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2021 AMC 10A Problems/Problem 4: Difference between revisions

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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>
<math>\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242</math>


==Solution 1==
==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 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>\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.


==Solution 2 (Answer Choices and Modular Arithmetic)==
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
From the <math>30</math>-term sum <cmath>5+12+19+26+\cdots</cmath> in the previous solution, taking modulo <math>10</math> gives <cmath>5+12+19+26+\cdots \equiv 3(0+1+2+3+4+5+6+7+8+9) = 3(45)\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 (Arithmetic Sequence but in a different way)==
==Solution 2 (Arithmetic Series)==
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>
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>
Or, <cmath>30 \cdot \dfrac{5+5+29(7)}{2} = 15 \cdot 213 = \boxed{\textbf{(D)} ~3195}.</cmath>
~BakedPotato66


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


~ North America Math Contest Go Go Go
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> (Because of the formula for the distance traveled in the <math>k</math>th one-second interval being <math>\Delta x_k = v_0 + a\left(k - \tfrac{1}{2}\right)</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 (Using Arithmetic Sequence) ==
== Video Solution by OmegaLearn ==
https://youtu.be/7NSfDCJFRUg
https://youtu.be/7NSfDCJFRUg


Line 24: Line 34:
https://youtu.be/qLDkSnxLvxM
https://youtu.be/qLDkSnxLvxM


Education, the Study of Everything
~ 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 12:15, 1 November 2025

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 $v_0=5-0.5\cdot7=1.5.$ (Because of the formula for the distance traveled in the $k$th one-second interval being $\Delta x_k = v_0 + a\left(k - \tfrac{1}{2}\right)$). 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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All AMC 10 Problems and Solutions

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