Arithmetic Aptitude: Percentage, Interest, Ratio, Speed & Work

Of the five Unit-5 questions in every UGC NET Paper-1 sitting, two or three are pure arithmetic — a percentage, an interest, a ratio, an average, a train or a pipe — worth 2 marks each with no negative marking. The first chapter of this subject dealt with series, codes and relations; this one is devoted entirely to the arithmetic half, and to the named slips that produce the wrong options. Every distractor a setter writes is a specific mistake: adding percentages that should be compounded, taking a discount on cost price instead of marked price, quoting the amount when the interest was asked, averaging two speeds arithmetically. Each worked example below shows every line of the correct working and names the slip beside it, so that in the hall you recognise a wrong option as a wrong method rather than merely a different number.

Fractions, Decimals and Percentage Change

Speed in this unit comes from refusing to do long multiplication. Multiplying by 5 is multiplying by 10 and halving: 348 × 5 = 3480 ÷ 2 = 1740. Dividing by 25 is multiplying by 4 and dividing by 100: 1375 ÷ 25 = 5500 ÷ 100 = 55. Squaring a number ending in 5: multiply the tens digit by the next integer and write 25 after it — 65² → 6 × 7 = 42 → 4225; 85² → 8 × 9 = 72 → 7225. Recurring decimals convert by writing the repeating block over as many 9s as it has digits: 0.3636… = 36/99 = 4/11; 0.1666… = (16 − 1)/90 = 15/90 = 1/6. And any decimal with three places is a fraction over 1000: 0.125 = 125/1000 = 1/8, 0.375 = 3/8, 0.625 = 5/8.

Percentage change as a multiplier and a fraction — the form that makes every question one multiplication
Change	Multiply by	As a fraction	Reverse change to undo it
+20%	1.20	6/5	−16⅔% (multiply by 5/6)
+25%	1.25	5/4	−20% (multiply by 4/5)
+33⅓%	1.333…	4/3	−25% (multiply by 3/4)
+12.5%	1.125	9/8	−11⅑% (multiply by 8/9)
−10%	0.90	9/10	+11⅑% (multiply by 10/9)
−37.5%	0.625	5/8	+60% (multiply by 8/5)
−50%	0.50	1/2	+100% (multiply by 2)

Successive percentage changes of a% and b% combine to a net change of a + b + ab/100, with signs carried. Worked example 1. A price rises 20% and then falls 10%: net = 20 + (−10) + (20 × −10)/100 = 10 − 2 = +8%. Check on ₹100: 100 → 120 → 108 ✓. The slip is to add and get +10%. Worked example 2. A rectangle's length increases 10% and breadth 20%: area changes by 10 + 20 + (10 × 20)/100 = +32%, not 30%. Worked example 3. A town's population grows 10% a year for two years: 10 + 10 + 1 = +21%; on 40,000 people that is 40,000 × 1.21 = 48,400. Worked example 4 — the reverse base. If A's salary is 25% more than B's, B's is less than A's by 100 × 25/(100 + 25) = 2500/125 = 20%, not 25% — because the base has switched from B to A. The general rule: a rise of R% is undone by a fall of 100R/(100 + R)%, and a fall of R% by a rise of 100R/(100 − R)%.

Worked example 5 — the pass-mark type with one candidate. A candidate needs 33% to pass, scores 125 marks and fails by 40. Failing by 40 means the pass mark is 125 + 40 = 165. Since 165 is 33% of the total, total = 165 ÷ 0.33 = 165 × 100/33 = 500 marks. Check: 33% of 500 = 165 ✓. Worked example 6 — x% of y. Because x% of y = xy/100 = y% of x, "8% of 150" is the same as "150% of 8" = 8 × 1.5 = 12 — switch whichever way is easier. And a percentage-point change is not a percentage change: if a party's vote share moves from 40% to 50%, that is a rise of 10 percentage points but a rise of 10/40 = 25 per cent in its share.

