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Tommy spent six lucky days in Las Vegas. On his first day, he won a net amount of $20, and on each of the following days, the daily net amount he won grew by $$\d$$. If Tommy won a total net amount of$1,620 during his stay in Las Vegas, how much did he win on the last day?
**Alternative Solution**: Instead, you could solve for $$d$$ by writing out the amount Tommy won on each day and then writing an equation where they add up to $1,620. - 1st day: $$\20$$ - 2nd day: $$\20 + \d$$ - 3rd day: $$\20 + \2d$$ - 4th day: $$\20 + \3d$$ - 5th day: $$\20 + \4d$$ - 6th day: $$\20 + \5d$$ Adding those up and setting them equal to$1,620 gives us > $$\20\cdot 6 + \15d = \1{,}620$$ > $$\120 + \15d = \1{,}620$$ > $$\15d = \1{,}500$$ > $$d = 100$$. That means that on the sixth day, Tommy made $520. Incorrect. [[Snippet]] Incorrect. [[Snippet]] Incorrect. [[Snippet]] Incorrect. [[Snippet]] Correct. [[Snippet]] Since the overall average amount he made is the average of the first and last days, it is > $$\displaystyle \text{Average} = \frac{x+\20}{2}$$. Putting that into the formula for finding the average, you get > $$\displaystyle \frac{x+\20}{2} = \frac{\ 1{,}620}{6}$$ > $$\displaystyle \frac{x+\20}{2} = \ 270$$ > $$\displaystyle x+\20 = \ 540$$ > $$\displaystyle x = \ 520$$. Hence, this is the correct answer.$330
$500$520
$540$620