Relationship between Revenue-Elasticity and Price-Elasticity I have attempted to solve this problem and would be grateful if someone looked over my solution

5% increase in price  results in 5% decrease in total revenue. I need to determine price elasticity of demand. I have done calculations but I am not sure whether the way I have solved it is true.

$$TR=PQ\mapsto Q=\frac{TR}{P}\rightarrow Q=\frac{0.8TR}{1.2 P}=2/3$$
And then I wrote $$PE=\frac{\frac{8}{12}}{\frac{12}{10}}=5/9$$

But this solution doesn't appear in the test answer options. Can someone please check it and tell what is wrong?

Thanks in advance!
Let's denote the elasticity of $X$ to $Y$ at $Y_0$ with $\eta_{X, Y}(Y_0)$ and solve for the demand ($Q$) elasticity in terms of the revenue elasticity ($R$).

We know that the elasticity is always given by

$$ \eta_{X,Y} = \frac{d X(Y)}{dY}\frac{Y}{X}$$ 

or equivalently (showing the equivalence is a useful exercise),

$$ \eta_{X,Y} = \frac{d \log X(Y))}{d \log Y}$$ 

Now, 

$$ R(p) = p Q(p) \\
\eta_{R, p}(p) = R'(p) \frac{p}{pQ(p)}\\
\eta_{R, p}(p) = (Q(p) + pQ'(p_0) ) \frac{p}{pQ(p)} \\
\eta_{R, p}(p) = 1 + \frac{pQ'(p_0)}{Q} \\
\eta_{R, p}(p) = 1 + \eta_{Q, p}(p) \\
$$ 

Given that the left-hand-side elasticity is $-1$ at the initial price $p_0$, the right-hand-side elasticity must be $-2$:

$$-1 = 1 + \eta_{Q, p}(p_0) \Rightarrow \eta_{Q, p}(p_0) = -2$$

Your mistake

You are being sloppy with the mathematics. First you state

$$ Q = \frac{TR}{P}$$ 

But then you claim

$$ Q = \frac{0.8TR}{1.2P}$$ 

These two equations just cannot be true at the same time.

Perhaps you want to look at initial $TR_0$, at total revenue after price change $TR_1$, and then state that $TR_1/TR_0 = 0.95$.