OG10 #251

oquiella wrote:On a certain street, there is an odd number of houses in a row. The houses in the row are painted alternatively white and green, with the first house painted white. If n is the total number of houses in the row, how many of the houses are painted white?

A. (n+1)/2
B. (n-1)/2
C. n/2 + 1
D. n/2 - 1
E. n/2

please break down

While maybe Brent's approach is the ideal approach for handling this question, here is an idea you can use for getting to the answer to this question or another like it.

One thing you could do if you are not sure what to do exactly is start eliminating answer choices. Choices C, D and E all involve dividing n by 2. n is an odd number. So dividing n by 2 will generate a fraction. We know that the houses are counted in integers not fractions, and dividing n by 2 or dividing n by 2 and adding an integer to the result will generate a fraction. None of C, D, and E can be the answer.

Now we have cleared the decks some and have only two similar answer choices to compare.

Since the row starts and ends with a white house and there are an odd number of houses, we know that there are more white houses than green houses. You could either write this down or picture it in your mind, using any number of houses.

If there are more white houses than green houses, then the number of white houses has to be more than half of n.

Choice B generates a number less than half of n.

Choice A generates a number more than half of n.

So A it is.