Time & Work, Pipes, Mixtures & Ages

These four topics look like four different chapters. They are not. Time and work, pipes and cisterns, mixtures, and ages are one idea wearing four costumes — in each of them something is being added at a fixed rate, and your only job is to find that rate and add it up. A labourer adds work per day. A pipe adds water per hour. A mixture adds price per litre. A person adds one year per year. Learn the rate habit once and all four collapse into simple arithmetic. In this chapter we will not hand you formulas to memorise — we will work out every single example on paper, number by number, so you can see where each digit came from.

Time & Work: Turn the Work into a Number

The textbook method is the unitary (rate) method: if a worker finishes a job in 12 days, then in one day he finishes 1/12 of it. Add the fractions of everyone working, and the reciprocal of the sum is the time taken together. It always works — but it drowns you in fractions with unfriendly denominators.

🎯 Why the LCM method exists

Nowhere does the question tell you how big the work is. So you are free to choose its size — and a clever choice removes every fraction. Choose the total work to be the LCM of the given days. Then each worker's one-day output becomes a whole number of units, and the entire sum becomes counting, not fraction addition. Same answer, one-third of the time. In an exam where you get roughly a minute per question, that is the whole game.

Worked example 1 — both methods side by side. A can finish a job in 12 days, B in 18 days. Working together, how long?

Unitary method: one day's work = 1/12 + 1/18. LCM of 12 and 18 is 36, so = 3/36 + 2/36 = 5/36. Time = 36/5 = 7.2 days (7 days and 1/5 of a day).

LCM method: let total work = LCM(12, 18) = 36 units. Then A does 36 ÷ 12 = 3 units/day, B does 36 ÷ 18 = 2 units/day. Together = 3 + 2 = 5 units/day. Time = 36 ÷ 5 = 7.2 days. Identical answer, and not one fraction was written until the last step.

Worked example 2 — three workers. Total work = LCM(10, 12, 15) = 60 units
Worker	Days alone	Units per day = 60 ÷ days
A	10	60 ÷ 10 = 6
B	12	60 ÷ 12 = 5
C	15	60 ÷ 15 = 4
A + B + C together	?	6 + 5 + 4 = 15 units/day

From that table: together they produce 15 units a day out of a total of 60 units. So time = 60 ÷ 15 = 4 days. Compare the fraction route — 1/10 + 1/12 + 1/15 = 6/60 + 5/60 + 4/60 = 15/60 = 1/4 → 4 days. Same numbers, but you had to build the 60 anyway. That is the proof that the LCM is the natural size of the work.

The four-step LCM drill — use it on every work question

Step 1 — Fix the total
Take total work = LCM of all the day-figures given in the question. Write it in the margin with a circle around it.
Step 2 — Get each rate
For each person, rate = total ÷ his days. These come out as clean whole numbers — that is the point of Step 1.
Step 3 — Add or subtract rates
Working together → add the rates. Someone leaving, or a leak, or an outlet pipe → subtract. You now have a single net rate.
Step 4 — Divide
Time = work remaining ÷ net rate. One division and you are done. Only here does a fraction ever appear.

🧠 One line to remember it

"LCM is the job, division is the man." The LCM of the days is how big the job is; dividing it by a man's days tells you how much of that job he does daily. Every work, pipe or cistern question in the objective paper yields to those seven words.

Worked example 3 — working backwards. A and B together finish a job in 12 days. A alone takes 20 days. How long does B alone take?

Total work = LCM(12, 20) = 60 units. Together they do 60 ÷ 12 = 5 units/day. A alone does 60 ÷ 20 = 3 units/day. So B's rate = 5 − 3 = 2 units/day, and B alone needs 60 ÷ 2 = 30 days. Notice Step 3 used subtraction — the same drill, run in reverse.

Three Classic Patterns: Together, Leaving Early, Alternate Days

Almost every work question the paper asks is one of three shapes. Once you can recognise the shape in five seconds, the arithmetic takes thirty more. Let us do all three fully.

Pattern 1 — one worker leaves partway. A can do a job in 20 days, B in 30 days. They start together, but A leaves after 8 days. In how many days in all is the job finished?