⚠️ Equal and opposite is never zero

A rise of x% followed by a fall of x% (or the reverse order) always leaves a net fall of x²/100 per cent, because the fall acts on a larger base. Up 20% then down 20%: 20 − 20 − 400/100 = −4% (100 → 120 → 96). Up 10% then down 10% = −1%. Up 50% then down 50% = −25%. The option reading "no change" is the setter's favourite distractor, and it is wrong every single time.

Profit, Loss, Discount and Marked Price

Three prices exist: the cost price (CP) the trader paid, the marked price (MP) written on the tag, and the selling price (SP) actually received. Two percentages exist, each on its own base: profit or loss per cent is always on CP, and discount per cent is always on MP. So SP = CP × (100 + profit%)/100 and SP = MP × (100 − discount%)/100, and the two expressions for SP are the bridge between the two worlds. The single most-tested slip is taking the discount on CP, or the profit on MP.

Worked example 1 — from tag to cost. An article is marked ₹1,200, sold at 15% discount, and the trader still gains 27.5%. Find CP. Discount acts on MP: SP = 1200 × 0.85 = ₹1,020. Profit acts on CP: 1020 = CP × 1.275, so CP = 1020 ÷ 1.275 = ₹800. Check: 800 × 1.275 = 1020 ✓. Named slips: stopping at ₹1,020 (forgetting the profit step); computing 1020 × 0.725 = ₹739.50 (subtracting the profit percentage from SP instead of dividing); computing 1200 ÷ 1.275 = ₹941.18 (ignoring the discount). Worked example 2 — the marked-up price. To gain 20% after allowing 20% discount, how much above cost must the trader mark? MP × 0.8 = CP × 1.2, so MP = CP × 1.2/0.8 = CP × 1.5 — a 50% mark-up, not 40%.

Profit-and-loss formulas with a verified number for each
Situation	Formula	Verified example
Profit % on cost	(SP − CP) × 100 / CP	CP ₹400, SP ₹460: 60 × 100/400 = 15%.
Discount % on marked price	(MP − SP) × 100 / MP	MP ₹2,500, SP ₹2,125: 375 × 100/2500 = 15%.
Two successive discounts a%, b%	a + b − ab/100	10% and 20%: 10 + 20 − 2 = 28%; on ₹5,000 the price is ₹3,600, not ₹3,500.
False weight (sells w g as 1000 g)	Gain % = (1000 − w) × 100 / w	w = 900: 100 × 100/900 = 11⅑% — on the 900 g actually given, not on 1000.
Same SP, one gains x%, other loses x%	Net loss of x²/100 %	x = 25: loss 6.25%. SP ₹1,500 each: CPs 1200 and 2000; total CP 3200, SP 3000, loss 200/3200 = 6.25% ✓.
Mark-up m% then discount d%, profit	m − d − md/100	Mark up 40%, discount 25%: 40 − 25 − 10 = 5% profit.

Worked example 3 — profit given in rupees on two prices. A trader sells a chair for ₹1,080 at a 10% loss. At what price must he sell it to gain 15%? The loss is on CP: 1080 = CP × 0.90, so CP = 1080 ÷ 0.9 = ₹1,200. For 15% gain: SP = 1200 × 1.15 = ₹1,380. The slip 1080 × 1.25 = ₹1,350 comes from adding the two percentages (10 + 15) and applying them to the wrong base. Worked example 4 — cost of the article in terms of selling price. A fruit-seller sells 12 mangoes for the cost price of 15. His gain per cent is (15 − 12)/12 × 100 = 3/12 × 100 = 25% — the gain is measured on the 12 he actually gave, i.e., on cost.

🧠 Read the base before you read the number

Before touching any figure in a profit-loss question, write three letters on your sheet — CP / MP / SP — and put each rupee amount in the question under one of them. Then write P on CP, D on MP. Half of all wrong options vanish the moment you know which base a percentage belongs to, because they were generated by a wrong base. In the ₹1,200 example the three named wrong options were all base errors, and none of them can survive the three-letter check.

