Geometry Puzzle – Challenge 71

Challenge:

The total weight of a box and the candies it contains is 12 pounds. After \(\frac{2}{3}\) of the candies are eaten, the box and the remaining candies weigh 5 pounds. What is the weight of the empty box in pounds?

A- \(\frac{2}{3}\)

B- 1

C- \(\frac{3}{2}\)

D- \(2\frac{1}{2}\)

E- 5

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Let “B” be the weight of the box and “C” be the weight of the candies. So:
B + C = 12
After \(\frac{2}{3}\) of the candies are eaten, the box and the remaining candies weigh 5 pounds. So,
B + \(\frac{1}{3}\) C = 5
Solve the system of two equations:
B + C = 12 → B = 12, C
B + \(\frac{1}{3}\) C = 5 → 12, C + \(\frac{1}{3}\) C = 5 → 12, \(\frac{2}{3}\) C = 5 → C = 10.5
B = 12, C → B = 12 to 10.5 = 1.5
The weight of the box is 1.5 pounds.

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Frequently Asked Questions

How do you set up a system of equations from a word problem?

Name each unknown with a letter, then turn each fact in the problem into one equation. In this puzzle, B is the box’s weight and C is the candy’s weight. “Together they weigh 12 pounds” becomes B + C = 12, and “after 2/3 of the candy is eaten they weigh 5 pounds” becomes B + (1/3)C = 5. Two facts, two unknowns, two equations.

How do you solve system of equations word problems?

After writing the two equations, eliminate one variable. Here, subtracting B + (1/3)C = 5 from B + C = 12 cancels B and leaves (2/3)C = 7, so C = 10.5 pounds of candy. Substitute back: B = 12 − 10.5 = 1.5 pounds. Finish by checking: 1.5 + 3.5 = 5, which matches the second fact.

How do you solve system of equations word problems using substitution?

Solve one equation for a variable and plug it into the other. From B + C = 12, write B = 12 − C. Substitute into B + (1/3)C = 5 to get 12 − C + (1/3)C = 5, so 12 − (2/3)C = 5 and (2/3)C = 7. That gives C = 10.5, and the empty box weighs 12 − 10.5 = 1.5 pounds.

How do you write a system of equations from a word problem?

Look for two separate statements that each describe the same unknowns. Key phrases help: “total” or “together” means add, “of” with a fraction means multiply. In the candy puzzle, “2/3 of the candies are eaten” means 1/3 of C is left, so the second equation uses (1/3)C, not (2/3)C. Mixing those up is the most common mistake here.

What are some examples of system of equations word problems?

Classic types include mixture, ticket-price, age, coin and weight problems. A ticket example: adult tickets cost $8, child tickets $5, 20 tickets sold for $130. Then a + c = 20 and 8a + 5c = 130, giving a = 10 and c = 10. The box-and-candy puzzle is a weight version, where one quantity (the box) stays fixed and the other (the candy) changes.

Related Lessons You May Like

• How to find the area of rectangles
• How to find the perimeter of rectangles
• How to find the area of triangles
• How to solve systems of equations
• How to solve multi-step word problems

If your student enjoys these puzzles, Geometry for Beginners works the same relationships inside a full curriculum. Pre-Algebra for Beginners covers the algebra foundations.