### Can you solve this tough math question?

1. For real numbers a, b, and c, ab = 1.5, bc = 6, and ac = 25. So abc =

(A) 9
(B) 12
(C) 15
(D) 100
(E) 225

Answer and explanation below the fold...

If I told you it was my birthday and I wanted a cake, and you liked me enough to provide one (pretty please?), you'd have a couple options. You could go to the grocery store and buy all the ingredients, and then go home and start baking. Or, you could go to the same store and go to a different aisle, buy a pre-made cake, and spend the time you would have been baking watching awesome videos on the internet. Listen, I'm glad you like me and all, but you don't really need to take all that time making me a cake. Internet videos. Get some.

What I mean to say is: when the SAT asks you to find an expression (by which I mean anything other than just a plain old variable like x) you should look for a way to solve directly for that expression.  Your mantra for a question like this: "How do I go from what they gave me to what they want?"

Often, you'll be adding or subtracting two equations. That doesn't do us any good with this problem, though. For this one, we're going to have to get a little more creative. Note that we are given three equations containing parts of what we want: abc. How do we go from what they gave us to what they want?

Try multiplying everything you have together:
(ab)(bc)(ac) = (1.5)(6)(25)
abbcac = 225
Now simplify a little bit:
aabbcc = 225
a2b2c2 = 225
(abc)2 = 225
And take the square root of both sides:
abc = 15.
That's answer choice (C).

#### 2 comments:

1. Well I guess I took the long way around...

I did a = 1.5/b, or 25/c.
1.5/b = 25/c. --> b = 1.5/25c. b = .06c, and 6/c. 6/c = .06c. --> c = 10.
10b = 6. b = .6
.6a = 1.5 --> a = 2.5
Then I multiplied.

2. ab=1.5
bc=6
ac=25
abc=15

ab/bc=1.5/6 the b's cross out
a/c X ac the c's cross out

therefore
a/\2 =6.25
sq. root. 2.5

2.5X bc=15

doing the algebra saved time for me.