Algebra Play · Lesson 9 of 10
Algebra in Money, Patterns, and Repeated Processes
“Apply algebra to profit models, odd-number identities, and repeated doubling-and-fee processes.”
• Build a simple profit model from fixed cost, variable cost, selling price, and quantity. • Solve for a required selling price or sales quantity from a target profit. • Explain a fraction pattern using the sum of consecutive odd numbers rather than repeated calculation. • Model repeated doubling-and-fee processes with a recurrence. • Work backwards through a repeated process and generalise the conditions for gaining or losing money.
Algebra is useful far beyond puzzles with hidden numbers. It can model business costs, explain numerical patterns, and track repeated changes. In each case the same habit matters: define the quantities, write the relationship, and interpret the result.
A Dosa Cart: Cost, Revenue, and Profit
Suppose the daily cart rent is ₹5000 and making one dosa costs ₹10. If n dosas are made and sold, the total daily cost is the fixed ₹5000 plus ₹10 for each dosa.
If each dosa sells for ₹p, revenue is pn. Profit is revenue minus cost.
Problem
If 100 dosas are sold, what price per dosa gives a ₹2000 profit?
- 1.For n=100, total cost is 5000+10(100)=6000.
- 2.To make ₹2000 profit, revenue must be 6000+2000=8000.
- 3.With 100 dosas, price = 8000÷100 = ₹80.
- 4.Check: revenue ₹8000 minus cost ₹6000 equals ₹2000.
Problem
Customers will pay ₹50 per dosa. How many dosas must be sold for a ₹2000 profit?
- 1.Profit per dosa before fixed rent is 50−10=₹40.
- 2.Use 40n−5000=2000.
- 3.Then 40n=7000, so n=175.
- 4.Check: revenue 175×50=₹8750; cost 5000+1750=₹6750; profit ₹2000.
Do not count the ₹5000 rent once per dosa. It is a fixed daily cost. The ₹10 ingredient-and-fuel amount is the variable cost that scales with n.
A Pattern Built from Odd Numbers
Consider 1/3, (1+3)/(5+7), and (1+3+5)/(7+9+11). Each fraction equals 1/3. Instead of checking one case at a time, use the fact that the sum of the first n odd numbers is n².
For the nth fraction, the numerator is the sum of the first n odd numbers, so it is n². The denominator contains the next n odd numbers. The sum of the first 2n odd numbers is (2n)²=4n². Remove the first n odd numbers, whose sum is n², and the remaining denominator is 3n².
Problem
What is the next fraction and why does it still equal 1/3?
- 1.The numerator uses the first four odd numbers: 1+3+5+7=16.
- 2.The denominator uses the next four odd numbers: 9+11+13+15=48.
- 3.16/48=1/3.
- 4.The general n² and 3n² argument explains why this continues.
Karim and the Genie
Karim begins with x coins. Each round doubles his current coins, then he pays a fixed fee f. If A_k is the amount after paying at the end of round k, then A_k=2A_{k-1}−f.
In the story the fee is 8 coins. On the third round, after doubling, Karim has exactly 8 coins—the amount he must pay—so after the payment he has 0. Working backwards is especially efficient.
Problem
Karim pays 8 coins after each doubling and is reduced to 0 after the third payment. How many coins did he start with?
- 1.Just before the third payment he had 8; before the third doubling he therefore had 4.
- 2.Just before the second payment he must have had 4+8=12; before that doubling he had 6.
- 3.Just before the first payment he must have had 6+8=14; before the first doubling he had 7.
- 4.So Karim started with 7 coins.
When Does Karim Gain Money?
After one round Karim has 2x−f. This exceeds his starting amount x exactly when 2x−f>x, or f<x. In fact, after three rounds the amount is 8x−7f. Comparing with x again gives the same condition f<x. A fee smaller than the starting amount lets the repeated doubling outrun the repeated fixed charge.
If the genie wants Karim to have exactly 0 after three paid rounds, set 8x−7f=0, so f=8x/7. For whole coins, the starting amount must make this fee integral if fractional fees are not allowed.
Always specify whether an amount is measured before or after the fee is paid. In the story, 'only 8 coins, exactly the number owed' refers to the amount after the third doubling but before the final payment.
Quiz
What is the total cost of selling n dosas when rent is ₹5000 and variable cost is ₹10 each?
At a selling price of ₹50, what is the contribution per dosa toward rent and profit?
What is the sum of the first n odd numbers?
Why is the denominator in the nth odd-number fraction equal to 3n²?
If Karim starts with x and pays fee f after one doubling, how much remains?
Practice Problems
- If 120 dosas are sold and the target profit is ₹2200, find the required selling price per dosa.
- At a selling price of ₹60, how many dosas must be sold to make ₹3000 profit?
- Write the fifth fraction in the odd-number pattern and simplify it.
- Use the n² identity to prove the odd-number fraction is always 1/3.
- If Karim starts with 10 coins and pays 6 coins after each doubling, find his amount after three rounds.
- For a starting amount x, find the fee f that leaves Karim with exactly 0 after three paid rounds.
Cost is 5000+1200=6200. Required revenue is 8400, so price is 8400/120=₹70.
Key Takeaways
• Profit equals revenue minus both fixed and variable costs. • A general algebraic model can solve either for price or for sales quantity. • The odd-number fraction pattern follows from the identity that the first n odd numbers sum to n². • Repeated doubling with a fixed fee can be modelled by A_k=2A_{k-1}−f. • Working backwards is often the fastest way to solve a repeated-process puzzle with a known final state.