Total work = LCM(20, 30) = 60 units. A = 60 ÷ 20 = 3 units/day; B = 60 ÷ 30 = 2 units/day; together = 5 units/day.
Work done in the first 8 days = 5 × 8 = 40 units.
Work left = 60 − 40 = 20 units.
B finishes it alone at 2 units/day → 20 ÷ 2 = 10 days.
Total = 8 + 10 = 18 days.

Pattern 2 — alternate days, and it divides evenly. A can do a job in 9 days, B in 18 days. They work on alternate days, A starting. When is the job finished?

Total work = LCM(9, 18) = 18 units. A = 18 ÷ 9 = 2 units/day; B = 18 ÷ 18 = 1 unit/day.
Do not think in single days — think in pairs of days. One pair (A's day + B's day) = 2 + 1 = 3 units in 2 days.
Number of pairs needed = 18 ÷ 3 = 6 pairs.
Days = 6 × 2 = 12 days.

Pattern 3 — alternate days with a leftover. A can do a job in 6 days, B in 8 days, working alternately with A starting.

Total work = LCM(6, 8) = 24 units. A = 24 ÷ 6 = 4 units/day; B = 24 ÷ 8 = 3 units/day. One pair = 4 + 3 = 7 units in 2 days.
24 ÷ 7 = 3 pairs with a remainder: 3 pairs = 21 units in 6 days, leaving 24 − 21 = 3 units.
Day 7 is A's turn, and A does 4 units in a full day. He needs only 3 units, so he needs 3/4 of that day.
Total = 6 + 3/4 = 6¾ days (6.75 days).

⚠️ In alternate-day sums, WHO starts changes the answer

Redo Pattern 3 with B starting. Pairs are still 7 units in 2 days, so after 6 days 21 units are done and 3 units remain — but day 7 now belongs to B, who does only 3 units a day. He needs a full day. Total = 7 days, not 6¾. The same two workers, the same job, a quarter-day of difference purely because of who went first. Always read the phrase "beginning with" before you compute.

Leaves partway → work done before leaving = (joint rate) × (days), then divide the remainder by the rate of whoever stays.
Alternate days → work in 2-day pairs. Divide the total by the pair-output; if there is a remainder, hand it to whoever's turn falls next.
"A did the first x days, B finished the rest" → identical to the leaving-partway pattern; only the joint rate is replaced by A's solo rate.
If a question gives you a fraction of the work ("A did 1/4 of it in 5 days"), first scale up to the whole: 1/4 in 5 days → the whole in 5 × 4 = 20 days.

Men, Days, Hours; Efficiency; and Wages

When a question mentions men, days and hours together, one relation covers everything:

M₁ × D₁ × H₁ ÷ W₁ = M₂ × D₂ × H₂ ÷ W₂

Here M = men, D = days, H = hours per day, W = amount of work. The logic is simple: men × days × hours is the total labour poured in, and if the work doubles you need double the labour. If the work is the same in both situations, drop W₁ and W₂ and just write M₁D₁H₁ = M₂D₂H₂.

Worked example 4 — same work. 10 men working 6 hours a day build a wall in 12 days. How many days will 8 men working 9 hours a day take to build the same wall?

M₁D₁H₁ = M₂D₂H₂
10 × 12 × 6 = 8 × D₂ × 9
720 = 72 × D₂
D₂ = 720 ÷ 72 = 10 days.

Worked example 5 — the work changes too. 20 men working 7 hours a day dig a 300 m canal in 12 days. In how many days will 25 men working 8 hours a day dig a 500 m canal? Now W must stay in the equation.

(20 × 12 × 7) ÷ 300 = (25 × D₂ × 8) ÷ 500
1680 ÷ 300 = 200 × D₂ ÷ 500
5.6 = 0.4 × D₂
D₂ = 5.6 ÷ 0.4 = 14 days.

Sanity check: the men and hours went up (more capacity, so fewer days) but the canal got much longer (much more work) — so the days rising from 12 to 14 is believable.

🧠 Efficiency and time are upside-down twins

More efficient means fewer days — always exactly inverted. If the efficiency ratio of A to B is 3 : 2, then the ratio of the days they take is 2 : 3. Flip the ratio, do not scale it. Students lose easy marks by writing the efficiency ratio where the time ratio belongs. Say it aloud once: "efficiency up, days down."

