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Compound Interest (Using Formula) — Flashcards

ICSE · Class 9 · Mathematics

25 flashcards for Compound Interest (Using Formula) (ICSE Class 9 Mathematics) to test yourself on key terms and facts.

37 questions25 flashcards4 formulas & key relations5 concepts

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A labeled diagram explaining the components of the compound interest formula A = P(1 + r/100)^n when interest is compounded yearly.
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25 Flashcards·
Amount when interest is compounded yearlyCompound interest from amountFinding principalFinding principal from compound interestFinding rateFinding rate and next year amountFinding timeSuccessive yearly rates
Card 1Amount when interest is compounded yearly

Find the amount on ₹7,500 in 2 years at 6% compounded annually.

Answer

Use A = P(1+r/100)^n. Step 1: P = 7,500, r = 6, n = 2. Step 2: A = 7,500(1 + 6/100)^2 = 7,500(1.06)^2. Step 3: A = 7,500 × 1.1236 = ₹8,427. Answer: Amount = ₹8,427.

Card 2Compound interest from amount

Find the compound interest on ₹18,000 for 2 years at 15% per annum.

Answer

Use A = P(1+r/100)^n. Step 1: P = 18,000, r = 15, n = 2. Step 2: A = 18,000(1.15)^2 = 18,000 × 1.3225 = ₹23,805. Step 3: C.I. = A - P = 23,805 - 18,000 = ₹5,805. Answer: Compound interest = ₹5,805.

Card 3Finding principal

What sum will amount to ₹3,630 in 2 years at 10% compound interest per annum?

Answer

Use A = P(1+r/100)^n. Step 1: A = 3,630, r = 10, n = 2. Step 2: 3,630 = P(1.10)^2 = P × 1.21. Step 3: P = 3,630 / 1.21 = ₹3,000. Answer: Principal = ₹3,000.

Card 4Finding principal from compound interest

A sum earns compound interest of ₹164 in 2 years at 5% per annum. Find the principal.

Answer

Use C.I. = P[(1+r/100)^n - 1]. Step 1: C.I. = 164, r = 5, n = 2. Step 2: 164 = P[(1.05)^2 - 1] = P(1.1025 - 1). Step 3: 164 = P(0.1025). Step 4: P = 164 / 0.1025 = ₹1,600. Answer: Principal = ₹1,600.

Card 5Finding rate

Find the rate if ₹2,000 amounts to ₹2,662 in 3 years at compound interest.

Answer

Use A = P(1+r/100)^n. Step 1: A = 2,662, P = 2,000, n = 3. Step 2: 2,662/2,000 = (1+r/100)^3. Step 3: 1.331 = (1+r/100)^3 = (11/10)^3. Step 4: 1+r/100 = 11/10. Step 5: r = 10%. Answer: Rate = 10% per …

Card 6Finding rate and next year amount

A sum of ₹10,000 becomes ₹11,200 after 1 year at compound interest. Find the rate and the amount after the second year.

Answer

Step 1: Use A = P(1+r/100)^n for 1 year. 11,200 = 10,000(1+r/100). Step 2: 11,200/10,000 = 1 + r/100. Step 3: 1.12 = 1 + r/100, so r = 12%. Step 4: For the second year, new principal = ₹11,200. A = 11…

Card 7Finding time

For ₹2,000 at 10% compound interest, how many years are needed to reach ₹2,662?

Answer

Use A = P(1+r/100)^n. Step 1: 2,662 = 2,000(1.10)^n. Step 2: 2,662/2,000 = 1.331. Step 3: 1.331 = 1.1^3. Step 4: So n = 3 years. Answer: Time = 3 years.

Card 8Successive yearly rates

₹12,000 is invested for 3 years at successive rates of 8%, 10%, and 15%. Find the amount.

Answer

Use A = P(1+r1/100)(1+r2/100)(1+r3/100). Step 1: P = 12,000. Step 2: A = 12,000 × 1.08 × 1.10 × 1.15. Step 3: 12,000 × 1.08 = 12,960. Step 4: 12,960 × 1.10 = 14,256. Step 5: 14,256 × 1.15 = ₹16,394.40…

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Frequently Asked Questions

What are the important topics in Compound Interest (Using Formula) for ICSE Class 9 Mathematics?
Key topics in Compound Interest (Using Formula) include Core idea and direct formula, Amount when rates are different in successive years, Finding principal, rate, and time, Age and sharing problems. Study these first, then practise questions on each for Class 9 exams.
How many flashcards are available for Compound Interest (Using Formula)?
There are 25 flashcards for Compound Interest (Using Formula) covering key definitions, facts and ideas. A few sample cards are shown on this page.
How should I revise Compound Interest (Using Formula) for Class 9 exams?
Learn the core ideas first, then work through the 37 practice questions on Compound Interest (Using Formula). Revise definitions regularly and use flashcards for quick recall before the exam.

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