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Profit-Loss, SI & CI

Profit-Loss, SI & CI is a frequently tested area in CUET General Test. Work through these free NTA-style sample questions with full answers and explanations, then attempt all 40 in a timed practice test to build exam-day speed.

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Snapshot

Part 1 — Profit & Loss

Everything is a percentage of the cost price (CP) unless stated otherwise.

Quantity Formula
Profit SP − CP
Profit % (Profit ÷ CP) × 100
Loss % (Loss ÷ CP) × 100
SP from CP and profit% CP × (100 + profit%) ÷ 100
Discount Marked Price (MP) − SP
Discount % (Discount ÷ MP) × 100

Two traps the exam loves: profit% is on CP but discount% is on MP — different bases; and a 20% profit then 20% loss is a net loss (the successive-change rule): 20 − 20 − 4 = 4% loss.

Successive discounts combine like successive change: two discounts of a% and b% give a single equivalent discount of a + b − (ab ÷ 100).

Dishonest dealer / false weights. A trader who sells at cost price but gives only 900 g for a kg makes a profit of (error ÷ true − error) = 100 ÷ 900 = 11.11%. The gain% = (true value − given value) ÷ given value × 100.

Part 2 — Simple vs Compound Interest

Simple Interest (SI) = P × R × T ÷ 100 — the same interest every year, so the amount grows in a straight line. Compound Interest: Amount = P × (1 + R/100)ᵀ, and CI = Amount − P. Each year's interest itself earns interest, so the amount curves upward and pulls away from SI:

₹10,000 at 10% — SI vs CI0200040006000800010000120001400016000Yr1Yr2Yr3Yr4Simple InterestCompound Interest

For ₹10,000 at 10%, year 1 SI and CI are equal (₹1,000), but by year 4 CI (₹14,641) is well clear of SI (₹14,000). The high-value shortcuts:

Part 3 — Speed techniques

  1. Anchor profit/loss to CP = 100 — then SP is just "100 + profit%", all mental.
  2. Use the successive formula for profit-then-loss and for two discounts (a + b − ab/100).
  3. For 2-year CI, use 2R + R²/100 instead of expanding the bracket.
  4. CI − SI (2 years) = P(R/100)² — recognise it on sight.
  5. Convert the rate to a fraction (12.5% = 1/8) to simplify interest sums.
  6. Match rate and time units before substituting (half-year ⇒ half the annual rate).

Part 4 — Worked examples

1. Profit%. Bought ₹400, sold ₹500. (100 ÷ 400) × 100 = 25%.

2. SP from profit%. CP ₹250, profit 18%. SP = 250 × 1.18 = ₹295.

3. Discount. MP ₹800, 15% discount. SP = 800 × 0.85 = ₹680.

4. Successive discount. Two discounts 10% and 20%. Single equivalent = 10 + 20 − (10×20÷100) = 28%.

5. Profit then loss. 20% profit then 20% loss on selling. Net = 20 − 20 − 4 = 4% loss.

6. False weight. A grocer sells at CP but uses a 900 g "kilogram". Gain% = 100 ÷ 900 × 100 = 11.11%.

7. Compound interest. CI on ₹5,000 at 10% for 2 years? 2-year rate 21% → CI = 21% of 5,000 = ₹1,050. (SI = ₹1,000; the ₹50 gap = P(R/100)².)

8. Find principal. SI for 3 years at 8% is ₹1,200. Principal? P = (1200 × 100) ÷ (8 × 3) = ₹5,000.

Part 5 — Common traps

Part 6 — How to use this page

Anchor every profit/loss question to CP = 100, memorise the 2-year CI shortcuts and the successive-discount formula, re-solve the eight examples closed-book, then attempt the practice set and the timed test.

One-line revision: profit/loss % sits on CP, discount % on MP, SI grows straight while CI curves, two discounts combine as a + b − ab/100, and for 2 years CI − SI = P(R/100)².

Practice questions

Now test yourself. 8 free sample questions with explanations. 32 more in the timed practice test.

Q1. Find the simple interest on ₹6400 at 7.5% per annum for 146 days. (Take 1 year = 365 days.)

▸ Show answer & explanation

Answer: B

$T=\frac{146}{365}=0.4$ year. SI $=\frac{6400\times7.5\times0.4}{100}=$ ₹192.

Q2. A retailer buys 100 pens for ₹500 and sells them at ₹6 each. What is the profit percentage?

▸ Show answer & explanation

Answer: C

CP per pen $=\frac{500}{100}=5$, SP $=6$. Profit% $=\frac{6-5}{5}\times100=20\%$.

Q3. A man sells an article at 5% profit. Had he sold it at 5% loss, he would have got ₹6 less. What is the cost price?

▸ Show answer & explanation

Answer: B

Difference $=5\%-(-5\%)=10\%$ of CP $=6$, so CP $=\frac{6}{0.10}=$ ₹60.

Q4. A shopkeeper sells a TV at 12% profit. If he had bought it for 10% less and sold it for ₹60 less, he would have gained 20%. What is the original cost price?

▸ Show answer & explanation

Answer: B

Let CP $=x$. New CP $=0.9x$, new SP at 20% gain $=1.2\times0.9x=1.08x$. Original SP $=1.12x$. Given $1.12x-1.08x=60\Rightarrow0.04x=60\Rightarrow x=$ ₹1500.

Q5. The marked price of a shirt is ₹1200. A customer pays ₹918 after two successive discounts, one of which is 15%. What is the other discount percentage?

▸ Show answer & explanation

Answer: B

After 15% discount price $=1200\times0.85=1020$. Second discount $=\frac{1020-918}{1020}\times100=\frac{102}{1020}\times100=10\%$.

Q6. A sum of ₹10000 is invested partly at 8% and partly at 10% simple interest. The total annual interest is ₹920. How much is invested at 8%?

▸ Show answer & explanation

Answer: B

Let ₹x at 8%. Then $0.08x+0.10(10000-x)=920\Rightarrow1000-0.02x=920\Rightarrow x=$ ₹4000.

Q7. An item is marked 50% above cost. To clear stock the seller offers a discount such that he just breaks even. What discount percent does he give?

▸ Show answer & explanation

Answer: B

Let CP $=100$, MP $=150$. To break even SP $=100$, discount $=\frac{50}{150}\times100=33.33\%$.

Q8. A dealer allows a discount of 10% and still gains 8%. If the cost price is ₹450, what is the marked price?

▸ Show answer & explanation

Answer: C

SP $=450\times1.08=486$. MP $=\frac{SP}{1-0.10}=\frac{486}{0.90}=$ ₹540.

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