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- ISEE Upper Level – Example Problems and Solutions
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## ISEE Upper Level – Example Problems and Solutions

Here are a couple of examples of problems you might see on the higher-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) infusion
(C) enriched
(D) exhausted**

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

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 use his vocabulary to choose the word from the answer options that has the closest meaning.

The phrase says that we are using our resources in large quantities; moreover, it says that we “continue” to use it in large quantities. So the logical conclusion would be that one day our supply will run out or run out. The word meaning “to exhaust” or “to exhaust” 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 sense. Obviously using large amounts of resources over a period of time won’t make them unlimited, that doesn’t make sense. Infused doesn’t make sense either, and enriched doesn’t fit the context of the sentence well enough to be an attractive answer either.

__MATHS__

**If y is directly proportional to ax, 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 ak times x, where k is a constant real number. The problem asks us to find y for a given value of x. To do this, we would plug x = 9 into the previous equation and see what the resulting y value is; however, we quickly see that we don’t know the value of k, so we need to find it first. The problem gives us another piece of information that will help us find the value of k. Plugging in the other values given by the problem (y = 20 when x = 6), we 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*

The answer is 30.

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