Profit, Interest, Time & Work, and Motion

This is the commercial and mechanical half of the arithmetic section, and it is where the marks are densest: profit and loss, discount, simple and compound interest, time and work, pipes and cisterns, time-speed-distance, trains, and boats and streams together supply eight to twelve of the twenty-five SSC CGL Tier-I quant questions and a large slice of Tier-II. Every one of these topics rests on the percentage and ratio machinery from Chapter 1 — profit is a percentage on cost, interest is a percentage on principal, work and speed are ratios in disguise. Because Tier-I penalises a wrong answer by −0.50, each section below ends with the shortcut that lets you finish in one line and a note on the trap that makes candidates lose the mark.

Cost Price, Selling Price & the One Golden Rule

Cost price (CP) is what the seller paid; selling price (SP) is what the buyer paid. If SP > CP there is a profit = SP − CP; if SP < CP there is a loss = CP − SP. The rule that decides half the marks in this chapter: profit% and loss% are always taken on the CP, never on the SP, unless the question explicitly says otherwise.

Worked example. A pair of shoes is bought for ₹640 and sold for ₹768. Profit = 768 − 640 = ₹128, and profit% = (128 ÷ 640) × 100 = 20%. Watch the denominator: 640, the CP. Dividing by 768 would give 16.67%, which is exactly the wrong option the examiner prints.

Profit-loss formula summary — six lines that cover the whole topic
You know	Formula	Worked value
CP and profit% → SP	SP = CP × (100 + p)/100	CP ₹640, p = 20 → 640 × 1.20 = ₹768
CP and loss% → SP	SP = CP × (100 − l)/100	CP ₹2,300, l = 15 → 2300 × 0.85 = ₹1,955
SP and profit% → CP	CP = SP × 100/(100 + p)	SP ₹1,242, p = 15 → 1242 × 100/115 = ₹1,080
SP and loss% → CP	CP = SP × 100/(100 − l)	SP ₹1,955, l = 15 → 1955 × 100/85 = ₹2,300
Profit% on CP → margin% on SP	100p/(100 + p)	p = 25 → 2500/125 = 20%
Margin% on SP → profit% on CP	100m/(100 − m)	m = 20 → 2000/80 = 25%

⚠️ Never subtract the percentage from the SP

Given SP = ₹1,242 at a 15% profit, the tempting move is 1242 − 15% of 1242 = 1242 − 186.30 = ₹1,055.70. Wrong. The 15% was charged on the CP of ₹1,080, not on ₹1,242. Always divide by (100 + p)/100. Both ₹1,055.70 and its rounded cousins appear as options in real papers.

Two selling prices, one bridge. An article sold for ₹1,080 causes a 10% loss. At what price should it be sold to gain 20%? Step 1 — the CP is the bridge: 1080 × 100/90 = ₹1,200. Step 2 — the new SP: 1200 × 120/100 = ₹1,440. Whenever a question mentions two selling prices, find the CP first; there is no shortcut that skips it.

Cost price from two scenarios. Two shapes recur.
(i) "The difference between selling an article at a 12% profit and at a 12% loss is ₹120." The gap between the two selling prices is 24% of the CP, so CP = 120 ÷ 0.24 = ₹500 (check: ₹560 and ₹440, difference ₹120 ✓).
(ii) "A man sells an article at a 10% loss; had he sold it for ₹90 more he would have gained 8%." The ₹90 spans 18% of the CP, so CP = 90 ÷ 0.18 = ₹500 (check: SP ₹450, and 450 + 90 = ₹540 = 500 × 1.08 ✓). In both, the trick is to see the given rupee amount as a percentage gap.

The Three Shortcuts the paper Tests Every Year

Shortcut 1 — two articles at the same selling price. A dealer sells two watches at ₹960 each; on one he gains 20% and on the other he loses 20%. Common sense says they cancel. Compute: CP of the first = 960 × 100/120 = ₹800; CP of the second = 960 × 100/80 = ₹1,200. Total CP = ₹2,000 against total SP = ₹1,920, so the loss is ₹80, that is 4%.

