Dosto, is chapter mein sawal aise hote hain: "A ek kaam 10 din mein karta hai, B 15 din mein — dono saath mein kitne din lenge?" Aur Pipes & Cisterns bhi bilkul yahi cheez hai — bas "kaam" ki jagah "tank bharna" hota hai!
CET Group D mein isse 1–2 questions aate hain. Ye topic students ko mushkil lagta hai, par ek LCM trick seekh lo to ye sabse aasan ban jata hai — fractions ka jhanjhat hi khatam.
Chapter 2 (LCM) yahan kaam aayega. Bas ek soch pakad lo: "1 din mein kitna kaam" — pura chapter isi par khada hai. Chalo! 💪
1. Base Idea — "1 din ka kaam"
Saath mein kaam karein to 1 din ke kaam JOD do, phir ulta (reciprocal) kar do.
Step 2: Pura kaam = ulta karo = 6 din ✓
2. LCM METHOD — ⭐ Is chapter ka asli hathiyar
Step 2: Har banda ki efficiency = Total work ÷ uske din (yaani 1 din mein kitne units)
Step 3: Efficiencies jodo/ghatao → Time = Total work ÷ total efficiency
Step 2: A = 30/10 = 3 units/din · B = 30/15 = 2 units/din
Step 3: Saath mein 5 units/din → 30 ÷ 5 = 6 din ✓
Fayda: Poore chapter mein ek bhi fraction nahi likhna padta!
Step 2: A = 3, B = 2 → saath = 5 units/din
Step 3: 36 ÷ 5 = 7.2 din (yaani 7⅕ din) ✓
3. "Ek ka time nikalna" (ulta sawal)
Step 2: (A+B) = 60/12 = 5 units/din · A = 60/20 = 3 units/din
Step 3: B = 5 − 3 = 2 units/din → B akela = 60 ÷ 2 = 30 din ✓
4. Efficiency (kaam karne ki taakat)
- Efficiency aur time ka ULTA rishta hai — jitna tez banda, utne kam din.
- Agar A, B se x guna tez hai → A ka time = B ka time ÷ x.
- Ratio ki bhasha mein: efficiency ratio a : b ho to time ratio b : a hota hai (Chapter 5 ka inverse proportion!).
Step 2: Time ratio ULTA = 4 : 5
Step 3: B = 20 din → 1 part = 4 din → A = 4 × 4 = 16 din ✓
5. Aadmi–Din–Kaam (M × D × H = Work)
Solved Example: 8 aadmi 12 din mein kaam karte hain. 6 din mein karna ho to kitne aadmi?
Step 1: 8 × 12 = M × 6 → Step 2: M = 96/6 = 16 aadmi ✓ (aadmi badhe → din ghate = inverse)
6. Pipes & Cisterns (nal aur tanki)
Step 2: A = 4, B = 3 → saath 7 units/ghanta
Step 3: 24 ÷ 7 = 24/7 = 3³⁄₇ ghante ✓
Step 2: Nal = +3, leak = −2 → net = 1 unit/ghanta
Step 3: 30 ÷ 1 = 30 ghante ✓ (leak ki wajah se 10 ki jagah 30 lag gaye!)
Step 1: A ne kiya = 5/20 = 1/4 kaam → bacha 3/4
Step 2: B ne 3/4 kaam 15 din mein kiya
Step 3: Pura kaam = 15 × 4/3 = 20 din ✓
Step 1: Efficiency A : B = 2 : 1 → total 3 parts
Step 2: Saath mein 12 din → total work = 3 × 12 = 36 units
Step 3: A = 2 units/din → 36 ÷ 2 = 18 din ✓ (B = 36 din)
Step 1: LCM(15,10) = 30 units → A = 2, B = 3 → saath 5 units/din
Step 2: 3 din mein = 15 units = 15/30 = 1/2 kaam
Step 3: Bacha = 1/2 ✓
Step 1: M₁D₁ = M₂D₂ → 10 × 15 = 15 × D
Step 2: D = 150/15 = 10 din ✓ (aadmi badhe to din ghate — inverse proportion)
Step 1: LCM(20,30) = 60 units → bharne wala +3, khali wala −2
Step 2: Net = +1 unit/min → 60 ÷ 1 = 60 minute ✓
⚠️ Agar khali karne wala tez hota (net minus), to tank kabhi bharta hi nahi — us option ko turant pakdo!
