Difference between revisions of "2002 AMC 8 Problems/Problem 14"

Revision as of 00:09, 9 July 2024

Problem

A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices. The total discount is

$\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$

Solution 1

Let's assume that each item is $100$ dollars. First we take off $30\%$ off of $100$ dollars. $100\cdot0.7=70$

Next, we take off the extra $20\%$ as asked by the problem. $70\cdot0.80=56$

So the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount. So the final answer is $44\%$ , which is answer choice $\boxed{(B) 44 \%}$ .

Solution 2

Let $x$ be the price of an item on sale. When the item is $30\%$ off, its new price is $(1-0.3)x=0.7x.$ When $20\%$ is taken off of that price, the item's final sales price is $(1-0.2)\cdot0.7x)=0.8\cdot0.7x=0.56x=(1-0.44)x.$ Therefore, the item was $44\%$ off, so the answer is

Video Solution

https://youtu.be/DUqszaQ01lM Soo, DRMS, NM

https://www.youtube.com/watch?v=UR2aLmJHoIs ~David

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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.

@@ Line 15: / Line 15: @@
 ==Solution 2==
-Let <math>x</math>
+Let <math>x</math> be the price of an item on sale. When the item is <math>30\%</math> off, its new price is
+<cmath>(1-0.3)x=0.7x.</cmath>
+When <math>20\%</math> is taken off of that price, the item's final sales price is
+<cmath>(1-0.2)\cdot0.7x)=0.8\cdot0.7x=0.56x=(1-0.44)x.</cmath>
+Therefore, the item was <math>44\%</math> off, so the answer is
 ==Video Solution==

Page

Toolbox

Search