🧠 It is ALWAYS a loss of (x/10)²%

Two articles sold at the same price, one at +x% and one at −x%, always give a net LOSS of (x/10)²% — never a profit and never a break-even. x = 10 → 1%; x = 15 → 2.25%; x = 20 → 4%; x = 30 → 9%. The reason: the loss falls on the bigger cost price (₹1,200) while the gain falls on the smaller one (₹800), so the loss wins. Any option saying "no profit no loss" in such a question is wrong by construction.

Shortcut 2 — profit on cost versus profit on selling price. Take CP = ₹100 with a 25% profit on cost: SP = ₹125 and profit = ₹25. Expressed on the SP, that same ₹25 is 25/125 = 1/5 = 20%. So "25% on cost" and "20% on selling price" describe one identical deal. The conversions are 100p/(100 + p) one way and 100m/(100 − m) the other, and both appear in the summary table above.

Shortcut 3 — the dishonest dealer with false weights. A grocer sells at his cost price but hands over only 900 g in place of a kilogram. Where does the profit hide? Suppose his goods cost ₹1 per gram: the customer pays ₹1,000 for "1 kg" but receives 900 g, which cost the grocer ₹900. Gain = 100/900 × 100 = 11 1/9%. The formula is gain% = error/(true weight − error) × 100. With an 800 g weight: 200/800 × 100 = 25%.

⚠️ False weight PLUS a markup — do not add the two

If the grocer also marks his price 10% above cost while using the 900 g weight, the gain is not 11 1/9 + 10. Work in cost units: he collects 1.10 × 1000 = 1,100 units of value for goods that cost him 900, so gain = 200/900 × 100 = 22 2/9%. The general form is gain% = (100 + markup%) × true weight/false weight − 100 = 110 × 1000/900 − 100 = 22 2/9%.

Marked Price, Discount & Equivalent Single Discount

Marked price (MP) — also called list price — is the tag on the article. Discount = MP − SP, and crucially discount% = (discount ÷ MP) × 100: the base here is the MP, not the CP. So this chapter now runs on two bases — profit sits on CP, discount sits on MP. Worked: an item marked ₹2,400 at a 15% discount sells for 2400 × 0.85 = ₹2,040.

Successive discounts never add. Two discounts of a% and b% together equal a single discount of a + b − ab/100. So 20% and 10% give 20 + 10 − 2 = 28%, not 30%. Verify on an MP of ₹2,000: 2000 × 0.80 = ₹1,600, then 1600 × 0.90 = ₹1,440, and (2000 − 1440)/2000 × 100 = 28% ✓. For three discounts apply the formula twice: 20% and 10% give 28%, then 28% with 5% gives 28 + 5 − 1.4 = 31.6%. The order never matters, because multiplication is commutative.

The markup-then-discount chain. A trader marks his goods 50% above cost and then allows a 20% discount. Take CP = ₹100: MP = ₹150, SP = 150 × 0.80 = ₹120 → a 20% profit. You can also read it as a successive change of +50% and −20%: 50 − 20 − 1000/100 = 30 − 10 = +20% — one line, no rupees. A tougher variant: markup 60%, then discounts of 20% and 25%. SP = 100 × 1.60 × 0.80 × 0.75 = ₹96, that is a 4% loss — proof that a big markup does not guarantee a profit once the discount chain gets long.

🧠 Fix the marked price to protect a target profit

Use MP = CP × (100 + profit%)/(100 − discount%). A shopkeeper whose CP is ₹1,200 wants a 20% profit even after a 25% discount: MP = 1200 × 120/75 = ₹1,920. Check: 1920 × 0.75 = ₹1,440 = 1200 × 1.20 ✓. Also worth knowing: "buy 3 get 1 free" is a 25% discount and "buy 2 get 1 free" is 33 1/3% — the free item is measured against the total handed over, not against the paid items.

