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16 Sep 2014, 01:49
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
55% (03:36) correct
45%
(03:41)
wrong
based on 84
sessions
History
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Time
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Not Attempted Yet
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. \(\frac{7}{12}\)
B. \(\frac{8}{41}\)
C. \(\frac{91}{348}\)
D. \(\frac{1}{8}\)
E. \(\frac{41}{91}\)
A. \(\frac{7}{12}\)
B. \(\frac{8}{41}\)
C. \(\frac{91}{348}\)
D. \(\frac{1}{8}\)
E. \(\frac{41}{91}\)
16 Sep 2014, 01:49
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Official Solution:
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. \(\frac{7}{12}\)
B. \(\frac{8}{41}\)
C. \(\frac{91}{348}\)
D. \(\frac{1}{8}\)
E. \(\frac{41}{91}\)
Here, we must determine the probability that a employee randomly selected from a pool of 348 employees is not uninsured, AND does not work part time. This survey divides employees into four groups: employees who are uninsured but are not part time, employees who work part time but who are not uninsured, part time employees who are also uninsured, and employees who neither are uninsured nor work part time. Assign variables accordingly:
We need to find the fraction \(\frac{z}{348}\).
The question tells us that the total number of uninsured employees is 104, so \(w + x = 104\). Similarly, the total number of those who work part time is 54, so \(x + y = 54\). We also know that \(12.5\) percent, or \(\frac{1}{8}\) of the employees who are uninsured also work part time: \(\frac{1}{8}(w + x) = x\). Finally, recall that all four quantities sum to 348, so \(w + x + y + z = 348\) and \(z = 348 - w - x - y\).
Use substitution to solve for each variable. First, solve \(\frac{1}{8}(w + x) = x\) for \(w\). Multiply both sides of \(\frac{1}{8}(w + x) = x\) by 8: \(w + x = 8x\). Combine like terms: \(w = 7x\). Now substitute into \(w + x = 104\) to get \(7x + x = 104\). Combine like terms and divide: \(8x = 104\), \(x = 13\).
Plug back into \(w = 7x\): \(w = 7(13) = 91\).
Next, substitute the value for \(x\) into \(x + y = 54\) to get \(13 + y = 54\). Subtract: \(y = 41\).
Finally, substitute and solve for \(z\): \(z = 348 - w - x - y = 348 - 91 - 13 - 41 = 203\).
Therefore, \(\frac{z}{348} = \frac{203}{348}\). This is not one of the answer choices, so we need to reduce the fraction. Factor a 29 out of the top and the bottom: \(\frac{7}{12}\).
Answer: A
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. \(\frac{7}{12}\)
B. \(\frac{8}{41}\)
C. \(\frac{91}{348}\)
D. \(\frac{1}{8}\)
E. \(\frac{41}{91}\)
Here, we must determine the probability that a employee randomly selected from a pool of 348 employees is not uninsured, AND does not work part time. This survey divides employees into four groups: employees who are uninsured but are not part time, employees who work part time but who are not uninsured, part time employees who are also uninsured, and employees who neither are uninsured nor work part time. Assign variables accordingly:
We need to find the fraction \(\frac{z}{348}\).
The question tells us that the total number of uninsured employees is 104, so \(w + x = 104\). Similarly, the total number of those who work part time is 54, so \(x + y = 54\). We also know that \(12.5\) percent, or \(\frac{1}{8}\) of the employees who are uninsured also work part time: \(\frac{1}{8}(w + x) = x\). Finally, recall that all four quantities sum to 348, so \(w + x + y + z = 348\) and \(z = 348 - w - x - y\).
Use substitution to solve for each variable. First, solve \(\frac{1}{8}(w + x) = x\) for \(w\). Multiply both sides of \(\frac{1}{8}(w + x) = x\) by 8: \(w + x = 8x\). Combine like terms: \(w = 7x\). Now substitute into \(w + x = 104\) to get \(7x + x = 104\). Combine like terms and divide: \(8x = 104\), \(x = 13\).
Plug back into \(w = 7x\): \(w = 7(13) = 91\).
Next, substitute the value for \(x\) into \(x + y = 54\) to get \(13 + y = 54\). Subtract: \(y = 41\).
Finally, substitute and solve for \(z\): \(z = 348 - w - x - y = 348 - 91 - 13 - 41 = 203\).
Therefore, \(\frac{z}{348} = \frac{203}{348}\). This is not one of the answer choices, so we need to reduce the fraction. Factor a 29 out of the top and the bottom: \(\frac{7}{12}\).
Answer: A
ConfusedAdmit
Joined: 28 Feb 2016
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21 Mar 2016, 18:00
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Bunuel, maybe a stupid question... how did you know to factor a 29?
