A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of 280 pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries.
How many cherries are there in the fruit salad?
Below is a table showing different possible amounts of each fruit and the total number of pieces of fruit:
| Blueberries | Raspberries | Cherries | Grapes | Total pieces of fruit |
|---|---|---|---|---|
| 1 | 2 | 8 | 24 | 35 |
| 2 | 4 | 16 | 48 | 70 |
| 4 | 8 | 32 | 96 | 140 |
| 8 | 16 | 64 | 192 | 280 |
When there are 280 pieces of fruit, there are 64 cherries. Notice that for the table we successively doubled the number of blueberries which also successively doubles the number of pieces of the other fruits. Students may notice, after the second line, that if they quadruple 70, this will give the desired 280 pieces of fruit, making the table a little shorter.
We are looking for the number of cherries in the fruit salad and will introduce a variable to relate the number of pieces of each fruit. If we let our variable denote the number of cherries then a some work is needed to set up our relationship because the first sentence in the problem deals with blueberries and raspberries. There are twice as many raspberries as blueberries so it is natural to let \(x\) denote the number of blueberries. Then we find the following:
| Fruit | Number of pieces |
|---|---|
| Blueberries | \(x\) |
| Raspberries | \(2x\) |
| Cherries | \(8x\) |
| Grapes | \(24x\) |
Adding up the total number of pieces of fruit gives 35\(x\). We have 280 pieces of fruit total so we want to solve $$ 35x = 280. $$ So \(x = 280 \div 35 = 8\). There are 8 times as many cherries as blueberries so there are 64 cherries.
Since we are looking for the number of cherries in the fruit salad, it is natural to let \(x\) denote the number of cherries.
| Fruit | Number of pieces |
|---|---|
| Blueberries | \(\frac18 x\) |
| Raspberries | \(\frac14 x\) |
| Cherries | \(x\) |
| Grapes | \(3x\) |
Adding up the total number of pieces of fruit gives \(4\frac38 x\). We have 280 pieces of fruit total so we want to solve $$ 4\frac38x = 280. $$ So $$x = 280 \div \left(4\frac38\right) = 280 \div \left(\frac{35}{8}\right)= 280 \times \left(\frac{8}{35}\right) = 64.$$ There are 64 cherries in the fruit salad.
Commentary
The purpose of this task is for students to solve a contextual problem where there is a multiplicative relationship between several quantities in the context. These relationships can either be represented in a ratio table or with a linear equation. Both approaches are valuable and the teacher can link the two by discussing what happens when a variable is used for the number of blueberries in the ratio table. For the algebraic approach, the student is faced with a dilemma: setting the variable equal to the number of cherries leads to a linear equation with fractions while using the number of blueberries as the variable gives a simpler equation. In either case, the students are setting up and solving an equation with positive rational coefficients of the form \(px=q\) as described in 6.EE.7.
This task provides opportunities for students to reason about their computations to see if they make sense. A teacher might start out by asking questions like "What is the most common fruit in the salad? What is the least common fruit in the salad?" in order to check that students are reading and interpreting the relationships correctly.
This task was adapted from problem #5 on the 2012 American Mathematics Competition (AMC) 12A Test.
For the 2012 AMC 12A, which was taken by 72,238 students, the multiple choice answers for the problem had the following distribution:
Of the 72,238 students, 28,268 (39%) were in 12th grade, 34,124 (47%) were in 11th grade 4,615 (6%) were in 10th grade and the remainder were below 10th grade.