Simple & Compound Interest

Simple interest is charged on the original principal every year: SI = PRT/100. Worked: ₹6,400 at 7.5% for 3 years gives SI = 6400 × 7.5 × 3/100 = ₹1,440. Compound interest is charged on the running amount, so the interest itself earns interest: A = P(1 + R/100)ᵗ and CI = A − P. Worked: ₹12,000 at 10% for 3 years gives A = 12000 × 1.331 = ₹15,972, so CI = ₹3,972 against an SI of only ₹3,600.

SI versus CI on the same money — and the difference formulae
Situation	Simple interest	Compound interest / difference
₹15,000, 8%, 2 years	15000 × 0.08 × 2 = ₹2,400	15000 × 1.08² − 15000 = ₹2,496; difference ₹96
₹10,000, 10%, 3 years	₹3,000	10000 × 1.331 − 10000 = ₹3,310; difference ₹310
CI − SI for 2 years	Always positive	P(R/100)² → 15000 × 0.0064 = ₹96 ✓
CI − SI for 3 years	Grows faster	P(R/100)² × (300 + R)/100 → 10000 × 0.01 × 3.10 = ₹310 ✓
Half-yearly compounding	Not applicable	Halve R, double t: ₹8,000 at 10% for 1 yr → 8000 × 1.05² = ₹8,820, CI ₹820
Quarterly compounding	Not applicable	Quarter R, quadruple t: ₹16,000 at 20% for 9 months → 16000 × 1.05³ = ₹18,522, CI ₹2,522

🎯 Why CI for 2 years = SI + interest on the interest

Take ₹15,000 at 8%. Year 1 interest = ₹1,200. Year 2 is charged on ₹16,200, so it is ₹1,200 plus 8% of that first ₹1,200 = ₹96. Total CI = 1200 + 1200 + 96 = ₹2,496, exactly SI (₹2,400) plus one interest-on-interest term (₹96). That is the whole content of the formula CI − SI = P(R/100)², and it is why the difference for 2 years never depends on anything else.

Doubling and multiplying. Under simple interest each additional principal takes the same time: if a sum doubles in 8 years then R × 8 = 100 → R = 12.5%, and becoming four times needs R × T = 300 → T = 24 years (2× in 8, 3× in 16, 4× in 24). Under compound interest the growth is multiplicative instead: if a sum doubles in 5 years it becomes 4 times in 10 years and 8 times in 15 years, because 2 × 2 = 4 and 2 × 2 × 2 = 8.

Equal instalments. Each instalment must be discounted back to the present. A loan is repaid in two equal annual instalments of ₹882 at 5% per annum compound interest; what was the loan? Loan = 882/1.05 + 882/(1.05)² = 840 + 800 = ₹1,640. Another: two equal instalments of ₹676 at 4% → 676/1.04 + 676/1.0816 = 650 + 625 = ₹1,275. Never simply add the instalments — ₹1,764 is the total paid, not the amount borrowed.

Population growth and depreciation are compound interest in disguise. A town of 20,000 growing at 5% a year holds 20000 × (1.05)² = 22,050 after two years — the same formula as CI. A machine bought for ₹1,00,000 that depreciates 10% a year is worth 100000 × (0.9)³ = ₹72,900 after three years; depreciation is simply compounding with a minus sign. If the rate differs year to year, multiply the factors one by one instead of using a power.

⚠️ Half-yearly is not "half the interest"

For half-yearly compounding halve the rate and double the number of periods; for quarterly, quarter the rate and quadruple the periods. ₹8,000 at 10% for one year compounded half-yearly is 8000 × (1.05)² = ₹8,820, giving CI ₹820 — more than the ₹800 of annual compounding, not less. More frequent compounding always yields more interest, so any option below the annual figure can be eliminated at once.

