3

Adding the numbers together won’t take long, but there is a shortcut to this problem. Match up the smallest number of miles (1) with the largest number of miles (12), and then the second smallest (2) with the second largest (11), etc. until all the numbers are matched up.

1 12
2 11
3 10
4 9
5 8
6 7

The sum of each of these 6 pairs is 13, which means you could multiply 13 by 6 = 78 to get the total number of miles. Next,find the prime factors of 78. We already know it’s divisible by 13 and 6 (because we had multiplied those together to get 78). Given that 13 is prime, you now need to find the prime factors of 6, which are 2 and 3. All together we have 3 prime factors at the bottom of the prime factor tree, so that’s our answer.

First, list all of the factors of 210.The easiest way to do this is in pairs, starting with 1 and 210 (remember, the number itself is considered one of its factors).Count up from one and check to see if each number you count is a factor of 210.

1 and 210
2 and 105
3 and 70
5 and 42
7 and 3
10 and 21
14 and 15

When you reach 14, you’ll see that you’ll just repeat 15 and 14 if you keep counting, so you know you’re finished.The question asked for the greatest and smallest positive difference between any two factors. (It’s helpful that the question specified positive difference so you know you don’t need to count negative differences, though once you look at the answer choices you can see that’s not what the question is asking for). The way you listed the factors makes this easy because as you look down the list, the difference between factors decreases. So the greatest difference is between 210 and 1, and the smallest difference is between 14 and 15. The final answer is 209 and 1.

The only number that is positive, even, and prime is 2.Because the number is divisible by 2, it must be even.

To solve this problem, Plug In for a and b, but don’t forget your restrictions. If a = −4, then a value of –¹⁄₂ for b would yield a product greater than 0 but less than |a|. Only choice (C) works.

41

Half of the integers from 1 to 100—inclusive—are even, so set X has 50 members. Set Y has 19 members, the integers divisible by 5 from 1 to 100 exclusive, so don’t include 100. Of the 19 members of set Y, 9 are even and therefore, in set X.The 50 members of set X minus the 9 members that \(\underline{are\: also\: in\: set}\) Y yields 50 − 9 = 41 members

Estimate that D is approximately 2.8 and A is approximately 0.8. So the answer is 2.8 ÷ 0.8, or 3.5, which is closest to choice (E).

D,E, and F

It’s an algebra question with numbers for answer choices, so set up your scratch paper to plug in the answers. Start with choice (C): If Emma is 17, then Merrick is 14, and Aliza is 22.That’s too young, so eliminate choices (A), (B), and (C). Now try choice (D): If Emma is 20, then Merrick is 17, and Aliza is 25.That’s old enough, so select choices (D), (E), and (F).

Try Plugging In. If a = 15, b = 30, and c = 60, Quantity A is 30 because c cannot divide into b even one time. Quantity B is 0 because 90 divided by 15 has no remainder.Eliminate choices (B) and (C).Try a new set of numbers to further narrow your choices. If a = 30, b = 45, and c = 120, Quantity A is 45, and Quantity B is 15.The answer is choice (A).

Plug in 12 for the number of people in the office, because the remainder is 0 when 12 is divided by 12.Because ³⁄₂ × 12 is 18, and the remainder when 18 is divided by 12 is 6, choice (E) must be correct.

Plug in values for a and b. If a = −2 and b = 2, then Quantity B is greater.Eliminate choices (A) and (C). If ¹⁄₂ and b = ¹⁄₄, then Quantity A is greater.Eliminate choice (B).