Profit, Interest & Time-Speed-Work for CLAT Quantitative Techniques
Three small toolkits cover most of CLAT's arithmetic-based data questions. Learn the on-cost-price convention, the SI-versus-CI gap and the speed–work formulas — then read them off a data passage in seconds.
CLAT Quantitative Techniques is data-interpretation in disguise. You are handed a short passage — a shopkeeper's accounts, a bank's interest table, a journey or a job — and asked questions that need nothing harder than class-10 arithmetic. Three toolkits do most of the heavy lifting: profit & loss, simple and compound interest, and time-speed-distance & time-work. Learn the handful of formulas behind each and these become some of the most reliable marks in the paper.
📌 One mindset for all three toolkits
Every question here is really about a relationship between three quantities. Profit links cost, selling price and gain; interest links principal, rate and time; speed links distance and time; work links rate and time. Once you know which two values a passage gives you, the third drops out of one formula. Since Quant is only about 10% of the paper, play for accuracy over ambition — clean marks others lose to the −0.25 negative marking.
Toolkit 1 — Profit & Loss
Profit and loss is shopkeeper arithmetic. Someone buys at one price and sells at another; the gap is the profit or loss. The two anchor terms are the cost price (CP) — what the seller paid — and the selling price (SP) — what the buyer paid. Everything else is built from these two.
✓Cost Price (CP) — the price at which an article is bought. The seller's outlay, including any cost added before sale.
✓Selling Price (SP) — the price at which the article is sold to the customer.
✓Profit — when SP > CP. Profit = SP − CP.
✓Loss — when SP < CP. Loss = CP − SP.
✓Marked Price (MP) — the listed or tagged price, before any discount is applied.
⚠️ The on-CP convention — profit % is on cost price, not selling price
The single most common CLAT slip. By convention, profit % and loss % are always calculated on the cost price, never the selling price — unless the question says otherwise. So Profit % = (Profit ÷ CP) × 100. Divide the profit by SP instead and you get a smaller, wrong percentage every time. Read 'profit of 20%' as 20% of what it cost.
Quantity
Formula
Profit
SP − CP
Loss
CP − SP
Profit %
(Profit ÷ CP) × 100
Loss %
(Loss ÷ CP) × 100
SP from CP & profit %
CP × (1 + Profit%/100)
CP from SP & profit %
SP ÷ (1 + Profit%/100)
The two recovery formulas in the last rows matter. CLAT often gives the SP and the profit % and asks for the CP — you must divide, not subtract a flat percentage, because the percentage sits on CP, the very number you are trying to find.
Discount, marked price and successive discounts
Shops rarely sell at the marked price. They mark up high, then offer a discount — always taken on the marked price, with what is left becoming the selling price.
Discount = MP − SP; Discount % = (Discount ÷ MP) × 100 — on the marked price.
SP after discount = MP × (1 − Discount%/100).
Successive discounts — 20% then 10% is not a single 30%. Apply them one after another; the second is taken on the already-reduced price.
💡 Successive discounts never simply add up
A '20% then 10%' offer is not 30% off. On ₹100 marked: 20% off leaves ₹80; 10% of ₹80 is ₹8, leaving ₹72 — a 28% effective discount. Quick formula for two discounts a% and b%: net = a + b − (ab/100) = 20 + 10 − 2 = 28%. Shops bank on customers adding them up; CLAT tests whether you don't.
🧩 Worked example
A stationery shop buys a fountain pen for 240 rupees. It marks the pen at 360 rupees, then sells it during a sale at a discount of 25 per cent on the marked price.
What is the shop's profit per cent on this pen?
A8.5%
B12.5%
C15%
D20%
▸ Show solution
Answer: B. Discount = 25% of ₹360 = ₹90, so SP = 360 − 90 = ₹270. Profit = 270 − 240 = ₹30. Apply the on-CP convention: Profit % = (30 ÷ 240) × 100 = 12.5%. The trap option A wrongly divides by the SP; D is the markup, not the profit. B is correct.
