One-third of the sum of two numbers is 12, twice their difference is 12. Find the numbers.

Analyze the question

We have two unknown numbers, so we form two equations from the two statements.

Let the numbers be x and y (x > y).

• "One-third of the sum is 12": (x + y)/3 = 12
• "Twice their difference is 12": 2(x − y) = 12

Concept

This is a pair of simultaneous linear equations. Simplify each equation, then add them to eliminate one variable.

Solution

1. From (x + y)/3 = 12: x + y = 36 ... (i)
2. From 2(x − y) = 12: x − y = 6 ... (ii)

Add (i) and (ii): 2x = 42, so x = 21

Substitute into (i): 21 + y = 36, so y = 15

Check

• Sum = 21 + 15 = 36, and one-third of 36 = 12 ✓
• Difference = 21 − 15 = 6, and twice 6 = 12 ✓

The numbers are 21 and 15, which is option C.

Checking the other options (a quick way to confirm in the exam):

• A. 22 + 14 = 36 but the difference is 8, so twice is 16 ✗
• B. 20 + 16 = 36 but the difference is 4, so twice is 8 ✗
• D. 23 + 13 = 36 but the difference is 10, so twice is 20 ✗

Tip: All four options have the same sum (36), so the difference is what separates them. Just solve x − y = 6 and look for the pair with a difference of 6. That takes only a few seconds.