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OMCC Nursing Program Entrance Test - PN

3. Randy drove the first third of his trip on a gravel road, the next 20 miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy's trip?

A. 30
B. [tex]\frac{400}{11} \approx 36.36[/tex]


Sagot :

To determine the total distance of Randy's trip, we need to break down the parts of the trip as described and then solve for the total distance. Here’s the detailed step-by-step solution:

1. Define the total trip distance:
Let the total distance of the trip be [tex]\( x \)[/tex] miles.

2. Distance on the gravel road:
Randy drove the first third of his trip on a gravel road. Therefore, the distance on the gravel road is [tex]\( \frac{x}{3} \)[/tex] miles.

3. Distance on the pavement road:
Randy drove the next 20 miles on pavement. So, the distance on the pavement road is 20 miles.

4. Distance on the dirt road:
Randy drove the remaining one-fifth of the trip on a dirt road. Thus, the distance on the dirt road is [tex]\( \frac{x}{5} \)[/tex] miles.

5. Total distance equation:
The sum of the distances on the gravel road, pavement road, and dirt road must equal the total trip distance [tex]\( x \)[/tex]. Setting up the equation:
[tex]\[ \frac{x}{3} + 20 + \frac{x}{5} = x \][/tex]

6. Combine the fractions:
To solve for [tex]\( x \)[/tex], we first need to combine the fractions on the left-hand side of the equation. To do this, find a common denominator, which is 15.

[tex]\[ \frac{5x}{15} + 20 + \frac{3x}{15} = x \][/tex]

Simplify the equation by adding the fractions:

[tex]\[ \frac{8x}{15} + 20 = x \][/tex]

7. Isolate [tex]\( x \)[/tex]:
To isolate [tex]\( x \)[/tex], we subtract [tex]\(\frac{8x}{15}\)[/tex] from both sides of the equation:

[tex]\[ 20 = x - \frac{8x}{15} \][/tex]

Now, express [tex]\( x \)[/tex] on a common denominator:

[tex]\[ 20 = \frac{15x}{15} - \frac{8x}{15} \][/tex]

Simplify the right-hand side of the equation:

[tex]\[ 20 = \frac{7x}{15} \][/tex]

8. Solve for [tex]\( x \)[/tex]:
Multiply both sides of the equation by 15 to solve for [tex]\( x \)[/tex]:

[tex]\[ 20 \times 15 = 7x \][/tex]
[tex]\[ 300 = 7x \][/tex]
[tex]\[ x = \frac{300}{7} \][/tex]

So, the total distance of Randy's trip is [tex]\( \frac{300}{7} \)[/tex] miles.

As a final note, [tex]\( \frac{300}{7} \)[/tex] approximates to around 42.86 miles.

Thus, the total distance of Randy's trip is [tex]\( \boxed{\frac{300}{7}} \)[/tex] miles.