Time and Work — SSC CGL Quantitative Aptitude

The textbook method drowns you in fractions: $\frac{1}{12}+\frac{1}{18}$, common denominators, inversions. The exam method assumes the total work = LCM of the given days, so every rate becomes a small integer of "units per day". Same mathematics, no fractions, half the time. Every section of this chapter is that one idea wearing different costumes.

1. What SSC actually asks

Tier 1: 1–2 Q · Tier 2: 2–3 Q. The costumes: two/three workers together, worker leaves midway, efficiency comparisons ("A is twice as good"), man–days scaling (the MDH formula), pipes and cisterns (a leak is a negative worker), wages split by work done, and alternate-day patterns.

2. The LCM method

A does a job in 12 days, B in 18. Together?

1. Total work = LCM(12, 18) = 36 units.
2. Rates: A = 36÷12 = 3 units/day, B = 36÷18 = 2 units/day.
3. Together: 5 units/day → $36÷5=7\frac{1}{5}$ days.

Every quantity is now an integer until the final division. The same frame handles "A leaves after 4 days" (subtract the units already done) and "how much does B get paid" (wages ∝ units contributed).

3. Efficiency language

• "A is twice as efficient as B" → rates 2 : 1 → A alone takes half the days B takes.
• "A is 50% more efficient than B" → rates 3 : 2 → days ratio 2 : 3 (days invert rates).
• Given "together they finish in $T$ days" with rate ratio $a:b$, total work $=(a+b)T$ units and each worker's solo time follows in one line.

4. The MDH formula (man–days scaling)

$\frac{M_1{D}_1{H}_1}{W_1}=\frac{M_2{D}_2{H}_2}{W_2}$

Men × days × hours per unit of work is constant. 15 men, 20 days, 8 h/day = 2400 man-hours; 20 men at 6 h/day need $2400÷120=20$ days. Include the $W$ terms only when the amount of work itself changes ("twice the wall").

Mixed workforces: "3 men or 5 women finish in 12 days" → 3m = 5w → convert everything to one currency (1 man = $\frac{5}{3}$ women) and count woman-days.

5. Pipes and cisterns — negative workers

An inlet is a worker; a leak or outlet is a worker with a negative rate.

A fills in 10 h, B in 15 h, C empties in 30 h; all open: LCM = 30 → rates $+3,+2,-1$ → net 4 units/h → $30÷4=7.5$ hours.

The one trap: if the tank starts full and the question asks about emptying, the target is 30 units in the negative direction — re-read which way the water goes before computing.

6. Solved PYQ-style examples

Q1. A and B together finish in 6 days; A alone in 9. B alone? Solution. LCM(6,9) = 18: together 3/day, A = 2/day → B = 1/day → 18 days.

Q2. A is twice as efficient as B; together they take 12 days. A alone? Solution. Rates 2:1 → together 3 units/day → work = 36 units → A alone: 36 ÷ 2 = 18 days.

Q3. A (12 days) and B (15 days) start together; A leaves after 4 days. How many more days does B need? Solution. LCM = 60: A = 5, B = 4. Four days together = 36 units. Remaining 24 at B's 4/day = 6 days.

Q4. A and B in 12 days, B and C in 15, C and A in 20. All three together? Solution. LCM(12,15,20) = 60: A+B = 5, B+C = 4, C+A = 3. Sum = 12 = 2(A+B+C) → A+B+C = 6 units/day → $60÷6=$ 10 days. (Each pair-sum counts every worker twice — halve it.)

Q5 (alternate days). A alone: 9 days; B alone: 12. They work alternate days, A starting. Total time? Solution. LCM = 36: A = 4, B = 3. Each 2-day cycle = 7 units → 5 cycles (10 days) = 35 units. Day 11 is A's: 1 unit left at 4/day = $\frac{1}{4}$ day. $10\frac{1}{4}$ days.

7. Exam protocol

1. First move, always: total work = LCM. Never start with $\frac{1}{x}+\frac{1}{y}$.
2. Translate efficiency phrases to rate ratios immediately; remember days invert rates.
3. Pair questions (A+B, B+C, C+A): add all three, halve, then subtract pairs for singles.
4. Pipes: write outlets with a minus sign as you read, and check whether the tank starts full or empty.
5. Wages divide by units contributed, not by days present — a common 3-mark trap.