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In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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Updated on: 12 Nov 2014, 09:43
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In a survey of 348 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured? (A) 7/12 (B) 8/41 (C) 9/348 (D) 1/8 (E) 41/91
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Originally posted by LalaB on 01 Oct 2011, 03:40.
Last edited by Bunuel on 12 Nov 2014, 09:43, edited 1 time in total.
RENAMED THE TOPIC.



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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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01 Oct 2011, 04:11
let a be the ones who are uninsured, c be those who work parttime and b be those who do both and d be neither. We have to find d/348 Given b+c=104 a+b=54 b=12.5(104)/100=13 so a=41 a+b+c=41+104=145 d=348145=203 d/348=203/348=7/12 or d/348>0.5 so eliminate b, c, d, and e
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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01 Oct 2011, 04:17
here is my approach
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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01 Oct 2011, 09:15
UINUITotal PT(12.5/100)*104 = 13 54 NPT10413 x294 Total104348
we have to find not part time and not uninsured . in other words not part time and insured = x/348 = (294104+13)/348
=203/348 = 7/12
Answer is A.



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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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02 Oct 2011, 09:45



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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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06 Oct 2011, 07:10
348 = 104+5413+n n (neither) =203 probability = 203/348 = 7/12
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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06 Oct 2011, 09:05



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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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08 Oct 2011, 04:51
catfreak wrote: Any shortcuts for reducing 203/348 > 7/12 . Took most of my time. look at the options...only one option is more than 50%...
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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09 Oct 2011, 21:42
GIVEN part time full time UNI A B 104 INS C X F Tot 54 E F=348104=244 E=34854=294 A=12.5 % of 54 = 13 A+B = 104 B = 91 A+C=54==>C=41 C+X=244===>X=203 part time full time UNI 13 91 104 INS 41 203 244 54 294 Probability ===>203/348 ==>7/12 ans (A)
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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09 Oct 2011, 22:58
Using Van diagram: if: U= #of uninsured P= Part Timer Common area between U and P = X= 13 U and P = 91+13+41= 145 neither U and P = 348145=203 Therefore, Probability = 203/348=7/12



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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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10 Oct 2011, 06:22
U = 104 (uninsured) P = 54 (part time) u U P = 104*12.5% = 13 (unisured and part time) I = 348104 = 244 (insured) I U F = 244 (5413) = 203 (insured and full time) Probability = 203/348 = 7/12
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In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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13 Nov 2014, 04:58
LalaB wrote: In a survey of 348 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured?
(A) 7/12 (B) 8/41 (C) 9/348 (D) 1/8 (E) 41/91
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File comment: the person will neither work part time nor be uninsured are 203 Total Number of persons are 348 So the probability will be 203/348 =7/12
SOL.png [ 2.65 KiB  Viewed 3989 times ]



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In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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25 Mar 2016, 18:35
LalaB wrote: In a survey of 348 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured?
(A) 7/12 (B) 8/41 (C) 9/348 (D) 1/8 (E) 41/91 i went with approximations on this one... insured total = 348  104 = 244 104 uninsured, 1/8 or 13 work part time. so 41 who work part time are insured. or 203 are insured and do not work part time. 203/348 that's way more than 1/2 B, C, D, and E can be eliminated right away. another way to look at this problem: work full time  294 out of this number, a very few are uninsured. since we are asked for those who are insured and those who work part time..it definitely needs to be >50%



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In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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13 Sep 2016, 18:10
Once we have found out that the probability looks like 203/348, it is better to look at the options. Before that, get an estimate of the ratio( 348 is less than double of 203) that we would be looking for. This would help in narrowing down the options that need to be simplified to look for an answer. (A) 7/12  I kept this for the last. Having checked the wide deviation of the other ratios from the possible correct one, I chose A. (B) 8/41  Denominator is more than 5 times the numerator. In other words, the ratio is more than 1:5. Incorrect (C) 9/348  Denominator is more than 40 times the numerator. In other words, the ratio is more than 1:40. Incorrect (D) 1/8  Denominator is 8 times the numerator. In other words, the ratio is more than 1:5. Incorrect (E) 41/91  Denominator is almost 2 times the numerator. In other words, the ratio is more than 1:2. Incorrect
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]
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