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"

m (→‎Solution 2 (Answer Choices and Modular Arithmetic))
m
Line 5: Line 5:
  
 
==Solution 1 (Arithmetic Series)==
 
==Solution 1 (Arithmetic Series)==
−
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 <math>30</math> 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>\mathrm{Distance}=\mathrm{Speed}\times\mathrm{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 <math>30</math> 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{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.</cmath> ~MRENTHUSIASM
  
 
==Solution 2 (Answer Choices and Modular Arithmetic)==
 
==Solution 2 (Answer Choices and Modular Arithmetic)==

Revision as of 18:29, 6 April 2021

Problem

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 1 (Arithmetic Series)

Since \[\mathrm{Distance}=\mathrm{Speed}\times\mathrm{Time},\] 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 \[\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.\] ~MRENTHUSIASM

Solution 2 (Answer Choices and Modular Arithmetic)

From the $30$-term sum \[5+12+19+26+\cdots\] in Solution 1, taking modulo $10$ gives \[5+12+19+26+\cdots \equiv 3(0+1+2+3+4+5+6+7+8+9) = 3(45)\equiv5 \pmod{10}.\] The only answer choices that are $5$ modulo $10$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimation, $\textbf{(A)}$ is too small, leaving us with $\boxed{\textbf{(D)} ~3195}.$

~MRENTHUSIASM

Solution 3

The distance (in inches) traveled within each 1-second interval is:

$5,5+1(7),5+2(7), \dots , 5+29(7).$

This is an arithmetic sequence so the total distance travelled, found by summing them up is: $\text{number of terms} \cdot \text{average of terms} = \text{number of terms} \cdot \dfrac{\text{first term}+\text{last term}}{2}.$

Or,

$30 \cdot \dfrac{5+5+29(7)}{2} = 15 \cdot 213 = \boxed{\textbf{(D)} ~3195}.$

~BakedPotato66

Video Solution (Simple and Quick)

https://youtu.be/qLDkSnxLvxM

~ Education, the Study of Everything

Video Solution (Arithmetic Sequence but in a different way)

https://www.youtube.com/watch?v=sJa7uB-UoLc&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=4

~ North America Math Contest Go Go Go

Video Solution (Using Arithmetic Sequence)

https://youtu.be/7NSfDCJFRUg

~ pi_is_3.14

Video Solution 4

https://youtu.be/aO-GklwkBfI

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/50CThrk3RcM?t=262

~IceMatrix

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=151156"