this week's MGRE Math Beast Challenge

From here:

Quantity A
The number of different ways all 9 letters in the word “TENNESSEE” can be arranged.

Quantity B
The number of different ways all 7 letters in the word “WYOMING” can be arranged.

(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

Go to it! My answer will eventually appear in the comments. This appears to be a permutations and combinations problem. Ugh. The obvious answer would seem to be (A), but that very obviousness is what makes me cautious.


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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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answer to last week's MGRE Math Beast Challenge

Yes! I got the answer to last week's MGRE Math Beast Challenge problem correct! The answer was indeed 3, 5, 41, and 43. Here's MGRE's explanation:

The prime numbers less than 12 are 2, 3, 5, 7, and 11. There are five possible ages for the two children, or 5!/2!3! = (5)(4)/(2)(1) = 10 possible combinations for the children’s ages.

The prime numbers between 40 and 52 are 41, 43, and 47. There are three possible ages for the two adults, or 3!/2!1! = 3 possible combinations for the adults’ ages.

In total, there are (10)(3) = 30 possible age combinations—far too many to test each scenario looking for a prime number average age.

An alternative is to start from the resulting average age. The minimum ages of the family members are 2, 3, 41, and 43, which average to 22.25. The maximum ages of the family members are 7, 11, 43, and 47, which average to 27. The only prime number in this range is 23, which implies an age sum of (4)(23) = 92.

If the adults are 43 and 47, the sum of their ages is 90. The sum of the children’s ages would need to be 92 – 90 = 2. The minimum sum of the children’s ages is 2 + 3 = 5, so no need to continue checking these possibilities.

If the adults are 41 and 47, the sum of their ages is 88. The sum of the children’s ages would need to be 92 – 88 = 4. The minimum sum of the children’s ages is 2 + 3 = 5, so again, no need to continue checking these possibilities.

If the adults are 41 and 43, the sum of their ages is 84. The sum of the children’s ages would need to be 92 – 84 = 8. This is only possible if the children are 3 and 5.

Check: (3 + 5 + 41 + 43) = 92, so the average is 92/4 = 23, which is prime.

The correct answers are 3, 5, 41, and 43.

Interesting deductive process! I think I arrived at my answer more intuitively.


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this week's MGRE Math Beast Challenge

From here:

"The Prime of Life"

In a family of four people, none of the people [has] the same age, but all are a prime number of years old. Two of the people are less than 12 years old, and the other two people are between 40 and 52 years old. If the average of their four ages is also a prime number, what are the ages of the family members?

Indicate four such ages (check 4 slots).

( ) 2
( ) 3
( ) 5
( ) 7
( ) 11
( ) 41
( ) 43
( ) 47

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


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answer to last week's MGRE Math Beast Challenge

The answer to last week's MGRE Math Beast Challenge was indeed (B). But MGRE shows how to arrive at this answer without using Heron's Formula. To wit:

To find the area of triangle PQR we need a base and a height. If we consider side PR the base of the triangle, then QS, which is at a right angle to PR, is the height. We know that both triangle PQS and triangle SQR are right triangles, but to find the height QS we’ll need to know the length of either PS or SR. Unfortunately we don’t know either one, so we’ll have to name variables for several legs of the triangle.

In this case we’ll let x represent the length of PS and y represent the height QS. Note that if PS has length x, then SR has length 21 – x, so we do not need to (and should not) name a variable for length SR. Updating our diagram yields the following:



Apply the Pythagorean Theorem to each right triangle.
From triangle PQS: x² + y² = 100
From triangle SQR: (21 – x)² + y² = 289

We solve each in terms of y²:
y² = 100 – x²
y² = 289 – (21 – x)²

Then set the two expressions equal to y² equal to each other:
289 – (21 – x)² = 100 – x²

Now simplify:
189 – (21 – x)² = -x²
189 = (21 – x)² – x²
189 = (441 – 42x + x²) – x²
189 = 441 – 42x
42x = 252
x = 6

Now that the value of x is known, solve for y:
100 = 36 + y²
64 = y²
8 = y
(Or just recognize that PQS is a 6 – 8 – 10 right triangle.)

Finally, we can solve for the area of triangle PQR:

(1/2)bh = (1/2)(21)(8) = 84

The correct answer is B.