Worked example 6 — efficiency given as a multiple. A is twice as efficient as B, and together they finish a job in 12 days. How long does each take alone?

Efficiency A : B = 2 : 1, so days A : B = 1 : 2. Let A take x days, then B takes 2x days.
1/x + 1/2x = 1/12
2/2x + 1/2x = 1/12
3/2x = 1/12
2x = 36, so x = 18.
A alone = 18 days, B alone = 36 days.
Check: 1/18 + 1/36 = 2/36 + 1/36 = 3/36 = 1/12. ✓

Efficiency given as a percentage — convert, then invert
The question says	Efficiency A : B	Days A : B (inverted)
A is 25% more efficient than B	125 : 100 = 5 : 4	4 : 5
A is 50% more efficient than B	150 : 100 = 3 : 2	2 : 3
A is twice as efficient (100% more)	200 : 100 = 2 : 1	1 : 2
A is 20% less efficient than B	80 : 100 = 4 : 5	5 : 4

Using row 1: if B alone takes 25 days and A is 25% more efficient, then days A : B = 4 : 5, so A takes 25 × 4/5 = 20 days. One line, no equation.

Worked example 7 — wages are shared in the ratio of EFFICIENCY, not of days. Total wage ₹1,800
Worker	Days alone	Units/day (of LCM 24)	Share of ₹1,800
A	6	24 ÷ 6 = 4	1800 × 4/9 = ₹800
B	8	24 ÷ 8 = 3	1800 × 3/9 = ₹600
C	12	24 ÷ 12 = 2	1800 × 2/9 = ₹400
Total	—	4 + 3 + 2 = 9 parts	800 + 600 + 400 = ₹1,800 ✓

💡 Think of it like this

Think of the units as bricks laid. A lays 4 bricks a day, B lays 3, C lays 2 — so of every 9 bricks in the wall, 4 are A's, 3 are B's and 2 are C's. Nobody would pay them by days spent standing there; they are paid per brick. That is the whole reason wages follow efficiency, and the wage ratio is simply 1/6 : 1/8 : 1/12 = 4 : 3 : 2.

Pipes & Cisterns — Work with a Minus Sign

There is no new theory here at all. A cistern is the work, an inlet pipe is a worker, and an outlet pipe or leak is a worker who undoes work — a negative worker. Set the tank's capacity to the LCM of the given hours, give every pipe its units per hour, put a plus on the fillers and a minus on the emptiers, and add.

The same chapter, two vocabularies
Time & Work	Pipes & Cisterns
The job to be done	Capacity of the tank
A worker (positive rate)	An inlet pipe — fills, so +
A worker who spoils the work	An outlet pipe or a leak — empties, so −
Days taken	Hours (or minutes) taken
Work already done	Water already in the tank — the classic trap

Worked example 8 — one inlet, one outlet. Pipe A fills an empty tank in 12 hours; pipe B empties a full tank in 18 hours. Both are opened together on an empty tank. When is it full?

Capacity = LCM(12, 18) = 36 units. A = 36 ÷ 12 = +3 units/hour. B = 36 ÷ 18 = −2 units/hour.
Net rate = 3 − 2 = +1 unit/hour.
Time = 36 ÷ 1 = 36 hours.

Worked example 9 — two inlets and one outlet. A fills in 10 hours, B fills in 15 hours, C empties in 12 hours. All three opened on an empty tank.

Capacity = LCM(10, 15, 12) = 60 units.
A = 60 ÷ 10 = +6; B = 60 ÷ 15 = +4; C = 60 ÷ 12 = −5.
Net = 6 + 4 − 5 = +5 units/hour.
Time = 60 ÷ 5 = 12 hours.

Worked example 10 — finding the leak's own time. A pipe can fill a tank in 6 hours, but because of a leak it actually takes 8 hours. How long would the leak alone take to empty a full tank?

The leak is the difference between what the pipe should do and what it does.
Leak's rate = 1/6 − 1/8 = 4/24 − 3/24 = 1/24 of the tank per hour.
So the leak alone empties the tank in 24 hours.

