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

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==Problem==
 
==Problem==
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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?
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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?
  
 
<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>
 
<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>
  
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==Solution==
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==Solution 1==
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First we need to find the bill for one person. Since 7 other friends payed <math>2.50 each, we get to know that Judy's part (and her friends's parts) is/are </math>17.50. To get the total bill, We multiply 17.50 by 8 to get the total bill which leaves with \boxed{\textbf{(C)} \ \text{\textdollar}140}$. ~avamarora
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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.
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==Solution 2==
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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>
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<cmath>\frac{m}{8}=17.5</cmath>
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<cmath>m=\boxed{\textbf{(C) } 140}.</cmath>
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~BananaBall00 ~joyoforigami (TOGFA mathematics) edits by ~megaboy6679
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==Video Solution==
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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 (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. AMC logo.png

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