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

(Created page with '== Problem == Let <math>a, b</math> and <math>c</math> be numbers with <math>0 < a < b < c</math>. Which of the following is impossible? <math>\mathrm{(A)} \ a + c < b \qquad …')
 
 
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impossible?
 
impossible?
  
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<math>\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</math>
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<math>\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</math>
  
 
== Solution ==
 
== Solution ==
  
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According to the given rules,  
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According to the given rules, every number needs to be positive. Since <math>c</math> is always 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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Every number needs to be positive.
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Therefore, the answer is <math>\boxed{\textbf{(A)}\ a+c<b}</math>
  
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Since <math>c</math> is always greater than <math>b</math>,
 
  
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adding a positive number (<math>a</math>) to <math>c</math> will always make it greater than <math>b</math>.
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==Solution 2==
  
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Therefore, the answer is <math>\boxed{A}</math>
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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) }a+c<b}</math> is correct.
 +
 
 +
—jason.ca
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/UdzJetT-XOY
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=_ZHS4M7kpnE
 +
 
 +
==Video Solution 2==
 +
https://youtu.be/GxR1giTQeD0  Soo, DRMS, NM
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=omFpSGMWhFc
 +
 
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2007|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 14:53, 2 July 2024

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

—jason.ca

Video Solution by WhyMath

https://youtu.be/UdzJetT-XOY

~savannahsolver

Video Solution

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

Video Solution 2

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

Video Solution by SpreadTheMathLove

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


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

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

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