Solutions

17

If the average of 11 numbers is 10, their sum is 11 × 10, which is 110. After one number is eliminated, the average is 9.3, so the sum of the 10 remaining numbers is 10 × 9.3, which is 93. The number eliminated is the difference between these sums: 110 − 93 = 17.
10.4
2

The mean of the listed terms is the sum of the numbers divided by the number of terms: 56 ÷ 7 = 8. The median is the middle number: 6. Thus, 8 is 2 greater than 6.
20

The sum of the other 7 numbers is 140 (168 − 28). So, the average of the numbers is 140/7 = 20.
8

If the average of seven terms is 6, then the sum of the terms is 7 × 6, which is 42. The listed terms have a sum of 34. Therefore, the remaining terms, x and y, must have a sum of 42 − 34, which is 8.
$360

Without x, Matt’s average sale is (300 + 40 + 140) ÷ 3, which is $160. With x, Matt’s average is $50 more, or $210. Therefore, the sum of (300 + 40 + 140 + x) = 4(210) = 840, and x = $360.
30

Elena’s score was within 2 standard deviations of the mean. The standard deviation is 15, so her score is no more than 30 points from the mean. The lowest possible score she could have received, then, is 60 − 30, or 30.
5

Before the $1,000 commission, Milo’s average commission was $250; this is expressed algebraically by the equation S = 250n.

After the sale, the sum of Milo’s sales increased by $1,000, the number of sales made increased by 1, and his average commission was $400. This is expressed algebraically by the equation:

Before the big sale, Milo had made 4 sales. Including the big sale, Milo has made 5 sales.
210

Grace wants to raise her average score by 10 percent. Because 10 percent of 150 is 15, her target average is 165. Grace’s total score is 150 × 6, which is 900. If, in 8 games, she wants to have an average score of 165, then she will need a total score of 165 × 8, which is 1,320. This is a difference of 1,320 − 900, which is 420. Her average score in the next two games must be 420 ÷ 2, which equals 210.
60

The sum of two numbers is twice their average. Therefore:

Substitute these expressions for z and x:

z − x = (160 − y) − (100 − y) = 160 − y − 100 + y = 160 − 100 = 60

Alternatively, pick Smart Numbers for x and y. Let x = 50 and y = 50 (this is an easy way to make their average equal 50). Because the average of y and z must be 80, z = 110. Therefore: z − x = 110 – 50 = 60.
3.76

If the range of the list is 7 and x > 0, then x² has to be the largest number in the list and x² − 1 = 7. Therefore, x² = 8 so . The average of the list is thus , which is approximately , or 3.76.
The list is given in order, therefore, you can see that the largest item in Quartile 2 is the eighth item in the list, which is 10. Furthermore the items in Quartile 4 are 23, 24, 25, and 26, and their average is , which equals 24.5. (Note that these numbers are an evenly spaced list, so the average equals the median or middle number.)

Thus, the ratio is .
If 2 percent of the observations are below −10, then −10 must approximately be 2 standard deviations from the mean. Thus the standard deviation is approximately , and thus roughly of the observations will fall between −5 and 5. Because normal variables are symmetric around the mean, half of that will be in the 0–5 range, so the correct answer is , which simplifies to .
(B)

This is a Weighted Average problem. The overall average score can be computed by assigning weights to the average scores of Poets and Bards that reflect the number of people in each subgroup. Because the ratio of Poets to Bards is 3 to 2, and collectively the two groups account for all students, the multiple ratio may be written as P:B:Total = 3:2:5.

This means that Poets constitute 3/5 of the students and Bards the remaining 2/5. Therefore, the overall average score is given by the weighted average formula:

Alternatively, you may argue as follows: if there were the same number of Poets as there were Bards, the overall average score would be 70. However, there are actually more Poets than Bards, so the overall average score will be closer to 60 than to 80 (i.e., less than 70). Therefore, Quantity B is greater.
(B)

Begin with the median. In a set with an odd number of terms, the median will be the middle term when the terms are put in ascending order. It is clear that x + 1 > x − 4. Moreover, because x > 2, 4x must be greater than x + 1. Therefore, the median is x + 1. Rewrite Quantity A:

Quantity A Quantity B

The median of x − 4, x + 1, and 4x = x + 1 The mean of x − 4, x + 1, and 4x

To compute the mean, add all three terms and divide by 3:

Rewrite Quantity B:

Quantity A Quantity B

The median of x − 4, x + 1, and 4x = x + 1 The mean of x − 4, x + 1, and 4x = 2x − 1

The comparison thus boils down to which is greater, x + 1 or 2x − 1. The answer is not immediately clear. Subtract x from both sides to try and isolate x:

Quantity A Quantity B

Now add 1 to both sides to isolate x:

Quantity A Quantity B

1 + 1 = 2 (x − 1) + 1 = x

The question stem states that x must be greater than 2, therefore, Quantity B is greater.
(C)

The sets in question are A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8, 10}. Each is a set of evenly spaced integers with an odd number of terms, such that the mean is the middle number. The deviations between the elements of the set and the mean of the set in each case are the same: −4, −2, 0, 2, and 4. Thus, the standard deviations of the sets must also be the same. Therefore, the two quantities are equal.

Quantity A	Quantity B
The median of x − 4, x + 1, and 4x = x + 1	The mean of x − 4, x + 1, and 4x