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# Distance Time | AIME I, 2012 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Distance Time. You may use sequential hints.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.

## Distance Time – AIME 2012

When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile.

• is 107
• is 279
• is 840
• cannot be determined from the given information

### Key Concepts

Time

Distance

Speed

AIME, 2012, Question 4

Problem Solving Strategies by Arther Engel

## Try with Hints

After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.

Then
$\frac{a}{6} + \frac{n-a}{4}$ = $\frac{n-a}{6} + \frac{2a}{5}$ implies that $a = \frac{5}{19}n$

Then integral value of n is 19 and a = 5 and $t = \frac{13}{3}$ hours that is 260 minutes. Then $19 + 260 = {279}$.