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

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

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.

@@ 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) }a+c<b}</math> is correct.
+—jason.ca
 ==Video Solution by WhyMath==
@@ Line 23: / Line 27: @@
 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}}

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