Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 10A Problems/Problem 4"
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
Since <cmath>\text{Distance}=\text{Speed}\times\text{Time},</cmath> we seek the sum <cmath>5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,</cmath> in which there are 30 addends. The last addend is <math>5+7(30-1)=208.</math> Therefore, the requested sum is <cmath>5+12+19+26+\cdots+208=\frac{(5+208)(30)}{2}=\boxed{\textbf{(D)} ~3195}.</cmath> Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely <cmath>\frac{\text{First}+\text{Last}}{2}\cdot\text{Count}.</cmath> ~MRENTHUSIASM
 
Since <cmath>\text{Distance}=\text{Speed}\times\text{Time},</cmath> we seek the sum <cmath>5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,</cmath> in which there are 30 addends. The last addend is <math>5+7(30-1)=208.</math> Therefore, the requested sum is <cmath>5+12+19+26+\cdots+208=\frac{(5+208)(30)}{2}=\boxed{\textbf{(D)} ~3195}.</cmath> Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely <cmath>\frac{\text{First}+\text{Last}}{2}\cdot\text{Count}.</cmath> ~MRENTHUSIASM
 +
 +
==Solution 2 (Answer Choices)==
 +
Since <cmath>\text{Distance}=\text{Speed}\times\text{Time},</cmath> we seek the sum <cmath>5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,</cmath> in which there are 30 addends. Taking modulo <math>10,</math> we get <cmath>5+12+19+26+\cdots \equiv 3(1+2+3+\cdots+10) = 3(55)\equiv5 \pmod{10}.</cmath> The only answer choices that are <math>5\mod{10}</math> are <math>\textbf{(A)}</math> and <math>\textbf{(D)}.</math> By a quick estimate, <math>\textbf{(A)}</math> is too small, leaving us with <math>\boxed{\textbf{(D)} ~3195}.</math> ~MRENTHUSIASM
  
 
== Video Solution (Using Arithmetic Sequence) ==
 
== Video Solution (Using Arithmetic Sequence) ==

Revision as of 08:18, 12 February 2021

Problem 4

A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$-second time interval, it travels $7$ inches more than during the previous $1$-second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?

$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$

Solution

Since \[\text{Distance}=\text{Speed}\times\text{Time},\] we seek the sum \[5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,\] in which there are 30 addends. The last addend is $5+7(30-1)=208.$ Therefore, the requested sum is \[5+12+19+26+\cdots+208=\frac{(5+208)(30)}{2}=\boxed{\textbf{(D)} ~3195}.\] Recall that to find the sum of an arithmetic series, we take the average of the first and last terms, then multiply by the number of terms, namely \[\frac{\text{First}+\text{Last}}{2}\cdot\text{Count}.\] ~MRENTHUSIASM

Solution 2 (Answer Choices)

Since \[\text{Distance}=\text{Speed}\times\text{Time},\] we seek the sum \[5(1)+12(1)+19(1)+26(1)+\cdots=5+12+19+26+\cdots,\] in which there are 30 addends. Taking modulo $10,$ we get \[5+12+19+26+\cdots \equiv 3(1+2+3+\cdots+10) = 3(55)\equiv5 \pmod{10}.\] The only answer choices that are $5\mod{10}$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimate, $\textbf{(A)}$ is too small, leaving us with $\boxed{\textbf{(D)} ~3195}.$ ~MRENTHUSIASM

Video Solution (Using Arithmetic Sequence)

https://youtu.be/7NSfDCJFRUg

~ pi_is_3.14

See Also

2021 AMC 10A (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_4&oldid=145942"