**Quantitative Comparisons**

Quantitative comparisons are the most common math questions. This is good news because they are the easiest problems to improve on.

Generally, quantitative comparison questions require much less calculating than do multiple-choice questions. But they are trickier.

Substitution is very effective with quantitative comparison problems. But you must plug in all five major types of numbers: positives, negatives, fractions, 0, and 1. Test 0, 1, 2, –2, and 1/2, in that order.

**General Principals for Solving Quantitative Comparisons**

The following principles can greatly simplify quantitative comparison problems.

**Strategy: **You Can Add or Subtract the Same Term (Number) from Both Sides of a Quantitative Comparison Problem.

**Strategy: **You Can Multiply or Divide Both Sides of a Quantitative Comparison Problem by the Same Positive Term (Number).

**(Caution:** This cannot be done if the term can ever be negative or zero.)

You can think of a quantitative comparison problem as an inequality/equation. Your job is to determine whether the correct symbol with which to compare the columns is < , = , > , or that it cannot be determined. Therefore, all the rules that apply to solving inequalities apply to quantitative comparisons. That is, you can always add or subtract the same term to both columns of the problem. If the term is __always positive,__ then you can multiply or divide both columns by it. (The term cannot be negative because it would then invert the inequality. And, of course, it cannot be zero if you are dividing.)

**Example 1:**

**Column A**

$\frac{1}{5}$+$\frac{1}{3}$+$\frac{1}{8}$

**Column B**

$\frac{1}{8}$+$\frac{1}{5}$+$\frac{1}{4}$

**Column A**

$\frac{1}{3}$

**Column B**

$\frac{1}{4}$

Now $\frac{1}{3}$ is larger than $\frac{1}{4}$ , so Column A is larger than Column B.

If there are only numbers (i.e., no variables) in a quantitative comparison problem, then “not enough information” cannot be the answer.

**Example 2:**

**Column A**

*y*^{3} + *y*^{4}

**Column B**

y^{4} - 2*y*^{2}

where *y* > 0

*y*

^{4}from both columns:

**Column A**

*y*

^{3}

**Column B**

-2

*y*

^{2}

where y > 0

Since

*y*> 0, we can divide both columns by

*y*

^{2}:

**Column A**

*y*

**Column B**

-2

where y > 0

Now, we are given that y > 0. Hence, Column A is greater.

**Example 3:**

**Column A**

$\frac{1}{x}$

**Column B**

$\frac{1}{x-1}$

where *x* > 1

*x*> 1,

*x*– 1 > 0. Hence, we can multiply both columns by x(x – 1) to clear fractions. This yields

**Column A**

*x*– 1

**Column B**

*x*

where

*x*> 1

Subtracting

*x*from both columns yields

**Column A**

-1

**Column B**

0

where

*x*> 1

In this form, it is clear that Column B is larger.

**The Answer is B**

**Example 4:**

**Column A**

$\frac{{n}^{2}}{x}$

**Column B**

*n*^{2}

where *n* is a positive integer and 0 < x < 1

**Column A**

$\frac{{n}^{2}}{x}$•$\frac{1}{{n}^{2}}$

**Column B**

*n*

^{2}• $\frac{1}{{n}^{2}}$

Reducing yields

**Column A**

$\frac{1}{x}$

**Column B**

1

We are also given that 0 <

*x*< 1. So we may multiply both columns by

*x*to get

**Column A**

1

**Column B**

*x*

But again, we know that 0 < x < 1. Hence, Column A is larger.

**You Must Be Certain That the Quantity You Are Multiplying or Dividing by Can Never Be Zero or Negative. **(There are no restrictions on adding or subtracting.)

The following example illustrates the __false__ results that can occur if you don’t guarantee that the number you are multiplying or dividing by is positive.

**Column A**

*x*^{3}

**Column B**

*x*^{2}

where 0 ≤* x* < 1

Solution (Invalid): Dividing both columns by *x*^{2} yields

**Column A**

*x*

**Column B**

1

We are given that *x* < 1, so Column B is larger. But this is a __false__ result because when *x* = 0, the two original columns are equal:

**Column A**

0^{3} = 0

**Column B**

0^{2} = 0

Hence, the answer is actually D, not-enough-information to decide.