Simple and Compound Interest

Simple interest is paid on the original principal only: SI = P × R × T / 100, and the amount is A = P + SI. Worked example 1. ₹6,000 at 7.5% per annum for 4 years: SI = 6000 × 7.5 × 4 / 100 = 6000 × 30 / 100 = ₹1,800; the amount is ₹7,800. Named slips: ₹450 is one year's interest only; ₹7,800 is the amount, which is wrong if the question asked for interest; ₹1,350 is three years. Compound interest re-invests each year's interest, so the amount after n years is A = P × (1 + R/100)ⁿ and CI = A − P. Worked example 2 — two years. ₹12,000 at 5% for 2 years: year 1 → 12,000 × 1.05 = 12,600; year 2 → 12,600 × 1.05 = 13,230; CI = 13,230 − 12,000 = ₹1,230. Shortcut: the effective two-year rate is 5 + 5 + 0.25 = 10.25%, and 10.25% of 12,000 = 1,230 ✓.

Worked example 3 — three years, line by line. ₹5,000 at 8% per annum compounded annually for 3 years. Year 1: interest = 8% of 5,000 = 400, amount = 5,400. Year 2: interest = 8% of 5,400 = 432, amount = 5,832. Year 3: interest = 8% of 5,832 = 466.56, amount = 6,298.56. So CI = 6,298.56 − 5,000 = ₹1,298.56. Cross-check with the effective three-year rate 3r + 3r²/100 + r³/10000 = 24 + 1.92 + 0.0512 = 25.9712%, and 25.9712% of 5,000 = 1,298.56 ✓. Named slips: ₹1,200 is the simple interest (5000 × 8 × 3/100); ₹832 is the two-year CI; ₹6,298.56 is the amount. Half-yearly compounding halves the rate and doubles the periods: ₹8,000 at 10% p.a. compounded half-yearly for 1 year = 8,000 × 1.05 × 1.05 = 8,820, CI ₹820, against ₹800 with annual compounding.

Compound multipliers (1 + R/100)ⁿ worth knowing by heart
Rate	2 years	3 years	Effective 2-year and 3-year rates
4%	1.0816	1.124864	8.16% and 12.4864%
5%	1.1025	1.157625	10.25% and 15.7625%
8%	1.1664	1.259712	16.64% and 25.9712%
10%	1.21	1.331	21% and 33.1%
20%	1.44	1.728	44% and 72.8%

The difference between CI and SI is the interest earned on interest. For 2 years it is exactly P × (R/100)²; for 3 years it is P × (R/100)² × (3 + R/100). Worked example 4. On ₹20,000 at 5% for 3 years: (R/100)² = 0.0025, and 3 + 0.05 = 3.05, so the difference = 20,000 × 0.0025 × 3.05 = 50 × 3.05 = ₹152.50. Verify the long way: CI = 20,000 × 1.157625 − 20,000 = 23,152.50 − 20,000 = 3,152.50; SI = 20,000 × 5 × 3/100 = 3,000; gap = 152.50 ✓. Reverse questions use the same formula: if the two-year gap is ₹64 at 8%, then P × 0.0064 = 64, so P = ₹10,000.

🧠 The rule of 72 and the doubling chain

At compound interest a sum doubles in roughly 72 ÷ R years. At 8% that is 9 years (1.08⁹ ≈ 1.999); at 12% it is 6 years (1.12⁶ ≈ 1.974); at 6% it is 12 years (1.06¹² ≈ 2.012). The rule is an approximation, good between about 4% and 15%, and is used to eliminate options, not to compute exact answers. Under simple interest doubling takes exactly 100 ÷ R years — 12.5 years at 8% — which is the distractor placed next to the rule-of-72 answer. And once a sum doubles in n years at CI, it becomes 4 times in 2n years and 8 times in 3n years: doubling in 4 years means eightfold in 12.

