## Tuesday, April 17, 2012

### 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!

_

#### 3 comments:

Kevin Kim said...

So! The people have been conveniently named A, B, C, and D.

2A = 9B; A/B = 9/2
2A = 7C; A/C = 7/2
2A = 15D; A/D = 15/2

Let's set up a compound ratio:

A : B : C : D

9 : 2
7 : x : 2
15: y : z : 2

What's the least common multiple (LCM) of 9, 7, and 15? Near as I can figure, it's 315. This means:

315/9 = 35.
315/7 = 45.
315/15 = 21.

Let's apply this to our compound ratio.

A : B : C : D

315 : 70 (factor of 35)
315 : x : 90 (factor of 45)
315 : y : z : 42 (factor of 21)

So the ratio of A:B:C:D is

315:70:90:42

Add those numbers together:

315 + 70 + 90 + 42 =

517.

The answer is (B).

_

Kevin Kim said...

I should note that everything depends on correctly determining the LCM of 9, 7, and 15. Sometimes it's possible to do this by multiplying the numbers together, but a better method exists. See Method #2 here.

For 9, 7, and 15, the above-linked method works like this:

1. Count the number of times each prime number appears in each of the factorizations.

For 9, the prime factor 3 appears twice.

For 7, the prime factor 7 appears once.

For 15, the prime factor 3 appears once, and the prime factor 5 appears once.

2. For each prime number, take the largest of these counts.

For 9: 3 (twice).
For 7: 7 (once).
For 15: 5 (once). (3 is already "spoken for" in 9 above. We don't count the 3 that's "inside" 15 because it doesn't appear TWICE in 15, as it did in 9.)

3. Write down that prime number as many times as you counted for it in step 2.

3 3 7 5

4. The least common multiple is the product of all the prime numbers written down.

3•3•7•5 = 315.

See how that works?

Kevin Kim said...

The dangers of multiplying three numbers together to get the LCM were cleverly considered in this problem. Had you multiplied 7, 9, and 15 together, you'd have gotten 945 (which is 3 times greater than 315). You would then have erroneously answered (D) after crafting your compound ratio on the basis of the incorrect LCM.