Question

1. sierra works in her parents sandwich shop. at the beginning of the year, a ham and cheese sandwich cost $2.45. by the end of the year, this kind of sandwich costs $2.80. what percent describes the change in price of a ham and cheese sandwich? explain your thinking.
2. in the 2011 - 2012 school year, the population at merry meadows middle school was 800 students. in the 2012 - 2013 school year, the population was 700 students. what percent describes the change in the population?
3. aida estimated that a line she drew was 12 cm. the line actually measures 11 cm. write a percent to show how the lines actual measurement compares with aidas estimate.
4. hassan went shopping for shoes. the price for his favorite type of shoes went from $78.95 to $65.79. what percent describes the change in the cost of the shoes?
5. josiah measured the volume of a container in chemistry class. he found that the volume was 65 ml. however, the actual measurement was 50 ml. what percent shows how the actual volume compares with josiahs measurement? explain your thinking.

Explanation:

Step1: Calculate the difference

For the first problem, the difference in price is \(2.80 - 2.45=\$0.35\).
For the second problem, the difference in population is \(800 - 700 = 100\) students.
For the third problem, the difference in length is \(12 - 11=1\) cm.
For the fourth problem, the difference in price is \(78.95 - 65.79=\$13.16\).
For the fifth problem, the difference in volume is \(65 - 50 = 15\) mL.

Step2: Use the percentage change formula

The formula for percentage change is \(\text{Percentage Change}=\frac{\text{Difference}}{\text{Original Value}}\times100\%\)

Problem 1

Original value \(=\$2.45\), \(\text{Percentage Change}=\frac{0.35}{2.45}\times 100\%=\frac{35}{2.45}\% \approx 14.29\%\)

Problem 2

Original value \( = 800\) students, \(\text{Percentage Change}=\frac{100}{800}\times100\% = 12.5\%\) (Since the value decreased, it is a \(- 12.5\%\) change)

Problem 3

Original value \(=12\) cm, \(\text{Percentage Change}=\frac{1}{12}\times100\% \approx 8.33\%\) (Since the actual is less than the estimate, it is a \(-8.33\%\) change)

Problem 4

Original value \(=\$78.95\), \(\text{Percentage Change}=\frac{13.16}{78.95}\times100\%=\frac{1316}{78.95}\% \approx 16.67\%\) (Since the price decreased, it is a \(-16.67\%\) change)

Problem 5

Original value \(=65\) mL, \(\text{Percentage Change}=\frac{15}{65}\times100\%=\frac{1500}{65}\% \approx 23.08\%\) (Since the actual is less than the measurement, it is a \(-23.08\%\) change)

Answer:

1. Approximately \(+14.29\%\) (the positive sign indicates an increase)
2. \(-12.5\%\) (the negative sign indicates a decrease)
3. Approximately \(-8.33\%\) (the negative sign indicates the actual is less than the estimate)
4. Approximately \(-16.67\%\) (the negative sign indicates a price decrease)
5. Approximately \(-23.08\%\) (the negative sign indicates the actual is less than the measurement)