7.2 Word Problems II: Age, Money, Percentage Change, and Interest

Applying Percentages to Everyday Problems

Chapter 6 covered how to compute a percentage; here we apply that skill to the money, age, and interest situations the CSC favors. Keep three master relationships ready: part = rate x whole; percentage change = (new - old) / old x 100; and for business items, profit or loss is measured against the stated base, which is almost always the cost.

Percentage Change, Discount, and Markup

A discount lowers a price; a markup raises it. A one-step multiplier beats doing two separate subtractions: an item at 25% off costs 75% of the original, so just multiply by 0.75.

Worked Example 1. A shirt priced at PHP 800 is 25% off. Sale price = 0.75 x 800 = PHP 600. The long way confirms it: 25% of 800 = 200, and 800 - 200 = 600.

Worked Example 2 (markup and profit). A vendor buys an item for PHP 250 and sells it for PHP 325. Profit = 325 - 250 = 75. Percentage profit on cost = 75 / 250 = 0.30 = 30%. Trap: dividing by the selling price 325 gives about 23%, the wrong base. Unless a problem says otherwise, profit and loss percentages use COST as the base.

Worked Example 3 (percentage increase). Enrollment rose from 250 to 300. Change = (300 - 250) / 250 = 50 / 250 = 1/5 = 20% increase.

Worked Example 4 (successive changes trap). A price rises 10% and then falls 10%. It does not return to the start: 100 -> 110 -> 99. A rise followed by an equal-percent fall always ends slightly lower, because the second percentage is taken on the larger amount.

Simple and Compound Interest

Simple interest uses I = P x R x T (principal x annual rate x years). The interest is identical every year because it is always based on the original principal.

Worked Example 5 (bank item). Invest PHP 10,000 at 5% simple interest for 3 years. I = 10,000 x 0.05 x 3 = 500 x 3 = PHP 1,500. The maturity value (principal plus interest) is 11,500.

Compound interest pays interest on the interest already earned: A = P x (1 + R) raised to the power T. With no calculator, expand year by year.

Worked Example 6 (compound, 2 years). PHP 10,000 at 5% compounded annually for 2 years. Year 1: 10,000 x 1.05 = 10,500. Year 2: 10,500 x 1.05 = 11,025. Interest = 1,025, which is PHP 25 more than the 1,000 simple interest over the same two years. That extra 25 is exactly 5% of the first year's 500 interest. A handy two-year shortcut is A = P(1 + 2R + R x R) = 10,000 x (1 + 0.10 + 0.0025) = 11,025.

Age Problems

Represent the present ages with a single variable and translate each clause carefully. The age gap between two people never changes, which is a fast way to check your work.

Worked Example 7. A father is 3 times as old as his son. In 10 years he will be twice as old. Let the son be x, so the father is 3x. In 10 years: 3x + 10 = 2(x + 10) -> 3x + 10 = 2x + 20 -> x = 10. The son is 10 and the father is 30. Check: in 10 years, 40 = 2 x 20. Correct.

Worked Example 8 (past tense). Ana is 30 now. Six years ago she was three times as old as her cousin was then. Ana six years ago was 24, so the cousin then was 24 / 3 = 8, making the cousin 14 now. Note the age gap of 16 years holds in every year.

Money and Coin Problems

Coin problems track both the number of coins and their total value. Value contributed = count x denomination.

Worked Example 9. A jar holds only PHP 1 and PHP 5 coins, 20 coins in all worth PHP 68. Let the number of five-peso coins be f, so the one-peso coins number 20 - f. Value equation: 5f + (20 - f) = 68 -> 4f + 20 = 68 -> 4f = 48 -> f = 12. There are 12 five-peso coins and 8 one-peso coins. Check: 60 + 8 = 68.

Worked Example 10 (budget split). A worker spends 1/3 of salary on rent and 1/4 on food, leaving PHP 5,000. The spent fraction is 1/3 + 1/4 = 4/12 + 3/12 = 7/12, so the remaining 5/12 equals 5,000, which makes the salary PHP 12,000. Rent is 4,000, food is 3,000, and 5,000 is left, matching the total.

Common Mistakes

The recurring errors are: using the selling price instead of cost as the base for profit; assuming compound interest equals simple interest (it only ties in year one, then pulls ahead); and setting up age equations without adding the SAME number of years to every person named. When a problem gives a future or past condition, adjust each person's age by the same amount before writing the equals sign.