GRE Word Problems and Strategy Chapter 3: Symbols

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GRE Word Problems Guide

		Ch 1: Word Problem Strategies

		Ch 2: 5-Step Method

		Ch 3: Functions & Symbols

		Ch 4: Progressions & Sequences

		Ch 5: Percentages

		Ch 6: Interest

		Ch 7: Ratio & Proportion

		Ch 8: Uniform Motion

		Ch 9: Work & Rate


		Ch 10: Grouping & Counting

		Ch 11: Data Interpretation

		Ch 12: Averages & Median

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The GRE uses ƒ(x) to describe a function

A function describes a relationship between one or more inputs and one output.

If ƒ(x) = 2x, then:
ƒ(2) = 4
ƒ(3) = 6

Note: For every ƒ(x) there is one unique value. So, for example, you can't have a ƒ(x) that corresponds to absolute value because ƒ(+2) would result in two answers: -2 and +2.

More interesting functions:

ƒ(x) = 2x/(4 - x)

What does ƒ(x +1) = ?

To solve this, simply plug in (x + 1) in for the function. This is just like Substitution for solving multiple variables.

	2(x + 1)	Substitute (x +1) in for x

	4 - (x + 1)

	2x + 2

	3 - x

Manipulating Functions

Manipulate functions as you would any number.

Note: that if you have f(x)/g(x) and g(x) = 0, then that isn't a real number because zero is in the denominator.

Given f(x) = 3x + 2 and g(x) = 4 - 5x, find:

f(x) - g(x) = [3x + 2] - [4 - 5x ] = 3x + 5x + 2 - 4 = 8x - 2
f (x) + g (x) = [3 x + 2] + [4 - 5x] = 3x - 5x + 2 + 4 = -2x + 6
f(x) × g(x) = (3x + 2)(4 - 5x) = 12x + 8 - 15x² - 10x = -15x² + 2x + 8

Compound Functions

RULE #1: go INSIDE, then go OUTSIDE

f(x) = x²

g(x) = x + 2

f(g(2)) = ?

First, take g(2) because it is on the inside
(2) + 2 = 4

Then take f(4), because it is the outside
f(4) = 16

The #$@#! Symbols

On some questions, the test will create new functions. You can identify these questions by the symbols that are used--(triangles, squares, ampersand, etc.). These questions are generally easy as long as you don't get confused by the symbols. Simply take the function and plug in the numbers.

Example (easy)

If a # b = a + b, then what is 2 # 3?

Solution

2 # 3 would equal 2 + 3, or 5.

Example (easy)

If a # b = a + b, then what is (2 # 3) # 2?

Solution

Solve inside of the parentheses first. 2 # 3 would equal 2 + 3, or 5. Then (5) # 2 = 5 + 2 or 7.

Example (hard)

If, for numbers x, y, z, the function # is defined as

x # y = xy - x

then

x # (y # z) =

Solution

The first step to solving x # (y # z) is to solve inside the parenthesis (y # z). After we have solved what is in the parenthesis, the second step is to do x # (what is in the parenthesis). The third step is to solve the equation using the symbol.

Step 1: (solve the parenthesis: y # z)

1a) if x # y = xy - x (as stated in the question stem)

1b) then y # z = yz - y ( you get this by substituting y for x and z for y)

Step 2: (insert the parenthesis value)

The original question asks x # (y # z), we have already solved y # z, which according to 1b) above

y # z = yz - y

So, in the original equation x # (y # z), substitute yz - y for y # z:

Now, x # (y # z) = x # (yz - y)

(NOTE: Remember order of operations when doing these.)

So, we are now dealing with:

x # (yz - y)

Step 3: (apply the # to the final equation)

The # symbol means x # y = xy - x.

(Essentially, take the first number--here x--, multiply it by the second number--here y--and then subtract the first number. Let's apply that to the equation at the end of step 2:

x # (yz - y) = x(yz - y) - x

then factor out the x's:
= xyz - xy - x
= x(yz - y - 1)

Help, I still don't get it!

To play with this question, try inserting numbers such as 1 and 3 for x and y to see how

x # y = xy - x

1 # 3 = 1(3)- 1

1 # 3 = 2

So now look at

x # (y # z) =

1. Set x = 1 , y = 3, z = 2 and plug them in:

1 # (3 # 2)

2. Do the parenthesis first (as is the rule always). To make a problem less intimidating, break it into smaller component parts:
(3 # 2) = (3)(2) - (3)
3 # 2 = 6 - 3
3 # 2 = 3

3. Go back to the original question and plug in the value you solved for the parenthesis.
1 # (3 # 2), plug in the value for (3 # 2) to get:
1 # (3) = (1)(3) - (1)
So, 1 # (3) = 2

4. Now that you have applied the function, you have the result:
x # (y # z) = 1 # (3 # 2) = 2

5. Set x = 1 , y = 3, z = 2 and plug them in and the result is 2.
Now, for fun, look at the answer we came up with before

x # (y # z) = x(yz - y - 1)

Now, plug in x = 1, y = 3, z = 2 for x(yz - y - 1), what do you get?
1 [(3)(2) - 3 - 1] = 2, so we can infer that we got the right answer.


	Double Checking When doing complex functions or algebra, Plug-In numbers to better understand the equation and to check if you got the right answer. This takes time, though, so use this judiciously.