By units: capacity = LCM(6, 8) = 24. Pipe alone = 24 ÷ 6 = 4/hour; with the leak the observed net = 24 ÷ 8 = 3/hour. Leak = 4 − 3 = −1/hour → 24 ÷ 1 = 24 hours. Same answer, no fractions.

⚠️ The tank that is already part-full

A tank is 3/5 full. Pipe A fills the whole tank in 10 minutes; pipe B empties the whole tank in 6 minutes. Both are opened. When does the tank become empty?

Net rate = 1/10 − 1/6 = 3/30 − 5/30 = −2/30 = −1/15 of the tank per minute, i.e. it is emptying at 1/15 per minute.
But only 3/5 of a tankful of water is present. Time = (3/5) ÷ (1/15) = (3/5) × 15 = 9 minutes.

The trap: most students answer 15 minutes — the time to empty a full tank. The rates are always given for the whole tank; the quantity to be moved is only what is actually there. Read the opening line of the question twice.

ℹ️ Note

If the net rate comes out negative on an empty tank, the tank will never fill — the outlet is stronger than the inlet. Example: A fills in 12 hours (+3 of 36 units) while B empties in 9 hours (−4 of 36); net = −1, so the tank only gets drier. If an option says "never" or "cannot be filled", check the sign of the net rate before dismissing it.

Mixtures, Alligation & Repeated Replacement

When you mix two things of different prices (or strengths, or purities), the result sits somewhere between the two — and exactly where it sits depends on how much of each you used. The price of one unit of the mixture is called the mean price. Alligation is nothing but a fast way of reading that "somewhere between".

The alligation rule. For two ingredients, a cheaper one at price C and a dearer one at price D, mixed to give a mean price M:

Quantity of cheaper : Quantity of dearer = (D − M) : (M − C)

The cross diagram, described in words: write C at the top left and D at the top right, and put M below, midway between them. Now draw two lines from M, one up-left to C and one up-right to D — this is the "cross". Along each line, subtract the smaller number from the bigger: the line to C gives M − C, the line to D gives D − M. Finally — and this is the whole trick — write each difference under the OTHER ingredient. So D − M lands under C, and M − C lands under D. Read the bottom row left to right and you have the ratio C : D.

Worked example 11 — finding the mixing ratio. In what ratio must rice at ₹30/kg be mixed with rice at ₹45/kg so that the mixture is worth ₹35/kg?

C = 30, D = 45, M = 35.
Cheaper : dearer = (D − M) : (M − C) = (45 − 35) : (35 − 30) = 10 : 5 = 2 : 1.

Verify: take 2 kg + 1 kg. Cost = 2 × 30 + 1 × 45 = 60 + 45 = ₹105 for 3 kg → 105 ÷ 3 = ₹35/kg. ✓

🧠 Distance from the mean, then swap

Two words carry the whole rule: "distance, swapped." Measure how far the mean is from each price, then give each ingredient the other one's distance. Why swapped? Because the ingredient whose price is nearer the mean must be present in the larger quantity — it is the one pulling the average towards itself. In example 11 the mean ₹35 is nearer ₹30 (distance 5) than ₹45 (distance 10), and indeed the ₹30 rice is the larger part, 2 out of 3. Nearer price, bigger share. If your answer ever violates that, you swapped in the wrong direction.

Worked example 12 — the classic milk-and-water sum. A milk seller buys milk at ₹40 per litre and wants to sell the mixture at cost price yet still earn, by adding water. If he wants the mixture to cost him only ₹32 per litre, in what ratio must he add water to milk?

Water is free, so its price is ₹0. Now C = 0 (water), D = 40 (milk), M = 32.
Water : Milk = (D − M) : (M − C) = (40 − 32) : (32 − 0) = 8 : 32 = 1 : 4.

Verify: 4 litres of milk cost 4 × 40 = ₹160; add 1 litre of water (₹0) → 5 litres for ₹160 → 160 ÷ 5 = ₹32/litre. ✓ So he adds 1 litre of water to every 4 litres of milk, i.e. water is 1/5 = 20% of the mixture.

