Solutions

  1. 9

    The primes from 10 to 41, inclusive, are: 11, 13, 17, 19, 23, 29, 31, 37, and 41. Note that the primes are NOT evenly spaced, so you have to list them and count them manually.

  2. No

    For any sequence of consecutive integers with an even number of items, the average is NEVER an integer. For example, if you pick 4, 5, 6, 7, 8, and 9:

    Since any set of consecutive integers is evenly spaced, the average is equal to the median. And if the set also has an even number of terms, the median is equal to the average of the two middle terms (i.e., exactly in between two consecutive integers). In the example above, the two middle terms are 6 and 7, giving you a median of 6.5, which is also the average of this evenly spaced set.

  3. 19.5

    Average . In this problem, you have as the average.

  4. 52

    Calculate 33 − (−18) = 51. Then add 1 before you are done: 51 + 1 = 52.

  5. (B)

    There are two ways to do this question. First, try to estimate the sum. Notice that there will be 11 terms in Set A, and one of them is 0. Even if every term in the set were 20, the sum would only be 220 (11 × 20 = 220). But half of the terms are less than 10. It is unlikely that Quantity A will be bigger than 150.

    To do it mathematically, use the equation for the sum of an evenly spaced sequence. You can find the median by adding up the first and last terms and dividing by 2:

    You can then find the number of terms by subtracting the first term from the last term, dividing by the interval (in this case, 2) and adding 1: and 10 + 1 = 11 terms.

    Finally, the sum of the terms will be the average value of the terms (10) times the number of terms (11): 10 × 11 = 110.

    Set A is comprised of all the even numbers between 0 and 20, inclusive.

    Quantity A Quantity A
    The sum of all the numbers in Set A = 110 150

    Therefore, Quantity B is greater.

  6. (C)

    Both quantities can be solved straightforwardly with the equations. The first multiple of 7 between 50 and 100 is 56, and the last is 98. Thus: 98 − 56 = 42; ; and 6 + 1 = 7.

    Another way to think about it is that 56 is the 8th multiple of 7, and 98 is the 14th multiple of 7. Now use the counting principle:

    14 − 8 = 6 + 1 = 7

    There are 7 multiples of 7 between 50 and 100.

    Similarly, the first multiple of 9 between 30 and 90 is 36, and the last is 90: 90 − 36 = 54; , and 6 + 1 = 7.

    Now note that 36 is the 4th multiple of 9, and 90 is the 10th multiple of 9:

    10 − 4 = 6 + 1 = 7

    There are 7 multiples of 9 between 30 and 90.

    Quantity A Quantity B
    The number of multiples of 7 between 50 and 100, inclusive = 7 The number of multiples of 9 between 30 and 90, inclusive = 7

    Therefore, the two quantities are equal.

  7. (A)

    The key here is to notice that Set A is an evenly spaced sequence, and thus you can easily solve for its sum. The median will be equal to

    Although there is a formula to figure out how many terms there are, it is easiest in this case to count. There are six terms in the set.

    The sum of the terms in the set is the average value of the terms (3x − 10) times the number of terms (6).

    (3x − 10) × 6 = 18x − 60

    Rewrite the quantities:

    Set A is comprised of the following terms: (3x), (3x − 4), (3x − 8), (3x − 12), (3x − 16), and (3x − 20)

    Quantity A Quantity B
    18x − 60 18x − 70

    Be careful. Quantity A is subtracting a smaller number (60) than is Quantity B (70), and so has a larger value. Therefore, Quantity A is greater.