{Incidentally, this problem is named for Heron’s Formula, which is an alternative to the (1/2)bh triangle area formula. For a triangle with side lengths a, b, and c, the semiperimeter is defined as s = (a + b + c)/2. The area of the triangle equals √(s(s-a)(s-b)(s-c)). Note that this doesn’t require a right triangle, nor does it require knowledge of any heights measured perpendicular to any base. You will NOT need to know this for the GRE, as the original solution above attests.}

I was glad to see the above explanation, because I was honestly stumped as to how to approach the problem. I had thought that Heron's Formula would be the only way to solve it.


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this week's MGRE Math Beast Challenge

From here:



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


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answer to last week's MGRE Math Beast Challenge

The answer to last week's challenge is indeed (A)! That means Charles and I are both right. But here's the thing: I had anticipated that MGRE would take Charles's tack and use the plug-and-chug method, but instead they went full-on algebra, as I did, and offered their own version of plug-and-chug only at the very end as an afterthought, and only as a way to check their algebra. To review, then-- here's what Charles had written in his comment:

Yeah, I got A, too, although my process was not nearly as detailed. I just took 15 as a possible number of women at the party, subtracted 8, and multiplied by 4 to get 28 men originally at the party. Since the question then says that 35 men left the party, I knew that the original number of women had to be greater than 15, so the answer was A.

It took me about five times as long to write the above paragraph as it did to work out the answer. I never did figure out how many women were originally at the party.

Disgustingly simple. My own approach, you may recall, went for the algebra:

Let m = original # of men.

Let w = original # of women.

1st phase: we have m and w.
[All men & women are present.]

2nd phase: we have m and (w - 8).
[Eight women have left.]

3rd phase: we have (m - 35) and (w - 8).
[Thirty-five men have left.]

Given (per what we know of the second phase, and what the word problem tells us):

m = 4(w - 8)

And for the third phase:

(w - 8) = 2(m - 35)

At this point, it's a matter of systems of equations.

m = 4w - 32 (2nd phase)

2m = w + 62 (3rd phase)

Multiply the first equation by 2:

2m = 8w - 64

Match it up with the other equation and solve:

2m = 8w - 64
-(2m = w + 62)

=

0 = 7w - 126

7w = 126

w = 18

The original number of women was 18, so Quantity A is greater.

I'm going with (A).

Not simple, but definitely thorough. And here, finally, is how MGRE tackled the problem:

This problem can be solved with a system of two equations.

First, “after 8 women leave, there are four times as many men as women.” Thus, once 8 is subtracted from the number of women, there is a 4 to 1 male/female ratio:

m/(w - 8) = 4/1

Cross-multiply and simplify:

m = 4(w – 8)
m = 4w – 32

Then, 35 men leave, and the 8 women don’t come back, resulting in a 1 to 2 male/female ratio:

(m - 35)/(w - 8) = 1/2

Cross-multiply and simplify:

2(m – 35) = (w – 8)
2m – 70 = w – 8

We now have two equations in two variables.
1st equation: m = 4w – 32
2nd equation: 2m – 70 = w – 8

Since the 1st equation is already solved for m, simply plug into the 2nd equation for m:

2(4w – 32) – 70 = w – 8
8w – 64 – 70 = w – 8
8w – 134 = w – 8
8w = w + 126
7w = 126
w = 18

Since 18 is more than 15, the correct answer is A.

Although we are not asked for the number of men, note that we could easily generate it by plugging w = 18 into either equation:

m = 4w – 32
m = 4(18) – 32
m = 40

This would allow us to check our answer. If we begin with 18 women and 40 men, and then 8 women leave, we would have 10 women and 40 men, which indeed would be a 1 to 4 ratio of women to men. If 35 men then leave, we would have 10 women and 5 men, which indeed would be a 2 to 1 ratio of women to men.

The correct answer is A.

sentence equivalence!

Here's a GRE Sentence Completion problem from Manhattan Prep's blog.

The exhibit is not so much a retrospective as a __________ ; the artist’s weaker early work is glossed over and any evidence of his ultimate dissolution is absent entirely.

Select two correct answers.

(A) paean
(B) philippic
(C) tirade
(D) panacea
(E) eulogy
(F) crescendo

In the revised GRE's Sentence Completion section, the object of the game is to select TWO words that are each capable of (1) completing the sentence correctly and (2) giving the sentence a similar meaning. In other words, the words you select need to be either synonyms or almost-synonyms. Two antonyms might conceivably complete the sentence, but this would violate criterion (2). To get around this problem, the GRE Sentence Completion questions are designed so that a pair of antonyms can't be selected without one or the other word in the selected pair sounding ridiculous in context.

