Difference between revisions of "2023 AMC 10B Problems/Problem 5"

Revision as of 12:00, 16 November 2023

Problem

Maddy and Lara see a list of numbers written on a blackboard. Maddy adds $3$ to each number in the list and finds that the sum of her new numbers is $45$ . Lara multiplies each number in the list by $3$ and finds that the sum of her new numbers is also $45$ . How many numbers are written on the blackboard?

$\textbf{(A) }10\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution

Let there be $n$ numbers in the list of numbers, and let their sum be $S$ . Then we have the following

$S+3n=45$

$3S=45$

From the second equation, $S=15$ , so $15+3n=45$ . Solving, we find $n=\boxed{\textbf{(A) }10}.$

~Mintylemon66

Solution 2

Let $x_1,x_2,x_3,...,x_n$ where $x_n$ represents the $n$ th number written on the board. Lara's multiplied each number by $3$ , so her sum will be $3x_1+3x_2+3x_3+...+3x_n$ . This is the same as $3\cdot (x_1+x_2+x_3+...+x_n)$ . We are given this quantity is equal to $45$ , so the original numbers add to $\frac{45}{3}=15$ . Maddy adds $3$ to each of the $n$ terms which yields, $x_1+3+x_2+3+x_3+3+...+x_n+3$ . This is the same as the sum of the original series plus $3 \cdot n$ . Setting this equal to $45$ , $15+3n=45 \Rightarrow n =\boxed{\textbf{(A) }10}.$

~vsinghminhas

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=SUnhwbA5_So

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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

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

@@ Line 7: / Line 7: @@
 ==Solution==
-Let there be <math>x</math> numbers in the list of numbers, and let their sum be <math>S</math>. Then we have the following
+Let there be <math>n</math> numbers in the list of numbers, and let their sum be <math>S</math>. Then we have the following
-<math>S+3x=45</math>
+<math>S+3n=45</math>
 <math>3S=45</math>
-From the second equation, <math>S=15</math>, so <math>15+3x=45</math>. Solving, we find <math>x=\boxed{\textbf{(A) }10}.</math>
+From the second equation, <math>S=15</math>, so <math>15+3n=45</math>. Solving, we find <math>n=\boxed{\textbf{(A) }10}.</math>

Page

Toolbox

Search