ICSE Class 9 Mathematics — Flashcards for Quick Revision

A: Step 1: √4 = 2, so it is rational and not a surd. Step 2: √2 cannot be written as a/b, so it is irrational. Step 3: ∛27 = 3, so it is rational and not a surd. Step 4: π is irrational, but it is not a

A: A rational number can be written as a/b, where a and b are integers and b ≠ 0. Example: 5/8 is rational because both numbers are integers and the denominator is not zero. Important check: b must never

A: Step 1: Check the denominator 8. Step 2: 8 = 2³, so it can be written only with factors 2 and 5. Step 3: Therefore 7/8 is a terminating decimal. Quick check: 7 ÷ 8 = 0.875. Answer: Yes, it is terminat

A: Step 1: First year interest = (8,000 × 5 × 1) / 100 = ₹400. Step 2: Amount at end of 1st year = ₹8,000 + ₹400 = ₹8,400. Step 3: Second year interest = (8,400 × 5 × 1) / 100 = ₹420. Step 4: Amount at e

A: Step 1: Half-yearly compounding means time for each conversion period = 1/2 year. Step 2: For 1st half-year, P = ₹10,000, R = 8%, T = 1/2. Interest = (10,000 × 8 × 1) / (100 × 2) = ₹400. Amount after

A: Compound interest is calculated on the growing principal. Example: If P = ₹1,000 and R = 10% per year: 1st year interest = (1,000 × 10 × 1) / 100 = ₹100. Amount after 1st year = ₹1,100. 2nd year inter

A: 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.

A: 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.

A: 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.