More Word Problems on Rational Algebraic Equation

Introduction

Now we will be studying word problems on rational equations using the concept of distance, rate, and time. These types of problems often appear in real-life situations such as travel, motion, and transportation.

By understanding the relationship among distance, rate, and time, we can translate word problems into rational equations and solve them systematically. This approach not only strengthens your algebraic skills but also enhances your ability to analyze and solve practical problems involving motion.

Before we proceed to solving such word problems, we have to be reminded of the key steps in solving word problems that involves rational algebraic equations as shown below:

Illustrative Example #1

George rode a bike for 32 km and walked for 2 \space km in 3 hours. If George’s speed in biking is 8 times his speed in walking, how fast did he bike?

Solution: Let \bm{x} represent George's speed in walking (km/h), and thus \bm{8x} her speed in biking.

With this, the required equation is

Note that the Least Common Denominator (LCD) of 8x \text{ and } x \text{ is } 8x.

To verify, substitute x = 2 into the crafted equation:

Thus, George’s walking speed is \bf 2 \space km/h, and his biking speed is \bf \boxed{2 \space km/h \times 8 = 16 \space km/h}.

Illustrative Example #2

It took Carla the same time to drive 120 \space kmas it took Dana to drive 90 \space km. If Carla’s speed is 10 \space km/hfaster than Dana’s, how fast did Dana drive?

Solution: Let \bm{x} represent Dana’s driving speed (km/h), and thus \bm{x + 10} represent Carla’s driving speed.

With this, the required equation is

Note that the Least Common Denominator (LCD) of x + 10 \text{ and } x \text{ is } x(x + 10).

To verify, substitute x = 30 into the crafted equation:

Thus, Dana's driving speed is \bf \boxed{ 30 \space km/h}, and Carla's is \bf 30+10=40 \space km/h.

Illustrative Example #3

Cindy biked for 3 hours longer than Allen. Cindy covered 90 km, and Allen covered 45 \space km. If they both maintained the same biking speed, how long did Cindy ride her bike?

Solution: Let x be the time (in hours) Allen biked; then Cindy biked for x + 3 hours.

Since their speeds are equal, we have this equation:

Note that the Least Common Denominator of x \text{ and } x + 3 \text{ is }x(x + 3).

Since x represents the number of hours Allen biked, then, Cindy’s time is

To verify if 6 hours is correct, we will substitute x = 3 to the crafted equation:

Hence, Cindy rode her bike for \boxed{\bm{6\textbf{ hours}}} .

Illustrative Example #4

Tom drove 60 \space km to his office, then took a train for 720 \space km to the capital city. His total travel time was 10 hours. If the train’s speed is 6 times faster than the car’s speed, how fast did Tom drive his car?

Solution: Let \bm{x} be Tom’s car speed (km/h), and thus \bm{6x} his train speed.

The equation modeling the total time is

Note that the Least Common Denominator of x \text{ and } 6x \text{ is } 6x.

To verify if the answer is correct, we will substitute x = 18 into the crafted equation:

Hence, Tom’s car speed is \bf{ \boxed{ 18 \space km/h}} (and the train’s speed is \bf{108} \space km/h).

Illustrative Example #5

A boat takes the same amount of time to go 150 \space km downstream as it does to go 100 \space km upstream. If the current flows at 8 \space km/h, what is the speed of the boat in still water?

Solution: Let \bm{x} be the boat’s speed in still water (km/h). Then

Hence

Setting the times equal gives the equation

Note that the Least Common Denominator of (x + 8) \text{ and } (x - 8) \text{ is }(x + 8)(x - 8).

To verify if the answer is correct, we will substitute x = 40 into the crafted equation:

Thus, the boat’s speed in still water is \bf \boxed{ 40 \space km/h}.