# Percentages Issue

I am having a problem with the following question could you guys tell me what I am doing wrong?

When the tires of a taxicab are under-inflated, the cab's odometer will read $10\%$ over the true mileage. If the odometer of a cab with under-inflated tires read $m$ miles, what is the actual distance driven? (answer = $10m/11$)

Here is how I am doing it: $$m = m - \frac{10m}{100}$$ $$m = \frac{9m}{10}$$

Could anyone tell me what I am doing wrong ?

-
You are using the same name for two different variables (the reading and the actual number of miles). – Alex Becker Jul 7 '12 at 22:22
I agree.. Just figured it out. Increase Value = Actual Value + increase – Rajeshwar Jul 7 '12 at 22:24
m = 11/10(ActualVAlue) – Rajeshwar Jul 7 '12 at 22:25

Let $x$ be the true mileage; then $$m=x+\frac{10x}{100}=x+\frac{x}{10}=\frac{10x+x}{10}=\frac{11x}{10}\;,$$ so $$x=\frac{10m}{11}\;.$$
What you wrote clearly cannot be right: $m$ cannot be nine-tenths of $m$ unless $m=0$. You seem to have tried to use $m$ to represent both the odometer reading and the true mileage; since those are different, this cannot work.
Let $T$ be the true mileage. Then $$m=T+(10\%)T=T+(0.1)T=(1.1)T=\frac{11}{10}\times T,$$ so that after multiplying both sides by $\frac{10}{11}$, $$\frac{10}{11}\times m=\frac{10}{11}\times \frac{11}{10}\times T=T.$$ Thus, we have that $T=\frac{10}{11} m$.