Have at it, then click the above link for the answer and explanation. I was able to answer correctly despite not knowing the meaning of "philippic," which is not a word I'd normally expect to see on the GRE.


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this week's MGRE Math Beast Challenge

From here:

Everyone at a party is either a man or a woman. After 8 women leave, there are four times as many men as women. After 35 men leave (and the 8 women do not return), there are twice as many women as men.

Quantity A
The number of women originally at the party

Quantity B
15

(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

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


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answer to last week's MGRE Math Beast Challenge

The correct answer was indeed (B)!

Here's MGRE's explanation:

This problem gives us an equilateral triangle and a circle, and tells us that the perimeter of the equilateral is 1.25 times the circumference of the circle.

This is license to plug in. Since both an equilateral and a circle are regular figures—that is, all equilaterals are in the same proportion as all other equilaterals, and all circles are in the same proportion as all other circles—we can be certain that we only need to plug in one set of values in order to be sure of the answer. Because we have regular figures and a way to relate them (triangle perimeter = 1.25 × circumference), we will not need to repeatedly try different values as we often do on Quantitative Comparisons.

We could say the radius of the circle is 2, so the circumference is 4π. Then, the perimeter of the equilateral triangle is (1.25)(4π) = 5π . This isn’t ideal, because then we are stuck with π in the calculations for the triangle, where it is unnecessarily awkward.

We could say circumference = 4 and therefore the perimeter of the triangle = (1.25)(4) = 5. But then the side of the equilateral triangle is 5/3, a non-integer, which is inconvenient. We can avoid this by using larger numbers (multiplied by a factor of 3).

So our smartest numbers are circumference = 12 and therefore the perimeter of the triangle = (1.25)(12) = 15, so each side of the triangle is 5. The height of an equilateral triangle is always (√3)/2 the side length. You can always derive this yourself by splitting the equilateral triangle into two 30–60–90 triangles, with known side ratio 1x : (√3)x : 2x. So, the area of the triangle is

(1/2)bh =

(1/2) • (5) • ((5)((√3)/2)) =

(25√3)/4 =

approx. 10.625. (Use the calculator.)

In the circle, circumference = 12 = 2πr, so

r = 12/(2π) = 6/π. The area of the circle is

π(r^2) = π((6/π)^2) = 36/π = approx. 11.46.

(Use the calculator and the approximation 3.14 for π.)

The correct answer is B.

I have to say... I like my method a lot better.


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this week's MGRE Math Beast Challenge

From here:

The perimeter of an equilateral triangle is 1.25 times the circumference of a circle.

Quantity A
The area of the equilateral triangle

Quantity B
The area of the circle

(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

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

By the way, sorry about the lack of posting last week. I think my sickness lingered on a bit longer than I had thought.


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answer to last week's MGRE Math Beast Challenge

Correct! The answer is indeed 14.286%. Strangely enough, MGRE's initial reasoning is rather tortured. Instead of following their own advice about solving a problem quickly through the plug-and-chug method, they went for the brute-force approach and solved the problem with variables. It's only at the end that the gurus discuss plug-and-chug as an afterthought. Here's what MGRE wrote:

When calculating a percent change “from (original) to (new),” be careful to use the ratio (change/original), not (change/new) or (new/original).

Create some variables:
x = attendance in 2010
y = attendance in 2011
z = attendance in 2012

In 2012, attendance was greater than in 2011, and even greater than it had been in 2010. So, x < y < z.

The question asks for the percent change from 2010 to 2011, or [(y - x)/x]•100%. This can be rewritten as [(y/x)-1]•100%, so what we really need to find is y/x.

We are given two percent changes from one year to another, but watch out! The “from (original)” year is different for each percent given.

“In 2012, attendance at an annual sporting event was 5% greater than it was in 2011”:
Note that “5% greater than” a number means “105% of” that number.
z = 105% of y
z = 1.05y

“In 2012, attendance at an annual sporting event was ... 20% greater than it was in 2010”:
Note that “20% greater than” a number means “120% of” that number.
z = 120% of x
z = 1.2x

To find y/x, we’ll first set both z equivalents equal:
z = 1.05y = 1.2x
y/x = 1.20/1.05

The answer is [(y/x) - 1]•100% = (1.14285714 - 1)•100% = 14.285714%. Rounded to the nearest 0.001%, the final answer is 14.286%.

