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In a survey of 348 employees, 104 of them are uninsured, 54 work part

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In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]

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New post Updated on: 12 Nov 2014, 09:43
3
7
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

75% (02:31) correct 25% (02:25) wrong based on 228 sessions

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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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New post 01 Oct 2011, 04:11
let a be the ones who are uninsured, c be those who work part-time 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=348-145=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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New post 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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New post 01 Oct 2011, 09:15
1
---------UI----------------NUI-------Total
PT----(12.5/100)*104 = 13----------- --54
NPT---104-13-------------- x--------294
Total--104----------------------------348

we have to find not part time and not uninsured . in other words not part time and insured = x/348 = (294-104+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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New post 02 Oct 2011, 09:45
Any shortcuts for reducing 203/348 --> 7/12 . Took most of my time.
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part [#permalink]

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New post 06 Oct 2011, 07:10
348 = 104+54-13+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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New post 06 Oct 2011, 09:05
104*12.5%=13
348=104+54-13+N
N=203
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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New post 08 Oct 2011, 04:51
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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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New post 09 Oct 2011, 21:42
GIVEN
part time full time

UNI A B 104


INS C X F
Tot 54 E

F=348-104=244
E=348-54=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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New post 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 = 348-145=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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New post 10 Oct 2011, 06:22
U = 104 (uninsured)
P = 54 (part time)
u U P = 104*12.5% = 13 (unisured and part time)
I = 348-104 = 244 (insured)
I U F = 244 -(54-13) = 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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New post 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
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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New post 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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New post 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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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part   [#permalink] 02 Mar 2018, 09:13
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