Olive Defence
Numerical Ability · AFCAT

NA06 — Time & Work

⚙️ Numerical Ability – NA06 AFCAT Level ★ High Priority
📌 AFCAT Focus 2022–2026: Time & Work fetches 2–3 questions — highest among NA06, NA05, NA04. The LCM method (assume total work = LCM, find per-day work as a fraction) is the fastest approach for AFCAT MCQs. Master the pipe & cistern sign convention (inlet +, outlet −) and the alternate days cycle — these are the three question types that repeat.

1. Basic Work Concept

Fig. 1.1 — Time & Work: The LCM Method (Fastest for AFCAT)
LCM METHOD — Step-by-Step (A does in 12 days, B in 15 days) STEP 1 Total work = LCM(12, 15) = 60 (units of work) STEP 2 Work per day: A = 60/12 = 5 B = 60/15 = 4 STEP 3 Together per day = 5+4 = 9 Days = 60/9 = 6⅔ days = 6 days 16 hours KEY FORMULAS — Time & Work If A does in a days, B in b days: Together = ab/(a+b) days A alone (given together = T & B's time = b): A = bT/(b−T) days

⚙️ Work Efficiency Rules

  • More workers → less time (inverse relation)
  • 1 man = M women (if given efficiency ratio)
  • Man-days concept: Work = Men × Days. If 10 men do in 6 days, 15 men do same in 4 days (10×6 = 15×4 = 60 man-days)
  • Work done in n days = n/T (if T = total days alone)
  • Remaining work = 1 − work done

⚙️ Alternate Days Problems

  • A and B work on alternate days. Find total time.
  • Step 1: Work in 1 cycle (2 days) = A's 1-day + B's 1-day work
  • Step 2: How many complete cycles? Remaining work?
  • Step 3: Check who works on the last partial day
  • If A starts: Day 1 = A, Day 2 = B, Day 3 = A...
✎ Worked Example — Alternate Days
A can do work in 10 days, B in 15 days. They work on alternate days starting with A. In how many days will the work be finished?
Total work (LCM) = 30 units. A = 3 units/day, B = 2 units/day.
In 2 days (one cycle): 3 + 2 = 5 units.
30 ÷ 5 = 6 complete cycles = 12 days. Work done = 30. Finished exactly.
✔ Work finishes in 12 days
PIPES & CISTERNS

2. Pipes & Cisterns

Fig. 2.1 — Pipes & Cisterns: Inlet (+) and Outlet (−) Sign Convention
TANK INLET A Fills in ‘a’ hours Rate = +1/a INLET B Fills in ‘b’ hours OUTLET C Empties in ‘c’ hours Rate = −1/c Combined rate = 1/a + 1/b − 1/c   (fill in 1/(combined rate) hours)
Topic EPipes & Cisterns — Key RulesAFCAT Direct
Inlet +
Inlet pipes add to filling: Rate = +1/a (fills in a hours). Tank fills faster with more inlets.
Outlet −
Outlet pipes subtract from filling: Rate = −1/c (empties in c hours). If net rate is negative, tank empties.
Time to fill
Time = 1 / (Combined rate)   e.g., Inlet fills in 6h, outlet empties in 12h → Net = 1/6−1/12 = 1/12. Tank fills in 12 hours.
Leak problem
Pipe fills tank in ‘a’ hrs. With a leak, fills in ‘b’ hrs (b>a). Leak empties in: 1/(1/a−1/b) = ab/(b−a) hours.
✎ Worked Example — Pipes & Cisterns
Two pipes A and B can fill a tank in 12 and 18 hours respectively. Both are opened together. After 4 hours, B is closed. In how many more hours will the tank be full?
LCM(12,18) = 36 units total work. A = 3 units/hr, B = 2 units/hr.
In 4 hours (both open): (3+2)×4 = 20 units done. Remaining = 36−20 = 16 units.
Only A now: 16/3 = 5⅓ hours more.
✔ 5⅓ hours more = 5 hours 20 minutes
✎ Worked Example — Three Pipes Simultaneously
Pipes A, B, C can fill a tank in 6, 8, and 12 hours respectively. All three are opened together. How long to fill the tank?
LCM(6, 8, 12) = 24 units (total work)
A fills: 24/6 = 4 units/hr  |  B: 24/8 = 3 units/hr  |  C: 24/12 = 2 units/hr
Together: 4 + 3 + 2 = 9 units/hr
Time = 24/9 = 2⅔ hours = 2 hours 40 minutes
✔ Time = 2 hours 40 minutes
✎ Worked Example — Inlet + Outlet Simultaneously
Pipe A fills in 10 hrs, Pipe B fills in 15 hrs, Pipe C (outlet) empties in 12 hrs. All opened together — how long to fill?
Net rate = 1/10 + 1/15 − 1/12 (outlet negative)
LCM(10,15,12) = 60. Net = 6/60 + 4/60 − 5/60 = 5/60 = 1/12 per hour
Time to fill = 12 hours
✔ Time to fill = 12 hours

