2007 AMC 8 Problems/Problem 15

Problem

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

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

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 $\boxed{\textbf{(A)}\ a+c<b}$

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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