Ratio, Proportion, Partnership and Averages

Worked example 1 — joining two ratios. If A : B = 2 : 3 and B : C = 4 : 5, B is 3 in one ratio and 4 in the other; make it the LCM, 12. Multiply the first ratio by 4 → A : B = 8 : 12; multiply the second by 3 → B : C = 12 : 15. So A : B : C = 8 : 12 : 15. Now divide ₹7,000 in that ratio: total parts = 8 + 12 + 15 = 35, one part = 7000 ÷ 35 = ₹200, so the shares are ₹1,600, ₹2,400 and ₹3,000 (sum 7,000 ✓). The slip is to write 2 : 3 : 5 or 2 : 4 : 5 by gluing the ratios without equalising B. Worked example 2 — proportion. The fourth proportional to 6, 9 and 14 satisfies 6 : 9 = 14 : x, so 6x = 126 and x = 21. The mean proportional between 4 and 25 is √(4 × 25) = √100 = 10. In inverse proportion the product is constant: if 15 men finish a wall in 20 days, 25 men need 15 × 20 ÷ 25 = 12 days.

Partnership shares profit in the ratio of capital × time, never capital alone. Worked example 3. A invests ₹25,000 for the whole year; B joins after 3 months with ₹40,000, so B's money works for 9 months. A's weight = 25,000 × 12 = 3,00,000; B's weight = 40,000 × 9 = 3,60,000. The ratio is 300 : 360 = 5 : 6. On a profit of ₹22,000, one part = 22,000 ÷ 11 = 2,000, so A gets ₹10,000 and B gets ₹12,000. Named slips: using 25 : 40 = 5 : 8 (capital only) gives B ₹13,538; using 12 : 9 (time only) gives B ₹9,429. If B had joined at the start with the same money the ratio would have been 5 : 8 — which is exactly why the months matter.

An average is a total in disguise: total = average × count. Worked example 4. A batsman averages 32 in 10 innings; how many must he score in the 11th to raise his average to 42? Present total = 32 × 10 = 320; required total = 42 × 11 = 462; needed = 462 − 320 = 142. The slip is 42 + (42 − 32) = 52, which forgets that the deficit of 10 must be made up on every one of the ten earlier innings. Worked example 5 — replacement. The average age of 8 committee members rises by 2 years when a 30-year-old member is replaced. The total rose by 8 × 2 = 16, so the newcomer is 30 + 16 = 46 years old. Worked example 6 — a weighted average. Section A (40 students) averages 62 and Section B (60 students) averages 72. Combined average = (40 × 62 + 60 × 72) ÷ 100 = (2,480 + 4,320) ÷ 100 = 68. The slip is the plain mean (62 + 72)/2 = 67; the answer must lie nearer to 72 because more students sit in B. By alligation, the two groups are in the ratio (72 − 68) : (68 − 62) = 4 : 6 = 2 : 3 = 40 : 60 ✓ — the same rule that mixes ₹30/kg rice with ₹45/kg rice to sell at ₹36/kg in the ratio (45 − 36) : (36 − 30) = 9 : 6 = 3 : 2.

Average of the first n natural numbers = (n + 1)/2; of consecutive numbers or any arithmetic sequence = (first + last)/2; of the first n even numbers = n + 1; of the first n odd numbers = n.
If every value is increased by k, the average rises by k; if every value is multiplied by k, the average is multiplied by k.
When a member leaves and the average falls by d, the leaver was above the new average by d × (new count) + … — always go back to totals rather than memorising this case.
A ratio is not a quantity: 3 : 5 tells you nothing about the actual numbers until a total or a difference is given — "the difference is 40" makes one part 20 and the values 60 and 100.

