Question

1. a basketball coach purchases bananas for the players on his team. the table shows the total price in dollars, p, of n bananas. which equation could represent the total price in dollars for n bananas?

number of bananas	total price in dollars
8	4.72
9	5.31
10	5.90

a. p = 0.59n
b. p = 5.90 - 0.59n
c. p = \frac{5.90}{n}
d. p = n + 0.59
(from unit 2, lesson 3.)

1. kiran is collecting dimes and quarters in a jar. he has collected $10.00 so far and has d dimes and q quarters. the relationship between the numbers of dimes and quarters, and the amount of money in dollars is represented by the equation 0.1d + 0.25q = 10.

select all the values (d, q) that could be solutions to the equation.
a. (100, 0)
b. (20, 50)
c. (50, 20)
d. (0, 100)
e. (10, 36)
(from unit 2, lesson 4.)

Explanation:

Step1: Find the unit - price for bananas

To find the unit - price of a banana, we can use the formula $unit\ price=\frac{total\ price}{number\ of\ bananas}$. For example, when $n = 7$ and $P=4.13$, the unit - price is $\frac{4.13}{7}=0.59$. When $n = 8$ and $P = 4.72$, the unit - price is $\frac{4.72}{8}=0.59$. When $n = 9$ and $P=5.31$, the unit - price is $\frac{5.31}{9}=0.59$. When $n = 10$ and $P = 5.90$, the unit - price is $\frac{5.90}{10}=0.59$. The total price $P$ is the unit - price times the number of bananas $n$. So the equation is $P = 0.59n$.

Step2: Check solutions for the dime - quarter equation

For the equation $0.1d+0.25q = 10$:

• Option A: Substitute $d = 100$ and $q = 0$ into the equation. We get $0.1\times100+0.25\times0=10 + 0=10$. So $(100,0)$ is a solution.
• Option B: Substitute $d = 20$ and $q = 50$ into the equation. $0.1\times20+0.25\times50=2 + 12.5=14.5

eq10$. So $(20,50)$ is not a solution.

• Option C: Substitute $d = 50$ and $q = 20$ into the equation. $0.1\times50+0.25\times20=5 + 5=10$. So $(50,20)$ is a solution.
• Option D: Substitute $d = 0$ and $q = 100$ into the equation. $0.1\times0+0.25\times100=0 + 25=25

eq10$. So $(0,100)$ is not a solution.

• Option E: Substitute $d = 10$ and $q = 36$ into the equation. $0.1\times10+0.25\times36=1+9 = 10$. So $(10,36)$ is a solution.

Answer:

1. A. $P = 0.59n$
2. A. $(100,0)$, C. $(50,20)$, E. $(10,36)$