Difference between revisions of "2020 AMC 10B Problems/Problem 6"

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So Megan drove <math>16061-15951=110</math> miles. Since this happened over <math>2</math> hours, she drove at <math>\frac{110}{2}=\boxed{\textbf{(B) }55}</math> mph. ~quacker88
 
So Megan drove <math>16061-15951=110</math> miles. Since this happened over <math>2</math> hours, she drove at <math>\frac{110}{2}=\boxed{\textbf{(B) }55}</math> mph. ~quacker88
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==Video Solution==
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https://youtu.be/OHR_6U686Qg
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~IceMatrix
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https://youtu.be/0bfIDBkfcZs
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 13:40, 29 June 2020

Problem

Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$-hour period?

$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70$

Solution

In order to get the smallest palindrome greater than $15951$, we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.

So we raise $9$ to the next largest value, $10$, but obviously, that's not how place value works, so we're in the $16000$s now. To keep this a palindrome, our number is now $16061$.

So Megan drove $16061-15951=110$ miles. Since this happened over $2$ hours, she drove at $\frac{110}{2}=\boxed{\textbf{(B) }55}$ mph. ~quacker88

Video Solution

https://youtu.be/OHR_6U686Qg

~IceMatrix

https://youtu.be/0bfIDBkfcZs

~savannahsolver

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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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