Question 1198078
### 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 \( 33.33\% \).

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### 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 \( \frac{1}{1 + r} \).