = 120

Answer: B
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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 25 Jun 2018, 12:04
Bunuel wrote:
From 2000 to 2003, the number of employees at a certain company increased by a factor of 1/4. From 2003 to 2006, the number of employees at this company decreased by a factor of 1/3. If there were 100 employees at the company in 2006, how many employees were there at the company in 2000 ?


A. 200

B. 120

C. 100

D. 75

E. 60



Letting x = the number of employees in 2000, we have:

x(1 + 1/4)(1 - 1/3) = 100

x(5/4)(2/3) = 100

x(10/12) = 100

x = 100(12/10)

x = 120

Answer: B
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From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 25 Jun 2018, 15:21
B is correct:

x+(1/4)x = (5/4)x
(5/4)x - (1/3)(5/4)x = (5/4)x[1 - 1/3] = (5/4)x.(2/3) = (5/6)x = 100

x = 120
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From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 25 Jun 2018, 19:46
Let no of employees in 2000 be x
2000 to 2003: increase of 25% (i.e. 1/4), 2003 to 2006 decrease of 33% (i.e. 1/3)
Hence,

x * 1.25 * (1-(1/3)) = 100
x = (100*3)/(2*1.25) = 120

Hence B
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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 25 Jun 2018, 20:09
Bunuel wrote:
From 2000 to 2003, the number of employees at a certain company increased by a factor of 1/4. From 2003 to 2006, the number of employees at this company decreased by a factor of 1/3. If there were 100 employees at the company in 2006, how many employees were there at the company in 2000 ?


A. 200

B. 120

C. 100

D. 75

E. 60


let x=number of employees in 2000
x*5/4*2/3=100
x=120
B
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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 23 Aug 2018, 06:22
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It is a very simple problem.. Its straightforward X=120..
Option B.
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Re: From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 14 Sep 2018, 13:36
Is it wise to use options here?
Like choose a middle option 'C' at first.

Option C: 100 so 100 + 1/4 of 100 = 125. Now 125- 125/3 = 125-41 = 84 (approx), but as of question 100 employes are there in 2006. So option C rejected.

Option B: 120, so 120+1/4 of 120 = 150. Now 150-150/3= 100. As of question, there are 100 employees in 2006. So Option B is correct.

Is it wise to solve the problem by this method of using options?
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From 2000 to 2003, the number of employees at a certain company increa  [#permalink]

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New post 16 Sep 2018, 06:10
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prav04 wrote:
Is it wise to use options here?
Like choose a middle option 'C' at first.

Option C: 100 so 100 + 1/4 of 100 = 125. Now 125- 125/3 = 125-41 = 84 (approx), but as of question 100 employes are there in 2006. So option C rejected.

Option B: 120, so 120+1/4 of 120 = 150. Now 150-150/3= 100. As of question, there are 100 employees in 2006. So Option B is correct.

Is it wise to solve the problem by this method of using options?



Hi prav04

Yes you can. That sometimes helps to solve questions quickly where forming equations and solving them is time consuming.
This approach is fine.

Hope this helps!
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