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Difference between revisions of "2005 AMC 8 Problems/Problem 11"

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==Solution==
 
==Solution==
 
The price Jack rings up is <math>\textdollar{(90.00)(1.06)(0.80)}</math>. The price Jill rings up is <math>\textdollar{(90.00)(0.80)(1.06)}</math>. By the commutative property of multiplication, these quantities are the same, and the difference is <math>\boxed{\textbf{(C)}\ \textdollar 0}</math>.
 
The price Jack rings up is <math>\textdollar{(90.00)(1.06)(0.80)}</math>. The price Jill rings up is <math>\textdollar{(90.00)(0.80)(1.06)}</math>. By the commutative property of multiplication, these quantities are the same, and the difference is <math>\boxed{\textbf{(C)}\ \textdollar 0}</math>.
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Reminder: Commutative states abc = bac= acb =..... and is the same in any order. This only works for addition and multiplication.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=10|num-a=12}}
 
{{AMC8 box|year=2005|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:01, 6 June 2021

Problem

The sales tax rate in Bergville is 6%. During a sale at the Bergville Coat Closet, the price of a coat is discounted 20% from its $90.00 price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $90.00 and adds 6% sales tax, then subtracts 20% from this total. Jill rings up $90.00, subtracts 20% of the price, then adds 6% of the discounted price for sales tax. What is Jack's total minus Jill's total?

$\textbf{(A)}\ \textdollar -1.06\qquad\textbf{(B)}\ \textdollar -0.53 \qquad\textbf{(C)}\ \textdollar 0\qquad\textbf{(D)}\ \textdollar 0.53\qquad\textbf{(E)}\ \textdollar 1.06$

Solution

The price Jack rings up is $\textdollar{(90.00)(1.06)(0.80)}$. The price Jill rings up is $\textdollar{(90.00)(0.80)(1.06)}$. By the commutative property of multiplication, these quantities are the same, and the difference is $\boxed{\textbf{(C)}\ \textdollar 0}$.


Reminder: Commutative states abc = bac= acb =..... and is the same in any order. This only works for addition and multiplication.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

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