Time and Work — The LCM Method

Forget fractions; work in units. If A finishes a job in 12 days and B in 18 days, take the total work as LCM(12, 18) = 36 units. Then A does 36/12 = 3 units a day and B does 36/18 = 2 units a day, so together they do 5 units a day and finish in 36/5 = 7.2 days = 7 1/5 days. The classical formula xy/(x + y) = 12 × 18/30 = 7.2 gives the same answer, but the unit method extends to three or more workers without any algebra.

Three workers, no formula needed. A, B and C alone take 10, 12 and 15 days. LCM = 60 units, so their daily rates are 6, 5 and 4 units — total 15 units a day → 4 days together. Note how the LCM turns an addition of fractions into an addition of small whole numbers.

The standard time-and-work shortcuts, each verified
Situation	Shortcut	Worked example
Two together	xy/(x + y) days	12 and 18 days → 216/30 = 7.2 days
Efficiency ratio	Times are in the inverse ratio of efficiencies	A twice as efficient as B, together 14 days → A alone 21, B alone 42
x% more efficient	Divide the time by (1 + x/100)	A 60% more efficient than B who takes 40 days → A takes 40/1.6 = 25 days
Men, days, hours, work	M₁D₁H₁/W₁ = M₂D₂H₂/W₂	12 men × 8 h × 15 d = 1440; 20 men × 9 h × D = 1440 → D = 8 days
Working alternately	Add a two-day cycle, then finish the remainder	A 12 d, B 15 d, A starts → 13 1/4 days
Wages	Divide in the ratio of one-day work	20 d and 30 d workers, ₹3,000 → 3 : 2 → ₹1,800 and ₹1,200

Men, women and boys equivalence. "4 men or 6 women can complete a work in 24 days. How long will 6 men and 6 women take?" First convert: 4 men = 6 women, so 1 man = 3/2 women, or 1 woman = 2/3 man. Total work = 4 × 24 = 96 man-days. Now 6 men + 6 women = 6 + 6 × (2/3) = 6 + 4 = 10 men, so the time is 96 ÷ 10 = 9.6 days. Always convert everything into a single unit before you divide.

Working alternately. A can do a job in 12 days and B in 15 days; they work on alternate days with A starting. Units: LCM = 60, A = 5/day, B = 4/day, so each two-day cycle yields 9 units. Six cycles = 12 days = 54 units. On day 13 A adds 5 → 59 units. Only 1 unit is left and B works at 4 units a day, so he needs 1/4 of a day. Total = 13 1/4 days. Track the cycle, then the leftover — never divide 60 by the average.

Leaving in between. (i) A and B can do a work in 20 and 30 days. They start together but A leaves after 6 days; in how many more days will B finish? Together they do 1/20 + 1/30 = 1/12 per day, so in 6 days half the work is done. B alone finishes the remaining 1/2 at 1/30 a day → 15 more days. (ii) The reverse phrasing: A and B alone take 15 and 10 days, and A leaves 5 days before completion — find the total time T. B works all T days and A works T − 5: (T − 5)/15 + T/10 = 1 → 2(T − 5) + 3T = 30 → 5T = 40 → T = 8 days.

🧠 Wages follow efficiency, not days worked

When two people finish a job together, the pay divides in the ratio of the work each did, which is the ratio of their one-day rates. Workers who alone take 20 and 30 days have rates 1/20 : 1/30 = 3 : 2, so ₹3,000 splits as ₹1,800 and ₹1,200. The instinct to divide 3000 equally, or in the ratio 20 : 30, is the trap — the ratio of times is the inverse of the ratio of work.

Pipes and Cisterns

Pipes and cisterns is time-and-work with a sign. A filling pipe contributes a positive rate; an emptying pipe or a leak contributes a negative one. Worked: pipe A fills a tank in 20 minutes and pipe B empties it in 30 minutes. LCM = 60 units, so A = +3 and B = −2 units per minute → net +1 unit per minute → the tank fills in 60 minutes. If the emptying rate were the larger of the two, the net would be negative and the tank could never fill — an option worth checking for.