Time, Speed and Distance; Time and Work

Everything here rests on distance = speed × time and one conversion: 1 km/h = 5/18 m/s. So 54 km/h = 54 × 5/18 = 15 m/s, 90 km/h = 25 m/s, 36 km/h = 10 m/s. Relative speed is the sum when two bodies move towards each other or in opposite directions, and the difference when they move the same way. A train crossing a pole travels its own length; crossing a platform or bridge it travels its own length plus the platform's. Worked example 1. A 180 m train at 54 km/h (= 15 m/s) passes a pole in 180 ÷ 15 = 12 s, and a 240 m platform in (180 + 240) ÷ 15 = 420 ÷ 15 = 28 s. Named slips: 16 s uses the platform length alone; 7.8 s divides 420 by 54 without converting units. Worked example 2. Trains of 150 m and 100 m run at 60 km/h and 30 km/h. Moving towards each other, relative speed = 90 km/h = 25 m/s, and they cross in (150 + 100) ÷ 25 = 10 s. Moving the same way, relative speed = 30 km/h = 25/3 m/s, so the faster overtakes in 250 × 3/25 = 30 s.

Average speed is total distance over total time, never the mean of the speeds. For equal distances at speeds x and y it is the harmonic mean 2xy/(x + y). Worked example 3. A cyclist rides to college at 30 km/h and returns at 60 km/h. Average speed = 2 × 30 × 60/(30 + 60) = 3600/90 = 40 km/h, not 45. Verify with a 60 km leg: going takes 2 h, returning 1 h, total 120 km in 3 h = 40 km/h ✓. For equal times the average is the plain mean — one hour at 30 and one hour at 60 does give 45. Boats and streams: downstream speed = boat + stream, upstream = boat − stream, so boat speed = (downstream + upstream)/2 and stream speed = (downstream − upstream)/2; a boat doing 12 km/h down and 8 km/h up has speed 10 km/h in still water and the stream flows at 2 km/h.

🎯 Why the LCM method beats fractions every time

The textbook method adds 1/12 + 1/18 and inverts; under time pressure that is where errors enter. Instead declare the total work to be the LCM of the days, so that every rate becomes a whole number of units per day. A finishes in 12 days and B in 18 → work = LCM(12, 18) = 36 units; A does 36/12 = 3 units/day, B does 36/18 = 2 units/day; together 5 units/day, so the job takes 36 ÷ 5 = 7.2 days (7 days and 4.8 hours). Emptying pipes and workers who leave are simply negative units or units removed from the total — the arithmetic never leaves the integers until the very last division.

Worked example 4 — pipes with an outlet. Pipe A fills a tank in 6 hours, pipe B in 8 hours, and outlet C empties it in 12 hours. All three are opened together. Work = LCM(6, 8, 12) = 24 units. A fills 24/6 = 4 units/h, B fills 24/8 = 3 units/h, C removes 24/12 = 2 units/h. Net = 4 + 3 − 2 = 5 units/h, so the tank fills in 24 ÷ 5 = 4.8 h = 4 h 48 min. Named slips: treating C as an inlet gives 24/9 = 2 h 40 min; ignoring C gives 24/7 ≈ 3 h 26 min; subtracting B as well gives 24/3 = 8 h. Worked example 5 — the chain rule. 12 men working 8 hours a day finish a road in 15 days. How many days will 10 men working 9 hours a day take? Man-hours are constant: 12 × 8 × 15 = 10 × 9 × D, so D = 1440 ÷ 90 = 16 days. Worked example 6 — a worker leaves. A and B can do a job in 12 and 24 days. They start together, and A leaves after 4 days. Work = 24 units; A = 2 units/day, B = 1 unit/day. In 4 days together they do 4 × 3 = 12 units; the remaining 12 units are B's alone at 1 unit/day → 12 more days, so the job takes 16 days in all.

Convert every speed to the unit of the lengths before dividing — km/h × 5/18 → m/s.
Decide whether the distance is the train's own length (pole, man, signal) or train plus object (platform, bridge, tunnel, another train).
Add speeds for opposite directions, subtract for the same direction.
For equal distances use 2xy/(x + y); for equal times use (x + y)/2.
For work, set total = LCM of the days, turn every worker or pipe into units per day, treat outlets as negative, and divide only at the end.

