2014 AMC 10B Problems/Problem 11


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

Video Solution



See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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 AMC 10 Problems and Solutions

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