Worked example 13 — mixing three quantities. Three varieties of rice cost ₹25, ₹30 and ₹40 per kg. Mix them in the ratio 2 : 3 : 5. What is the mean price?

Mean price = (2 × 25 + 3 × 30 + 5 × 40) ÷ (2 + 3 + 5)
= (50 + 90 + 200) ÷ 10 = 340 ÷ 10 = ₹34/kg.

And now the reverse — how would you have found 2 : 3 : 5 if only ₹34 were given? With three ingredients you pair the dearest with each of the others and apply alligation twice:
- Pair ₹25 with ₹40 at mean 34 → 25 : 40 = (40 − 34) : (34 − 25) = 6 : 9 = 2 : 3.
- Pair ₹30 with ₹40 at mean 34 → 30 : 40 = (40 − 34) : (34 − 30) = 6 : 4 = 3 : 2.

Now collect: ₹25 appears once, with share 2. ₹30 appears once, with share 3. ₹40 appears in both pairs, so add its shares: 3 + 2 = 5. Hence 25 : 30 : 40 = 2 : 3 : 5 — exactly the ratio we started from. ✓

Repeated replacement — the one formula worth memorising. From a vessel of total volume V, you draw out x units of liquid and pour back x units of water. Repeat this n times. Then:

Original liquid left = V × (1 − x/V)ⁿ

Worked example 14. A vessel holds 40 litres of pure milk. 4 litres is drawn out and replaced by water; this operation is done twice in all. How much milk remains?

Milk left = 40 × (1 − 4/40)² = 40 × (9/10)² = 40 × 81/100 = 32.4 litres.
Water = 40 − 32.4 = 7.6 litres. Ratio milk : water = 32.4 : 7.6 = 81 : 19.

Verify it by hand, one step at a time:
- Operation 1: remove 4 L of pure milk → 36 L milk. Add 4 L water → 36 milk + 4 water.
- Operation 2: the 40 L is now 90% milk, so the 4 L you remove is 3.6 L milk + 0.4 L water. Left: 36 − 3.6 = 32.4 L milk and 4 − 0.4 = 3.6 L water. Add 4 L water → 3.6 + 4 = 7.6 L water.
- Total 32.4 + 7.6 = 40 L ✓, and the milk figure matches the formula exactly.

⚠️ Two things the replacement formula demands

(1) V must be the total volume of the vessel, not the amount removed. Writing (1 − 4/36) instead of (1 − 4/40) is the single commonest error here — the fraction is removed ÷ total, and the total never changes because you always pour back exactly what you took out.
(2) n is the number of operations, not the number of litres. "The process is repeated twice more" after one operation means n = 3, not 2. Read the wording with care.

One more for practice: a container has 250 litres of milk; 50 litres is drawn out and replaced by water, three times. Milk left = 250 × (1 − 50/250)³ = 250 × (4/5)³ = 250 × 64/125 = 128 litres; water = 250 − 128 = 122 litres; ratio = 64 : 61.

Problems on Ages

Age problems are pure translation. English (or Hindi) goes in, one linear equation comes out. There is only one discipline you must never break: always name your unknown as the PRESENT age. Then "5 years ago" is simply (x − 5) and "after 8 years" is (x + 8). Students who set the unknown at some past year end up subtracting twice and lose the sum.

Translating the words into an equation
The question says	Write this
A is 4 years older than B	A = B + 4
A is thrice as old as B	A = 3B
The sum of their present ages is 50	A + B = 50 (so B = 50 − A — use one letter)
Five years ago A was twice as old as B	(A − 5) = 2 × (B − 5) — subtract from both
The ratio of their present ages is 3 : 4	Ages = 3k and 4k (one unknown, not two)
After 6 years the ratio will be 4 : 5	(3k + 6) ÷ (4k + 6) = 4/5, then cross-multiply

Worked example 15 — present and future. A father's present age is 3 times his son's. After 12 years, the father's age will be twice the son's. Find their present ages.