Toolkit 2 — Simple & Compound Interest
Interest is the rent charged for using money. You borrow or deposit a principal (P), and after a time (T) at a rate (R) per cent per year, you owe or earn interest on top. CLAT loves a small bank table and a question on the difference between the two ways interest can grow.
Simple interest
With simple interest, the interest is calculated only on the original principal, year after year — it never changes. The formula to memorise cold:
SI = (P × R × T) ÷ 100
P = principal; R = rate per cent per annum; T = time in years.
Amount (A) = P + SI — the total you finally have or owe.
Compound interest
With compound interest, each year's interest is added to the principal, and the next year's interest is charged on that larger amount. Interest earns interest — which is why real-world savings and loans almost always compound. The amount after T years is:
A = P × (1 + R/100) raised to the power T, and CI = A − P
Read it as: multiply the principal by (1 + R/100) once for each year. For 2 years at 10%, A = P × 1.1 × 1.1 = 1.21P, and the compound interest is the amount minus the original principal.
For the first year they are identical — both charge interest on the same starting principal. From the second year onward, CI charges interest on the interest too, so it pulls ahead and the gap widens every year. Over the same P, R and T (for T > 1 year), CI is always greater than SI. The longer the time, the bigger the gap.
Feature
Simple Interest
Compound Interest
Interest charged on
Original principal only
Principal + interest accumulated so far
Formula
SI = PRT/100
A = P(1 + R/100)^T ; CI = A − P
Year-on-year interest
Same every year
Grows each year
Size (for T > 1 year)
Smaller
Always larger
Real-world use
Rare — some short loans
Banks, deposits, EMIs
ℹ️ The CI–SI difference for 2 years has a shortcut
Over 2 years, the difference between compound and simple interest on the same principal equals P × (R/100)². On ₹10,000 at 10% for 2 years, that is 10000 × (0.1)² = ₹100 — exactly the 'interest on the first year's interest' that CI charges and SI does not. One line, a full calculation saved.
🧩 Worked example
A bank offers a fixed deposit of 20,000 rupees at 10 per cent per annum for 2 years. A customer compares the interest under simple interest with that under compound interest, compounded annually.
By how much does the compound interest exceed the simple interest over the 2 years?
A100 rupees
B200 rupees
C400 rupees
DBoth are equal
▸ Show solution
Answer: B. SI = (P × R × T)/100 = (20000 × 10 × 2)/100 = ₹4,000. For CI, A = 20000 × 1.1 × 1.1 = ₹24,200, so CI = 24,200 − 20,000 = ₹4,200. The difference is 4,200 − 4,000 = ₹200. Shortcut check: P × (R/100)² = 20000 × 0.01 = ₹200. B is correct; D ignores that CI overtakes SI after year one.
Lock these formulas with real data sets
10 drills, 150 questions — CLAT-style passages on shop accounts, bank tables and journeys, with full working in every answer.
The third toolkit is two close cousins. Time-speed-distance is about movement; time-work is about jobs getting done. Both turn on the same idea — a steady rate acting over time. The master relationship is the one every commuter knows: speed = distance ÷ time. Rearrange it for whatever the question hides.
Want
Formula
Speed
Distance ÷ Time
Distance
Speed × Time
Time
Distance ÷ Speed
⚠️ Average speed is NOT the mean of the two speeds
Drive there at 40 km/h and back at 60 km/h and the average is not (40 + 60)/2 = 50. You spend more time at the slower speed, so the average is pulled below 50. Correct average speed = total distance ÷ total time; for equal distances at a and b it is the harmonic mean 2ab ÷ (a + b) = (2 × 40 × 60)/100 = 48 km/h. CLAT plants 50 as the tempting wrong option.
Unit conversion — km/h and m/s
Passages mix units on purpose. A train's speed is in km/h but the platform length is in metres and the time in seconds. Convert before you compute.
