Answer to Question #237439 in Algebra for Mr. Lee

Question

A is 8% taller than B, while B is 9% taller than C. by how much

in percent is A taller than C?

Answer 1

Let

$A =[\frac{100+8}{100}]B=1.08B$

$B =[\frac{100+9}{100}]C=1.09C$

$A=(1.08)1.09C=1.1772C$

Hence $(1.1772×100)-100=17.72$

A taller than C by 17.72%