VIII Algebra & Word Problems

Distance, Rate & Time

$Distance = Rate \times Time$

Example 1

$D=RT$, $R=\frac D T$, $T=\frac D R$

Example: On a jog you walk at 5mph for a while and then at 8mph. In total you run 7 miles in 1.1 hours. How long did you run each walk?

This can be interpreted as:
$5t = x$ and $8t = y$ resolve to $x + y = 7$. This leads to the equation $5t + 8(1.1-t) = 7$ which can be solved for $t$ where $t$ represents the time taken for one walk and $1.1-t$ represents the time taken for the other walk.

Example 2

You travel in a car from town A to town B at an average speed of 64km/h. On the return journey your average speed is 80km/h. You take 9 hours in total.

How far is it from A to B?

$D = \frac R T$ implies $T = \frac D R$. Since we know the rates of both trips and their total time, an equation can be derived that has the distance as its only unknown:

$\frac {x}{64} + \frac {x}{80} = 9$

Percentage

$x \frac {n}{100}$ equals to $x$% of $n$.

Example 1

A number plus 4% of itself is 41.6. Find that number. This implies:

$x = 4 \frac {n}{100}$, $41.6 = n - x$: $x = 4 \frac {41.6 - x}{100}$

Celsius & Fahrenheit

Given a known Fahrenheit value, its Celsis equivalent can be found by solving the formula for $x$. For 86 degrees Fahrenheit to Celsius:

$86 = \frac {9x}{5} + 32$

Misc

Various word problems involving algebra.

Example 1

A man is 4 times as old as his son. In 4 years time he will be three times as old. How old are both now?

This problem is somewhat ambiguous. In 4 years time the man will be three times as old as what? Plugging in values for their ages helps finding the correct equations.

Age of man: $x = 4y$

Age of man in 4 years: $x + 4 = 3(y + 4)$

Example of Problem involving Quadratic Equations

A watermelon falls from a 225ft tall building. How long does it take the melon to hit the ground level? The height of the watermelon is given by the following formula:

$h(T) = -16T^2+225$

The goal is to find $T$ when $h = 0$. This implies the quadratic equation:

$-(16T^2-225) = 0$

That can be solved for $T=\frac {15}{4}$ and $T=\frac{-15}{4}$. Given the context $T$ cannot be a negative, therefore $3\frac 3 4sec$ is the solution to the problem.