The leak question. A cistern is normally filled in 8 hours, but because of a leak it takes 10 hours. How long would the leak alone take to empty a full cistern? 1/8 − 1/L = 1/10 → 1/L = 1/8 − 1/10 = (5 − 4)/40 = 1/40 → 40 hours. Note the subtraction: the leak removes from the filling rate, and its own rate is what is left over.

Two more standard shapes. (i) Two pipes fill a tank in 15 and 20 hours. Both are opened together but the second is closed after some time, and the tank fills in 12 hours. When was it closed? The first ran all 12 hours, contributing 12/15 = 4/5, so the second supplied 1/5 → t/20 = 1/5 → t = 4 hours. (ii) Two taps fill in 10 and 15 minutes and a waste pipe empties; with all three open the tank fills in 18 minutes. How long does the waste pipe alone take? 1/10 + 1/15 − 1/w = 1/18 → 1/6 − 1/w = 1/18 → 1/w = 1/6 − 1/18 = 2/18 = 1/9 → 9 minutes.

💡 A tank is a bank account

Filling pipes are deposits, emptying pipes and leaks are standing debits. All you ever compute is the net monthly balance, and then you ask how long the target takes at that net rate. That one image tells you instantly that a leak lengthens the filling time and that two leaks bigger than the inflow mean the tank never fills — no formula involved.

Time, Speed and Distance

Distance = speed × time, and everything follows. Convert units first, always: km/h → m/s means multiply by 5/18, and m/s → km/h means multiply by 18/5. So 72 km/h = 72 × 5/18 = 20 m/s, 54 km/h = 15 m/s, 63 km/h = 17.5 m/s, and 25 m/s = 25 × 18/5 = 90 km/h. Mixing units is the single largest source of wrong answers in the trains section.

Speed conversion and relative-speed rules
Rule	Statement	Verified example
km/h to m/s	Multiply by 5/18	54 × 5/18 = 15 m/s
m/s to km/h	Multiply by 18/5	15 × 18/5 = 54 km/h
Same direction	Relative speed = difference	45 and 27 km/h → 18 km/h = 5 m/s
Opposite directions	Relative speed = sum	45 and 27 km/h → 72 km/h = 20 m/s
Same distance	Ratio of speeds = inverse ratio of times	Speeds 3 : 4 → times 4 : 3
Average over equal distances	2xy/(x + y), a harmonic mean	60 and 40 → 4800/100 = 48 km/h

⚠️ Equal distances, not equal times

Going 60 km/h and returning 40 km/h gives an average of 48 km/h, not 50 — the harmonic mean, because more time is spent at the slower speed. But one hour at 40 followed by one hour at 60 genuinely averages 50 km/h. Read the question for what is equal: equal distance → 2xy/(x + y); equal time → (x + y)/2. Examiners set both in the same paper precisely to catch a memorised formula.

Early and late arrival. A boy walking at 4 km/h reaches school 5 minutes late; at 5 km/h he arrives 10 minutes early. How far is the school? The two journeys differ by 15 minutes = 1/4 hour, so d/4 − d/5 = 1/4 → d(1/20) = 1/4 → d = 5 km. The shortcut: distance = (product of speeds × time difference)/(difference of speeds) = (4 × 5 × 1/4)/1 = 5 km ✓. Always convert the minutes to hours first — this is where marks leak.

The fraction-of-usual-speed type. Covering a distance at 3/4 of his usual speed, a man is 20 minutes late. What is his usual time? Speed and time are inversely proportional, so 3/4 of the speed means 4/3 of the time; the extra 1/3 of the usual time is the 20 minutes → usual time = 60 minutes. Similarly, at 5/6 of the usual speed the time becomes 6/5, and an extra 1/5 of the usual time of 8 minutes gives a usual time of 40 minutes.

