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A school employs math teachers and physics teachers. If 4 of

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A school employs math teachers and physics teachers. If 4 of

by Max@Math Revolution » Mon May 14, 2018 12:29 am

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[GMAT math practice question]

A school employs math teachers and physics teachers. If 4 of these teachers are selected randomly, what is the probability that at least one math teacher is selected?

1) The ratio of the number of physics teachers to the number of math teachers is 2 to 1.
2) The sum of the number of physics teachers and the number of math teachers is 24.
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Source: — Data Sufficiency |
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by Max@Math Revolution » Wed May 16, 2018 2:05 am

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Let m be the number of math teachers and p be the number of physics teachers. Since we have 2 variables and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2):
Condition 1) yields p : m = 2 : 1 => p = 2m.
Condition 2) gives m + p = 24.
Combining these equations gives m + p = m + 2m = 3m = 24.
So, m = 8 and p = 16.
The probability that at least one math teacher is selected is 1 - pC4 / m+pC4 = 1 - 16C4 / 24C4.
Both conditions together are sufficient.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.


Condition 1) (p = 2m)
If m = 8 and p = 16, the probability is 1 - pC4 / m+pC4 = 1 - 16C4 / 24C4.
If m = 4 and p = 8, the probability is 1 - pC4 / m+pC4 = 1 - 8C4 / 12C4.
These values are different. Since we don't have a unique solution, condition 1) is not sufficient.

Condition 2) (m + p = 24)
If m = 8 and p = 16, the probability is 1 - pC4 / m+pC4 = 1 - 16C4 / 24C4.
If m = 12 and p = 12, the probability is 1 - pC4 / m+pC4 = 1 - 12C4 / 24C4.
These values are different. Since we don't have a unique solution, condition 2) is not sufficient.

Therefore, C is the answer.

Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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