Difference between revisions of "2013 AMC 8 Problems/Problem 4"

(Solution)
(Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
Since Judi's 7 friends had to pay <math>2.50 extra each to cover the total amount that Judi should have paid, we can multiply </math>2.50 by 7, which is equal to <math>17.50. We can now tell that </math>17.50 is the amount that each of her friends needed to pay ( such that Judi didn't forget her money). So, we multiply <math>17.50 by 8 and get </math>140 is the total amount of the meal, which is C
+
Since Judi's 7 friends had to pay <math>2.50 extra each to cover the total amount that Judi should have paid, we can multiply </math>2.50 by 7, which is equal to <math>17.50. We can now tell that </math>17.50 is the amount that each of her friends needed to pay, such that Judi didn't forget her money. So, we multiply <math>17.50 by 8 and get </math>140 is the total amount of the meal, which is C
  
 
==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 21:47, 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?

$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$

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 (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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png