Interesting Numerical Reasoning Question 
Q: Which period saw the greatest decrease in water prices?
How do I calculate the percentage decrease between period $y_0$ and $y_1$ for example? My initial instinct was to simply calculate: $\frac {y_0-y_1}{y_0}\times 100 = \frac{100-95}{100} \times100= 5\%$. However this is not correct. Any ideas how it's supposed to be calculated? 
This is the proposed solution


*

*y0-y1 1-(100/95)=-5.3%

*y1-y2 1-(95/90)=-5.6%

*y2-y3 1-(90/85)=-5.8%


y2-y3 had the greatest decrease.
 A: Each of the years $y_1$, $y_2$, $y_3$ saw a decrease of 5 from the previous year; so there is no year that saw the "greatest decrease". 
I suspect, they want percentage decrease, as JavaMan surmises.  Then you compute
$$
{5\over100}, {5\over95}, {5\over 90}
$$ for the decimal representation of the percentage decrease for, respectively, the years $y_1$, $y_2$, and $y_3$. Then select the largest.

To find the percentage decrease, $D$, from one year, $y_{old}$ to the next year, $y_{new}$, solve
$$
y_{old}-\underbrace{\textstyle{D\over 100}y_{old}}_{\text{change in }\atop{\text{value}}}=y_{new} 
$$
for $D$.
This gives $$D=100\cdot{y_{old}-y_{new}\over y_{old}}$$ (as you calculated).
 In the "solution", it seems that they are giving the percentage decrease from one year to the previous year. This will allow you to select the year with the greatest percentage decrease; but it's a strange way to do it in my opinion. 

I suspect the solution is in error, and they meant for the fractions therein to be "flipped". This would give the formula above (the percentage decrease from one year to the next year).