Key takeaways

Successive changes of a% and b% net to a + b + ab/100; equal and opposite changes always lose x²/100 per cent, and a rise of R% is undone by a fall of 100R/(100 + R)%.
Profit and loss sit on cost price, discount on marked price; the same SP with x% gain and x% loss is a net loss of x²/100%, and a false weight of w grams gains (1000 − w)/w × 100%.
CI amount = P(1 + R/100)ⁿ; the CI − SI gap is P(R/100)² for two years and P(R/100)²(3 + R/100) for three; a sum doubles in about 72/R years at CI and exactly 100/R years at SI.
Join ratios through the LCM of the shared term; partnership profit follows capital × time; an average is a total in disguise, and a weighted average is pulled towards the larger group (alligation gives the weights).
Trains: convert with 5/18, add the platform length, add speeds when opposite; equal-distance average speed is 2xy/(x + y); time-and-work is solved with total work = LCM of the days and outlets as negative units.

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.

The price of a commodity falls by 20% and then rises by 20%. Compared with the original price, the final price is
1. unchanged
2. 4% lower
3. 4% higher
4. 2% lower
Show answer
Answer: B — 4% lower
Use a + b + ab/100 with a = −20 and b = +20: −20 + 20 + (−20 × 20)/100 = 0 − 4 = −4%. Check on ₹100: after the fall 80, after a 20% rise on 80 the price is 96 — a 4% fall. The value "unchanged" is the slip of adding −20 and +20 and forgetting that the rise acts on the smaller base of 80; the value 4% higher has the right size but the wrong sign — equal and opposite percentage changes always produce a loss of x²/100 per cent, never a gain; the value 2% lower halves the cross term for no reason.
What is the simple interest on ₹6,000 at 7.5% per annum for 4 years?
1. ₹1,800
2. ₹450
3. ₹7,800
4. ₹1,350
Show answer
Answer: A — ₹1,800
SI = P × R × T / 100 = 6,000 × 7.5 × 4 / 100. Multiply 7.5 × 4 = 30 first, then 6,000 × 30 / 100 = 6,000 × 0.30 = ₹1,800. The value ₹450 is the interest for one year only (6,000 × 7.5/100), the slip of dropping T. The value ₹7,800 is the amount — principal plus interest — which is the answer to a different question; a setter places it because candidates who compute correctly then add the principal by habit. The value ₹1,350 is three years' interest, a miscount of the period.
Find the compound interest on ₹5,000 at 8% per annum, compounded annually, for 3 years.
1. ₹1,200
2. ₹1,298.56
3. ₹832
4. ₹6,298.56
Show answer
Answer: B — ₹1,298.56
Compound year by year. Year 1: 5,000 × 1.08 = 5,400. Year 2: 5,400 × 1.08 = 5,832. Year 3: 5,832 × 1.08 = 6,298.56. CI = 6,298.56 − 5,000 = ₹1,298.56; equivalently the effective three-year rate 24 + 1.92 + 0.0512 = 25.9712% of 5,000. The value ₹1,200 is the simple interest 5,000 × 8 × 3/100 — the slip of not compounding at all. The value ₹832 is the compound interest for only two years (5,832 − 5,000), a period miscount. The value ₹6,298.56 is the amount, not the interest, the slip of forgetting to subtract the principal.
An article marked at ₹1,200 is sold at a discount of 15%, and the seller still makes a profit of 27.5%. The cost price of the article is
1. ₹1,020
2. ₹739.50
3. ₹800
4. ₹941.18
Show answer
Answer: C — ₹800
