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Difference between revisions of "2001 AMC 12 Problems/Problem 1"

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numbers?
 
numbers?
  
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<math>\text{(A)}\ 2S + 3\qquad \text{(B)}\ 3S + 2\qquad \text{(C)}\ 3S + 6 \qquad\text{(D)}\ 2S + 6 \qquad \text{(E)}\ 2S + 12</math>
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<math>\textbf{(A)}\ 2S + 3\qquad \textbf{(B)}\ 3S + 2\qquad \textbf{(C)}\ 3S + 6 \qquad\textbf{(D)}\ 2S + 6 \qquad \textbf{(E)}\ 2S + 12</math>
  
 
== Solution ==
 
== Solution ==
 
Suppose the two numbers are <math>a</math> and <math>b</math>, with <math>a+b=S</math>.
 
Suppose the two numbers are <math>a</math> and <math>b</math>, with <math>a+b=S</math>.
 
Then the desired sum is
 
Then the desired sum is
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<math>2(a+3)+2(b+3)=2(a+b)+12=2S +12</math>, which is answer <math>\boxed{\text{(E)}}</math>.
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<math>2(a+3)+2(b+3)=2(a+b)+12=2S +12</math>, which is answer <math>\boxed{\textbf{(E)}}</math>.
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==Video Solution by Daily Dose of Math==
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https://youtu.be/FxFb_QALttI?si=qUQVUkuBeK1cRbtS
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~Thesmartgreekmathdude
  
 
== See also ==
 
== See also ==
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{{AMC10 box|year=2001|num-b=2|num-a=4}}
 
{{AMC10 box|year=2001|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Algebra Problems]]

Latest revision as of 16:12, 18 October 2025

The following problem is from both the 2001 AMC 12 #1 and 2001 AMC 10 #3, so both problems redirect to this page.

Problem

The sum of two numbers is $S$. Suppose $3$ is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

$\textbf{(A)}\ 2S + 3\qquad \textbf{(B)}\ 3S + 2\qquad \textbf{(C)}\ 3S + 6 \qquad\textbf{(D)}\ 2S + 6 \qquad \textbf{(E)}\ 2S + 12$

Solution

Suppose the two numbers are $a$ and $b$, with $a+b=S$. Then the desired sum is $2(a+3)+2(b+3)=2(a+b)+12=2S +12$, which is answer $\boxed{\textbf{(E)}}$.

Video Solution by Daily Dose of Math

https://youtu.be/FxFb_QALttI?si=qUQVUkuBeK1cRbtS

~Thesmartgreekmathdude

See also

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by
First question 		Followed by
Problem 2
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
2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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