✓km/h to m/s — multiply by 5/18. (60 km/h = 60 × 5/18 = 16.67 m/s.)
✓m/s to km/h — multiply by 18/5. (10 m/s = 10 × 18/5 = 36 km/h.)
✓Remember why — 1 km = 1000 m and 1 hour = 3600 s, so 1 km/h = 1000/3600 = 5/18 m/s.
🧩 Worked example
A train 180 metres long is running at a speed of 72 kilometres per hour. It needs to cross a railway platform that is 220 metres long.
How many seconds does the train take to completely cross the platform?
A16 seconds
B18 seconds
C20 seconds
D22 seconds
▸ Show solution
Answer: C. To fully cross a platform, the train travels its own length plus the platform: distance = 180 + 220 = 400 m. Convert the speed: 72 km/h × 5/18 = 20 m/s. Time = 400 ÷ 20 = 20 seconds. The classic trap is forgetting the train's own length (using only 220 m gives 11 s — a signal you missed a step). C is correct.
Toolkit 3b — Time & Work
Time-work uses the same engine with one trick: think in rate of work per day. If a person finishes a job in n days, in one day they do 1/n of it. Add the daily rates of everyone working together to find how fast the job goes.
✓Work rate — if a job takes n days, one day's work = 1/n of the whole.
✓Combined work — A does 1/a per day, B does 1/b per day; together they do (1/a + 1/b) per day. Time taken together = 1 ÷ (1/a + 1/b).
✓Pipes & cisterns — exactly the same idea. A filling pipe is positive work (+1/a per hour); a draining pipe (a leak or outlet) is negative work (−1/b per hour). Add them with signs to get the net fill rate.
ℹ️ Pipes & cisterns is just time-work with a leak
Do not learn pipes and cisterns separately. A tap filling a tank in 6 hours does +1/6 per hour; an outlet emptying it in 12 hours does −1/12 per hour. Open together, the net rate is 1/6 − 1/12 = 1/12 per hour, so the tank fills in 12 hours. Same rates, same addition — one sign is just negative.
🧩 Worked example
Two friends are painting a wall. Working alone, Asha would finish the wall in 12 hours and Ravi would finish the same wall in 6 hours. They decide to paint it together.
Working together, how long will they take to paint the whole wall?
A3 hours
B4 hours
C6 hours
D9 hours
▸ Show solution
Answer: B. Work in rates per hour. Asha does 1/12 per hour; Ravi does 1/6 = 2/12 per hour. Together: 1/12 + 2/12 = 1/4 of the wall each hour, so the whole wall takes 1 ÷ (1/4) = 4 hours. The trap option D averages the times (12 + 6)/2 = 9 — but rates add, times do not. B is correct.
🧩 Worked example
A shopkeeper sold two watches for 1,200 rupees each. On the first watch he made a profit of 20 per cent, and on the second he suffered a loss of 20 per cent.
On the two watches taken together, what was his overall result?
ANo profit, no loss
BA loss of 100 rupees
CA profit of 100 rupees
DA loss of 50 rupees
▸ Show solution
Answer: B. Find each cost price using the on-CP convention — divide, never subtract. Watch 1: CP = 1200 ÷ 1.20 = ₹1,000. Watch 2: CP = 1200 ÷ 0.80 = ₹1,500. Total CP = ₹2,500; total SP = ₹2,400, a loss of ₹100. Option A is the classic trap — equal +20%/−20% on the same SP is not a wash, because the percentages sit on different cost prices. B is correct.
Reading a CLAT Quant data passage
CLAT does not ask bare sums — it gives a passage (a paragraph, a small table or a chart) and a set of questions hanging off it. Run the same method every time so the arithmetic feels routine under the clock.
1
Read for the numbers and their labels
Note every value and what it measures — is ₹360 the cost price or the marked price? Is 72 in km/h or m/s? Mislabelling a number is the commonest error here.
