Printable

2012 AMC 10A Problem 6

2012 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Systems of Equations · difficulty 2 of 5 · about 2 min · position P1-10

The product of two positive numbers is $9$ . The reciprocal of one of these numbers is $4$ times the reciprocal of the other number. What is the sum of the two numbers?

A)
$\frac{10}{3}$
B)
$\frac{20}{3}$
C)
$7$
D)
$\frac{15}{2}$
E)
$8$

Answer

Key insight

The reciprocal condition 1/x = 4/y just says y = 4x; substitute into xy = 9 to get x = 3/2, y = 6.

Solution

Call the numbers $x$ and $y$ with $xy = 9$ and $\dfrac{1}{x} = 4 \cdot \dfrac{1}{y}$ .

Cross-multiplying the second equation gives $y = 4x$ . Substituting into the first:
$x \cdot 4x = 9 \quad\Longrightarrow\quad x^2 = \frac{9}{4} \quad\Longrightarrow\quad x = \frac{3}{2},$
taking the positive root. Then $y = 4x = 6$ .

Check: $\frac{3}{2} \cdot 6 = 9$ and $\frac{1}{3/2} = \frac{2}{3} = 4 \cdot \frac{1}{6}$ . The sum is $\frac{3}{2} + 6 = \frac{15}{2}$ .

The answer is $\boxed{\textbf{(D)}\ \frac{15}{2}}$ .

Why this works

A statement about reciprocals is a statement about the numbers themselves: $\frac1x = \frac4y$ means $y$ is four times $x$ . Once one variable is a multiple of the other, the product equation becomes a single-variable quadratic. Always verify the final pair against both original conditions.

The trap

Turning 1/x = 4 * (1/y) into x = 4y (the wrong way round), which gives 6 and 3/2 swapped but also tempts the wrong sum for other setups.

Common mistakes

Turning 1/x = 4 * (1/y) into x = 4y (the wrong way round), which gives 6 and 3/2 swapped but also tempts the wrong sum for other setups.
Answering with $x$ or $y$ alone, or with the product, instead of the sum.

Techniques

Set up the equation/formula and compute; no special trick needed · Substitute to simplify (u = x+1/x, shifting, scaling)