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 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.

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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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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rock and roll! got last week's MGRE problem correct!

I admit I felt shaky about my solution to last week's MGRE Math Beast Challenge problem, but I turned out to be correct. Here's MGRE's explanation:

Let’s put the words into equation form:

(Total Income – Exclusion)(Tax Rate) = Income Tax.

The question asks about total income, so we’ll solve the equation for Total Income:

Total Income – Exclusion = [Income Tax / Tax Rate]
Total Income = [Income Tax / Tax Rate] + Exclusion

To maximize total income, we’ll minimize Tax Rate (smaller denominator→larger value) and maximize the Exclusion:

Maximum Total Income = [$8700/0.15] + $9800 = $58,000 + $9,800 = $67,800.

To minimize total income, we’ll maximize Tax Rate (larger denominator→smaller value) and minimize the Exclusion:

Minimum Total Income = [$8700/0.35] + $5200 = $24,857.14 + $5,200 = $30,057.14.

The correct answers are C, D, E, and F.

The letters C, D, E, and F correspond to the values I had selected. MGRE arrived at the exact same range that I had arrived at, too: roughly $30,057 at the bottom end, and $67,800 at the top end.

Triomphe!


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this week's MGRE Math Beast Challenge

From here:

For Jack, income tax is between 15 and 35 percent of total income after an “exclusion” amount has been subtracted (that is, Jack does not have to pay any income tax on the exclusion amount, only on the remainder of his total income). If the exclusion amount is between $5200 and $9800, and Jack’s income tax was $8700, which of the following could have been Jack’s total income?

(Choose all that are appropriate)

$13,000

$23,100

$33,200

$43,300

$53,400

$63,500

$73,600

Go to it! I'll leave my response in the comments.


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