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 – 10.5 = 1.5
The weight of the box is 1.5 pounds.

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Frequently Asked Questions

How do I set up the system of equations?

Use B for the box weight and C for the candy weight. The two facts become B + C = 12 (original) and B + (1/3)C = 5 (after eating 2/3).

Why is the remaining candy (1/3)C?

After eating 2/3 of the candy, 1 – 2/3 = 1/3 of the candy remains. So the remaining candy weighs (1/3)C.

How do I solve the system?

Subtract the second equation from the first: (B + C) – (B + (1/3)C) = 12 – 5, so C – (1/3)C = 7, which gives (2/3)C = 7. Then C = 7 / (2/3) = 21/2 = 10.5.

How is the box weight calculated?

From B + C = 12 with C = 10.5: B = 12 – 10.5 = 1.5 pounds.

Why does subtracting the equations eliminate B?

Because both equations have B with coefficient 1. Subtracting cancels B and leaves an equation in just C.

Can I check the answer?

Yes. B + C = 1.5 + 10.5 = 12 ✓. B + (1/3)C = 1.5 + 3.5 = 5 ✓. Both check out.

What if 3/4 of the candy were eaten instead?

Remaining candy = (1/4)C. Equation: B + (1/4)C = 5. With B + C = 12, subtract: (3/4)C = 7, so C = 28/3 ≈ 9.33 lbs, B ≈ 2.67 lbs.

Why is this a useful kind of problem?

It introduces systems of two linear equations in a friendly context. Mixture problems, age problems, and rate problems all use the same structure.

What grade level?

Late middle school or early high school (Grades 7-9). The substitution and elimination methods are both Algebra I staples.

Where does this kind of math appear in real life?

Recipe scaling, inventory accounting (what is the container weight vs the contents), cost analysis (fixed vs variable costs), and any setting that separates a constant amount from a varying one.

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.