VIII Algebra & Word Problems

Distance, Rate & Time

Distance=Rate×TimeDistance = Rate \times Time

Example 1

D=RTD=RT, R=DTR=\frac D T, T=DRT=\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=x5t = x and 8t=y8t = y resolve to x+y=7x + y = 7. This leads to the equation 5t+8(1.1t)=75t + 8(1.1-t) = 7 which can be solved for tt where tt represents the time taken for one walk and 1.1t1.1-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=RTD = \frac R T implies T=DRT = \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:

x64+x80=9\frac {x}{64} + \frac {x}{80} = 9


xn100x \frac {n}{100} equals to xx% of nn.

Example 1

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

x=4n100x = 4 \frac {n}{100}, 41.6=nx41.6 = n - x: x=441.6x100x = 4 \frac {41.6 - x}{100}

Celsius & Fahrenheit

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

86=9x5+3286 = \frac {9x}{5} + 32


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=4yx = 4y

Age of man in 4 years: x+4=3(y+4)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)=16T2+225h(T) = -16T^2+225

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

(16T2225)=0-(16T^2-225) = 0

That can be solved for T=154T=\frac {15}{4} and T=154T=\frac{-15}{4}. Given the context TT cannot be a negative, therefore 334sec3\frac 3 4sec is the solution to the problem.