Difference between revisions of "2007 AMC 10A Problems/Problem 4"

(solution)
 
(Problem: The previous problem was the actual problem)
(4 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
The larger of two consecutive odd integers is three times the smaller. What is their sum?
 
The larger of two consecutive odd integers is three times the smaller. What is their sum?
 
<math>\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20</math>
 
  
 
== Solution ==
 
== Solution ==
Let the two consecutive odd integers be <math>a</math>, <math>a+2</math>. Then <math>a+2 = 3a</math>, so <math>a = 1</math> and their sum is <math>4\ \mathrm{(B)}</math>.
+
Let the two consecutive odd integers be <math>a</math>, <math>a+2</math>. Then <math>a+2 = 3a</math>, so <math>a = 1</math> and their sum is <math>4\ \mathrm{(A)}</math>.
  
 
== See also ==
 
== See also ==
{{AMC10 box|year=2007|ab=A|num-b=3|num-a=5}
+
{{AMC10 box|year=2007|ab=A|num-b=3|num-a=5}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 12:11, 30 December 2020

Problem

The larger of two consecutive odd integers is three times the smaller. What is their sum?

Solution

Let the two consecutive odd integers be $a$, $a+2$. Then $a+2 = 3a$, so $a = 1$ and their sum is $4\ \mathrm{(A)}$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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. AMC logo.png