Statistics: Averages
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1. In a typical day at the TSA (Testing School of America), Sally takes five tests. Her score on the first test is x. On each test after the first, her score is 4 points lower than her score on the previous test. What is her average score for all five tests?

(A) x–8
(B) x–10
(C) x–12
(D) x–15
(E) x–16

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Sally's test scores are: x, x–4, x–8, x–12, and x–16. Add them up to get the total: 5x–40. Divide by 5 to get the average: x–8. The answer is (A).

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2. Sally is scheduled to take her Nth test at the TSA. She has an average of A for her first N–1 tests. She is finally fed up and doesn't bother to show up for the Nth test. She receives a zero for this test. How many points lower than A is Sally's average for all N tests?

(A) A/N
(B) A/(N–1)
(C) AN/(N–1)
(D) A(N–1)/N
(E) A/(N(N–1))

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The zero on the last test cost Sally A points from her total. These are divided evenly amongst the N tests so her new average is A – A/N. This isn't the answer though. Sally's average is A/N points lower than it was before so the answer is (A).

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3. The average of four positive numbers is 70. Which of the following is true?

I. If one number is 60, one number must be 80.
II. At least one number is greater than 70.
III. The largest number cannot be 300.

(A) none
(B) I and II, only
(C) II, only
(D) III, only
(E) II and III, only

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Four numbers that average to 70 add up to 280. They could be 60, 70, 71, and 79 for instance. They could also be 70, 70, 70, 70. So (I) and (II) are NOT true. However, since all the numbers are positive, you can't have a 300 so (III) is true and the answer is (D).

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4. The average of five positive integers is 50. One of the five integers is 26. The largest possible value of one of the other four numbers is:

(A) 212
(B) 215
(C) 218
(D) 221
(E) 224

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The five numbers have to add up to 250. The smallest possible first four numbers, all positive integers, are 26, 1, 1, and 1 (the question didn't say they had to be different). The fifth number would be 221. The answer is (D).

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5. The average of a, b, c, and d is equal to y and the value of d is 12 less than the value of y. The average of a, b, and c must be:

(A) 12 more than y
(B) 8 more than y
(C) 6 more than y
(D) 4 more than y
(E) 3 more than y

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The three numbers a, b, and c have to make up the 12 (d=y–12). If a, b, and c are each y+4 or if they average to y+4 that will make up the 12. So the answer is (D).

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6. One group of people (group I) has an average height of 62 inches. Another group (group II) has an average height of 70 inches. The average height of all the people in both groups together is 67 inches. If there are 120 people in group I, how many are in group II?

(A) 45
(B) 72
(C) 75
(D) 180
(E) 200

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The average is just another way of giving you the total. The total height of group I is 120·62. The total height of group II is N·70. The total height of all the people in both groups is (120+N)·67. To get N solve: 120·62 + N·70 = (120+N)·67. You get N=200 or (E).

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7. Jack is an independent contractor working for a radioactive isotope production company. Jack earns 30 dollars per hour for routine checks and 45 dollars per hour for work in radioactive areas. On one particular day he earns d dollars for routine checks and another d dollars for work in a radioactive area. He doesn't take any breaks. What is the average number of dollars per hour (dollars earned divided by hours worked) Jack earns that day?

(A) 35
(B) 36
(C) 37.5
(D) 38.5
(E) cannot be determined

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Jack earned 2d dollars. Jack worked d/30 + d/45 hours. Jack's overall dollars per hour is 2d / (d/30 + d/45) = 2 / (5/90) = 36. His average pay rate is closer to his rate for routine checks because he spent longer doing the routine checks. The answer is (B).

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8. The average of a, b, c, and d is x. If a = x/2, b = 2x, and c = x–10, then d =

(A) x/2 + 10
(B) x/2 – 10
(C) x + 10
(D) x – 10
(E) (x – 10) / 2

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Since the average is x, the sum is 4x. You'll need an x/2 to get the 4x and a +10 to cancel out the –10. So d = x/2 + 10 or (A).

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9. The set of different integers a, b, c, and d has an average equal to x. The set of integers a2, b2, c2, and d2 has an average equal to y. Which of the following statements could be true?

I. y < x
II. y = x
III. y < 2

(A) none
(B) I, only
(C) II, only
(D) III, only
(E) I, II, and III

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Squares (of integers) are bigger so y has to be bigger than x so (I) is not true. You could have two of the numbers be 0 and 1 but the four integers are different so the other two will produce bigger numbers when squared so y can't be equal to x and (II is not true). To make y as small as possible use, -1, 0, 1, and 2. The avearage of the squares is 1.5 so (III) could be true (the question did NOT say "must be true") and the answer is (D).

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10. The average of 5 different integers is 20. The largest of the 5 integers is 23. The smallest of the five integers is N. How many possible values of N are there?

(A) 3
(B) 4
(C) 5
(D) 6
(E) more than 6

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You can pick 23, 22, 21, 20, and 14. The smallest is 14 and that's as low as you can go. You could also do 23, 22, 21, 19, and 15. You could do 23, 22, 21, 18, and 16. You could do 23, 22, 20, 18, and 17. If you try to make the lowest one 18, you "run out of room." The 23 is 3 above 20 and an 18 and a 19 make 3 below 20. But you still have two more numbers to pick and you can't keep the average equal to 20. So there are four possible smallest integers: 14, 15, 16, and 17. The answer is (B).

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