Let the son be x now; then the father is 3x.
After 12 years: son = x + 12, father = 3x + 12.
3x + 12 = 2(x + 12)
3x + 12 = 2x + 24
3x − 2x = 24 − 12
x = 12.
So the son is 12 and the father is 36.
Check: after 12 years they are 24 and 48, and 48 = 2 × 24. ✓

Worked example 16 — the ratio-multiples shortcut. The ratio of the present ages of A and B is 4 : 5. Eight years hence the ratio will be 6 : 7. Find their present ages.

Do not write two letters. Write ages as 4k and 5k — that single k already encodes the ratio, halving your work.
(4k + 8) ÷ (5k + 8) = 6/7
Cross-multiply: 7(4k + 8) = 6(5k + 8)
28k + 56 = 30k + 48
56 − 48 = 30k − 28k
8 = 2k → k = 4.
So A = 4 × 4 = 16 years, B = 5 × 4 = 20 years.
Check: after 8 years, 24 and 28 → 24 : 28 = 6 : 7. ✓

⚠️ The ratio changes; the difference never does

Both people age by exactly the same number of years, so the difference between two ages is fixed for life. In example 16 the gap is 20 − 16 = 4 years, and eight years later it is still 28 − 24 = 4 years. Two uses for this:
(a) it is a free 5-second check on any answer you produce;
(b) many questions can be solved by it directly. And the matching trap — never assume the difference of the ratio terms is the difference of the ages. Going from 4 : 5 to 6 : 7 does not mean the gap grew from 1 to 1 in years; the k is different in the two ratios. Only actual ages may be subtracted.

Worked example 17 — a sum condition with a past condition. The sum of the present ages of a mother and her daughter is 70 years. Five years ago the mother was 5 times as old as her daughter. Find both present ages.

Let the daughter be d and the mother be m now, with d + m = 70.
Five years ago: m − 5 = 5(d − 5)
m − 5 = 5d − 25
m = 5d − 20.
Substitute into the sum: d + (5d − 20) = 70
6d = 90 → d = 15, and m = 70 − 15 = 55.
So the daughter is 15 and the mother is 55.
Check: five years ago they were 10 and 50, and 50 = 5 × 10. ✓

Worked example 18 — the average age changes when someone joins. The average age of 20 students in a class is 12 years. When the class teacher is included, the average becomes 13 years. Find the teacher's age.

Long way: total age of students = 20 × 12 = 240 years. With the teacher there are 21 people averaging 13, so their total = 21 × 13 = 273 years. Teacher = 273 − 240 = 33 years.

Fast way: the newcomer must carry the new average for himself, plus enough extra to lift each of the 20 old members by 1 year. So teacher = new average + (number of old members × rise in average) = 13 + (20 × 1) = 33 years. Same answer in one line — and this shortcut works for a cricket captain joining a team, a baby joining a family, anything.

💡 Think of it like this

Picture the average as the water level in a row of equal glasses. Adding a person is pouring in one more glass. If the level rose by 1 cm across 21 glasses, the newcomer had to bring his own 13 cm plus 1 cm for each of the 20 glasses already there — 13 + 20 = 33. If the level had fallen by 1 cm instead, he would have brought 13 − 20, i.e. he is younger than the average and drags it down. The sign of the change tells you whether to add or subtract.

Spotting Which Method to Use

In the exam hall you will not have time to try two methods. The good news is that these questions announce themselves — a single phrase in the stem tells you exactly which tool to reach for. Learn the table below and your first ten seconds go into recognition instead of panic.

Decision table — read the phrase, pick the method
The phrase in the question	Method to use	First thing to write
"A in x days, B in y days, together in ...?"	LCM unit method	Total work = LCM(x, y)
"... but A left after k days" / "B finished the rest"	LCM units, then subtract done from total	Units done = joint rate × k
"working on alternate days, beginning with A"	Pair up 2 days, then hand the remainder to the next turn	Pair output = rateA + rateB
"men ... hours a day ... days" (numbers change)	M₁D₁H₁/W₁ = M₂D₂H₂/W₂	Line up the two situations, cross-multiply
"x% more/less efficient" / "twice as efficient"	Efficiency ratio, then INVERT for days	Efficiency 5 : 4 → days 4 : 5
"total wages of ₹P are to be divided"	Split P in the ratio of the daily units	1/a : 1/b : 1/c → whole-number ratio
"pipe empties" / "leak" / "outlet"	LCM units with a MINUS sign	Net rate = fillers − emptiers
"the tank is already 2/3 full"	Net rate as usual, but divide only the water present	Time = quantity present ÷ net rate
"in what ratio must ... be mixed" (two prices, one mean)	Alligation	C : D = (D − M) : (M − C)
"x litres drawn out and replaced by water, n times"	Replacement formula	Left = V × (1 − x/V)ⁿ
"the ratio of their ages is ... and after k years ..."	Ratio multiples in one letter k	Ages = ak and bk, then form the equation
"the average age changes when X joins/leaves"	Total-age method (or the one-line shortcut)	X = new average ± (old count × change)

