Simple Interest — Solutions

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1. Calculate the simple interest

   1. (a) I = 1000×5×3÷100 = $150
   2. (b) I = 2500×6×4÷100 = $600
   3. (c) I = 800×3×2÷100 = $48
   4. (d) I = 4500×8×5÷100 = $1 800
   5. (e) I = 600×2.5×2÷100 = $30
   6. (f) I = 10 000×4.5×3÷100 = $1 350
   7. (g) T = 18÷12 = 1.5 yr: I = 350×7×1.5÷100 = $36.75
   8. (h) T = 30÷12 = 2.5 yr: I = 12 000×3.5×2.5÷100 = $1 050

2. Total amount A = P + I

   1. (a) I = $240. A = $2 000 + $240 = $2 240
   2. (b) I = $600. A = $5 000 + $600 = $5 600
   3. (c) I = $150. A = $750 + $150 = $900
   4. (d) I = $1 200. A = $8 000 + $1 200 = $9 200
   5. (e) T = 1.5 yr: I = 1200×9×1.5÷100 = $162. A = $1 200 + $162 = $1 362
   6. (f) I = 6500×4×2.5÷100 = $650. A = $6 500 + $650 = $7 150

3. Find the unknown (P, R, or T)

   1. (a) R: R = 360×100÷(3000×3) = 36 000÷9 000 = 4% p.a.
   2. (b) R: R = 450×100÷(5000×2) = 45 000÷10 000 = 4.5% p.a.
   3. (c) T: T = 480×100÷(4000×6) = 48 000÷24 000 = 2 years
   4. (d) T: T = 700×100÷(2000×5) = 70 000÷10 000 = 7 years
   5. (e) P: P = 560×100÷(7×4) = 56 000÷28 = $2 000
   6. (f) P: P = 900×100÷(6×5) = 90 000÷30 = $3 000

4. True or False

   1. True — since I = PRT ÷ 100, doubling P doubles I (all else equal).
   2. True — I = P × 2R × (T/2) ÷ 100 = PRT ÷ 100. The two changes cancel exactly.
   3. False — simple interest is calculated on the original principal only; it does not earn further interest. (That is compound interest.)
   4. False — 18 months must be converted to years first: T = 18 ÷ 12 = 1.5 years.
   5. True — I = 2000×5×3÷100 = $300. A = $2000 + $300 = $2300. ✓
   6. True — A = P + I and I is always positive (the interest earned or charged is added to P), so A > P.

5. Apply simple interest in context

   1. Jasmine: I = 6000×7×3÷100 = $1 260 interest. Total repayment = $6 000 + $1 260 = $7 260
   2. Aiden: T = 540×100÷(4500×4.8) = 54 000÷21 600 = 2.5 yr. He needs 3 full years (at T=3, I = $648 > $540).
   3. Sam: P = 800×100÷(5×4) = 80 000÷20 = $4 000
   4. Car loan:
      1. I = 15 000×9×4÷100 = $5 400
      2. A = $15 000 + $5 400 = $20 400
      3. Monthly repayment = $20 400 ÷ 48 = $425/month

6. Compare investments (P = $5 000)

   1. Scenario 1: A = 5000×6×2÷100 = $600. B = 5000×4.5×3÷100 = $675. Option B is better.
   2. Scenario 2: A = 5000×8×1÷100 = $400. B = 5000×5×2÷100 = $500. Option B is better.
   3. Scenario 3: A = 5000×3×5÷100 = $750. B = 5000×7×2÷100 = $700. Option A is better.

7. Interest timeline ($4 000 at 5% p.a.)

   Annual interest = $4 000 × 5% = $200 per year (same every year with simple interest).

   Year	Interest this year	Total interest	Balance
   1	$200	$200	$4 200
   2	$200	$400	$4 400
   3	$200	$600	$4 600
   4	$200	$800	$4 800
   5	$200	$1 000	$5 000

   1. The interest is $200 every year — it is constant. This is because simple interest is always calculated on the original principal ($4 000), never on the accumulated balance.
   2. The balance reaches exactly $5 000 at Year 5. It first exceeds $5 000 after Year 5 is complete (or would first exceed during Year 6).

8. Find unknowns in context

   1. Zara: I = $11 200 − $8 000 = $3 200. T = 3200×100÷(8000×5) = 8 years
   2. P = 1750×100÷(7×5) = 175 000÷35 = $5 000
   3. I = $10 350 − $9 000 = $1 350. R = 1350×100÷(9000×3) = 135 000÷27 000 = 5% p.a.
   4. Raj:
      1. $7 500 − $6 000 = $1 500 more needed
      2. T = 1500×100÷(6000×3.6) = 150 000÷21 600 ≈ 6.94 yr. 7 full years (at T=7: I = 6000×3.6×7÷100 = $1512 > $1500 ✓)

9. Extended real-world problems

   1. Sophie’s savings club. Each $200 earns 4% p.a. for the remaining years:
      Year 1 contribution: I = 200×4×5÷100 = $40
      Year 2 contribution: I = 200×4×4÷100 = $32
      Year 3 contribution: I = 200×4×3÷100 = $24
      Year 4 contribution: I = 200×4×2÷100 = $16
      Year 5 contribution: I = 200×4×1÷100 = $8
      Total interest = $40 + $32 + $24 + $16 + $8 = $120
   2. Priya: I = 10 000×8×3÷100 = $2 400. Marcus: I = 10 000×6×5÷100 = $3 000. Marcus earns more by $600.
   3. Lena’s loan comparison ($12 000):
      1. Lender X: I = 12 000×6×4÷100 = $2 880. Total = $14 880. Lender Y: I = 12 000×8×3÷100 = $2 880. Total = $14 880
      2. Both cost the same total: $14 880. Neither is cheaper overall.
      3. Lender X monthly: $14 880 ÷ 48 = $310/month. Lender Y monthly: $14 880 ÷ 36 = $413.33/month. Lender Y has higher monthly repayments.

10. Investigation — Effect of changing variables

    Change	P	R	T	New Interest	Change
    Base	$1 000	5%	4	$200	—
    Double P	$2 000	5%	4	$400	+$200 (doubled)
    Halve P	$500	5%	4	$100	−$100 (halved)
    Double R	$1 000	10%	4	$400	+$200 (doubled)
    Halve T	$1 000	5%	2	$100	−$100 (halved)
    Double R, halve T	$1 000	10%	2	$200	No change

    2. Doubling the principal doubles the simple interest (direct proportion).
    3. Doubling R and halving T leaves interest unchanged. The ×2 from R and the ÷2 from T cancel exactly, because I = P×R×T÷100 and 2R×(T/2) = RT.
    4. “Simple interest is proportional to the principal, the rate, and the time.”