2021 AMC 10A Problems/Problem 4

Revision as of 18:09, 14 February 2021 by Timz2005 (talk | contribs) (Solution 2 (Answer Choices and Modular Arithmetic))

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

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

Solution 2 (Answer Choices and Modular Arithmetic)

From the $30$-term sum \[5+12+19+26+\cdots\] in the previous solution, 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 $5\mod{10}$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimate, $\textbf{(A)}$ is too small, leaving us with $\boxed{\textbf{(D)} ~3195}.$ ~MRENTHUSIASM

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

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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