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## ISEE Upper Level – Example Problems and Solutions

Here are some examples of problems you might see on the top level 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) exhausted

This is an end-of-sentence question that tests the student’s vocabulary and ability to understand the context of the sentence. The correct answer is (D).

In order to solve a sentence completion problem, the student must use the context of the sentence to infer the meaning of the missing word; then, the student must draw on his vocabulary to choose the word among the answer choices that has the closest meaning.

The sentence indicates that we use our resources in large quantities; moreover, it is said that we “continue” 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 that means “exhausted” or “exhausted” is exhausted, which is answer choice (D).

If, by chance, the student does not know the definition of “exhausted” but knows the definitions of the other three words, it is still possible to answer the question by eliminating the rest of the answers because they do not make any sense. Obviously, using large amounts of resources over a period of time won’t make them unlimited – it just doesn’t make sense. Infused also doesn’t make sense, and enriched doesn’t fit well enough in the context of the sentence to be an appealing 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 a constant real number. The problem asks us to find y for some value of x. To do this, we would plug x = 9 into the equation above and see what value y results from; however, we quickly see that we don’t know the value of k, so we need to find it 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 of 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

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