# Math Formulas You Need To Know For The Act ISEE Upper Level – Example Problems and Solutions

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## ISEE Upper Level – Example Problems and Solutions

Here are a few examples of problems you may see on the ISEE Advanced Level and their solutions.

VERBAL

If we continue to use our resources in such abundance, one day our supply will be—-.

(A) limit

(B) is included

(C) enriched

(D) tired

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 the sentence completion problem, the student must use the context of the sentence to find the meaning of the missing word; After that, the student must draw from his vocabulary to choose a word from the answer choices that has the closest meaning.

The sentence says that we use our resources in abundance; in addition, it says that we “continue” to use it in abundance. Therefore, the logical conclusion would be that one day, our supply will end or be used up. A word that means “to expire” or “to use up” is expired, which is answer choice (D).

If, by chance, the student does not know the definition of the word “finished” but knows the definitions of the other three words, it is possible to answer the question by removing all the answers because they do not make sense. Obviously, using large amounts of resources over time won’t cause them to run out of limits — that’s absurd. Reposting doesn’t make sense, and reposting doesn’t fit the context of the sentence well enough to be an attractive response.

STATISTICS

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 linear variables. If one variable is directly proportional to the other, 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 uk is a real constant. The problem asks us to find y for some value of x. To do this, we will plug in x = 9 in the equation above and see what the results are for the value of y; however, we quickly see that we don’t know the value of k, so we need to find that first. The problem gives us some information that will help us find the value of k. Plugging in the other values ​​the problem gives us (y = 20 when x = 6), we get the following:

y = kx

20 = k*6

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

k = 20/6 = 10/3

Now that we know uk, 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

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