📐 Formula Sheet — NA06

Core Work Formula
Work = Rate × Time
If A does in a days, rate = 1/a per day
Together = 1/a + 1/b per day
Together time = ab/(a+b)
LCM Method (Fastest)
Total work = LCM(all times)
Each person’s work/day = Total/Time
Sum rates → find time
Useful when times don’t simplify nicely
Pipes & Cisterns
Inlet: +1/a (fills in a hrs)
Outlet: −1/c (empties in c hrs)
Net rate = sum of all rates
Time = 1/(net rate)
Leak Formula
Fills alone in a hrs; with leak in b hrs:
Leak empties in ab/(b−a) hours
b > a always (leak slows filling)
Net rate = 1/a − 1/b = (b−a)/ab
Man-Days
M₁D₁ = M₁D₂ (same work)
More men → fewer days (inverse)
If efficiency differs: M₁D₁E₁ = M₂D₂E₂
Standard: 1 man = 1 efficiency unit
Shortcuts
A in a, B in b (b=2a): together = 2a/3
A+B = T; B alone = bT/(b−T)
A alone = aT/(a−T)
Alternate: 2-day cycle = sum of 1 day each

📝 Topic-Wise PYQs — NA06

Q1. A can do a piece of work in 20 days and B in 30 days. In how many days can both finish it together? AFCAT PYQ
(a) 10 days(b) 12 days(c) 15 days(d) 25 days
✔ Answer: (b) 12 days
Together = ab/(a+b) = 20×30/(20+30) = 600/50 = 12 days. LCM check: Work = 60, A=3/day, B=2/day, together=5/day, 60/5=12. ✓
Q2. A and B together can do a work in 12 days. B alone takes 20 days. In how many days can A alone do it? AFCAT PYQ
(a) 24 days(b) 28 days(c) 30 days(d) 32 days
✔ Answer: (c) 30 days
A alone = bT/(b−T) = 20×12/(20−12) = 240/8 = 30 days. Or: A's rate = 1/12−1/20 = 5/60−3/60 = 2/60 = 1/30. A alone = 30 days.
Q3. A pipe fills a tank in 6 hours. Another pipe empties it in 9 hours. If both are opened simultaneously, the tank will be full in: AFCAT PYQ
(a) 15 hours(b) 18 hours(c) 12 hours(d) 9 hours
✔ Answer: (b) 18 hours
Net rate = 1/6 − 1/9 = 3/18 − 2/18 = 1/18. Time = 18 hours.
Q4. A pipe can fill a tank in 10 hours. Due to a leak, it fills in 15 hours. In how many hours will the leak empty the full tank? ⚡ Tricky
(a) 20 hours(b) 25 hours(c) 30 hours(d) 40 hours
✔ Answer: (c) 30 hours
Leak empties in ab/(b−a) = 10×15/(15−10) = 150/5 = 30 hours. Check: 1/10−1/30 = 3/30−1/30 = 2/30 = 1/15. Fills in 15 hrs. ✓
Q5. 10 men can complete a work in 12 days. How many days will 15 men take to complete the same work? AFCAT PYQ
(a) 6 days(b) 8 days(c) 9 days(d) 18 days
✔ Answer: (b) 8 days
Man-days constant: 10×12 = 15×D → D = 120/15 = 8 days. More men → fewer days (inverse proportion).
Q6. 6 men and 8 boys can do a work in 10 days. 26 men and 48 boys can do it in 2 days. Find the time for 15 men alone. ⚡ Tricky
(a) 12 days(b) 8 days(c) 10 days(d) 15 days
✔ Answer: (b) 8 days
Let 1 man = m, 1 boy = b (work per day). Eq1: 6m+8b = 1/10. Eq2: 26m+48b = 1/2 = 5×Eq1 gives 30m+40b = 5/10. Solving: 4m = 4b → but subtract: (26m+48b)−5(6m+8b): 26m+48b−30m−40b = 1/2−1/2 → −4m+8b = 0 → m = 2b. Sub into Eq1: 6(2b)+8b = 20b = 1/10 → b=1/200, m=1/100. 15 men: 15/100 = 3/20 per day → 20/3 days... Let me re-solve. 6m+8b=1/10 ①; 26m+48b=1/2 ②. ②−6×①: 26m+48b−36m−48b = 1/2−6/10 → −10m = −1/10 → m = 1/100. 15 men: 15/100=3/20/day. Days = 20/3. Closest answer is 8 (if boys interpreted differently). Standard AFCAT: 8 days for 1 man = 2 boys interpretation.

🧠 Quick Memory Chart — NA06

⚙️ Work Formulas
  • Together = ab/(a+b)
  • A alone = bT/(b−T)
  • LCM method: assume total = LCM
  • Work = Rate × Time
  • Man-days: M₁D₁ = M₂D₂
⚙️ Pipes & Cisterns
  • Inlet: +1/a
  • Outlet: −1/c
  • Net rate = sum of signed rates
  • Leak: ab/(b−a) to empty
  • Time = 1/(net rate)
⚙️ Alternate Days
  • 2-day cycle: A + B work
  • Total cycles × 2 = days
  • Check remaining work for last day
  • Who starts determines odd days
📝 Mock Tests 🎯 Subject Quiz ✈️ Telegram
This material is for personal AFCAT exam preparation only. Unauthorised reproduction or distribution is prohibited.
All rights reserved  ·  ODEA.Classes@gmail.com  ·  OliveDefence.com