Alternatively, we could pick numbers. A smart number for z would be a multiple of 120 and 105 (reflecting 5% and 20% increases from an easy base of 100).

z = (105)(120)
y = (100)(120)
x = (105)(100)

The answer is [(y/x) - 1]•100% = {[(100•120)/(105•100)] - 1}•100% = 14.285714%. Rounded to the nearest 0.001%, the final answer is 14.286%

For what it's worth, my own method took me 45 seconds with an on-screen calculator.


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this week's MGRE Math Beast Challenge

From here:

In 2012, attendance at an annual sporting event was 5% greater than it was in 2011 and 20% greater than it was in 2010. What was the percent increase in attendance from 2010 to 2011?

Give your answer to the nearest 0.001%.

Go to it! I'm no longer sick, so my answer will definitely appear in the comments sometime during the next 36 hours. As you can tell, this is a "grid-in" problem, i.e., you have to write the correct answer, not select an answer from multiple choices.


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answer to last week's MGRE Math Beast Challenge

I was still sick last week, so I failed to answer this rather complicated question. Here is MGRE's answer:

Rebecca began with 288 friends, evenly divided among 12 months. Thus, she had 24 friends with birthdays in each month.
Let’s make a simple chart:

[chart 1]

Now we’ll simply calculate and record all the changes. First, “the number of Rebecca’s friends with birthdays in the last quarter of the year increased by 25%.”

Thus, October, November, and December’s totals collectively increased by 25%, so from 3(24) = 72 to 72(1.25) = 90. We don’t actually know if each month increased by 25% (from 24 to 30) or if their total simply increased by 25% (for instance, maybe the entire increase of 18 occurred in one month, bringing that month’s total to 42, and the number of people with birthdays in the other two months remained at 24). The new total is 90 either way, and this question is ultimately about the total. However, one of the later constraints in this problem mentions “the month with the largest number of birthdays,” so let’s put the increase all in one month, as it might ultimately be the relevant month.

[chart 2]

Next, “the number of friends with birthdays in each month beginning with “J” increased by one-third.” To increase a number by one-third, multiply by one and one-third (this is faster than multiplying by one-third and then adding it back to the original): 24 (4/3) = 32

[chart 3]

Next, “the number of people with birthdays in February was increased by 12.5%.” Since 12.5% is 0.125, multiply by 1.125 to ADD 12.5% percent to the original number in one step: 24(1.125) = 27

[chart 4]

Next, “the number of people with birthdays in March became 166.6666...% of the new number of people with birthdays in February.”

166.6666...% of 27 is simply one hundred percent of the number, plus another two-thirds. Since 2/3 of 27 is 18, the new total for March is 45. Or, in the calculator: 27(1.666666666...) = 45. (Actually, putting this in the calculator will yield 44.9999999...., since you didn’t actually type in infinity 6’s. This is fine! The answer is 45.)

[chart 5]

Next, “the number of people with birthdays in April became five less than 75% of the new number of people with birthdays in February and March combined.”

February + March = 27 + 45 = 72
75% of 72 = 72(0.75) = 54.
We need the number 5 less than that: 54 – 5 = 49.

[chart 6]

Now, “the number of people with birthdays in May increased by 1, and the number of people with birthdays in August became one less than 20% greater than the new number of people with birthdays in May.”

May is now simply 24 + 1 = 25.
August is one less than 20% greater than 25. In the calculator: 25(1.2) = 30, then one less, or 29.

[chart 7]

Finally, “September’s total increased to 6% less than one more than the new total for the month with the largest number of birthdays.”

The month with the largest number of birthdays is April, with 49. Remember that even if the 25% increase in the total for the last quarter of the year occurred in a single month, that month (October in our chart) would only have 42 people.

One more than 49 is 50.

September’s total is 6% less than 50. To decrease a number by 6%, take 94% of it (this is faster than finding 6% and subtracting it from the original): 50(0.94) = 47

[chart 8]

To calculate the final answer, simply add the “AFTER” row of the chart:

32 + 27 + 45 + 49 + 25 + 32 + 32 + 29 + 47 + 42 + 24 + 24 = 408

The correct answer is B.

I don't have time to do it right now, but I'll be adding the charts later tonight.


