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In an office, 40 percent of the workers have at least 5

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In an office, 40 percent of the workers have at least 5 [#permalink] New post 31 Mar 2008, 18:38
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In an office, 40 percent of the workers have at least 5 years of service, and a total of 16 workers have at least 10 years of service. If 90 percent of the workers have fewer than 10 years of service, how many of the workers have at least 5 but fewer than 10 years of service?

(A) 48
(B) 64
(C) 50
(D) 144
(E) 160
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Re: PS: workers [#permalink] New post 31 Mar 2008, 19:15
And going for the trifecta..

What we know
employees >= 5 = 40% -- 1
employees <10 = 90 %
This implies that 10% of the employees have >= 10yrs experience. -- 2
We are also given that 16 employees have greater than 10 yrs experience.

If n is the total number of employees. 0.1n=16 or n = 160

out of these 30% are in the range, [ 5 < yrs of experience < 10 ]
(remember that >5 includes the employees who have greater than 10yrs of work ex)

Thus answer is 0.3n = 48 (A).


Man it took 3 minutes longer to type the answer than it took to solve the problem.

Last edited by daszero on 31 Mar 2008, 19:20, edited 1 time in total.
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Re: PS: workers [#permalink] New post 31 Mar 2008, 19:18
W = total workers

at least 5: W * .4
at least 10: 16
<10: W * .9

W = W * .9 + 16
.1W = 16
W = 160

5 <= x < 10
x = W * .4 - 16 = 48
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Re: PS: workers [#permalink] New post 31 Mar 2008, 21:51
daszero wrote:
And going for the trifecta..

What we know
employees >= 5 = 40% -- 1
employees <10 = 90 %
This implies that 10% of the employees have >= 10yrs experience. -- 2
We are also given that 16 employees have greater than 10 yrs experience.

If n is the total number of employees. 0.1n=16 or n = 160

out of these 30% are in the range, [ 5 < yrs of experience < 10 ]
(remember that >5 includes the employees who have greater than 10yrs of work ex)

Thus answer is 0.3n = 48 (A).


Man it took 3 minutes longer to type the answer than it took to solve the problem.


I used the table to solve it, but I lost in the gem. Could you figure out the boldface and colored above? I still not get it. Thank you!
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Re: PS: workers [#permalink] New post 01 Apr 2008, 11:21
sondenso wrote:
daszero wrote:
And going for the trifecta..

What we know
employees >= 5 = 40% -- 1
employees <10 = 90 %
This implies that 10% of the employees have >= 10yrs experience. -- 2
We are also given that 16 employees have greater than 10 yrs experience.

If n is the total number of employees. 0.1n=16 or n = 160

out of these 30% are in the range, [ 5 < yrs of experience < 10 ]
(remember that >5 includes the employees who have greater than 10yrs of work ex)

Thus answer is 0.3n = 48 (A).


Man it took 3 minutes longer to type the answer than it took to solve the problem.


I used the table to solve it, but I lost in the gem. Could you figure out the boldface and colored above? I still not get it. Thank you!


Let me rephrase the answer, lets see if this conveys it better :).

The problem says that 40% of the employees have work experience of at least 5 years

The problem also says that 16 employees have work experience of at least 10 years.

If the total number of employees = n,

the number of people who are between 5 and 10 years of experience = ((40/100) * n) - 16 -- 1

The problem also says that 90% of employees have < 10 years experience. From this statement we see that the 16 employees who have more than 10 years of experience, constitute only 10% of the work force.

Thus (10/100) * n =16

Solving for this we get n=160. Substitute this in 1 and the answer works out to 48..
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Re: PS: workers [#permalink] New post 01 Apr 2008, 12:10
it is a Straight A ->

If 90 percent of the workers have fewer than 10 years of service =>a total of 16 workers have at least 10 years of service represent 10%

thus 10% =16=> 100% is 160

how many of the workers have at least 5 but fewer than 10 years of service=> Given 40 percent of the workers have at least 5 years of service thus 40% = 64 BUT the 40% also involves the 16 folks who have more than 10 yrs of service thus 64-16 =48
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Re: PS: workers [#permalink] New post 01 Apr 2008, 12:25
daszero wrote:
And going for the trifecta..

What we know
employees >= 5 = 40% -- 1
employees <10 = 90 %
This implies that 10% of the employees have >= 10yrs experience. -- 2
We are also given that 16 employees have greater than 10 yrs experience.

If n is the total number of employees. 0.1n=16 or n = 160

out of these 30% are in the range, [ 5 < yrs of experience < 10 ]
(remember that >5 includes the employees who have greater than 10yrs of work ex)

Thus answer is 0.3n = 48 (A).


Man it took 3 minutes longer to type the answer than it took to solve the problem.


I can't understand how did you come up with 30%? Please clarify this :
out of these 30% are in the range, [ 5 < yrs of experience < 10 ]
(remember that >5 includes the employees who have greater than 10yrs of work ex)

Thus answer is 0.3n = 48 (A).
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Re: PS: workers [#permalink] New post 02 Apr 2008, 11:16
el1981 wrote:
daszero wrote:
And going for the trifecta..

What we know
employees >= 5 = 40% -- 1
employees <10 = 90 %
This implies that 10% of the employees have >= 10yrs experience. -- 2
We are also given that 16 employees have greater than 10 yrs experience.

If n is the total number of employees. 0.1n=16 or n = 160

out of these 30% are in the range, [ 5 < yrs of experience < 10 ]
(remember that >5 includes the employees who have greater than 10yrs of work ex)

Thus answer is 0.3n = 48 (A).


Man it took 3 minutes longer to type the answer than it took to solve the problem.


I can't understand how did you come up with 30%? Please clarify this :
out of these 30% are in the range, [ 5 < yrs of experience < 10 ]
(remember that >5 includes the employees who have greater than 10yrs of work ex)

Thus answer is 0.3n = 48 (A).


> =5 yrs = 40%
>=10 yrs = (16/160) *100 = 10%

Between 5 and 10 yrs = 40-10 = 30%
Re: PS: workers   [#permalink] 02 Apr 2008, 11:16
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