🎯 Why this chapter is worth the effort

Arithmetic of this kind appears in the General Studies paper of the objective paper and again in the General Studies Paper II of the Mains, and it is one of the few areas where the answer is provably right — no opinion, no revised figure, no "as per latest data". Polity and current affairs can surprise you; 36 ÷ 5 cannot. Every question here is a mark you can guarantee to yourself. And because the objective paper carries negative marking, guaranteed marks are worth far more than probable ones.

Underline the day-figures / hour-figures in the stem. Their LCM is your total.
Look for a minus word — leak, outlet, empties, leaves, spoils. Every one of them means subtract.
Ask "how much work is actually left to do?" — not always the whole. Part-full tanks and partly done jobs live here.
If a ratio is given, use one letter k, never two unknowns.
Spend the last five seconds putting your answer back into the question. Age differences must match; totals must add up; rates must have the right sign.

⚠️ Three careless losses to avoid

(1) Units. Pipe questions mix hours and minutes freely — convert before you add rates.
(2) What was asked. A question may compute cleanly to "18 days total" while asking only for "days B worked alone". Reread the last line before you shade the bubble.
(3) Efficiency written where time belongs. If your answer says the more efficient worker takes more days, you have inverted nothing — go back one line.

Key takeaways

LCM unit method: total work = LCM of the given days; each worker's rate = total ÷ his days. Add rates to work together, subtract for anyone leaving. Time = work left ÷ net rate.
Alternate days → work in 2-day pairs and give the remainder to whoever's turn is next. Who starts first can change the answer (A-first 6¾ days vs B-first 7 days for the 6-and-8-day pair).
M₁D₁H₁/W₁ = M₂D₂H₂/W₂ for men-days-hours-work. Efficiency and time are exactly inverse — efficiency 5 : 4 means days 4 : 5. Wages are split in the ratio of daily units (1/6 : 1/8 : 1/12 = 4 : 3 : 2).
Pipes = work with a minus sign. Inlet is +, outlet or leak is −. Leak's rate = (pipe alone) − (pipe with leak); a 6-hour pipe taking 8 hours means a leak that empties in 24 hours. If the tank is part-full, divide only the water actually present.
Alligation: cheaper : dearer = (D − M) : (M − C) — distance from the mean, then swapped. The price nearer the mean always has the bigger share. Water counts as price ₹0.
Repeated replacement: original liquid left = V × (1 − x/V)ⁿ, where V is the vessel's full volume and n the number of operations. 40 L milk, 4 L replaced twice → 40 × (9/10)² = 32.4 L milk, milk : water = 81 : 19.
Ages: always let the unknown be the present age; write a given ratio as ak and bk (one letter). The difference of two ages never changes — use it as a free check. When one person joins a group, newcomer = new average + (old count × rise in average).