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this week's Math Beast Challenge problem

From here:

Rebecca had 288 Facebook friends, and noticed that an equal number of these friends had birthdays in each of the twelve months of the year. Then, Rebecca approved many friend requests at once. After doing so, the number of Rebecca’s friends with birthdays in the last quarter of the year increased by 25%, the number of friends with birthdays in each month beginning with “J” increased by one-third, the number of people with birthdays in February was increased by 12.5%, the number of people with birthdays in March became 166.6666...% of the new number of people with birthdays in February, the number of people with birthdays in April became five less than 75% of the new number of people with birthdays in February and March combined, the number of people with birthdays in May increased by 1, and the number of people with birthdays in August became one less than 20% greater than the new number of people with birthdays in May. Finally, September’s total increased to 6% less than one more than the new total for the month with the largest number of birthdays. Assuming no one de-friended her, after approving all her friend requests, how many Facebook friends did Rebecca then have?

(A) 396

(B) 408

(C) 453

(D) 512

(E) 696

Good Lord. Given the time it takes to read this problem, you'd run out of time to do the rest of the Quant section if this were an actual GRE! Anyway, go to it! I'm still sick, so my answer may or may not appear in the comments section below.


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answer to last week's MGRE Math Beast Challenge

Having gotten disgustingly sick on Monday, I never answered last week's MGRE Math Beast Challenge. Here's MGRE's explanation, which-- strangely enough-- doesn't seem to require a very deep knowledge of how standard deviations work.

The main challenge in this problem is working through the math language to figure out what the question is really asking.

First, we are given a function: f(x) = 0.27(-3.12x – 4)

Let’s distribute: f(x) = -0.8424x – 1.08

The ugliness of these numbers is a good clue that this is more of a logic problem than a straight math problem.

We are then told that Set P consists of n distinct values that are inputted into f(x). Keep in mind that n here is just the number of numbers in the set. So, if Set P were 10, 11, 12, n would simply be 3. We are also told that the values are distinct (so the set could not be 1, 1, 1, 1, 1, for instance).

Set Q consists of all the results you get from plugging the values in Set P into the function.

Let’s try a simple example. What if Set P = {1, 2}?

f(1) = -0.8424(1) – 1.08
f(1) = -1.9224

f(2) = -0.8424(2) – 1.08
(2) = - 2.7648

Thus, if Set P = {1, 2}, then Set Q = {-1.9224, -2.7648}. Note that the values in Set P are further apart (exactly 1 apart), while the values in Set Q are closer together (less than 1 apart). Thus, the standard deviation of Set P is greater. But will this always be true?

At this point the answer is either A or D. We could try other possibilities – Set P could be nearly anything, after all. But a bit of logic might prove helpful.

The standard deviation of a distinct set increases when every item in the set is multiplied by a value > 1 or < -1.

The standard deviation of a distinct set decreases when every item in the set is multiplied by a value between -1 and 1, not inclusive.

The standard deviation of a distinct set does not change when every item in the set has the same value added to it (or subtracted from it).

Thus, in the function f(x) = -0.8424x – 1.08, x undergoes two changes:

It is multiplied by a number between -1 and 1.
It has a value subtracted from it.

When you perform both these changes to every item in Set P, the first change will cause the standard deviation to decrease – that is, the numbers get closer together. The second change makes no difference to the standard deviation.

Thus, no matter what numbers you pick, Set P will always have a greater standard deviation than Set Q (said another way, running at least two distinct numbers through this particular function yields output numbers that are closer to one another than the input numbers were).

Notice that the problem specified “distinct” values? If that one word were removed from the problem, the answer would become D. Why? Without the word “distinct,” Set P could be something like {10, 10, 10}, which has a standard deviation of zero. Putting {10, 10, 10} through the function would yield {-9.504, -9.504, -9.504}, which also has a standard deviation of zero. Since it would then be possible for Quantity A to be larger but also possible for the quantities to be equal, the answer would become D. Watch out for the “distinct trap” in standard deviation problems!

The correct answer is A.

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this week's MGRE Math Beast Challenge problem

From here:

f(x) = 0.27(-3.12x – 4)

Set P consists of n distinct values, where n > 1. Set Q consists of the n values that result from inputting the n distinct values from Set P into f(x).

Quantity A
The standard deviation of Set P

Quantity B
The standard deviation of Set Q

(A) Quantity A is greater

(B) Quantity B is greater

(C) The two quantities are equal.

(D) The relationship cannot be determined from the information given.

Go to it! My own answer will eventually appear in the comments... but first, I really need to review standard deviations!

(Some help can be found here.)


