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

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<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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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 which is, (C) 140, the total amount of the meal.
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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.
  
 
==Solution 2==
 
==Solution 2==
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Say <math>m</math> is the total amount needed to be paid. At the beginning, each person was supposed to pay <math>\$\frac{m}{8}</math>. But since Judi forgot her money, we have to add <math>\$\frac{m}{8}</math> to <math>\$2.50</math> to get the amount of money each of the <math>7</math> friends have to pay. Multiply <math>\$\frac{m}{8} + 2.50</math> by <math>7</math> to get <math>\$\frac{7m}{8} + 140</math>. We can set that equal to <math>m</math> because after we multiply by <math>7</math> we have the total amount needed to be paid. Solving for <math>m</math> we have our answer, <math>\boxed{\text{(C) } 140}</math>.
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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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~BananaBall00
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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
  
 
==Video Solution==
 
==Video Solution==

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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