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 and B together can complete a piece of work in 12 days. A alone can complete it in 20 days. In how many days can B alone complete it?
1. 24 days
2. 30 days
3. 32 days
4. 36 days
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Answer: B — 30 days
Take total work = LCM(12, 20) = 60 units. Together they do 60 ÷ 12 = 5 units/day; A alone does 60 ÷ 20 = 3 units/day. So B's rate = 5 − 3 = 2 units/day, and B alone needs 60 ÷ 2 = 30 days.
If 15 men working 8 hours a day can build a wall in 24 days, in how many days will 20 men working 9 hours a day build the same wall?
1. 16 days
2. 18 days
3. 20 days
4. 24 days
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Answer: A — 16 days
The work is the same, so M₁D₁H₁ = M₂D₂H₂. Thus 15 × 24 × 8 = 20 × D₂ × 9 → 2880 = 180 × D₂ → D₂ = 2880 ÷ 180 = 16 days. More men and longer hours mean fewer days, so 16 (less than 24) is the sensible direction.
A can do a work in 12 days and B in 18 days. They begin the work together, but A leaves after 6 days. In how many days in all is the work completed?
1. 8 days
2. 9 days
3. 10 days
4. 12 days
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Answer: B — 9 days
Total work = LCM(12, 18) = 36 units. A = 36 ÷ 12 = 3 units/day, B = 36 ÷ 18 = 2 units/day, together 5 units/day. In 6 days they do 5 × 6 = 30 units, leaving 36 − 30 = 6 units. B finishes those alone at 2 units/day → 6 ÷ 2 = 3 days. Total = 6 + 3 = 9 days.
A, B and C can individually complete a work in 8, 12 and 24 days respectively. They work together and are paid ₹1,680 in all. Consider the following statements:
1. Working together, they complete the work in 4 days.
2. B's share of the money is ₹480.
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: A — Only 1
Total work = LCM(8, 12, 24) = 24 units. A = 3, B = 2, C = 1 unit per day, so together 6 units/day → 24 ÷ 6 = 4 days. Statement 1 is correct. Wages follow the efficiency ratio 3 : 2 : 1 (6 parts), so one part = 1680 ÷ 6 = ₹280, giving A ₹840, B ₹560 and C ₹280. B gets ₹560, not ₹480, so statement 2 is wrong.
A pipe can fill a tank in 9 hours, but because of a leak at the bottom it takes 12 hours. How long will the leak alone take to empty a full tank?
1. 21 hours
2. 30 hours
3. 36 hours
4. 48 hours
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Answer: C — 36 hours
The leak's rate is the difference between what the pipe should do and what it actually does: 1/9 − 1/12 = 4/36 − 3/36 = 1/36 of the tank per hour. So the leak alone empties the full tank in 36 hours. (Option 21 is the wrong habit of adding 9 and 12.)
A tank is 2/3 full of water. Inlet pipe A can fill the whole tank in 12 minutes and outlet pipe B can empty the whole tank in 8 minutes. If both are opened together, in how many minutes will the tank be empty?
1. 24 minutes
2. 16 minutes
3. 12 minutes
4. 8 minutes
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Answer: B — 16 minutes
Net emptying rate = 1/8 − 1/12 = 3/24 − 2/24 = 1/24 of the tank per minute. But only 2/3 of a tankful is present, so time = (2/3) ÷ (1/24) = (2/3) × 24 = 16 minutes. The trap answer is 24 minutes — that is the time to empty a completely full tank.
Consider the following two statements:
1. To make a mixture worth ₹28 per kg from wheat costing ₹24 per kg and wheat costing ₹36 per kg, the two must be mixed in the ratio 2 : 1.
2. If 6 litres is drawn out of a vessel containing 60 litres of pure milk and replaced by water, and the same operation is performed once more, 48.6 litres of milk remain.
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: C — Both 1 and 2
Statement 1: by alligation, cheaper : dearer = (36 − 28) : (28 − 24) = 8 : 4 = 2 : 1. Check — 2 kg at ₹24 plus 1 kg at ₹36 = ₹84 for 3 kg = ₹28/kg. Correct. Statement 2: milk left = 60 × (1 − 6/60)² = 60 × (9/10)² = 60 × 0.81 = 48.6 litres. Correct. Hence both are right.
The sum of the present ages of a mother and her daughter is 70 years. Five years ago, the mother was 5 times as old as her daughter. What is the daughter's present age?
1. 10 years
2. 12 years
3. 14 years
4. 15 years
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Answer: D — 15 years
Let the daughter be d and the mother m now, with d + m = 70. Five years ago: m − 5 = 5(d − 5) → m − 5 = 5d − 25 → m = 5d − 20. Substituting, d + 5d − 20 = 70 → 6d = 90 → d = 15, so m = 55. Check: five years ago they were 10 and 50, and 50 = 5 × 10.