Difference between revisions of "2021 AMC 10A Problems/Problem 4"

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==Problem==
 
==Problem==
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Nam libero justo laoreet sit amet. Eget arcu dictum varius duis at consectetur. Amet venenatis urna cursus eget nunc. Feugiat nisl pretium fusce id velit. Quam elementum pulvinar etiam non quam lacus. Feugiat nisl pretium fusce id velit. Felis eget nunc lobortis mattis aliquam faucibus purus. Iaculis urna id volutpat lacus laoreet non curabitur gravida arcu. Et ultrices neque ornare aenean euismod elementum nisi quis eleifend. Tellus id interdum velit laoreet id donec ultrices. Nascetur ridiculus mus mauris vitae ultricies. Ut placerat orci nulla pellentesque dignissim enim sit. Varius vel pharetra vel turpis. Aliquam id diam maecenas ultricies mi eget mauris. At tempor commodo ullamcorper a lacus vestibulum sed arcu non. Nulla malesuada pellentesque elit eget gravida cum sociis. Auctor augue mauris augue neque. Risus ultricies tristique nulla aliquet enim
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A cart rolls down a hill, travelling <math>5</math> inches the first second and accelerating so that during each successive <math>1</math>-second time interval, it travels <math>7</math> inches more than during the previous <math>1</math>-second interval. The cart takes <math>30</math> seconds to reach the bottom of the hill. How far, in inches, does it travel?
== Solution ==  
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Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Nam libero justo laoreet sit amet. Eget arcu dictum varius duis at consectetur. Amet venenatis urna cursus eget nunc. Feugiat nisl pretium fusce id velit. Quam elementum pulvinar etiam non quam lacus. Feugiat nisl pretium fusce id velit. Felis eget nunc lobortis mattis aliquam faucibus purus. Iaculis urna id volutpat lacus laoreet non curabitur gravida arcu. Et ultrices neque ornare aenean euismod elementum nisi quis eleifend. Tellus id interdum velit laoreet id donec ultrices. Nascetur ridiculus mus mauris vitae ultricies. Ut placerat orci nulla pellentesque dignissim enim sit. Varius vel pharetra vel turpis. Aliquam id diam maecenas ultricies mi eget mauris. At tempor commodo ullamcorper a lacus vestibulum sed arcu non. Nulla malesuada pellentesque elit eget gravida cum sociis. Auctor augue mauris augue neque. Risus ultricies tristique nulla aliquet enim
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<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math>
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==Solution 1 (Arithmetic Series)==
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Since <cmath>\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},</cmath> we seek the sum <cmath>5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,</cmath> in which there are <math>30</math> terms.
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The last term is <math>5+7\cdot(30-1)=208.</math> Therefore, the requested sum is <cmath>5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\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: <cmath>\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.</cmath> ~MRENTHUSIASM
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==Solution 2 (Arithmetic Series)==
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The distance (in inches) traveled within each <math>1</math>-second interval is: <cmath>5,5+1(7),5+2(7), \dots , 5+29(7).</cmath>
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This is an arithmetic sequence so the total distance travelled, found by summing them up is:
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<cmath>\text{number of terms} \cdot \text{average of terms} = \text{number of terms} \cdot \dfrac{\text{first term}+\text{last term}}{2}.</cmath>
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Or, <cmath>30 \cdot \dfrac{5+5+29(7)}{2} = 15 \cdot 213 = \boxed{\textbf{(D)} ~3195}.</cmath>
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~BakedPotato66
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==Solution 3 (Answer Choices and Modular Arithmetic)==
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From the <math>30</math>-term sum <cmath>5+12+19+26+\cdots</cmath> in Solution 1, taking modulo <math>10</math> gives <cmath>5+12+19+26+\cdots \equiv 3\cdot(5+2+9+6+3+0+7+4+1+8) = 3\cdot45\equiv5 \pmod{10}.</cmath> The only answer choices congruent to <math>5</math> modulo <math>10</math> are <math>\textbf{(A)}</math> and <math>\textbf{(D)}.</math> By a quick estimation, <math>\textbf{(A)}</math> is too small, leaving us with <math>\boxed{\textbf{(D)} ~3195}.</math>
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~MRENTHUSIASM
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==Solution 4 (Motion With Constant Acceleration)==
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This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases <math>7 \ \text{in/s}</math> per second. So, the acceleration <math>a=7 \ \text{in/s\textsuperscript{2}}.</math> The average speed of the first second is <math>5 \ \text{in/s}.</math> We can know the initial velocity <math>v_0=5-0.5\cdot7=1.5.</math> The displacement at <math>t=30</math> is <cmath>s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{\textbf{(D)} ~3195}.</cmath>
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~Bran_Qin
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== Video Solution by OmegaLearn ==
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https://youtu.be/7NSfDCJFRUg
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~ pi_is_3.14
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==Video Solution (Simple and Quick)==
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https://youtu.be/qLDkSnxLvxM
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~ Education, the Study of Everything
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== Video Solution (Arithmetic Sequence but in a Different Way)==
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https://www.youtube.com/watch?v=sJa7uB-UoLc&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=4
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~ North America Math Contest Go Go Go
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==Video Solution==
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https://youtu.be/aO-GklwkBfI
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~savannahsolver
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==Video Solution by TheBeautyofMath==
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https://youtu.be/50CThrk3RcM?t=262
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~IceMatrix
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==Video Solution by The Learning Royal==
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https://youtu.be/slVBYmcDMOI
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==See Also==
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{{AMC10 box|year=2021|ab=A|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 06:59, 13 November 2023

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 \[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\] we seek the sum \[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\] in which there are $30$ terms.

The last term is $5+7\cdot(30-1)=208.$ Therefore, the requested sum is \[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{\textbf{(D)} ~3195}.\] 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: \[\mathrm{Sum}=\frac{\mathrm{First}+\mathrm{Last}}{2}\cdot\mathrm{Count}.\] ~MRENTHUSIASM

Solution 2 (Arithmetic Series)

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

Solution 3 (Answer Choices and Modular Arithmetic)

From the $30$-term sum \[5+12+19+26+\cdots\] in Solution 1, taking modulo $10$ gives \[5+12+19+26+\cdots \equiv 3\cdot(5+2+9+6+3+0+7+4+1+8) = 3\cdot45\equiv5 \pmod{10}.\] The only answer choices congruent to $5$ 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 4 (Motion With Constant Acceleration)

This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases $7 \ \text{in/s}$ per second. So, the acceleration $a=7 \ \text{in/s\textsuperscript{2}}.$ The average speed of the first second is $5 \ \text{in/s}.$ We can know the initial velocity $v_0=5-0.5\cdot7=1.5.$ The displacement at $t=30$ is \[s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{\textbf{(D)} ~3195}.\] ~Bran_Qin

Video Solution by OmegaLearn

https://youtu.be/7NSfDCJFRUg

~ pi_is_3.14

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

https://youtu.be/aO-GklwkBfI

~savannahsolver

Video Solution by TheBeautyofMath

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

~IceMatrix

Video Solution by The Learning Royal

https://youtu.be/slVBYmcDMOI

See Also

2021 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

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