2007 AMC 8 Problems/Problem 15

Revision as of 15:50, 26 March 2010 by YottaByte (talk | contribs) (Created page with '== Problem == Let <math>a, b</math> and <math>c</math> be numbers with <math>0 < a < b < c</math>. Which of the following is impossible? <math>\mathrm{(A)} \ a + c < b \qquad …')
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Let $a, b$ and $c$ be numbers with $0 < a < b < c$. Which of the following is impossible?

$\mathrm{(A)} \ a + c < b  \qquad \mathrm{(B)} \ a * b < c \qquad \mathrm{(C)} \ a + b < c \qquad \mathrm{(D)} \ a * c < b \qquad \mathrm{(E)}\frac{b}{c} = a$

Solution

According to the given rules,

Every number needs to be positive.

Since $c$ is always greater than $b$,

adding a positive number ($a$) to $c$ will always make it greater than $b$.

Therefore, the answer is $\boxed{A}$