Many quantitative comparison questions come with parameters. Parameters are basically a few ground rules that are listed above Column A and Column B.

x is a positive integer

Column A | Column B |
---|---|

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

In this question *x is a positive integer* is a parameter. The parameter is essential information. That is you must take into account if you want to get the correct answer.

The answer to the question above is (D). If x = 1, then the columns are equal. For all other values (B) is bigger.

To see how a simple tweak in a parameter can change a question have a look at a similar problem.

x is an integer greater than 1

Column A | Column B |
---|---|

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

Let’s now look at two questions that are very similar save for a word or two. Both questions are much more difficult so you may want to grab some scratch paper.

1. Isosceles right triangle ABC and Square EFGH have the same area.

Column A | Column B |
---|---|

Perimeter of ABC | Perimeter of EFGH |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

2. Quadrilateral EFGH and right triangle ABC have the same perimeter

Column A | Column B |
---|---|

Area of ABC | Area of EFGH |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

## Solution for #1:

Hint: The answers are different.

In this question, we have an isosceles right triangle, aka a 45-45-90 triangle. This is fixed triangle—meaning you can’t endlessly manipulate it. Two sides are equal and it is a right triangle, so the sides must be in the ratio of x: x: x√2.

But let’s back up a moment. A square has the same area as our 45-45-90 triangle. At this point, I like to use a convenient number (but if you are predisposed to using ‘x’ then go ahead and do so). Let’s say the area of the square is 8. That means each side of the triangle is 4 (Area of triangle: (4)(4)/2 = 8).

The three sides of the triangle added together will be 4 + 4 + 4√2 = 8 + 4√2.

If the square has an area of 8, each side is equal to √8 = 2√2. The perimeter of the square is 4 x 2√2 = 8√2.

Now we can compare these two numbers:

Column A | Column B |
---|---|

Be very careful here! First off do not add 8 + 4√2. You cannot add an integer to a number expressed as a radical. One solution:

√2 = 1.4 (approximately)

4 x 1.4 = 5.6

5.6 + 8 = 13.6

8√2 = 8 x 1.4 = 11.2

Column A | Column B |
---|---|

Therefore (A).

Another solution:

8√2 = 4√2 + 4√2

4√2 + 8 vs. 4√2 + 4√2 (Subtract 4√2 from both sides)

8 vs. 4√2

(4)(2) vs (4)(√2) so (A).

## Solution for #2

A slight (but devilish) change in the details gives us question #2. Note that the square has changed to a quadrilateral and the triangle is now simply a right triangle. Does this change anything?

Imagine that quadrilateral is a square. Imagine that the triangle is an isosceles. We have the first question. How does this help us? Well, in the first question (A) was bigger. If we find an instance in which (A) is not bigger, then the answer is (D).

Imagine the quadrilateral is very skinny, let’s say length is 9 and width is 1. The area would only be 9, but the perimeter would be 20. A triangle with area nine could have sides 3 and 6. Just to equal 20, the triangle would have to have a hypotenuse of 11, which would be impossible: longest possible side of triangle with legs 3 and 6 is < 9.

And like that we’ve found a case where (B) is larger. So the answer is (D).

## Takeaway

On Quantitative Comparison a simple word can change an entire answer.