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Difference between revisions of "2014 AMC 10B Problems/Problem 11"

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(3) a <math>25\%</math> discount followed by a <math>5\%</math> discount
 
(3) a <math>25\%</math> discount followed by a <math>5\%</math> discount
  
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What is the possible positive integer value of <math>n</math>?
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What is the smallest possible positive integer value of <math>n</math>?
  
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<math> \textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}}\ 31\qquad\textbf{(E)}\ 33 </math>
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<math> \textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33 </math>
  
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==Solution==
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==Solution 1==
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Let the price be <math>x</math>. Then, for option <math>1</math>, the discounted price is <math>(1-.15)(1-.15)x = .7225x</math>. For option <math>2</math>, the discounted price is <math>(1-.1)(1-.1)(1-.1)x = .729x</math>. Finally, for option <math>3</math>, the discounted price is <math>(1-.25)(1-.05) = .7125x</math>. Therefore, the discount must be greater than <math>\max(x - .7225x, x-.729x, x-.7125x)</math>. Thus, the discount must be greater than <math>.2875</math>. We multiply this by <math>100</math> to get the percent value, and then round up because <math>x</math> is the smallest integer that provides a greater discount than <math>28.75</math>, leaving us with the answer of <math>\boxed{\textbf{(C) } 29}</math>
+
Let the original price be <math>x</math>. Then, for option <math>1</math>, the discounted price is <math>(1-.15)(1-.15)x = .7225x</math>. For option <math>2</math>, the discounted price is <math>(1-.1)(1-.1)(1-.1)x = .729x</math>. Finally, for option <math>3</math>, the discounted price is <math>(1-.25)(1-.05) = .7125x</math>. Therefore, <math>n</math> must be greater than <math>\max(x - .7225x, x-.729x, x-.7125x)</math>. It follows <math>n/100</math> must be greater than <math>.2875</math>. We multiply this by <math>100</math> to get the percent value, and then round up because <math>n</math> is the smallest integer that provides a greater discount than <math>28.75</math>, leaving us with the answer of <math>\boxed{\textbf{(C) } 29}</math>
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 +
==Solution 2 (a bit easier)==
 +
Assume the original price was <math>100</math> dollars. Thus, after a discount of <math>n\%</math>, the price will be <math>100-n</math> dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: <math>100-n</math>, <math>\frac{289}{4}</math>, <math>\frac{9^3}{10}</math>, and <math>\frac{15\cdot19}{4}</math>. Simplify these to get <math>100-n</math>, <math>72</math> (rounded down), <math>72.9</math>, and <math>71</math> (rounded down). To have the greatest discount, we need the least price, which is <math>71</math> dollars. Now we get the original fractional value of this, and the discount of this is <math>100-\frac{285}{4}</math>, and we round this down to <math>28</math>. Now, it's pretty easy. The integer value greater than this is <math>\boxed{\textbf{(C) } 29}</math>
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 +
~solution by sakshamsethi
 +
 
 +
~edited by sosiaops
 +
 
 +
==Solution 3 (Straightforward)==
 +
 
 +
Assume WLOG that the original price was <math>\$100</math>. Then, option 1 would cost <math>100 \cdot \frac{17}{20} \cdot \frac{17}{20} = \$ 72.25</math>. Option 2 would cost <math>100 \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} = \$72.90</math>. Option 3 would cost <math>100 \cdot \frac{3}{4} \cdot \frac{19}{20} = \$71.25</math>. Therefore, the largest integer smaller than all of these three is <math>71</math>, so <math>100-71= \boxed{\textbf{(C) } 29}</math>.
 +
 
 +
~MrThinker
 +
 
 +
==Video Solution==
 +
https://youtu.be/TVk6fRdTFJU
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2014|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:03, 11 August 2023

Problem

For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:

(1) two successive $15\%$ discounts

(2) three successive $10\%$ discounts

(3) a $25\%$ discount followed by a $5\%$ discount

What is the smallest possible positive integer value of $n$?

$\textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33$

Solution 1

Let the original price be $x$. Then, for option $1$, the discounted price is $(1-.15)(1-.15)x = .7225x$. For option $2$, the discounted price is $(1-.1)(1-.1)(1-.1)x = .729x$. Finally, for option $3$, the discounted price is $(1-.25)(1-.05) = .7125x$. Therefore, $n$ must be greater than $\max(x - .7225x, x-.729x, x-.7125x)$. It follows $n/100$ must be greater than $.2875$. We multiply this by $100$ to get the percent value, and then round up because $n$ is the smallest integer that provides a greater discount than $28.75$, leaving us with the answer of $\boxed{\textbf{(C) } 29}$

Solution 2 (a bit easier)

Assume the original price was $100$ dollars. Thus, after a discount of $n\%$, the price will be $100-n$ dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: $100-n$, $\frac{289}{4}$, $\frac{9^3}{10}$, and $\frac{15\cdot19}{4}$. Simplify these to get $100-n$, $72$ (rounded down), $72.9$, and $71$ (rounded down). To have the greatest discount, we need the least price, which is $71$ dollars. Now we get the original fractional value of this, and the discount of this is $100-\frac{285}{4}$, and we round this down to $28$. Now, it's pretty easy. The integer value greater than this is $\boxed{\textbf{(C) } 29}$

~solution by sakshamsethi

~edited by sosiaops

Solution 3 (Straightforward)

Assume WLOG that the original price was $$100$. Then, option 1 would cost $100 \cdot \frac{17}{20} \cdot \frac{17}{20} = $ 72.25$. Option 2 would cost $100 \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} = $72.90$. Option 3 would cost $100 \cdot \frac{3}{4} \cdot \frac{19}{20} = $71.25$. Therefore, the largest integer smaller than all of these three is $71$, so $100-71= \boxed{\textbf{(C) } 29}$.

~MrThinker

Video Solution

https://youtu.be/TVk6fRdTFJU

~savannahsolver

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AMC 10 Problems and Solutions

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