**Watch out when canceling.**

Some people are tempted to cancel the *x*^{2}*s* from both columns of the following problem:

**Column A**

*x*

^{2}+ 4

*x*-6

**Column B**

6 + 4

*x*-

*x*

^{2}

You cannot cancel the

*x*

^{2}

*s*from both columns because they do not have the same sign. In Column A,

*x*

^{2}is positive. Whereas in Column B, it is negative.

**You Can Square Both Sides of a Quantitative Comparison Problem to Eliminate Square Roots.**

**Example 1:**

**Column A**

√3+√5

**Column B**

√8

**Column A**

(√3 + √5 )

^{2}

**Column B**

( √8 )

^{2}

or

**Column A**

3 + 2√3√5 + 5

**Column B**

8

Reducing gives

**Column A**

8 + 2√3√5

**Column B**

8

Now, clearly Column A is larger.

**Example 2:**

**Column A**

$\frac{\sqrt{2}}{3}$

**Column B**

$\frac{2}{5}$

**Column A**

5√2

**Column B**

6

Squaring both columns yields

**Column A**

25•2

**Column B**

36

Performing the multiplication in Column A yields

**Column A**

50

**Column B**

36

Hence, Column A is larger.

**Substitution (Special Cases)**

We already studied this method in the section Substitution. Here, we will practice more and learn a couple of special cases.

**In a problem with two variables, say, x and y, you must check the case in which x = y.** (This often gives a double case.)

**Example 1:**

**Column A**

Average of

*x*and

*y*

**Column B**

Average of

*x*

^{3}and

*y*

^{3}

where

*x*and

*y*are positive.

Let *x = y *= 1. Then Column A becomes $\frac{1+1}{2}$ = 1. And Column B becomes $\frac{{1}^{3}+{1}^{3}}{2}$ = 1. In this case, the columns are equal. But if *x* = *y* = 2, then Column A becomes $\frac{2+2}{2}$ = 2 and Column B becomes $\frac{{2}^{3}+{2}^{3}}{2}$ = 8. In this case, the columns are unequal. This is a double case and therefore the answer is D.

**Example 2:**

**Column A**

2^{x+ y}

**Column B**

2^{x} + 2^{y}

where *x* and *y* are integers greater than or equal to 1.

*x*≠

*y*, then Column A is larger than Column B. (Plug in a few numbers until you are convinced.) But if

*x = y*= 1, then the columns are equal: 2

^{x+ y}= 2

^{1+1}= 2

^{2}= 4 and 2

^{x}+ 2

^{y}= 2

^{1}+ 2

^{1}= 4. Hence, there is not enough information to decide.

**When you are given x < 0, you must plug in negative whole numbers, negative fractions, and –1.** (Choose the numbers –1, –2, and –1/2, in that order.)

**Example:**

**Column A**

*k*

^{2}(k+$\frac{1}{2}$ )

^{2}

**Column B**

0

where

*k*< 0

*k*

^{2}(k+$\frac{1}{2}$ )

^{2}

= (-$\frac{1}{2}$ )

^{2}(-$\frac{1}{2}$ + $\frac{1}{2}$)

^{2}=$\frac{1}{4}$•0=0

Hence, there is not enough information to decide.

**Strategy: Sometimes you have to plug in the first three numbers (but never more than three) from a class of numbers.**

**Example:**

*x*is both an integer and greater than 1. Let [

*x*] stand for the smallest positive integer factor of

*x*not equal to 1.

**Column A**

[

*x*]

**Column B**

[

*x*

^{3}]

*x*= 2, 3, and 4. If

*x*= 2, then [

*x*] = 2 and [x

^{3}] = [8] The smallest factor of 8 is 2 which is equal to 2. So for this choice of

*x*, the two columns are equal. If

*x*= 3, then [

*x*] = 3 and [

*x*

^{3}] = [27] the smallest factor of 27 is 3, again the columns are equal. Finally, If

*x*= 4, then the smallest factor of [

*x*] is 2 and [

*x*

^{3}] = [64] then then smallest factor is also 2, still again the columns are equal. Note, there is no need to check

*x*= 5. The writers of the GRE do not change the results after the third number.