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

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~savannahsolver
 
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==Video Solution==
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https://www.youtube.com/watch?v=_ZHS4M7kpnE
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=14|num-a=16}}
 
{{AMC8 box|year=2007|num-b=14|num-a=16}}
 
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Revision as of 22:56, 9 September 2022

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

Video Solution by WhyMath

https://youtu.be/UdzJetT-XOY

~savannahsolver

Video Solution

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

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 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

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