Problem 2 · AMC 8 Stretch
Stretch
Algebra & Patterns
Fractions, Decimals & Percents
work-backwardasking-key-questions
Two numbers add up to \(12\), and when you multiply them you get \(4\). Without finding the two numbers, find the sum of their reciprocals (one over each number added together).
Show answer
Answer: 3
Show hints
Hint 1 of 3
You do NOT need to find the two numbers. Give them friendly names: call them \(x\) and \(y\). You are told \(x+y=12\) and \(x\cdot y=4\).
Still stuck? Show hint 2 →
Hint 2 of 3
You want \(\frac{1}{x}+\frac{1}{y}\). To add two fractions, put them over a common denominator. What single fraction do you get?
Still stuck? Show hint 3 →
Hint 3 of 3
Adding gives \(\frac{1}{x}+\frac{1}{y} = \frac{y}{xy}+\frac{x}{xy} = \frac{x+y}{xy}\). Now plug in the two numbers you already know.
Show solution
Approach: Asking key questions — combine the reciprocals into sum-over-product
- Call the numbers \(x\) and \(y\). We are told the sum \(x+y=12\) and the product \(xy=4\).
- We want \(\frac{1}{x}+\frac{1}{y}\). Adding fractions means a common denominator, which is \(xy\): \(\frac{1}{x}+\frac{1}{y} = \frac{y}{xy}+\frac{x}{xy} = \frac{x+y}{xy}\).
- The top is the SUM and the bottom is the PRODUCT, both of which we already know, so \(\frac{1}{x}+\frac{1}{y} = \frac{12}{4} = 3\). We never had to find the two messy numbers themselves.
Mark:
· log in to save