Are there more girls than boys at a school?

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[GMAT math practice question]

Are there more girls than boys at a school?

1) 3/7 of the number of girls is more than 1/3 of the number of boys
2) 1/3 of the number of girls is more than 2/5 of the number of boys

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by GMATGuruNY » Mon Apr 02, 2018 3:05 am
Max@Math Revolution wrote:[GMAT math practice question]

Are there more girls than boys at a school?

1) 3/7 of the number of girls is more than 1/3 of the number of boys
2) 1/3 of the number of girls is more than 2/5 of the number of boys
Statement 1:
(3/7)G > (1/3)B
G > (7/3)(1/3)B
G > (7/9)B.
Case 1: B=9
Substituting B=9 into G > (7/9)B, we get:
G > (7/9)(9)
G > 7.
If G=8, then there are more boys than girls, so the answer to the question stem is NO.
If G=10, then there are more girls than boys, so the answer to the question stem is YES.
INSUFFICIENT.

Statement 2:
(1/3)G > (2/5)B
G > (6/5)B.
Since the value of G more than 6/5 the value of B, there are more girls than boys.
SUFFICIENT.

The correct answer is B.
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by Max@Math Revolution » Wed Apr 04, 2018 12:14 am
=>

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.

Since we have 2 variables (b for boys and g for girls) 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) together:
From condition 1:
(3/7)g > (1/3)b
=> 9g > 7b

From condition 2:
(1/3)g > (2/5)b
=> 5g > 6b
=> 6g > 5g > 6b
=> g > b

Both conditions together are sufficient.

Since this question is an inequality 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):

(3/7)g > (1/3)b
⇔ 9g > 7b

If g = 10 and b = 8, then the answer is 'yes'.
If g = 8 and b = 8, then the answer is 'no'.

Thus, condition 1) is not sufficient on its own.


Condition 2):

(1/3)g > (2/5)b
=> 5g > 6b
=> 6g > 5g > 6b
=> g > b

Thus, condition 2) is sufficient on its own.

Therefore, B is the answer.

Answer: B

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.