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 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

## 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.

I solved Q1 in the following manner:

If Area of triangle = Area of square,

1/2a^2 = l^2

=> a = √2L

Therefore, hypotenuse of this triangle, b = √2 x ( √2L ) = 2L

Plug this into our equations for perimeters:

Perimeter of triangle V/S Perimeter of square, and using Logic method on quantities A and B,

√2L + √2L + 2L vs. 4L

2√2L + 2L vs. 4L

2√2L vs. 2L (subtracting 2L from both sides)

and hence, A > B

The second question was challenging, took a while to get the concept. But it’s a great concept. It means developing a conscious thought about the math logic we just performed in the 1st question, which is, for a triangle to have its area equal to that of a square, it will have a perimeter (and thus the size) greater than that of the square. Great example, thank you!

Hey Chris,

Could you explain this problem :

|x+3| = 4x

Qty A: x

Qty B : 1

Answer is C

Also could you throw some insight into these types where we have mod or absolute values.

Thanks !

Hi Ajay,

Here it goes. Just solve for x using negative and positive (x+3) which will give us answer of 1 and -3/5. But do not be tempted to choose D as -3/5 is actually a extraneous root. For example if you plug it in equation you will see makes an absolute value equal to a negative number which is of course impossible.

Seems like I missed the initial comment from Ajay :(. Thanks so much Sultan for providing an excellent explanation :).

x=(2/3)r r=(12/5)y y=(5/2)z

Column A: x

Column B: 4z

Hello, Chris

I’am having hard times answering this question, could you plez see look at it.

Thank you in advanced.

NB: The answer given is “C”

Hi Desland,

You can just look at the proportion between each. Meaning. to go from ‘x’ to ‘z’ you have to go through r and y. Working one step at a time, you can figure out the relationships:

If r = 12/5y and 2/3 of x is r, then x = (2/3)(12/5) —> 24/15 y, (24/15)(5/2) = 120/30 = 4. Therefore z is four times x.

Hope that helps!

Yes, if EFGH is a square, then there is no way a triangle with equal perimeter could ever have a greater area. So if EFGH has to be a square, then the answer would be (B).

Hi,

With regard to Q2, If EFGH was a square instead of a quadilateral, then Column B would be larger right??