CUET UG General Test Quantitative Reasoning Percentage Profit and Loss Simple and Compound Interest

Must‑Know (15–20 detailed bullets)

• Percentage change = (Change / Original) × 100; if a shirt increases from ₹800 to ₹960, percentage increase = (160/800) × 100 = 20%.
• If A is x% more than B, then B is less than A by [x/(100 + x)] × 100%; e.g., if A is 25% more than B, B is (25/125) × 100 = 20% less than A.
• If the price of sugar increases by 20%, to keep expenditure constant, consumption must be reduced by (20/120) × 100 = 16.67%.
• Profit = SP – CP; if CP = ₹500, SP = ₹600, profit = ₹100.
• Loss = CP – SP; if CP = ₹800, SP = ₹720, loss = ₹80.
• Profit % = (Profit / CP) × 100; on CP ₹500 and profit ₹100, profit % = (100/500) × 100 = 20%.
• Loss % = (Loss / CP) × 100; on CP ₹800 and loss ₹80, loss % = (80/800) × 100 = 10%.
• SP = CP × (1 + P%/100) when profit; CP = ₹600, profit 10%, SP = 600 × 1.10 = ₹660.
• SP = CP × (1 – L%/100) when loss; CP = ₹500, loss 12%, SP = 500 × 0.88 = ₹440.
• CP = SP / (1 + P%/100) if profit; SP = ₹1100, profit 10%, CP = 1100 / 1.10 = ₹1000.
• CP = SP / (1 – L%/100) if loss; SP = ₹900, loss 10%, CP = 900 / 0.90 = ₹1000.
• Overhead expenses are added to CP; if CP = ₹1200, repair = ₹80, total CP = ₹1280.
• Simple Interest (SI) = (P × R × T)/100; ₹5000 at 8% p.a. for 3 years: SI = (5000 × 8 × 3)/100 = ₹1200.
• Amount in SI = P + SI = P(1 + RT/100); ₹10,000 at 5% for 4 years: A = 10000(1 + 0.05×4) = ₹12,000.
• Compound Interest (CI) = P[(1 + R/100)^T – 1]; ₹10,000 at 10% p.a. compounded annually for 2 years: CI = 10000[(1.1)^2 – 1] = ₹2100.
• For half-yearly compounding, rate = R/2%, time = 2T; ₹8000 at 10% p.a. half-yearly for 1 year: A = 8000(1 + 5/100)^2 = ₹8820.
• For quarterly compounding, rate = R/4%, time = 4T; ₹4000 at 8% p.a. quarterly for 6 months: A = 4000(1 + 2/100)^2 = ₹4161.60.
• Depreciation uses CI formula with negative rate; machine ₹20,000 depreciates 10% p.a., value after 2 years = 20000(0.9)^2 = ₹16,200.
• Population growth uses CI; town population 10,000 grows at 5% p.a., population after 2 years = 10000(1.05)^2 = 11,025.
• If interest compounded annually, CI > SI for T > 1 year; for P = ₹10,000, R = 10%, T = 2 years: SI = ₹2000, CI = ₹2100 → difference ₹100.

Difficulty Level

Intermediate — combines direct formulas with application-based scenarios involving multiple steps, common in CUET.

Common CUET Traps

• Trap: Using SI formula for CI questions when compounding is mentioned.
  Avoid: Always check if interest is compounded; use CI formula or conversion of time/rate for half-yearly/quarterly.

• Trap: Calculating profit % on SP instead of CP.
  Avoid: Profit % is always calculated on CP unless stated otherwise.

• Trap: Ignoring time conversion (e.g., months to years) in interest problems.
  Avoid: Convert time into years; 6 months = 0.5 year, 9 months = 0.75 year before applying formula.

Practice MCQs

Q1. A shopkeeper sells a book for ₹450 and incurs a loss of 10%. What was the cost price?
A. ₹480
B. ₹490
C. ₹500
D. ₹520
Answer: C
Explanation: CP = SP / (1 – L%/100) = 450 / 0.90 = ₹500.
Why others fail: Option A is tempting if student adds 10% of SP to SP (450 + 45).

Q2. What is the simple interest on ₹7,500 at 12% per annum for 5 years?
A. ₹3,600
B. ₹4,500
C. ₹4,800
D. ₹5,000
Answer: B
Explanation: SI = (P × R × T)/100 = (7500 × 12 × 5)/100 = ₹4,500.
Why others fail: Option A results from using 9.6% or miscalculating multiplication.

Q3. A sum amounts to ₹13,310 in 3 years at 10% p.a. compound interest compounded annually. What is the principal?
A. ₹10,000
B. ₹9,000
C. ₹11,000
D. ₹10,500
Answer: A
Explanation: A = P(1 + R/100)^T → 13310 = P(1.1)^3 → P = 13310 / 1.331 = ₹10,000.
Why others fail: Option B arises if student subtracts 30% of amount as interest.

Q4. The price of a mobile increases by 20%, then decreases by 20%. What is the net percentage change?
A. 4% increase
B. 4% decrease
C. 2% decrease
D. No change
Answer: B
Explanation: Net change = +20 – 20 – (20×20)/100 = –4%, i.e., 4% decrease.
Why others fail: Option D is tempting if student assumes increase and decrease cancel out.

Q5. A machine depreciates at 10% per annum. If its present value is ₹8,100, what was its value 2 years ago?
A. ₹10,000
B. ₹9,500
C. ₹9,000
D. ₹8,500
Answer: A
Explanation: Value 2 years ago = 8100 / (0.9)^2 = 8100 / 0.81 = ₹10,000.
Why others fail: Option C results from subtracting 20% from current value instead of reversing depreciation.

Last‑Minute Revision

• ⚠️ Profit % = (Profit / CP) × 100 — never on SP unless specified.
• ⚠️ Loss % = (Loss / CP) × 100 — CP always base.
• ⚠️ SP = CP × (100 + P%) / 100 — use in profit cases.
• ⚠️ SP = CP × (100 – L%) / 100 — use in loss cases.
• ⚠️ CP = 100 × SP / (100 + P%) — when profit given.
• ⚠️ CP = 100 × SP / (100 – L%) — when loss given.
• ⚠️ SI = PRT/100 — R in % per annum, T in years.
• ⚠️ Amount in SI = P(1 + RT/100).
• ⚠️ CI = P[(1 + R/100)^T – 1] — T must be in compounding periods.
• ⚠️ For half-yearly: R → R/2, T → 2T.
• ⚠️ For quarterly: R → R/4, T → 4T.
• ⚠️ Depreciation: Value = Initial × (1 – R/100)^T.
• ⚠️ Population growth: Final = Initial × (1 + R/100)^T.
• ⚠️ Net % change after x% increase and y% decrease: x – y – (xy/100).
• ⚠️ If A is r% more than B, B is [r/(100 + r)] × 100% less than A.
• ⚠️ If price increases r%, consumption must decrease [r/(100 + r)] × 100% to keep expenditure same.
• ⚠️ CI > SI for T > 1 year at same rate — difference = P(R/100)^2 for 2 years.
• ⚠️ Overhead costs added to CP — included in total cost.
• ⚠️ "Per annum" means yearly — adjust for compounding frequency.
• ⚠️ Use multiplication factor: 10% increase → ×1.10, 10% decrease → ×0.90.