SOLUTION: You must use the constant product princivle to solve these problems. Answer al questions in the space Provided. Suppose you drive the same route to work every day, Each morning

Question 1198078: You must use the constant product princivle to solve these problems. Answer al questions in the space
Provided.
Suppose you drive the same route to work every day, Each morning you leave home at 8:20
and arrive at work at 9:00. However, today you are late and do not leave until 8:30.
a.What IS the fractional decrease in time So that you will arrive at 9:00?
b. by What fraction must your speed change so that you'll to get to work at 9:00?
2. Each time I stop at the gas station I put exactly $60 worth of gas into my car
a. This week the gas prices have risen by 5%. Does this mean that I will get 5% less gas for
my $60? Explain.
b. How much gas Will I get?
c. Suppose gas prices rise by a fraction r. What is the effect on the amount of gas my $60 will buy?
Answer by onyulee(41) (Show Source): You can put this solution on YOUR website!
### Problem 1: Driving to Work
#### Part a: Fractional Decrease in Time
- Original time: $ 40 \, \text{minutes} $ (from 8:20 to 9:00).
- New time: $ 30 \, \text{minutes} $ (from 8:30 to 9:00).
- Fractional decrease in time is given by:
\[
\text{Fractional Decrease} = \frac{\text{Original Time} - \text{New Time}}{\text{Original Time}} = \frac{40 - 30}{40} = \frac{10}{40} = 0.25
\]
Thus, the fractional decrease in time is $ 0.25 $ or $ 25\% $.
#### Part b: Fractional Increase in Speed
Using the constant product principle ($ \text{Speed} \times \text{Time} = \text{Distance} $):
- Original speed: $ v $, original time: $ 40 \, \text{minutes} $.
- New speed: $ v_{\text{new}} $, new time: $ 30 \, \text{minutes} $.
- Since the distance is the same:
\[
v \times 40 = v_{\text{new}} \times 30
\]
\[
v_{\text{new}} = \frac{40}{30} v = \frac{4}{3} v
\]
Fractional increase in speed is:
\[
\text{Fractional Increase} = \frac{v_{\text{new}} - v}{v} = \frac{\frac{4}{3}v - v}{v} = \frac{4}{3} - 1 = \frac{1}{3}
\]
Thus, the fractional increase in speed is $ \frac{1}{3} $ or $ 33.33\% $.
---
### Problem 2: Gas Station Purchase
#### Part a: Does a 5% Price Increase Reduce Gas by 5%?
No. The relationship between price per gallon and the quantity of gas purchased is inverse. A 5% increase in price does not correspond to a 5% decrease in gas but will decrease the gas by slightly less.
#### Part b: How Much Gas Will I Get?
Let:
- Original price per gallon: $ p $.
- New price per gallon: $ p_{\text{new}} = 1.05p $.
- Gas obtained: $ g $, where $ g = \frac{60}{p_{\text{new}}} = \frac{60}{1.05p} = \frac{60}{p} \times \frac{1}{1.05} $.
- Original gas purchased: $ g_{\text{original}} = \frac{60}{p} $.
The new amount of gas is reduced by a factor of $ \frac{1}{1.05} $:
\[
g = g_{\text{original}} \times \frac{1}{1.05}
\]
#### Part c: Effect of Fractional Price Increase $ r $
If gas prices increase by a fraction $ r $, the new price is:
\[
p_{\text{new}} = (1 + r)p
\]
The amount of gas purchased becomes:
\[
g = \frac{60}{p_{\text{new}}} = \frac{60}{(1 + r)p} = \frac{g_{\text{original}}}{1 + r}
\]
Thus, the amount of gas decreases inversely to $ 1 + r $. For every fraction $ r $ increase in price, the quantity of gas decreases by a factor of $ \frac{1}{1 + r} $.
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