Question

practice distributed and integrated
*1. multi-step the regular price of a cell phone was $89.99. the price was reduced by 25%.
a. find the amount saved. then find the new price of the cell phone. round to the nearest hundredth.
b. if the new price is later reduced by $15, find the percent change from the original price to the lowest price at which the cell phone is sold. round to the nearest percent.
*2. error analysis two students analyzed the data table shown at right. one student determined that the relationship between the variables is an inverse variation, while the other determined that it is a direct variation. which student is correct? what error did the other student make?

Explanation:

1.

Step1: Calculate the amount saved

The amount saved is \(25\%\) of the original price. The formula for finding a percentage \(p\) of a number \(x\) is \(p\times x\). Here, \(p = 0.25\) and \(x=89.99\).
\(0.25\times89.99 = 22.4975\approx22.50\)

Step2: Calculate the new price

The new price \(N\) is the original price \(O\) minus the amount saved \(S\). So \(N=O - S\).
\(N=89.99 - 22.50=67.49\)

Step3: Calculate the lowest price

The new - price is later reduced by \(15\). So the lowest price \(L\) is \(L = 67.49-15=52.49\)

Step4: Calculate the percent change

The formula for percent change is \(\text{Percent Change}=\frac{\text{Original}-\text{New}}{\text{Original}}\times100\%\)
Here, the original price \(O = 89.99\) and the new price (lowest price) \(L = 52.49\)
\(\text{Percent Change}=\frac{89.99 - 52.49}{89.99}\times100\%=\frac{37.5}{89.99}\times100\%\approx42\%\)

For inverse variation, the product of the two variables \(Q\times P\) is a constant. For direct variation, the ratio \(\frac{Q}{P}\) is a constant.
Student A calculated \(Q\times P\) for each pair of values:
For \(Q = 1000\) and \(P=90\), \(Q\times P=1000\times90 = 90000\)
For \(Q = 900\) and \(P = 100\), \(Q\times P=900\times100=90000\)
For \(Q = 750\) and \(P = 120\), \(Q\times P=750\times120 = 90000\)
For \(Q = 600\) and \(P = 150\), \(Q\times P=600\times150=90000\)
For \(Q = 500\) and \(P = 180\), \(Q\times P=500\times180=90000\)
Since \(Q\times P\) is a constant (\(k = 90000\)), the relationship is an inverse variation (\(Q\times P=k\)).
Student B made an error. Just because \(Q\div P\) gives the same numerical value (\(90000\) when calculated as \(Q\div P\) in their table, but the operation \(Q\div P\) is not the correct test for direct variation. For direct variation, \(\frac{Q}{P}=k\) (constant), but here \(Q\times P\) is constant.

Answer:

a. The amount saved is \(\$22.50\) and the new price is \(\$67.49\)
b. The percent change is approximately \(42\%\)

2.