# 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

## Solution 2

We can test numbers into the inequality we’re given. The simplest is $0<1<2<3$. We can see that $3+1>2$, so $\boxed{\textbf{(A) }a+c is correct.

—jason.ca

~savannahsolver

## Video Solution 2

https://youtu.be/GxR1giTQeD0 Soo, DRMS, NM

## See Also

 2007 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.