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# Tip #3 to Get a Top SAT Math Score (page 2)

By McGraw-Hill Professional
Updated on Sep 10, 2011

This is one of my favorite SAT strategies. It's true, I have favorite SAT strategies, but it's just such a good one and makes average questions so much easier.

The "average" of a list of numbers is found by adding them and dividing by how many there are.

This is very simple, and the SAT knows that you have been figuring out your grade point average, test average, or bowling average for years. Therefore, they do not usually ask you only to find the average of a list of numbers; they say, "Given the average, find the sum of the numbers." This gets kids, but if you know the strategy, it's easy! Just know this formula:

Sum = (average) × (number of items)

This is the key to unlock almost any average question on the SAT. As soon as you see the word "average" on the test, consider using this formula. By the way, anytime the SAT prints the word "average," they follow it with "(arithmetic mean)." This is simply a clarification and a synonym for "average." They are just referring to the normal average that you are used to, so ignore "(arithmetic mean)."

Let's look at this question:

Solution: For an average question, we use either the average formula or the sum formula. How do you know which? The question asks for one or the other! This question asks for the sum, so set up the sum formula:

### Medium

1. If the average (arithmetic mean) of m and 4m is 30, what is the value of m ?
1. 4
2. 6
3. 8
4. 10
5. 12
2. The average (arithmetic mean) of the scores of 21 students on an exam is k. In terms of k, what is the sum of the scores of all 21 students?
1. 21 + k
2. k – 21
3. 21k
3. The average (arithmetic mean) of m and n is 7, and the average of m, n, and p is 11. What is the value of p ?
1. 31
2. 29
3. 21
4. 19
5. 15
4. If the perimeters of two shapes have a sum of 9, what is the average (arithmetic mean) of the perimeters of the two shapes?
1. 0
2. 10
5. The average (arithmetic mean) of the test scores of a class of 20 students is 80, and the average of the scores of a class of 30 students is 92. What is the average of both classes combined?
1. 25
2. 86
3. 87.2
4. 89.5
5. 92
6. The average (arithmetic mean) of m, n, o, and p is 55. If the average of m, n, o, p, and q is 70, what is the value of q ?
7. ### Hard

8. Each of 5 judges had a scorecard on which they wrote a different positive integer. If the average (arithmetic mean) of these integers is 7, what is the greatest possible integer that could be on one of the cards?

1. E   Set up the average formula: (m + 4m)/2 = 30. Then use algebra to solve for m or "Use the Answers" to see which one works.
2. E   Most kids get this one wrong, but with the sum formula it's easy! Set up the sum formula: Sum = (k)(21). Therefore, Sum = 21k. EASY!
3. D   Set up the sum formula: (m + n) = 7 × 2. Thus, m + n equals 14. Then use the sum formula for the second set of data: (m + n + p) = 11 × 3. So their sum is 33. Since m + n + p equals 33 and m + n equals 14, p must equal 33 – 14 = 19.
4. D   This question is a medium because many people fear fractions. Don't! They are easy, you can even turn them into decimals on your calculator if you don't like them (just divide the top number by the bottom number). With the average formula this question is very easy. Average = sum / # items. So average = 9/2.
5. C   Set up the sum formula to find the sum for each class, then add the sums and divide by the total number of students in the classes to arrive at the combined average:
6. 130   This question follows our strategy perfectly. If the average of m, n, o, and p is 55, then their sum is 55 × 4 = 220; and if m, n, o, p, and q average to 70, then their sum is 70 × 5 = 350. Therefore the value of q is the difference between the sums: 350 – 220 = 130.
7. 25   If 5 judges average 7, then using our sum formula, sum = average × # items, the sum is 35. If we want one judge's number to be as big as possible, then we want the other judges' numbers to be as low as possible. Since they must be "different integers," the first 4 numbers would be 1, 2, 3, 4, which are the lowest different positive integers. These numbers add up to 10, so we have 35 – 10 = 25. The last judge can give a whopping score of 25.

Go to: Tip #4