Problems on Trains

A train is not a point — it has length, and that length is part of the distance covered. Crossing a pole, a man, or a tree: distance = length of the train. Crossing a platform, bridge or tunnel: distance = length of train + length of platform. Worked: a 240 m train at 54 km/h = 15 m/s crosses a pole in 240/15 = 16 seconds, and crosses a 360 m platform in 600/15 = 40 seconds.

Two trains crossing each other. The distance is the sum of the two lengths, and the speed is the relative speed. Trains of 120 m and 180 m travel at 45 km/h and 27 km/h. In the same direction the relative speed is 18 km/h = 5 m/s, so they take 300/5 = 60 seconds. In opposite directions it is 72 km/h = 20 m/s, so they take 300/20 = 15 seconds — four times faster, which is exactly why the direction must be read carefully.

A train crossing a walking man. The man has no length, so the distance is just the train length, but his speed still counts. A 150 m train at 63 km/h overtakes a man walking at 3 km/h in the same direction: relative speed = 60 km/h = 50/3 m/s, so the time = 150 ÷ 50/3 = 150 × 3/50 = 9 seconds. Had the man been walking towards the train, the relative speed would be 66 km/h and the time about 8.18 seconds — smaller, as it must be.

🧠 Two crossing times give both length and speed

A train crosses a pole in 15 seconds and a 100 m platform in 25 seconds — find its length and speed. The speed is constant, so L/15 = (L + 100)/25 → 25L = 15L + 1500 → L = 150 m, and the speed = 150/15 = 10 m/s = 36 km/h. Even faster: the extra 10 seconds were spent covering the extra 100 m, so the speed is 100/10 = 10 m/s at once, and the length is 10 × 15 = 150 m.

Boats and Streams

Let b be the speed of the boat in still water and s the speed of the stream. Downstream speed = b + s (the current helps) and upstream speed = b − s (the current resists). Reading it backwards, b = (down + up)/2 and s = (down − up)/2 — the average and the half-difference. Worked: a boat covers 30 km downstream in 2 hours and returns in 3 hours. Down = 15 km/h, up = 10 km/h, so b = 12.5 km/h and s = 2.5 km/h.

Round-trip time and average speed. A boat rows at 15 km/h in still water and the stream runs at 3 km/h. For a 36 km trip each way: downstream 36/18 = 2 hours, upstream 36/12 = 3 hours, total 5 hours. Average speed = 72/5 = 14.4 km/h, which the formula (b² − s²)/b confirms: (225 − 9)/15 = 216/15 = 14.4 ✓. Notice it is below the still-water speed of 15 — a current always costs you time on a round trip, never saves it.

The classic two-condition question. A man rows to a place 48 km away and comes back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against it. Find the speed of the stream. The second sentence gives the ratio down : up = 4 : 3, so write down = 4k and up = 3k. Then 48/4k + 48/3k = 14 → 12/k + 16/k = 28/k = 14 → k = 2. Hence down = 8 km/h, up = 6 km/h, so b = 7 km/h and s = 1 km/h. Verify: 48/8 + 48/6 = 6 + 8 = 14 hours ✓.

📖 The escalator, the wind and the river are one question

A man on a moving escalator, an aeroplane in a tailwind, and a boat in a current are the same problem wearing different clothes: a base speed plus or minus a medium speed. Walking down a moving escalator is "downstream"; walking up against it is "upstream". So if you can do boats and streams, you can already do escalator and wind questions — and the paper does set all three.

