# A Closer Look at GMAT Function Questions Last week, we looked at the basics of how to handle function questions. Today, let’s look at a couple of questions. We will start with an easier one and then go on to a slightly tougher one.

Question 1: If f(x) = 343/x^3, what is the value of f(7x)* f(x/7) in terms of f(x)?

(A)  f(x^2)

(B) (f(x))^2

(C) f(x^3)

(D) (f(x))^3

(E) f(343x)

Solution: We discussed that to get f(a) given f(x), all you need to do is substitute x with a.

f(x) = 343/x^3

f(7x) = 343/(7x)^3 = 1/x^3

f(x/7) = 343/(x/7)^3 = 343*343/x^3

So we get f(7x) * f(x/7) = (1/x^3) * (343*343/x^3) = (343/x^3)^2

But we know that 343/x^3 = f(x)

So, f(7x) * f(x/7) = (f(x))^2

There are other ways of solving this too:

Say x = 1, then f(1) = 343

f(7x)* f(x/7) = f(7)*f(1/7) = (343/7^3) * (343/(1/7)^3 = (343)^2

So f(7x)* f(x/7) = (f(1))^2

We hope you see that the question was not difficult to solve. Once you get over your fear of symbols, it is quite straight forward.

Now, let’s take a question similar to an official question from the GMAT paper tests:

Question 2: The function f is defined for each positive three-digit integer n by f(n) = 2^x * 3^y * 5^z, where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive integers such that f(m) = 25f(v), then m-v=?

(A) 2
(B) 9
(C) 18
(D) 20
(E) 100

Solution: The question may seem a bit difficult to understand first so let’s take one sentence at a time:

f is defined for each positive three-digit integer n by f(n) = 2^x * 3^y * 5^z, where x, y and z are the hundreds, tens, and units digits of n

Let’s take an example to see how to make sense of it: say 146 is a three digit positive integer. So f(146) = 2^1 * 3^4 * 5^6

In the same way, f(283) = 2^2 * 3^8 * 5^3

If m and v are three-digit positive integers such that f(m) = 25f(v)

So f(m) = 5^2 * f(v)

If m is represented as abc and v as def, then (2^a * 3^b * 5^c) = 5^2 * (2^d * 3^e * 5^f)

Note that for the left hand side to be equal to right hand side, a = d, b = e and c = 2 + f.

So the units digit of m is 2 more than the units digit of v but their tens and hundreds digits are the same.

So m – v = 2.