Difference between revisions of "2013 AMC 8 Problems/Problem 4"
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Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill? | Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill? | ||
| − | <math>\textbf{(A)}\ | + | <math> \textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160 </math> |
==Solution== | ==Solution== | ||
| − | Judi's | + | Since Judi's 7 friends had to pay 2.50 extra each, to cover the total amount that Judi should have paid, we can multiply 2.50 by 7, which is equal to 17.50. We can now tell that 17.50 is the amount that each of her friends needed to pay, such that Judi didn't forget her money. So, we multiply 17.50 by 8 and get 140 is the total amount of the meal, {which is C.} |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=3|num-a=5}} | {{AMC8 box|year=2013|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:56, 3 November 2020
Problem
Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
Solution
Since Judi's 7 friends had to pay 2.50 extra each, to cover the total amount that Judi should have paid, we can multiply 2.50 by 7, which is equal to 17.50. We can now tell that 17.50 is the amount that each of her friends needed to pay, such that Judi didn't forget her money. So, we multiply 17.50 by 8 and get 140 is the total amount of the meal, {which is C.}
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
