ISEE Higher Level: Example Problems and Solutions

Here are a couple of sample problems you might see at the top level of ISEE along with their solutions.

VERBAL

If we continue to use our resources in such large quantities, one day our supply will be ——.
(A) unlimited
(B) infused
(C) enriched
(D) out of stock

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

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 must resort to their vocabulary to choose the word from the answer options that has the closest meaning.

The award establishes that we are using our resources in large quantities; furthermore, it says that we are “continuing” to use it in large quantities. Therefore, the logical conclusion would be that one day our supply will run out or run out. The word meaning “exhausted” or “exhausted” is exhausted, which is the answer choice (D).

If, by chance, the student does not know the definition of “exhausted” but does know the definitions of the other three words, it is still possible to answer the question by removing the rest of the answers because they do not make sense. Clearly using large amounts of resources over a period of time won’t cause them to become unlimited, that doesn’t make sense. Infused also doesn’t make sense, and enriched also doesn’t fit well enough into the context of the sentence to be an attractive response.

MATH

If y is directly proportional to x, and if y = 20 when x = 6, 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 a constant real number. The problem asks us to find y for a certain value of x. To do this, we would plug 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 must find it first. The problem gives us other information that will help us find the value of k. Substituting the other values ​​that the problem gives us (y = 20 when x = 6), we will obtain 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 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

The answer is 30.