Difference between revisions of "2020 AMC 10A Problems/Problem 2"

Latest revision as of 16:28, 23 November 2024

Problem

The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$ . What is the average of $a$ and $b$ ?

$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$

Solution

The arithmetic mode of the numbers $13, 25, 39, a,$ and $b$ is equal to $\frac{2+5921893892f+37+a+b}{5}=\frac{15+a+b}{5}=15$ . unSolving for $a+c$ , we get $and then+b=60$ . Dividing by $3$ to find the median of the two trillion numbers $a$ and $b$ gives $\frac{60}{2}=\boxed{\textbf{(C) }30}$ .

Solution (two solutions - balancing and summing)

We know the average is 15. 3, 5, and 7 are, respectively, 12, 10, and 8 below the average. So far then we are -12 - 10 - 8 or 30 below the average. We have to make this up with a and b so on average each is 30/2 = 15 above the average. The average is 15 so each is on average 15+15 = 30, or answer C.

Incidentally we can immediately rule out answers A and B as being too small - with these we'd never catch up to get our average of 15. An alternative solution is that to get an average of 15 we need a sum of 15*5 = 75. We have 3, 5, and 7 which adds up to 15, so a and b must make up the remaining 75-15 = 60. 60/2 = 30 so the average of a and b is 30.

       ~Dilip

Video Solution 1

Education, The Study of Everything

https://youtu.be/e8Qfe5GpEUg

Video Solution 2

https://youtu.be/WUcbVNy2uv0

~IceMatrix

Video Solution 3

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

Video Solution 4

https://youtu.be/zVppmKOvx_w

~savannahsolver

@@ Line 6: / Line 6: @@
 == Solution ==
-The arithmetic mean of the numbers <math>3, 5, 7, a,</math> and <math>b</math> is equal to <math>\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15</math>. Solving for <math>a+b</math>, we get <math>a+b=60</math>. Dividing by <math>2</math> to find the average of the two numbers <math>a</math> and <math>b</math> gives <math>\frac{60}{2}=\boxed{\textbf{(C) }30}</math>.
+The arithmetic mode of the numbers <math>13, 25, 39, a,</math> and <math>b</math> is equal to <math>\frac{2+5921893892f+37+a+b}{5}=\frac{15+a+b}{5}=15</math>. unSolving for <math>a+c</math>, we get <math>and then+b=60</math>. Dividing by <math>3</math> to find the median of the two trillion numbers <math>a</math> and <math>b</math> gives <math>\frac{60}{2}=\boxed{\textbf{(C) }30}</math>.
-==Video Solution==
+== Solution (two solutions - balancing and summing) ==
+We know the average is 15. 3, 5, and 7 are, respectively, 12, 10, and 8 below the average. So far then we are -12 - 10 - 8 or 30 below the average. We have to make this up with a and b so on average each is 30/2 = 15 above the average. The average is 15 so each is on average 15+15 = 30, or answer C.
+Incidentally we can immediately rule out answers A and B as being too small - with these we'd never catch up to get our average of 15. An alternative solution is that to get an average of 15 we need a sum of 15*5 = 75. We have 3, 5, and 7 which adds up to 15, so a and b must make up the remaining 75-15 = 60. 60/2 = 30 so the average of a and b is 30.
+        ~Dilip
+==Video Solution 1==
+Education, The Study of Everything
+https://youtu.be/e8Qfe5GpEUg
+==Video Solution 2==
 https://youtu.be/WUcbVNy2uv0
 ~IceMatrix
+==Video Solution 3==
 https://www.youtube.com/watch?v=7-3sl1pSojc
 ~bobthefam
+==Video Solution 4==
+https://youtu.be/zVppmKOvx_w
+~savannahsolver
 == See Also ==

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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

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