Algebra Play · Lesson 8 of 10
Algebraic Story Puzzles
“Model repeated processes, animals, ages, and sharing puzzles with equations and context-aware checks.”
• Translate story conditions into variables and equations. • Solve repeated-process puzzles by choosing between forward and backward reasoning. • Model heads-and-legs, age, and sharing situations with equations. • Compare an algebraic solution with a non-algebraic reasoning method. • Check whether a numerical solution satisfies every condition in the original story.
Word problems become easier when we separate the story from the mathematical relationships hidden inside it. A variable names an unknown quantity; each sentence then becomes a condition that the variable must satisfy. The equation is not an extra step added to the story—it is a compact record of the story.
Magical Ponds and Equal Offerings
A person starts with x flowers. At each magical pond the number of flowers doubles, then the same number y of flowers is placed at the nearby shrine. After the third pond, all remaining flowers are placed at the third shrine. Because each shrine receives the same amount, every offering is y.
| Stage | Flowers remaining |
|---|---|
| Start | x |
| After pond 1 and shrine 1 | 2x − y |
| After pond 2 and shrine 2 | 4x − 3y |
| After pond 3 before shrine 3 | 8x − 6y |
The amount before the third offering must equal y, because all remaining flowers go to shrine 3. So 8x−6y=y, giving 8x=7y. The smallest positive whole-number solution is x=7 and y=8.
The equation does not determine one unique whole-number pair. If x=7k and y=8k for any positive whole number k, the same equal-offering pattern works. The pair 7 and 8 is the smallest positive solution; larger solutions such as 14 and 16 are scaled versions of the same structure.
Problem
Verify that starting with 7 flowers allows 8 flowers at each shrine.
- 1.Start with 7; the first pond doubles this to 14. Offer 8, leaving 6.
- 2.The second pond doubles 6 to 12. Offer 8, leaving 4.
- 3.The third pond doubles 4 to 8. Offer all 8.
- 4.Each shrine receives 8, so the conditions are satisfied.
Horses and Hens
Suppose there are h horses and n hens. Each animal contributes one head, so h+n=55. Horses have 4 legs and hens have 2, so 4h+2n=150.
Problem
Find the numbers of horses and hens.
- 1.Use h+n=55 and 4h+2n=150.
- 2.Double the heads equation: 2h+2n=110.
- 3.Subtract from the legs equation: 2h=40, so h=20.
- 4.Then n=55−20=35.
- 5.Check: 20+35=55 heads and 80+70=150 legs.
There is also a clever non-algebraic route. Pretend all 55 animals are hens: that would give 110 legs. The actual total has 40 extra legs. Replacing one hen by one horse adds 2 legs, so 40÷2=20 animals must be horses. This method and the equation method express the same difference idea.
Age Relationships
Age statements involve quantities that change together over time. If a daughter is d years old now and her mother is five times as old, the mother's age is 5d. Six years later both ages increase by 6, not by the same multiplicative factor.
Problem
A mother is 5 times her daughter's age. In 6 years she will be 3 times her daughter's age. Find the daughter's current age.
- 1.Let the daughter's current age be d; the mother's is 5d.
- 2.In 6 years their ages are d+6 and 5d+6.
- 3.Use 5d+6=3(d+6).
- 4.Then 5d+6=3d+18, so 2d=12 and d=6.
- 5.The mother is 30 now. In 6 years they will be 12 and 36, and 36=3×12.
When time passes, add the same number of years to both people. Do not multiply both present ages by the same factor.
Sharing Cows
Gauri says Naina has twice as many cows as she does. If Gauri has g, Naina has 2g. Naina then says that giving 3 cows to Gauri would make the numbers equal. The transfer changes both holdings but keeps the total number of cows unchanged.
Problem
Find how many cows each friend has.
- 1.Let Gauri have g cows. Naina has 2g.
- 2.After Naina gives 3 cows, Gauri has g+3 and Naina has 2g−3.
- 3.Set them equal: g+3=2g−3.
- 4.So g=6 and Naina has 12.
- 5.After the transfer, both have 9.
Choosing Forward or Backward Reasoning
Repeated processes are often easier to undo from the end. Ordinary relationship problems are often easier to write as simultaneous or single-variable equations. Good algebra is partly about choosing a representation that reduces unnecessary work.
- Name the unknown quantity or quantities.
- Translate each sentence into a mathematical relationship.
- Look for a relationship that can eliminate one unknown or express one variable in terms of another.
- Solve the resulting equation carefully.
- Return to the story and check every condition, including whole-number or positivity requirements.
Solve the magical-pond problem backwards from the final offering of 8: before the third pond there must have been 4, before the second offering 12, and so on. Compare this reasoning with the equation 8x=7y.
Quiz
In the horses-and-hens problem, which equation represents heads?
If all 55 animals were hens, how many legs would there be?
If a daughter is d now, what is her mother's age now when the mother is five times as old?
Six years later, what is the mother's age?
If Gauri has g cows and Naina has twice as many, which equation represents equality after Naina gives 3 cows to Gauri?
Practice Problems
- Solve the magical-pond puzzle and explain why 7 starting flowers and 8 flowers per shrine is the smallest positive whole-number solution.
- A farm has 40 animals, all goats or chickens, and 112 legs. Find the number of each.
- A father is 4 times his son's age. In 8 years he will be twice his son's age. Find their current ages.
- Riya has three times as many stickers as Aman. If Riya gives Aman 10 stickers, they have equal numbers. Find their original amounts.
- For the horses-and-hens problem, explain the 'pretend all are hens' method and connect it to subtracting equations.
- Write your own story problem that can be represented by one linear equation, solve it, and verify the answer in context.
From 8x=7y, the smallest positive integer pair is x=7,y=8. Direct simulation confirms equal offerings of 8.
Key Takeaways
• A story equation is a compact translation of the relationships stated in words. • Repeated doubling-and-subtracting processes can often be solved either algebraically or backwards. • Heads-and-legs problems can be solved by equations or by comparing with an all-one-animal scenario. • Age problems add the same number of years to each person's current age. • A solution is complete only after it is checked against every condition in the original situation.