Difference between revisions of "2007 AMC 8 Problems/Problem 15"

Revision as of 21:07, 7 January 2023

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

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 $\boxed{\textbf{(A)} }$ is correct.

—jason.ca

Video Solution by WhyMath

https://youtu.be/UdzJetT-XOY

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=_ZHS4M7kpnE

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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

@@ Line 14: / Line 14: @@
 ==Solution 2==
+We can test numbers into the inequality we’re given. The simplest is <math>0<1<2<3</math>. We can see that <math>3+1>2</math>, so <math>\boxed{\textbf{(A)} }</math> is correct.
+—jason.ca
 ==Video Solution by WhyMath==

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