Work = LCM(dino ka)
10 aur 15 din → work = 30 units, efficiency 3 aur 2. Ab sab kuch jodna-ghatana poore numbers mein — fraction ek bhi nahi!
T = (a × b)/(a + b)
A 10, B 15 → (10×15)/(10+15) = 150/25 = 6 din. Sirf DO bandon par lagta hai — teen ho to LCM method hi karo.
Efficiency 5:4 → Time 4:5
"25% zyada efficient" → efficiency 5:4 → time ratio ulta. Ratio palatna bhoolna sabse badi galti hai.
Bharna +, khali karna −
Bas ye ek badlaav — baaki pura Time-Work jaisa. Leak wale sawal mein hamesha net efficiency nikalo.
M₁D₁ = M₂D₂ (kaam same ho)
10×15 = 15×10. Aadmi aur din ka guna hamesha same rahega — 5 second mein answer.
| Cheez | Formula / Rule |
|---|---|
| 1 din ka kaam | 1/n |
| Do bandon ka time | ab/(a+b) |
| LCM method | Work = LCM, eff = work/din |
| Time = ? | Total work ÷ total efficiency |
| Efficiency vs Time | Ulta (inverse) |
| Eff ratio a:b | Time ratio b:a |
| Aadmi–din | M₁D₁ = M₂D₂ |
| Bharne wala nal | +efficiency |
| Khali karne wala | −efficiency |
| A10 + B15 | 6 din |
| Fill 10 + leak 15 | 30 ghante |
| Net minus ho to | Tank kabhi nahi bharega |
Q1. A can do a piece of work in 10 days and B in 15 days. Working together, they will finish it in:
Q2. A can complete a work in 12 days and B in 18 days. Together they will complete it in:
Q3. A and B together can do a work in 12 days. A alone can do it in 20 days. B alone will take:
Q4. 10 men can complete a work in 15 days. In how many days will 15 men complete it?
Q5. A is twice as good a workman as B. Together they finish a work in 12 days. A alone can finish it in:
Q6. A can do a work in 20 days. He works for 5 days and leaves; B finishes the remaining work in 15 days. B alone can do the whole work in:
Q7. Pipe A fills a tank in 6 hours and pipe B in 8 hours. If both are opened together, the tank will be full in:
Q8. A pipe can fill a tank in 10 hours, but a leak empties it in 15 hours. With both working, the tank fills in:
Q9. A tap fills a tank in 20 minutes while another empties it in 30 minutes. If both are opened together, the tank will be filled in:
Q10. If 8 men can do a piece of work in 12 days, how many men are needed to do it in 6 days?
Q11. A man completes 1/3 of a work in 5 days. He will complete the whole work in:
Q12. A is 25% more efficient than B. If B takes 20 days to complete a work, A will take:
Q13. A can do a work in 15 days and B in 10 days. If they work together for 3 days, the fraction of work left is:
Q14. Two pipes can fill a tank in 15 hours and 20 hours respectively. If both are opened together, the tank will be full in:
Pura chapter ek soch par hai: "1 din mein kitna kaam". Sabse tez tarika LCM method — dino ka LCM = total work, efficiency = work ÷ din, phir time = work ÷ total efficiency (fraction ek bhi nahi!). Sirf do banda ho to shortcut ab/(a+b). Efficiency aur time hamesha ulta chalte hain (eff 5:4 → time 4:5). Aadmi-din mein M₁D₁ = M₂D₂. Aur Pipes & Cisterns bilkul yahi hai — bas khali karne wale nal ki efficiency minus lagti hai (fill 10 + leak 15 → 30 ghante). 🚀