Estonian Mathematical Olympiad

Since the new model of watch A is more expensive than the old model of watch A and the new model of watch B is cheaper than the old model of watch B, but the difference in price is still $p\%$, the old model of watch A must be $p\%$ cheaper than the old model of watch B, and the new model of watch A must be $p\%$ more expensive than the new model of watch B.

Let the price of the old model of watch B be $x$. Then the price of the new model of watch B is $(1-\frac{q}{100})x$. The old model of watch A then costs $(1-\frac{p}{100})x$, while the new model of watch A costs $(1+\frac{q}{100})(1-\frac{p}{100})x$. Since from the previous paragraph the price of the new model of watch A is $(1+\frac{p}{100})(1-\frac{q}{100})x$, we have $$\left(1+\frac{q}{100}\right)\left(1-\frac{p}{100}\right)x=\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right)x.$$ After cancelling $x$ from both sides, opening the parentheses and simplifying, we get $p=q$.

a. Since $p=q$, the prices of the new model of watch B and of the old model of watch A are both $(1-\frac{p}{100})x$.

b. Since $p=q$, the price of the new model of watch A is $(1-\frac{p^2}{10000})x$ which is less than the price of the old model of watch B.