Skip to lesson content

Lesson 1 of 13

Finding the Unknown · Lesson 1 of 13

Unknown Weights and Balanced Scales

“Use balanced branches to discover unknown weights before writing equations.”

Learning Objectives

• Interpret balance, tilt, and the total weight of a hanging scale. • Find unknown weights from equal branches and smaller balances. • Remove equal weights from both sides without disturbing balance. • Separate a combined unknown weight into equal individual weights.

What a balance tells us

Imagine hanging two groups of objects from opposite ends of a light rod. If the rod is horizontal, the two groups weigh the same. If one end hangs lower, that group is heavier. In these puzzles we ignore the weight of rods and strings, and treat matching objects as having equal weights.

A balanced total of 4 can have 2 on each branch. A total of 7 split as 4 and 3 tilts towards the 4-unit branch because the weights are unequal.

A number written above a hanging scale tells you the total weight of everything beneath it. It is not the weight of a single branch. A balanced scale of total weight 16 has 8 on each branch. If another small rod hangs from one branch, that smaller rod also balances its own two groups. Work from the total towards the smaller groups, keeping these two kinds of information separate.

Definition
Unknown weight

A weight that has not yet been given. Matching objects in one puzzle represent the same unknown weight.

Total weight: 163 + 3 + unknownBranch total: 8One flowerBranch total: 8Equal branches share the total equally.
Total weight and equal branches— Half of 16 belongs to each branch; the left unknown fills the gap between 6 and 8.
One total, two unknowns

Problem
A balanced scale weighs 16 in total. The left branch has two leaves weighing 3 each and one blue object. The right has one flower.

  1. 1.Each branch weighs 16 ÷ 2 = 8. Therefore the flower weighs 8.
  2. 2.The leaves together weigh 3 + 3 = 6. The blue object must weigh 8 − 6 = 2.
  3. 3.Check: the left branch weighs 3 + 3 + 2 = 8, equal to the right; together they weigh 16.
Read a balance inside a balance

Problem
A hanging scale has total weight 8. One branch holds a book; the other holds a balanced smaller rod with one identical note-shaped object at each end.

  1. 1.The book and the complete smaller branch each weigh 8 ÷ 2 = 4.
  2. 2.The smaller branch shares its 4 equally between two matching objects, so each weighs 2.
  3. 3.Check: 2 + 2 = 4 balances the book of weight 4. The whole scale weighs 8.

Remove the same weight from both sides

Sometimes the same object appears on both branches. You can remove one matching object from each branch: the two sides lose equal weight, so the remaining groups still balance. This gives a simpler puzzle with the same answer. Remove equal amounts, rather than equal numbers of objects that may have different weights.

Known crosses and unknown rings

Problem
Four crosses weighing 4 each balance one cross and two equal rings.

  1. 1.The four crosses weigh 4 × 4 = 16. Remove one cross of weight 4 from each side.
  2. 2.Three crosses remain on the left, weighing 12. Two rings remain on the right.
  3. 3.Each ring weighs 12 ÷ 2 = 6. Check the original: 16 = 4 + 6 + 6.
A sack on each branch

Problem
Two identical sacks balance one identical sack, a 10 kg weight, and a 4 kg weight.

  1. 1.Remove one sack from each side. One sack now balances 10 kg + 4 kg.
  2. 2.The sack weighs 14 kg. Originally the left weighs 28 kg and the right weighs 14 + 10 + 4 = 28 kg.
  3. 3.The removal did not assume the sack’s weight in advance; it used only the fact that matching sacks weigh equally.

Find one item from a group

After equal removals, you may know the total weight of several identical items. Divide by the number of items to find one item. The group total and the weight of one item are different quantities. This distinction is the bridge from balance puzzles to equations involving a number multiplied by an unknown.

Many sacks, the same reasoning

Problem
90 identical sacks and 50 kg balance 60 identical sacks and 500 kg.

  1. 1.Remove 60 sacks from each side: 30 sacks + 50 kg balance 500 kg.
  2. 2.Remove 50 kg from each side: 30 sacks balance 450 kg.
  3. 3.One sack weighs 450 ÷ 30 = 15 kg. Check: 90 × 15 + 50 = 1400 kg; 60 × 15 + 500 = 1400 kg.
Common mistake

A top total must be shared between the branches. Also, removing one heavy object from one branch and one light object from the other does not preserve balance merely because you removed one object on each side.

Quiz

Quick check

A balanced hanging scale has total weight 24. What is the weight of each branch?

Quick check

Three breads weighing 2 each balance two identical eggs. What does one egg weigh?

Quick check

A watermelon of weight 10 balances an orange of weight 4 and two equal bananas. What does each banana weigh?

Quick check

Five sacks balance two matching sacks and 21 kg. What does each sack weigh?

Quick check

Which action certainly keeps an already balanced scale balanced?

Practice Problems

Practice Problems
  1. A total of 24 is split equally. The left branch has two stars weighing 2 each and two equal blue fish; the right has one grey fish. Find both fish weights.
  2. A scale of total 18 has a sun weighing 5 and four equal clouds on the left, and three equal lightning shapes on the right. Find a cloud and a lightning shape.
  3. A scale of total 40 has one crown and five equal hexagons on the left and four matching crowns on the right. Find both unknown weights.
  4. A sack and 2 kg balance 10 kg and 2 kg. Explain a removal that finds the sack’s weight.
  5. Five sacks balance two matching sacks, two 10 kg weights, and 1 kg. Find one sack and check the original balance.
  6. Draw a balance puzzle of your own. Give enough information to find one unknown, and explain your reasoning.

Key Takeaways

Key Takeaways

• Horizontal branches represent equal weights. • A top total includes both branches. • Matching objects in a puzzle have equal weights. • Removing equal weights from both sides preserves balance. • Divide the remaining group weight by the number of matching objects to find one.