answer to last week's MGRE Math Beast Challenge

MGRE has this to say about last week's Math Beast Challenge:

We are told that the 11th grade girls at Stumpville High School have an average GPA of 3.1, and the overall 11th grade average GPA is 3.05. Fortunately, the 11th grade has the same number of boys and girls, so rather than using the weighted average formula, we can simply conclude that the boys’ average GPA must be 3.0. Write on your paper something like:

11th grade boys average GPA = 3.0

(If you’re not sure about our quick inference, try this example: If a dozen people in a room each have an average of $10 and another dozen people each have an average of $20, then the average amount of money each person has is exactly $15, since the $10 group and the $20 group are the same size. Similarly, if this example had told you that a dozen people have an average of $10, another dozen people have x dollars, and the overall average is $15, then – since 15 is exactly halfway between 10 and 20 – you could confidently conclude that the other dozen people have an average of $20.)

We are told that all of the boys enrolled in Honors Chemistry are in 11th grade. From the first chart, add up the total number of boys: 46 + 52 + 52 + 50 = 200. From the bottom chart, we can see that 6% of boys take Honors Chemistry. 6% of 200 is 12, so write on your paper something like:

11th grade boys in Honors Chem = 12

We are told that these 12 boys have an average GPA of 3.8. And yet the average GPA for boys in 11th grade is only 3.0 – thus, we are expecting the rest of the boys’ GPAs to be much lower than the Honors Chemistry boys’ GPAs.

However, we CANNOT do the kind of “quick logic” we did above and assume that, since the Honors Chem 11th grade boys have an average GPA of 3.8 and the 11th grade boys in general have an average GPA of 3.0, therefore the rest of the boys have an average GPA of 2.2 (since 3.0 is exactly in the middle of 2.2 and 3.8). THIS IS A TRAP! We cannot conclude that the answer is 2.2, because the number of Honors Chem 11th grade boys and the number of other 11th grade boys are NOT THE SAME.

We must calculate a weighted average (to review Weighted Averages, see Manhattan Prep’s GRE Word Problems Strategy Guide). Remember that there are 12 boys in the 11th grade who are in Honors Chem and 40 who are not in Honors Chem:

[12(3.8) + 40(x)]/52 = 3.0

12(3.8) + 40x = 156

45.6 + 40x = 156

40x = 110.4

x = 2.76

The correct answer is B.

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this week's MGRE Math Beast Challenge

From here:

The 11th-grade girls at Stumpville High School have an average GPA of 3.1, and the overall 11th-grade average GPA is 3.05. If all of the boys enrolled in Honors Chemistry are in the 11th grade and those boys have an average GPA of 3.8, what is the average GPA of all the 11th-grade boys who are not enrolled in Honors Chemistry?

(A) 2.2
(B) 2.76
(C) 2.96
(D) 3.05
(E) 3.16

Go to it! My own attempted solution will appear in the comments.


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right again!

The official answer to last week's MGRE Math Beast Challenge problem is indeed (E)!

Here is MGRE's explanation:

Interest on debt comprises 2% of total expenses for each division. So if the pharmaceutical division spends 4 times as much as the chemical division does on interest, then total expenses for the pharmaceutical division must be 4 times total expenses for the chemical division. The best thing to do is pick some smart numbers:

Total expenses for pharma = 400
Total expenses for chem= 100

{We might pause here to check our earlier logic. With these numbers, chem would spend 2 for interest on debt and pharma would spend 8. This agrees with what the problem told us about interest on debt.}

Payroll expenses for pharma = 26% of pharma total = (0.26)(400) = 104
Payroll expenses for chem = 38% of chem total = (0.38)(100) = 38

The question asks “What percent of the chemical division's payroll expense is the pharmaceutical division's payroll expense?”
With our numbers, this question is “What percent of 38 is 104?”
Rephrasing a bit: “104 is what percent of 38?”

We can see that 104 is more than 100% of 38, so (A) and (B) can be eliminated.
Actually, 104 is more than twice 38 (i.e. 104 > 76), so (C) and (D) can also be eliminated.

We can solve to prove that (E) is the answer by translating the percent question into an equation:
104 is what percent of 38?
104 = (x/100)(38), where x is the answer.
x = (104)(100)/38
x = 273.6842....

To the nearest whole percent, 104 is 274% of 38.

The correct answer is E.

I like MGRE's explanation, which involves, arguably, a little less math than my own explanation does. Many math problems on both the SAT and the GRE can be solved in this way: through logical deduction instead of hardcore algebra. Remember to eliminate possibilities as you work: by narrowing the number of plausible answers, you increase your chances of guessing correctly-- if you find yourself needing to make a guess. Of course, if you're able to use deduction to eliminate four out of five possibilities, then you're golden!


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this week's MGRE Math Beast Challenge

My answer to last week's challenge, regarding the honeycombed hexagons, was indeed correct: Quantity A is greater. MGRE says:

Regular hexagons are very special—divide the hexagon with three diagonals (running through the center) and you will get six equilateral triangles. Why are the triangles equilateral? Since the sum of the angles in any polygon is (n – 2)(180), the sum for a hexagon is 720. Divide by 6 to get that each angle is 120. When you divide the hexagon into triangles, you split each 120 to make two 60 degree angles for each triangle. Any triangle that has two angles of 60 must have a third angle of 60 as well, since triangles always sum to 180.

Since the triangles are equilateral, we know that all of them have all sides equal to 2√3. For each triangle, we know that the height will always equal half the side times √3. (This is a good fact to simply memorize; however, had you not memorized this, you could divide each equilateral into two 30–60–90 triangles and use the side ratios of 1 : √3 : 2 for a 30–60–90 triangle.)

Therefore, the height of each triangle is simply √3 * √3 = 3.

Since the area of a triangle is [(1/2)bh], the area of each equilateral triangle is: (1/2)*(2√3)*3 = 3√3.

Since there are six such equilateral triangles in each hexagon, the area of each hexagon is 18.

Since there are six hexagons in the honeycomb, the total area of the figure is 108√3 = 187.061487....

The correct answer is A.

This week, we've got another Data Interpretation challenge:




Go to it! My own answer will eventually appear in the comments.


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vindication

I got last week's MGRE Math Beast Challenge problem right. MGRE's reasoning:

The chemical division’s legal expenses are between 1/3 and 1/2 of the $720,000 spent by the pharmaceutical division on legal expenses. Thus, the chemical division spends between $240,000 and $360,000 on legal expenses.

The chemical division’s legal expenses are also given to us as 15 percent of the division’s total expenses.
$240,000 < 0.15c < $360,000, where c represents the total expenses of the chemical division. Solve for c by dividing by 0.15 (remember to do so on ALL sides of the inequality): $240,000/0.15 < c < $360,000/0.15 $1,600,000 < c < $2,400,000 The only answer choice between $1.6 million and $2.4 million is $1,855,100. The correct answer is D.

Pretty close to what I said in the comments.


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this week's MGRE Math Beast Challenge problem

Here it is:



Give it a go! My own attempted answer will appear in the comments; MGRE's official answer will appear next week.


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