A price goes up and then down

A price $P$ is a positive real number.

First the price increases by $a\%$ , then the new price decreases by $a\%$ , where $0 \le a \le 100$ .

Is the final price higher, lower, or the same as the original price?

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Solution by Ancient Tree

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Increasing $P$ by $a\%$ gives the new price

$P_2=P+\frac{a}{100}P=\frac{100+a}{100}P.$

Then the final price is

$P_3=\frac{100-a}{100}P_2=\frac{(100-a)(100+a)}{100^2}P.$

But

$(100-a)(100+a)=100^2-a^2\le 100^2.$

Therefore $P_3\le P$ . Equality happens only when $a=0$ ; if $a>0$ , the final price is strictly lower than the original price.

Solution by Ancient TreeFR

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Augmenter $P$ de $a\%$ donne le nouveau prix
$P_2=P+\frac{a}{100}P=\frac{100+a}{100}P.$ Alors, le prix final est
$P_3=\frac{100-a}{100}P_2=\frac{(100-a)(100+a)}{100^2}P.$ Mais
$(100-a)(100+a)=100^2-a^2\le 100^2.$ Par conséquent $P_3\le P$ . L’égalité se produit uniquement lorsque $a=0$ ; si $a>0$ , le prix final est strictement inférieur au prix initial.

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