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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
(PS15957)
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A. 200
B. 120
C. 100
D. 75
E. 60
(PS15957)
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Bunuel
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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Bunuel
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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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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General Discussion
Posts: 28
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
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
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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In 2003 employees then 5x/4
In 2006 employees then (5x/4-5x/12)=10x/12
10x/12=100
X=120
Sent from my iPhone using GMAT Club Forum
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
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
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?
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?
Posts: 6,727
Bunuel
From 2000 to 2003, the number of employees at a certain company increased by a factor of 1/4
Let x = the number of employees in 2000
So, the number of employees in 2003 = x + (1/4)x
= (5/4)x
From 2003 to 2006, the number of employees at this company decreased by a factor of 1/3
(5/4)x = the number of employees in 2003
So, the number of employees in 2006 = (5/4)x - (1/3)((5/4)x
= (5/4)x - (5/12)x
= (15/12)x - (5/12)x
= (10/12)x
= (5/6)x
If there were 100 employees at the company in 2006, how many employees were there at the company in 2000 ?
We can write: (5/6)x = 100
Multiply both sides by 6 to get; 5x = 600
Solve: x = 600/5 = 120
Answer: B
Cheers,
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Bunuel
Hello Experts,
EMPOWERgmatRichC, VeritasKarishma, IanStewart, Bunuel, chetan2u, ArvindCrackVerbal, GMATGuruNY, AaronPond, GMATinsight
We can cancel choice D (75) because if 1/4 of 75 is not an integer-people has to be integer. Can we also cancel choice E (60) by this way?
Also, appreciating your help if you put an insight so that we can visually cancel the most choices.
Thanks__
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Asad
Yes, you can eliminate many choices, and it is always useful to keep your eyes and mind open to such details..
As you have mentioned that 75 does not give an integer when divided by 4.
Similarly, we also have to decrease the total by 3, so our answer has to be divisible by 3. So, we can eliminate A, and C. However, if the increase was by 1/2 then we would have got a multiple of 3, as we would have had the new number 1+1/2 or 3/2 of initial .
Next you can also apply logic, wherein there is an increase of 1/4 over a smaller number, say x, as compared to decrease of 1/3( which is >1/4) of a greater number ( x+1/4 of x), so there is a overall decrease as we come from 2000 to 2006. Our answer, therefore, has to be > than what was there in 2006 or 100.
Now, you can eliminate except A and B.
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Asad
Hi Asad
Yes you are right till the point that you can eliminate 75 (Option D) cause it's NOT divisible by 4 (whereas the population of 2000 must be divisible by 4 as per the language of the question
But we can cancel Option E due to the same reason cause it's divisible by 4. However, this option may be eliminated based on a little calculation that 60, when increased by a factor 1/4, becomes 75 which when decreased by 1/3 becomes 50 which actually should have been 100 as per given info.
100 could be eliminated too cause it in 2003 must becomes 100--->(5/4)*100=125 which is NOT divisible by 3 to decrease by a factor 1/3
120 (in 2000) -------> (5/4)*120=150 (in 2003)----------->(2/3)*150=100 BINGO!!!
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Asad
Hello Asad,
IMHO, you should never eliminate answer options based on how they look. Looks can be deceptive.
What you should look at is to eliminate answer options based on your understanding of the constraints in the question, like how you have done with 75. The reason that you have provided to eliminate 75 is a sound one as well.
However, there’s much more to this question. You are increasing the value in 2000 by\(\frac{ 1}{4}\)th and then reducing the resultant value by \(\frac{1}{3}\)rd. This means that the resultant value should be a multiple of 3 and 4 both i.e. it should be a multiple of 12. This is a stronger constraint. Based on this, you can eliminate answer options A, C and D.
Of the two options left i.e. B and E, 60 can be eliminated based on its magnitude – increasing 60 by\(\frac{ 1}{4}\)th takes us to 75 and reducing it by \(\frac{1}{3}\)rd takes us down to 50. Answer option E can be eliminated.
The correct answer option has to be B.
From the above, we see that after doing the two percentage change operations, we got 50. But, we are supposed to get 100 which is exactly double of 50. Therefore, double of 60 i.e. 120 should be our answer, which is indeed answer option B.
Hope that helps!
Asad
Yes, you can cancel off 75 using this logic but not 60.
1/4 of 75 is not an integral value but 1/4 of 60 is 15.
Alternatively, all you need to see is this:
x * (5/4) * (2/3) = 100
x = 120
Answer (B)
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Bunuel
30 seconds solution
Use multiplying factor
Assume no of employees
In 2000=y ; 2003=x
and given 2006=100
Now , x*2/3=100
=> x =150
Also given
y*5/4=x
=> y = 120. (Use x=150)
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Bunuel
Answer: Option B
Video solution by GMATinsight
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Posts: 43
Here is my approach :
Here let us assume total number of employees in 2000 be 12 (factor of 4 and 3)
Now from 2000 -2003 , they increased by 1/4 = 1+1/4 = 5/4 * 12 = 15 , so the increase is by 3 employees
from 2003 -2006 , they decreased by 1/3 = 1-1/3 = 2/3 * 15 = 10 , so the decrease is by 5 employees
If 12 employees --> dec to 10 overall , if there were 100 employees by 2006
so, there will be same factor 10 * 10 = 100 so 12*10 = 120 in the beginning of 2000
Ans B
Here let us assume total number of employees in 2000 be 12 (factor of 4 and 3)
Now from 2000 -2003 , they increased by 1/4 = 1+1/4 = 5/4 * 12 = 15 , so the increase is by 3 employees
from 2003 -2006 , they decreased by 1/3 = 1-1/3 = 2/3 * 15 = 10 , so the decrease is by 5 employees
If 12 employees --> dec to 10 overall , if there were 100 employees by 2006
so, there will be same factor 10 * 10 = 100 so 12*10 = 120 in the beginning of 2000
Ans B
Dealing with fractions in this question is the most time saving strategy
Approach:
Let the no of employees in 2000 be N
According to the Q,
N*(4/4+1/4)(3/3-1/3)=100
N*5/4*2/3=100
N=120
Answer (B)
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Approach:
Let the no of employees in 2000 be N
According to the Q,
N*(4/4+1/4)(3/3-1/3)=100
N*5/4*2/3=100
N=120
Answer (B)
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Expert reply
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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No need to plug-in numbers. Prefer fractions rather than decimals. Stop trying tricks and establish a pattern in basic math specially to arithmetic questions.
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