🎯 Guessing strategy under −0.50

With four options and −0.50 for a wrong answer, a blind guess has an expected value of (1/4)(+2) + (3/4)(−0.5) = 0.5 − 0.375 = +0.125, so a pure guess is very slightly positive — but a guess after eliminating one option is worth (1/3)(2) + (2/3)(−0.5) = +0.33, and after eliminating two it is (1/2)(2) + (1/2)(−0.5) = +0.75. That is the real argument for learning the elimination checks in this chapter: units, order of magnitude, and the sanity rules (a round trip is always slower than the still-water speed, a leak always lengthens the filling time, equal-percentage same-price sales are always a loss).

Key takeaways

Profit% and loss% always sit on the CP, discount% always sits on the MP. To go backwards from SP, divide by (100 ± p)/100 — never subtract the percentage from the SP.
Two articles at the same selling price with +x% and −x% always give a net loss of (x/10)²%: 4% for x = 20, 2.25% for x = 15. A false weight gives gain% = error/(true weight − error) × 100 — so a 900 g weight is 11 1/9%.
Successive discounts a% and b% equal a single a + b − ab/100, and markup-then-discount is the same formula with a plus and a minus: +60% then −20% then −25% is a 4% loss.
CI − SI = P(R/100)² for 2 years and P(R/100)²(300 + R)/100 for 3 years. Half-yearly compounding means halve R and double t; more frequent compounding always yields more interest.
For time and work take the total as the LCM of the individual times, convert everyone into units per day, and add. Wages divide in the ratio of one-day work, which is the inverse of the ratio of days.
Pipes are work with signs: fillers positive, leaks negative. A tank filled in 8 h but taking 10 h with a leak means the leak alone empties it in 40 h.
Convert km/h to m/s with 5/18 before touching a train question. Crossing a pole uses the train length; crossing a platform adds the platform; crossing another train adds both lengths at the relative speed.
Boats: b = (down + up)/2 and s = (down − up)/2, and the round-trip average speed is (b² − s²)/b, always less than b. Escalator and wind questions are the same structure.
Under −0.50 negative marking a blind guess is worth only +0.125 marks, but a guess after eliminating two options is worth +0.75 — so learn the sanity checks, not just the formulae.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A shopkeeper sells two articles at ₹1,020 each. On one he gains 15% and on the other he loses 15%. What is his overall result?
1. No profit, no loss
2. Loss of 2.25%
3. Profit of 2.25%
4. Loss of 15%
Show answer
Answer: B — Loss of 2.25%
When two articles are sold at the same price, one at +x% and one at −x%, the result is always a loss of (x/10)²%. Here x = 15, so the loss = (15/10)² = 1.5² = 2.25%. The longer route confirms it: CP₁ = 1020 × 100/115 ≈ ₹886.96 and CP₂ = 1020 × 100/85 = ₹1,200, total ≈ ₹2,086.96 against SP ₹2,040, a loss of about ₹46.96 which is 2.25% of ₹2,086.96. The loss falls on the bigger cost price, so it can never be a profit or a break-even.
A trader marks his goods 60% above the cost price and then allows two successive discounts of 20% and 25%. His net result is:
1. A profit of 15%
2. A profit of 4%
3. No profit, no loss
4. A loss of 4%
Show answer
Answer: D — A loss of 4%
Take CP = ₹100. MP = ₹160. After 20% off: 160 × 0.80 = ₹128. After a further 25% off: 128 × 0.75 = ₹96. Since 96 < 100 he makes a 4% loss. As a chain of successive changes: +60%, −20%, −25% gives 1.60 × 0.80 × 0.75 = 0.96, that is −4%. This is the standard warning that a large markup does not guarantee a profit once the discount chain is long — the two discounts together equal a single 40% discount, and 160 × 0.60 = 96 as well.
The difference between the compound interest and the simple interest on a certain sum for 2 years at 10% per annum is ₹150. Find the sum.
