combination

You say 5x5 board, I'm assuming that there's 5 squares along each edge, but the length of square is not defined here.

If I assume a length of 1, then the area of the board is 25 square feet (arbitrarily picking a unit). The factors of 6 square feet are

3x2 = 2 rectangles
6x1 = 0 rectangles (since sides of board are only 5 feet long)

Since the answer choices are much higher, I'm assuming the problem is missing information, or I misunderstood problem.

winnerhere wrote:How many rectangles of area=6 are there in a 5 x 5 square grid(like chess board but 5 x 5 )

1. 12

2. 18

3. 24

4. 30

Here's what the question is really asking:

An area of the xy plane is bounded by (0,0), (0,5), (5,5) and (5,0). How many rectangles with an area of 6 units can be formed within this section of the xy plane if the vertices of the rectangles must lie on points with integer values.

As cbenk121 notes, to get an area of 6 we can have 1*6 or 2*3 rectangles. 1*6 would lie outside the bounded area, so we know that our rectangle must be 2*3.

So, how many different 2*3 rectangles can we create within the bounded area, given that our vertices must all lie on points with integer coordinates?

You could draw it out and brute force it, or you can use counting methods.

First, we have to recognize that we can orient our rectangle two different ways: our 2 side can be parallel to either the x- or y-axis.

Second, once we've split the question into two parts (one for each orientation), we need to recognize that once we place the lower left corner, we create 1 unique rectangle.

If our 2 side is parallel to the y-axis, it fits within the bounded area if the lower left corner of the rectangle begins at y=0, 1, 2, or 3. So, there are 4 possible y-coordinates.

Since our 3 side is parallel to the x-axis, it fits within the bounded area if the lower left corner begins at x = 0, 1 or 2. So, there are 3 possible x-coordinates.

Therefore, there are 4*3 = 12 possible rectangles in this first case.

Similarly, if we orient the rectangle with the side of length 3 parallel to the y-axis, we get 3 possible y co-ordinates and 4 possible x-coordinates for the lower left corner. Accordingly, there are 12 possible rectangles in the second case.

So, there are 12 + 12 = 24 possible rectangles.

What you say makes complete sense.

But how did you know that it was asking for all the possible places for a 3x2 rectangle to sit, versus how many could sit in at one time?

In case a similiar question appears in my practice or on the GMAT, I'd like to be able to recognize what the question is really asking for.

cbenk121 wrote:What you say makes complete sense.

But how did you know that it was asking for all the possible places for a 3x2 rectangle to sit, versus how many could sit in at one time?

In case a similiar question appears in my practice or on the GMAT, I'd like to be able to recognize what the question is really asking for.

I only knew because I've seen similar questions before. On the actual GMAT, questions will never be worded as poorly as the one was in this post. Notice that there are also only 4 answer choices in this question, which confirms that it's not really from the GMAT (or reliable GMAT practice materials).

If, on the exam, you see a question that you think has multiple interpretations, go with common sense (and use the answer choices to help you determine which interpretation must be correct).