Uninsured part time

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j_shreyans: Legendary Member; Posts: 510; Joined: Thu Aug 07, 2014 2:24 amFollowed by: 5 members; Thanked: 3 times

by j_shreyans » Wed Sep 10, 2014 10:48 pm

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)91/348
D)1/8
E)41/91

OAA

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Source: Beat The GMAT — Problem Solving | Wed Sep 10, 2014 10:48 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Sep 11, 2014 2:53 am

j_shreyans 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)91/348
D)1/8
E)41/91

We can use the following equation:

Total = Group 1 + Group 2 - Both + Neither

The big idea with overlapping groups is to SUBTRACT THE OVERLAP.
When we count everyone in Group 1 (uninsured) and everyone in Group 2 (part-time), the employees in BOTH groups (those who are BOTH uninsured and part-time) get counted twice.
So that we don't double-count the employees who belong to both groups, we SUBTRACT THE OVERLAP from the total.

In the problem above:
Total = 348.
Uninsured = 104.
Part-time = 54.
Both = 12.5% of uninsured = (1/8)(104) = 13.
N = neither.

Plugging these values in the equation in bold, we get:
348 = 104 + 54 - 13 + N
N = 203.

Thus:
P(neither) = N/total = 203/348 = more than 1/2.
Of the 5 answer choices, only 7/12 is greater than 1/2.

The correct answer is A.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Sep 12, 2014 10:20 am

j_shreyans 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)91/348
D)1/8
E)41/91

We can also use the Double Matrix Method here. This technique can be used for most questions featuring a population in which each member has two characteristics associated with it.

Here, we have a population of 348 employees, and the two characteristics are:
- uninsured or not uninsured (i.e., insured)
- work part time or don't work part time

So, we can set up our diagram as follows:

Note: I placed a star in the bottom right box to remind me that this is the value we are trying to determine (employees who neither work part time nor are uninsured).

Now, if there are 348 employees and 104 are UNinsured, then the other 244 employees are INsured.
Similarly, if 54 employees work part time, then the other 294 employees do NOT work part time.
So, we can add that information to the diagram.

12.5 percent of employees who are uninsured work part time.
12.5% = 1/8
1/8 of 104 = 13, so 13 of the 104 uninsured employees work part time.
When we add this information to our diagram, we get the following:

Since we know the sums of the rows and columns, we can determine the values in every box to get the following:

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Sep 12, 2014 10:32 am

j_shreyans 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)91/348
D)1/8
E)41/91

OAA

ASIDE: On test day, if you didn't know how to tackle this question, you could still ELIMINATE 3 answer choices, and leave yourself a 50% chance of guessing correctly.

Since there are 348 employees, P(selected person neither works part time nor is uninsured) = (some integer)/348

So, the correct answer will EITHER be a fraction with 348 in the denominator OR be a reduced fraction.

Since 348 = (2)(2)(3)(29), the DENOMINATOR of the reduced fraction will be some product of the two 2's, 3 and/or 29.
For example, the reduced fraction could be something/2, something/4, something/6, something/12, something/29 etc.

Now check the answer choices.
A) 7/12 Since 12 can be written as (2)(2)(3), answer A is possible. KEEP
B) 8/41 Since 41 cannot be written as the product of the two 2's, 3 and/or 29, ELIMINATE B.
C) 91/348 KEEP
D) 1/8 Since 8 cannot be written as the product of the two 2's, 3 and/or 29, ELIMINATE D.
E) 41/91 Since 91 cannot be written as the product of the two 2's, 3 and/or 29, ELIMINATE E.

Now guess either A or C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Sun Sep 14, 2014 10:30 pm

j_shreyans 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)91/348
D)1/8
E)41/91

OAA

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