Question 6

To find the total number of hours Dan was driving, we can use the formula:

\[ \text{Total hours} = \frac{\text{Total gas used}}{\text{Gas used per hour}} \]

Here, Dan used a total of \(5 \frac{3}{4}\) gallons of gas and used \(\frac{5}{6}\) gallons of gas per hour. First, we need to convert \(5 \frac{3}{4}\) into an improper fraction.

Convert \(5 \frac{3}{4}\) to an improper fraction:

\[ 5 \frac{3}{4} = \frac{5 \times 4 + 3}{4} = \frac{20 + 3}{4} = \frac{23}{4} \]

Now we have:

• Total gas used = \(\frac{23}{4}\) gallons
• Gas used per hour = \(\frac{5}{6}\) gallons

Now plug these values into the formula:

\[ \text{Total hours} = \frac{\frac{23}{4}}{\frac{5}{6}} \]

Dividing by a fraction is the same as multiplying by its reciprocal:

\[ \text{Total hours} = \frac{23}{4} \times \frac{6}{5} \]

Now multiply the fractions:

\[ \text{Total hours} = \frac{23 \times 6}{4 \times 5} = \frac{138}{20} \]

Next, we simplify \(\frac{138}{20}\). First, find the greatest common divisor (GCD) of 138 and 20.

The prime factorization of 138 is \(2 \times 69\) (which is \(3 \times 23\)) and the prime factorization of 20 is \(2^2 \times 5\). The GCD is \(2\).

Now divide both the numerator and the denominator by \(2\):

\[ \frac{138 \div 2}{20 \div 2} = \frac{69}{10} \]

Now we convert \(\frac{69}{10}\) into a mixed number. Divide \(69\) by \(10\):

• The whole number part is \(6\) (since \(10\) goes into \(69\) six times).
• The remainder is \(9\) (since \(69 - 60 = 9\)).

Thus, we can write \(\frac{69}{10}\) as:

\[ 6 \frac{9}{10} \]

So the total number of hours Dan was driving is:

\[ \boxed{6 \frac{9}{10}} \]