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

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adding a positive number (<math>a</math>) to <math>c</math> will always make it greater than <math>b</math>.
 
adding a positive number (<math>a</math>) to <math>c</math> will always make it greater than <math>b</math>.
  
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Therefore, the answer is <math>\boxed{A}</math>
+
Therefore, the answer is <math>\boxed{\textbf{(A)}\ a+c<b}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=14|num-a=16}}
 
{{AMC8 box|year=2007|num-b=14|num-a=16}}

Revision as of 12:21, 9 December 2012

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

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
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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