Problem Solving

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:


So, of the 348 employees, 203 NEITHER work part time NOR are uninsured (i.e., are insured)
So, the probability = 203/348 = [spoiler]7/12[/spoiler]

Answer: A
-------------------------------------
NOTE: This type of question is VERY COMMON on the GMAT, so be sure to master the technique.

To learn more about the Double Matrix Method, watch our free video: https://www.gmatprepnow.com/module/gmat- ... ems?id=919

Once you're familiar with this technique, you can attempt these additional practice questions:

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Medium Problem Solving questions
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Difficult Problem Solving questions
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- https://www.beatthegmat.com/mba/2011/05/ ... question-2

Easy Data Sufficiency questions
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- https://www.beatthegmat.com/ds-quest-t187706.html

Medium Data Sufficiency questions
- https://www.beatthegmat.com/sets-matrix-ds-t271914.html
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- https://www.beatthegmat.com/a-manufacturer-t270331.html
- https://www.beatthegmat.com/in-costume-f ... 69355.html
- https://www.beatthegmat.com/mba/2011/05/ ... question-1

Difficult Data Sufficiency questions
- https://www.beatthegmat.com/double-set-m ... 71423.html
- https://www.beatthegmat.com/sets-t269449.html
- https://www.beatthegmat.com/mba/2011/05/ ... question-3

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