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last week's MGRE Math Beast Challenge: correct!

Here's what MGRE has to say about last week's Math Beast Challenge problem:

Let’s start by translating this into algebra.

Abe has k ketchup packets, Beata has m mustard packets, Cruz has s soy sauce packets, and Dion has b barbecue sauce packets. We are trying to find the smallest possible value for k + m + s + b.

We know that 2k = 9m = 7s = 15b. In other words: 2k, 9m, 7s and 15b each equal the same integer number, but what is that number? It would have to be cleanly divisible by 2, 9, 7, and 15. In order to minimize the number of packets owned by the group as a whole, we would want to find the smallest such number. That would be least common multiple of 2, 9, 7, and 15.

First, we find the factors of each number:
2 = 2
9 = 3 * 3
7 = 7
15 = 3 * 5

Then we multiply only the necessary factors together:
2 * 3 * 3 * 7 * 5 = 630.

Note that we leave out a 3, compared to the list of all factors above. Remember that we only include the factors needed to build each of the starting numbers individually. With two 3’s and a 5, we could make either 9 or 15. That’s good enough for a least common multiple.

Now we know that 2k, 9m, 7s, and 15b each equals 630. Let’s find how many packets each person has.

2k = 630, so k = 315 9m = 630, so m = 70 7s = 630, so s = 90 15b = 630, so b = 42

To finish, just add together the individual number of packets (k + m + s + b) = 315 + 70 + 90 + 42 = 517.

It is worth analyzing how the other choices all represent a possible mistake, a quality that makes this question harder than it would be with different answer choices:

(A) is a trap. It’s just the number in the problem added together (2 + 9 + 7 + 15).

(B) CORRECT.

(C) is a trap for those who stop at 630 and don’t remember that it’s just a step on the way to finding the individual number of packets each person has.

(D) is the sum of the packets if starting with 1890 instead of 630 as the common multiple. 2k = 1890 (k = 945). 9m = 1890 (m = 210). 7s = 1890 (s = 270). 15b = 1890 (b = 126). Sum: 945 + 210 + 270 + 126 = 1551.

(E) is another possible multiple of 2, 9, 7 and 15. (2)(9)(7)(15) = 1890. But it’s not the least common multiple. It’s the trap for those who don’t omit a redundant 3, and who also forget to finish the solving. (The value of 2k = 9m = 7s = 15b is not the answer, k + m + s + b is the answer.)

The correct answer is B.

Woo-hoo! But I actually think my own method, with the compound ratio, is quicker.


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this week's MGRE Math Beast Challenge

From here:

Abe, Beata, Cruz, and Dion each collect a different type of condiment packet, and each person only collects one type. Twice the number of ketchup packets in Abe’s collection is 9 times the number of mustard packets Beata has, 7 times the number of soy sauce packets Cruz has, and 15 times the number of barbecue sauce packets possessed by Dion. If each collector owns at least one packet and only whole packets, what is the fewest possible number of packets owned by all four people?

(select only one)

A. 33
B. 517
C. 630
D. 1551
E. 1890

Go to it! My own answer will appear in the comments. Hint: the easiest way to solve this problem is probably via compound ratios. Ever worked with those? I started using them only a few months ago myself! Compound ratios basically look like multi-tiered fractions, because that's exactly what they are. Here's an example of such ratios in action:

In Arkansas, for every 2 snakes there are 7 gerbils; for every gerbil there are 5 mice. On Farmer Brown's many-acred property, there are 6125 mice. How many snakes are on Farmer Brown's property?

Start building a compound ratio by first assembling the data you have:

Let snakes = s; let gerbils = g; let mice = m.

s : g : m

2 : 7 : x

y : 1 : 5

Solve for x by making a proportion: 7/x = 1/5. X therefore equals 35.

s : g : m

2 : 7 : 35

No need to solve for y! We now have our compound ratio, and we can derive other ratios from the above information. To wit:

s/g = 2/7 (given)
g/m = 7/35 = 1/5 (given)
s/m = 2/35 (we'll need this info)

Now that we know the basic ratio of snakes to gerbils to mice, we can apply the compound ratio to the problem.

s : g : 6125

2 : 7 : 35

The snakes-to-mice ratio is 2 : 35. Set up a proportion:

2/35 = s/6125

Solve for s:

35s = 6125•2

s = (6125•2)/35 = 350

Farmer Brown's looking at 350 snakes on his property.

And that's how compound ratios work!



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