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

m
 
(28 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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?
+
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)}\ &#036;120 \qquad \textbf{(B)}\ &#036;128 \qquad \textbf{(C)}\ &#036;140 \qquad \textbf{(D)}\ &#036;144 \qquad \textbf{(E)}\ &#036;160</math>
+
<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 1==
Judi's seven friends each paid &#036;2.50 for her, so the total amount of money that was paid for her was <math>7\cdot &#036;2.50 = &#036;17.50</math>. This is Judi's portion of the bill; what she was supposed to pay.
+
Since Judi's 7 friends had to pay \$2.50 extra each to cover the total amount that Judi should have paid, we multiply <math>2.50\cdot7=17.50</math> is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply <math>17.50\cdot8=\boxed{\textbf{(C) } 140}</math> to find the total the 8 friends paid.
  
Judi was supposed to pay <math>\frac{1}{8}</math> of the total bill, so the total bill must be <math>8\cdot &#036;17.50 = \textbf{(C)}\ &#036;140</math>.
+
==Solution 2==
 +
Let <math>m</math> be the total bill the 8 friends split up; each person was supposed to pay <math>\$\frac{m}{8}</math>. Since Judi forgot her money, we have to add <math>\$\frac{m}{8}+2.50</math> to get the amount of money each of the <math>7</math> friends have to pay. After multipling by 7, we have <math>7(\frac{m}{8}+2.50)=\frac{7m}{8} + 17.5</math>. This is equal to <math>m</math> because both sides represent the total bill. Solving for <math>m</math>:<cmath>\frac{7m}{8} + 17.5=m</cmath>
 +
<cmath>\frac{m}{8}=17.5</cmath>
 +
<cmath>m=\boxed{\textbf{(C) } 140}.</cmath>
 +
 
 +
~BananaBall00 ~joyoforigami (TOGFA mathematics) edits by ~megaboy6679
 +
 
 +
==Video Solution==
 +
https://youtu.be/lvaWrFwW6BM ~savannahsolver
  
 
==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}}

Latest revision as of 22:57, 1 February 2023

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 1

Since Judi's 7 friends had to pay $2.50 extra each to cover the total amount that Judi should have paid, we multiply $2.50\cdot7=17.50$ is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply $17.50\cdot8=\boxed{\textbf{(C) } 140}$ to find the total the 8 friends paid.

Solution 2

Let $m$ be the total bill the 8 friends split up; each person was supposed to pay $$\frac{m}{8}$. Since Judi forgot her money, we have to add $$\frac{m}{8}+2.50$ to get the amount of money each of the $7$ friends have to pay. After multipling by 7, we have $7(\frac{m}{8}+2.50)=\frac{7m}{8} + 17.5$. This is equal to $m$ because both sides represent the total bill. Solving for $m$:\[\frac{7m}{8} + 17.5=m\] \[\frac{m}{8}=17.5\] \[m=\boxed{\textbf{(C) } 140}.\]

~BananaBall00 ~joyoforigami (TOGFA mathematics) edits by ~megaboy6679

Video Solution

https://youtu.be/lvaWrFwW6BM ~savannahsolver

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