1999 CEMC Pascal Problems/Problem 8

Problem

The average of four test marks was $60$ . The first three marks were $30$ , $55$ , and $65$ . What was the fourth test mark?

$\text{ (A) }\ 40 \qquad\text{ (B) }\ 55 \qquad\text{ (C) }\ 60 \qquad\text{ (D) }\ 70 \qquad\text{ (E) }\ 90$

Solution 1

Let $x$ be the fourth test mark. We can then set up an equation using the four numbers and the current average:

$\frac{30 + 55 + 65 + x}{4} = 60$

$\frac{x + 150}{4} = 60$

$x + 150 = 60 \times 4 = 240$

$x = 240 - 150 = \boxed {\textbf {(E) } 90}$

~anabel.disher

Solution 2 (answer choices)

We can try different answer choices to see if the average is too high or too low.

If $60$ was the fourth test mark, then the average would be:

$\frac{30 + 55 + 65 + 60}{4} = 60$

$\frac{210}{4} = 52.5$

This is too low, meaning that only answer choices that can be the answer are $70$ and $90$ .

Trying $70$ , we get $\frac{220}{4} = 55$ . This is still not the answer, so the only answer choice remaining is $\boxed {\textbf {(E) } 90}$ .

~anabel.disher

Solution 3 (answer choices)

We can try different answer choices to see if the average is too high or too low.

If $60$ was the fourth test mark. The rest of the numbers would've been $30$ less than $60$ , $5$ less than $60$ , the same as $60$ , and $5$ more than $60$ . Since these numbers are not centered around the average of $60$ , this is too low.

Trying $90$ as the fourth test mark, one number is $30$ less than the average, $5$ less than the average, $5$ more than the average, and $30$ more than the average. This balances out, so the fourth test mark is $\boxed {\textbf {(E) } 90}$ .

~anabel.disher

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1999 CEMC Pascal Problems/Problem 8

Contents

Problem

Solution 1

Solution 2 (answer choices)

Solution 3 (answer choices)