Difference between revisions of "2021 AMC 10A Problems/Problem 4"
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==Solution 1 (Arithmetic Series)== | ==Solution 1 (Arithmetic Series)== | ||
− | Since <cmath>\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},</cmath> we seek the sum <cmath>5 | + | 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. |
− | + | 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 | |
− | |||
− | ~ | + | ==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 | ||
− | ==Solution 3== | + | ==Solution 3 (Answer Choices and Modular Arithmetic)== |
− | The | + | 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 | ||
− | ~ | + | ~ pi_is_3.14 |
==Video Solution (Simple and Quick)== | ==Video Solution (Simple and Quick)== | ||
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~ Education, the Study of Everything | ~ Education, the Study of Everything | ||
− | == Video Solution (Arithmetic Sequence but in a | + | == Video Solution (Arithmetic Sequence but in a Different Way)== |
https://www.youtube.com/watch?v=sJa7uB-UoLc&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=4 | https://www.youtube.com/watch?v=sJa7uB-UoLc&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=4 | ||
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~ North America Math Contest Go Go Go | ~ North America Math Contest Go Go Go | ||
− | == Video Solution | + | ==Video Solution== |
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https://youtu.be/aO-GklwkBfI | https://youtu.be/aO-GklwkBfI | ||
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~IceMatrix | ~IceMatrix | ||
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+ | ==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 07:59, 13 November 2023
Contents
- 1 Problem
- 2 Solution 1 (Arithmetic Series)
- 3 Solution 2 (Arithmetic Series)
- 4 Solution 3 (Answer Choices and Modular Arithmetic)
- 5 Solution 4 (Motion With Constant Acceleration)
- 6 Video Solution by OmegaLearn
- 7 Video Solution (Simple and Quick)
- 8 Video Solution (Arithmetic Sequence but in a Different Way)
- 9 Video Solution
- 10 Video Solution by TheBeautyofMath
- 11 Video Solution by The Learning Royal
- 12 See Also
Problem
A cart rolls down a hill, travelling inches the first second and accelerating so that during each successive
-second time interval, it travels
inches more than during the previous
-second interval. The cart takes
seconds to reach the bottom of the hill. How far, in inches, does it travel?
Solution 1 (Arithmetic Series)
Since we seek the sum
in which there are
terms.
The last term is Therefore, the requested sum is
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:
~MRENTHUSIASM
Solution 2 (Arithmetic Series)
The distance (in inches) traveled within each -second interval is:
This is an arithmetic sequence so the total distance travelled, found by summing them up is:
Or,
~BakedPotato66
Solution 3 (Answer Choices and Modular Arithmetic)
From the -term sum
in Solution 1, taking modulo
gives
The only answer choices congruent to
modulo
are
and
By a quick estimation,
is too small, leaving us with
~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 per second. So, the acceleration
The average speed of the first second is
We can know the initial velocity
The displacement at
is
~Bran_Qin
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution (Simple and Quick)
~ 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
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/50CThrk3RcM?t=262
~IceMatrix
Video Solution by The Learning Royal
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.