# ISEE Upper Level – Example Problems and Solutions

Here are a couple examples of problems you might see on the Upper Level ISEE along with their solutions.

VERBAL

If we continue to use our resources in such large quantities, someday our supply will be ——.

(A) unlimited

(B) infused

(C) enriched

(D) exhausted

This is a sentence completion question that tests the student’s vocabulary and ability to understand sentence context. The correct answer is (D).

In order to solve a sentence completion problem, the student needs to use the context of the sentence to deduce the meaning of the missing word; then, the student needs to draw on his or her vocabulary to choose the word from the answer choices that has the closest meaning.

The sentence states that we are using our resources in large quantities; additionally, it says that we are “continuing” to use it in large quantities. Therefore, the logical conclusion would be that someday, our supply will run out or be used up. The word that means “run out” or “used up” is exhausted, which is answer choice (D).

If, by chance, the student doesn’t know the definition of “exhausted” but knows the definitions of the other three words, it’s still possible to answer the question by eliminating the rest of the answers because they make no sense. Clearly, using large amounts of resources over a period of time will not cause them to become unlimited — that makes no sense. Infused also makes no sense, and enriched also does not fit well enough in the context of the sentence to be an attractive answer.

MATH

If y is directly proportional to x, and if y = 20 when x = 6, then what is the value of y when x = 9?

This problem is an algebra problem that tests the student’s knowledge of direct variation. If one variable is directly proportional to another, then it follows the general formula (by definition):

y = kx

This means: as x increases, y increases at a rate proportional to k times x, where k is some constant real number. The problem asks us to find y for a certain value of x. In order to do this, we would plug in x = 9 into the above equation and see what y-value results; however, we quickly see that we don’t know the value of k, so we have to find that first. The problem gives us other information that will help us find the value of k. Plugging in the other values that the problem gives us (y = 20 when x = 6), we will get the following:

y = kx

20 = k*6

Now we can solve for k by dividing both sides of the equation by 6:

k = 20/6 = 10/3

Now that we know k, we know that the general equation is:

y = (10/3) x

This means that as x increases, y increases at a rate proportional to 10/3 that of x. If x increases by 1, y increases by 10/3; if x increases by 3, y increases by 10. Using our new equation, we can find the answer to the question by plugging in x = 9:

y = (10/3) * (9)

y = 30