1. ₹7,500
2. ₹12,000
3. ₹15,000
4. ₹18,000
Show answer
Answer: C — ₹15,000
For 2 years, CI − SI = P(R/100)². So 150 = P × (10/100)² = P × 0.01 → P = 150 ÷ 0.01 = ₹15,000. Full check: CI = 15000 × 1.21 − 15000 = ₹3,150 and SI = 15000 × 0.10 × 2 = ₹3,000, difference exactly ₹150 ✓. The reason the formula works: the whole difference is one year of interest on the first year of interest.
A and B can complete a piece of work in 20 days and 30 days respectively. They start together, but A leaves after 6 days. In how many more days will B finish the remaining work?
1. 15 days
2. 12 days
3. 18 days
4. 10 days
Show answer
Answer: A — 15 days
Use the LCM method: total work = LCM(20, 30) = 60 units, so A does 3 units a day and B does 2, together 5 units a day. In 6 days they complete 30 units, leaving 30 units. B alone works at 2 units a day, so he needs 30 ÷ 2 = 15 more days. Cross-check with fractions: together 1/20 + 1/30 = 1/12 per day, so 6 days finishes half the job, and B alone does the remaining half at 1/30 a day → 15 days ✓.
A cistern is normally filled by a tap in 8 hours, but because of a leak in the bottom it takes 10 hours. How long would the leak alone take to empty a full cistern?
1. 18 hours
2. 20 hours
3. 25 hours
4. 40 hours
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Answer: D — 40 hours
The tap fills 1/8 of the tank per hour, and with the leak the net rate is only 1/10 per hour. So the leak removes 1/8 − 1/10 = (5 − 4)/40 = 1/40 of the tank per hour, meaning it empties a full cistern in 40 hours. Option 18 comes from wrongly adding 8 and 10, and option 40 is confirmed by a sanity check: the leak must be slower than the tap, otherwise the tank would never fill at all.
A train 150 metres long, running at 63 km/h, overtakes a man walking at 3 km/h in the same direction. How long does the train take to pass him completely?
1. 9 seconds
2. 10 seconds
3. 8.18 seconds
4. 12 seconds
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Answer: A — 9 seconds
A man has no length, so the distance to cover is just the 150 m of train. Both move in the same direction, so use the relative speed 63 − 3 = 60 km/h = 60 × 5/18 = 50/3 m/s. Time = 150 ÷ (50/3) = 150 × 3/50 = 9 seconds. Option 10 s is what you get by ignoring the man and using 15 m/s; option 8.18 s is the answer for the opposite direction, where the relative speed would be 66 km/h.
A man rows to a place 48 km away and comes back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against it. The speed of the stream is:
1. 3 km/h
2. 2 km/h
3. 1 km/h
4. 1.5 km/h
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Answer: C — 1 km/h
Equal times for 4 km downstream and 3 km upstream means downstream : upstream = 4 : 3, so write them as 4k and 3k. Then 48/(4k) + 48/(3k) = 14 → 12/k + 16/k = 28/k = 14 → k = 2. So downstream = 8 km/h and upstream = 6 km/h, giving boat speed b = (8 + 6)/2 = 7 km/h and stream speed s = (8 − 6)/2 = 1 km/h. Verify: 48/8 + 48/6 = 6 + 8 = 14 hours ✓.
A car travels from town P to town Q at 60 km/h and returns along the same road at 40 km/h. Consider the following statements:
1. Its average speed for the whole journey is 50 km/h.
2. Its average speed for the whole journey is 48 km/h.
Which is/are correct?
1. Only 1
2. Only 2
3. Both 1 and 2
4. Neither 1 nor 2
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Answer: B — Only 2
The two distances are equal, so the average speed is the harmonic mean 2xy/(x + y) = 2 × 60 × 40/(60 + 40) = 4800/100 = 48 km/h, making statement 2 correct and statement 1 wrong. Verify with a concrete distance of 120 km each way: 120/60 = 2 hours out and 120/40 = 3 hours back, so 240 km in 5 hours = 48 km/h. The arithmetic mean 50 would be correct only if equal times were spent at each speed, not equal distances.