2
Decide which toolkit applies
Shop and prices → profit & loss. Bank, principal, rate → interest. Journey or job → speed/work. Name it before you compute.
3
Fix units first
Convert km/h↔m/s, years↔months, before plugging in. Mixed units are deliberate bait.
4
Apply the formula, then check the option trap
Most questions need a single formula. Then glance for the planted wrong answer — the averaged speed, the on-SP percentage, the un-added train length.
💡 Let negative marking guide your pace
CLAT marks +1 correct, −0.25 wrong, 0 unattempted. Do not bleed time on a stubborn data set — solve the clean ones for sure marks, and if a value will not resolve, leave it. A zero beats a −0.25.
🎯 Profit, interest & TSW in a nutshell
Profit/loss % is always on the cost price, not the selling price; discount is on the marked price, and successive discounts never simply add.
SI = PRT/100; CI = P(1 + R/100)^T − P. CI always beats SI for more than one year because interest earns interest.
The 2-year CI − SI difference = P × (R/100)² — a one-line shortcut.
Speed = distance ÷ time; convert km/h to m/s by × 5/18, and back by × 18/5.
Average speed = total distance ÷ total time — never the mean of the two speeds.
Time-work runs on rates: a job in n days = 1/n per day; combined rates add; pipes & cisterns is the same with a negative sign for draining.
Common mistakes to stop making
✓Taking profit % on the selling price instead of the cost price, or adding successive discounts (20% + 10% ≠ 30%).
✓Assuming SI and CI are equal, or forgetting that CI overtakes SI from the second year.
✓Averaging two speeds for average speed instead of using total distance ÷ total time.
✓Forgetting to convert units, or to add the train's own length when it crosses a platform.
✓Averaging times in a time-work problem instead of adding the work rates.
Is profit percentage calculated on the cost price or the selling price in CLAT?
Always on the cost price, unless the question says otherwise: Profit % = (Profit ÷ Cost Price) × 100. Many students wrongly divide by the selling price and get a smaller, incorrect figure. Read '20% profit' as 20% of what the article cost the seller.
Why is compound interest always more than simple interest?
In the first year both charge interest on the same starting principal, so they are equal. From the second year, compound interest charges interest on the interest already earned too, while simple interest stays on the original principal. So for any period over one year, CI exceeds SI, and the gap widens with time.
What is the quick way to find the CI–SI difference for two years?
Use the shortcut: the difference equals P × (R/100)², where P is the principal and R the annual rate. On ₹20,000 at 10% for 2 years, that is 20000 × (0.1)² = ₹200. It is the extra interest CI charges on the first year's interest, and it saves a full calculation.
Why isn't average speed just the average of the two speeds?
Because you spend more time at the slower speed, so it weighs more. The correct method is total distance ÷ total time. For equal distances at speeds a and b, average speed is the harmonic mean 2ab ÷ (a + b) — for 40 and 60 km/h that is 48, not 50.
How do I convert between km/h and m/s?
From km/h to m/s, multiply by 5/18; from m/s to km/h, multiply by 18/5. This comes from 1 km being 1000 m and 1 hour being 3600 seconds, so 1 km/h equals 5/18 m/s. Always convert to a single unit first, as CLAT passages mix units deliberately.
How do time-and-work and pipes-and-cistern problems relate?
They are the same idea. Think in rates: a job done in n days means 1/n per day, and combined rates simply add. Pipes and cisterns is identical — a filling pipe adds a positive rate, a draining pipe subtracts a negative one. Take the reciprocal of the net rate for the time.
How much does Quantitative Techniques count for in CLAT, and how should I approach it?
Quant is about 10% of the 120-question CLAT UG paper — roughly two data sets — and uses only class-10 maths. Because of the −0.25 negative marking, prioritise accuracy: solve the clean questions and leave any value that won't resolve rather than guessing. Sure marks here are easier than in the longer sections.
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