Confusing Workers

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Confusing Workers

by smclean23 » Wed Jul 30, 2008 4:02 pm
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







Answer is A
Source: — Problem Solving |

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by VP_RedSoxFan » Thu Jul 31, 2008 7:42 am
If 90 percent of the office has less than 10 years service, then 10% has at least 10 years. We know that there are 16 people with at least 10 years and that represents 10 percent of the workforce. Now we can solve for the number of employees, x, as follows:

.1x=16

x=160

If 40 percent have at least 5 years, then that represents 64 workers (160 * 0.4). The question asks for "at least 5 years, but less than 10" so it's 64 - 16 = 48

Hope this helps.
Last edited by VP_RedSoxFan on Thu Jul 31, 2008 7:56 am, edited 1 time in total.
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by sudhir3127 » Thu Jul 31, 2008 7:50 am
i did the same as ryan did..
total 160
16 are >10 and
60% less than 5

hence 160-16-96 = 48.

hope its clear..

I am sure its a typo by Ryan its 64-16 = 48 and not 64- 18.

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by VP_RedSoxFan » Thu Jul 31, 2008 7:58 am
Thanks for the catch. I usually do the problems on paper first and then transcribe. Must be my Homer-Simpson-like fat fingers... :)
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