Advanced Word Problems • Topic 4 of 5

Growth Problems

Growth problems apply a percentage change to a starting amount. A rise of r percent multiplies the old value by (1 + r/100); a fall multiplies by (1 − r/100). To go the other way — finding the percent change from old to new — divide the change by the original and multiply by 100: (new − old)/old × 100. Repeated growth compounds: doubling each period multiplies by 2 each time, so after n periods the factor is 2ⁿ. Identify the starting value, the rate, and whether you want the new amount or the percent change, then apply the matching form.

A quantity growing by a percent: old value to new valuenew = old × (1 + r)oldnew+ r%percent change = (change ÷ original) × 100

✅ Solved examples

1. A town of 2,000 grows 10% in a year. New population?
2000 × 1.10 = 2,200.
2. Sales rose from 80 to 100 units. Percent increase?
(100 − 80)/80 × 100 = 25%.
3. 500 bacteria double every hour. After 3 hours?
500 × 2³ = 500 × 8 = 4,000.
4. A $400 price rises 25%. New price?
400 × 1.25 = $500.

✏️ Practice — try these, take hints as needed

1. A city of 5,000 grows 20% in a year. New population?
old × (1 + r).
5000 × 1.20.
—
6,000.
2. A score rose from 50 to 60. Percent increase?
(60 − 50)/50 × 100.
10/50 × 100.
—
20%.
3. 300 cells double every hour. After 2 hours?
300 × 2².
300 × 4.
—
1,200.
4. A $200 item rises 50%. New price?
200 × 1.5.
—
—
$300.
5. Membership fell from 400 to 360. Percent decrease?
(400 − 360)/400 × 100.
40/400 × 100.
—
10%.

📝 Topic test — 8 questions

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