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SAT Average Questions
1. If the average (arithmetic mean) of m and 4m is 30, what is the value of m ?
a. 4
b. 6
c. 8
d. 10
e. 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?
a. 21 + k
b. k – 21
c.
d.
e. 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 ?
a. 31
b. 29
c. 21
d. 19
e. 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?
a.
0
b.
c.
d.
e. 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?
a. 25
b. 86
c. 87.2
d. 89.5
e. 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. 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?
SAT Average Questions
Answers
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 + nequals 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.