Lower Elementary
Question: Rachel found three types of candy she liked at the store. The first type of candy was 35 cents each, the second type was 6 for $2.00, and the third type was 4 for $1.50. Which type of candy should Rachel buy to get the best deal?
Answer: 6 for $2.00 (the second type)
Note: There are several way to figure out the problem. To solve this problem without division, you can start by comparing the first and third types of candy. If you buy 4 pieces of the first type then it is only $1.40 (4 x .35), while the third type was 4 for $1.50, so the first is a better deal than the third. Then compare the first type to the second type: if you buy 6 pieces of the first type it will cost $2.10 (6 x .35), compared to the second type which was 6 for $2.00, so the second type is a better buy than the first.

Upper Elementary
Question: The numbers 1 through 9 are individually written on 9 identical marbles. The marbles are placed into a bag. Without looking, what is the probability of pulling out a marble that is both even and prime number from the bag?
Answer: 1/9
Note: 2 is the only even prime number.

Middle School
Question: Mark has an average of 80 for the first four tests in his math class. If he gets a 100 on his next test, what will his average be for all five tests?
Answer: 84
Note:  To find the answer we must first find Mark’s total points for the first 4 tests. If Mark has an average of 80 on the first 4 tests, that means the total points earned for all 4 tests is 320 points ( 80 x 4). Then we add the points from his fifth test which gives his him 420 points (320 + 100).  To find the new average we divide by the new number of tests which is 5.  420 / 5 = 84.

Algebra and Up
Question: In a class of 15 high school students, 9 students are taking French, 4 students are taking an art class, and 5 students are taking neither. How many students are taking both French and art?
Answer: 3 students
Note: It is helpful to draw a Venn Diagram to show the different groups and how they overlap. Since 5 students take neither French nor art, that leaves 10 people taking French and/or art. These 10 students represent the union of both French students and art students. This includes: French only students, art only students, and students taking both French and art (the intersection).

Another way to look at it is that these 10 students are equal to all the students taking French plus all the students taking art minus the number of students who take both French and art. If we let n = number of students taking both French and art. 10 = (9 + 4) – n => 10 = 13 – n => n = 3