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

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==See Also==
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Revision as of 03:47, 12 February 2021

Problem 4

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

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

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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