Discount acts on the marked price: SP = 1,200 × 0.85 = ₹1,020. Profit acts on the cost price: 1,020 = CP × 1.275, so CP = 1,020 ÷ 1.275 = ₹800 (check: 800 × 1.275 = 1,020). The value ₹1,020 is the selling price — the slip of stopping after the discount and never applying the profit. The value ₹739.50 comes from 1,020 × 0.725, i.e., subtracting 27.5% of the selling price instead of dividing by 1.275, which puts the profit on the wrong base. The value ₹941.18 is 1,200 ÷ 1.275, the slip of ignoring the discount and treating the marked price as the selling price.
A train 180 m long is running at 54 km/h. How long will it take to cross a platform 240 m long?
1. 12 seconds
2. 16 seconds
3. 7.8 seconds
4. 28 seconds
Show answer
Answer: D — 28 seconds
Convert the speed: 54 km/h × 5/18 = 15 m/s. To clear a platform the train must cover its own length plus the platform: 180 + 240 = 420 m. Time = 420 ÷ 15 = 28 seconds. The value 12 seconds is 180 ÷ 15 — the time to pass a pole, the slip of forgetting the platform length. The value 16 seconds is 240 ÷ 15 — the slip of using the platform alone and forgetting the train's own length. The value 7.8 seconds is 420 ÷ 54, which divides metres by kilometres per hour without converting units.
In a college, Section A has 40 students with an average score of 62 and Section B has 60 students with an average score of 72. The average score of all 100 students is
1. 68
2. 67
3. 66
4. 70
Show answer
Answer: A — 68
A combined average is total marks over total students: (40 × 62 + 60 × 72) ÷ 100 = (2,480 + 4,320) ÷ 100 = 6,800 ÷ 100 = 68. Alligation confirms it: 68 splits 62 and 72 in the ratio (72 − 68) : (68 − 62) = 4 : 6 = 2 : 3, which is exactly 40 : 60. The value 67 is the plain mean (62 + 72)/2, the slip of ignoring that the sections are of different sizes; the answer must lie nearer to 72 because B is larger. The value 66 comes from swapping the weights — (60 × 62 + 40 × 72)/100 — and pulls the average the wrong way. The value 70 is 72 − 2, a guess that has no arithmetic behind it.
Pipe A can fill a tank in 6 hours and pipe B in 8 hours, while an outlet pipe C can empty the full tank in 12 hours. If all three pipes are opened together on an empty tank, in how much time will the tank be full?
1. 3 hours 26 minutes
2. 8 hours
3. 4 hours 48 minutes
4. 2 hours 40 minutes
Show answer
Answer: C — 4 hours 48 minutes
Take the tank as LCM(6, 8, 12) = 24 units. A fills 24/6 = 4 units/h, B fills 24/8 = 3 units/h, and C removes 24/12 = 2 units/h. Net inflow = 4 + 3 − 2 = 5 units/h, so the tank fills in 24 ÷ 5 = 4.8 h = 4 hours 48 minutes. The value 2 hours 40 minutes is 24 ÷ 9, the slip of treating the outlet C as a third filling pipe (4 + 3 + 2). The value 3 hours 26 minutes is 24 ÷ 7, the slip of ignoring the outlet altogether. The value 8 hours is 24 ÷ 3, the slip of subtracting B as well as C (4 − 3 + 2), i.e., misreading which pipes fill and which empties.
A starts a business with ₹25,000. After 3 months B joins with ₹40,000. At the end of the year the profit is ₹22,000. B's share of the profit is
1. ₹13,538
2. ₹10,000
3. ₹9,429
4. ₹12,000
Show answer
Answer: D — ₹12,000
Partnership profit is shared in the ratio of capital × time. A: 25,000 × 12 months = 3,00,000. B: 40,000 × 9 months (joined after 3 months) = 3,60,000. Ratio = 300 : 360 = 5 : 6, so 11 parts = ₹22,000 and one part = ₹2,000. B receives 6 × 2,000 = ₹12,000; A receives ₹10,000. The value ₹10,000 is A's share, the slip of reading the wrong partner. The value ₹13,538 comes from the capital ratio alone, 25 : 40 = 5 : 8, giving 22,000 × 8/13 — the slip of ignoring that B's money worked for only nine months. The value ₹9,429 comes from the time ratio alone, 12 : 9, giving 22,000 × 9/21 — the slip of ignoring the capitals.