Difference between revisions of "2014 AMC 10A Problems/Problem 10"

Latest revision as of 03:36, 1 November 2024

The following problem is from both the 2014 AMC 12A #9 and 2014 AMC 10A #10, so both problems redirect to this page.

Problem

Five positive consecutive integers starting with $a$ have average $b$ . What is the average of $5$ consecutive integers that start with $b$ ?

$\textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7$

Solution 1

Let $a=1$ . Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$ . Our next set starting with $3$ is $\{3,4,5,6,7\}$ . Our average is $25\div 5=5$ .

Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{\textbf{(B)}\ a+4}$ .

Solution 2

We are given that $b=\frac{a+a+1+a+2+a+3+a+4}{5}$ $\implies b =a+2$

We are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is $\frac{a+2+a+3+a+4+a+5+a+6}5=a+4$

Thus, the answer is $\boxed{\textbf{(B)}\ a+4}$

Solution 3 (fast)

We know from experience (or logical reasoning) that the average of $5$ consecutive numbers is the $3^\text{rd}$ one or the $1^\text{st} + 2$ . With the logic, we find that $b=a+2$ . Therefore, $b+2=(a+2)+2=\boxed{a+4}$ .

~MathFun1000

Solution 4

The list of numbers is $\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}$ so $b=a+2$ . The new list is $\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}$ and the average is $a+4 \Longrightarrow \boxed{\textbf{(B) } a+4}$ .

~JH. L

Video Solution (CREATIVE THINKING)

https://youtu.be/GonWHjrROzI

~Education, the Study of Everything

Video Solutions

Video Solution 1

https://youtu.be/wBdD6Ge8FuE

~savannahsolver

Video Solution 2

https://youtu.be/rJytKoJzNBY

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 AMC 10 Problems and Solutions

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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 AMC 12 Problems and Solutions

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #9]] and [[2014 AMC 10A Problems|2014 AMC 10A #10]]}}
 ==Problem==
 Five positive consecutive integers starting with <math>a</math> have average <math>b</math>. What is the average of <math>5</math> consecutive integers that start with <math>b</math>?
-<math> \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}}\ a+6\qquad\textbf{(E)}\ a+7</math>
+<math> \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7</math>
-==Solution==
+==Solution 1==
+Let <math>a=1</math>. Our list is <math>\{1,2,3,4,5\}</math> with an average of <math>15\div 5=3</math>. Our next set starting with <math>3</math> is <math>\{3,4,5,6,7\}</math>. Our average is <math>25\div 5=5</math>.
+Therefore, we notice that <math>5=1+4</math> which means that the answer is <math>\boxed{\textbf{(B)}\ a+4}</math>.
+==Solution 2==
+We are given that <cmath>b=\frac{a+a+1+a+2+a+3+a+4}{5}</cmath>
+<cmath>\implies b =a+2</cmath>
+We are asked to find the average of the 5 consecutive integers starting from <math>b</math> in terms of <math>a</math>. By substitution, this is <cmath>\frac{a+2+a+3+a+4+a+5+a+6}5=a+4</cmath>
+Thus, the answer is <math> \boxed{\textbf{(B)}\ a+4} </math>
+==Solution 3 (fast)==
+We know from experience (or logical reasoning) that the average of <math>5</math> consecutive numbers is the <math>3^\text{rd}</math> one or the <math>1^\text{st} + 2</math>. With the logic, we find that <math>b=a+2</math>. Therefore, <math>b+2=(a+2)+2=\boxed{a+4}</math>.
+~MathFun1000
+==Solution 4==
+The list of numbers is <math>\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}</math> so <math>b=a+2</math>. The new list is <math>\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}</math> and the average is <math>a+4 \Longrightarrow \boxed{\textbf{(B) } a+4}</math>.
+~JH. L
+==Video Solution (CREATIVE THINKING)==
+ https://youtu.be/GonWHjrROzI
+~Education, the Study of Everything
+==Video Solutions==
+===Video Solution 1===
+https://youtu.be/wBdD6Ge8FuE
+~savannahsolver
+===Video Solution 2===
+https://youtu.be/rJytKoJzNBY
 ==See Also==
 {{AMC10 box|year=2014|ab=A|num-b=9|num-a=11}}
+{{AMC12 box|year=2014|ab=A|num-b=8|num-a=10}}
 {{MAA Notice}}
+[[Category: Introductory Algebra Problems]]

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