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Difference between revisions of "2021 AMC 10A Problems/Problem 4"
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==Problem 4==
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==Problem==
 
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?
 
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?
  
 
<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math>
 
<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==
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==Solution 1 (Arithmetic Series)==
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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
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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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==Solution 2 (Answer Choices)==
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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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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
 
  
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== Video Solution (Using Arithmetic Sequence) ==
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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:
 +
<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
 +
 
 +
==Solution 3 (Answer Choices and Modular Arithmetic)==
 +
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>
 +
 
 +
~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 <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>
 +
~Bran_Qin
 +
 
 +
== Video Solution by OmegaLearn ==
 
https://youtu.be/7NSfDCJFRUg
 
https://youtu.be/7NSfDCJFRUg
  
 
~ pi_is_3.14
 
~ 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==
 
==See Also==
 
{{AMC10 box|year=2021|ab=A|num-b=3|num-a=5}}
 
{{AMC10 box|year=2021|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{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 (Problems • Answer Key • Resources)
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Problem 5
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