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Alternate Working Days

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Introduction

Alternate Working Days pattern तब आता है जब दो या अधिक workers बारी-बारी से काम करते हैं (Day 1 पर A, Day 2 पर B, Day 3 पर फिर A, और इसी तरह)। यह pattern इसलिए important है क्योंकि कई exam-type Time & Work problems में alternating schedule दिया होता है - अगर आप एक पूरा cycle (आमतौर पर 2-day या multi-day) को एक single work unit की तरह treat करना सीख लेते हैं, तो ये questions बहुत आसान हो जाते हैं।

Key idea: एक पूरा alternation cycle को एक unit मानें (जिसमें उस cycle के सभी days का कुल work शामिल हो), फिर cycle-counting + partial cycle logic से पूरा work complete करें।

Pattern: Alternate Working Days

Pattern: Alternate Working Days

Key concept: पूरा alternation cycle में किया गया work निकालें, फिर integer cycles + बचा हुआ partial cycle मिलाकर total work = 1 पूरा करें।

संक्षेप में steps:
1. हर worker का daily rate (work per day) निकालें।
2. एक alternation cycle में आने वाले days के rates को जोड़ें (जैसे A + B = 2-day cycle)।
3. देखें कि कितने full cycles से काम लगभग पूरा होता है, फिर बचे हुए काम को day-by-day पूरा करें।

Step-by-Step Example

Question

A एक काम 10 days में और B वही काम 15 days में करता है। वे alternate days पर काम करते हैं और शुरुआत A से होती है (Day 1 A, Day 2 B, Day 3 A, ...). काम कुल कितने दिनों में पूरा होगा?

Solution

  1. Step 1: Daily rates पहचानें:

    A = 1/10; B = 1/15.

  2. Step 2: 2-day full cycle का work निकालें (A + B):

    Work in 2 days = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6.

  3. Step 3: देखिए कितने full cycles से काम 1 के करीब पहुँचता है:

    एक 2-day cycle = 1/6 work। 6 cycles = 6×(1/6) = 1 पूरा काम, लेकिन 6 cycles = 12 days। पहले देखें कि 5 cycles क्या देते हैं: 5×(1/6) = 5/6; remaining = 1 - 5/6 = 1/6.

  4. Step 4: Remaining work को दिन-प्रतिदिन पूरा करें (next turn A की होती है):

    Day 11: A करता है 1/10 = 3/30 = 1/10 ≈ 0.1; remaining था 1/6 ≈ 0.1667, इसलिए काम पूरा नहीं होता। Remaining after Day 11 = 1/6 - 1/10 = (5/30 - 3/30) = 2/30 = 1/15. Day 12: B करता है 1/15, जो remaining के बराबर है - काम Day 12 को end में पूरा हो जाता है।

  5. Final Answer:

    काम 12 days में पूरा होगा।

  6. Quick Check:

    6 cycles = 12 days = 6×(1/6) = 1 पूरा काम ✔️ - हमारा stepwise result match करता है।

Quick Variations

1. अगर शुरुआत B से होती है - cycle order बदल दें।

2. Multi-person alternation (A → B → C → A → …): cycle length = number of people, cycle-work = तीनों के daily rates का sum।

3. Unequal shifts (A 2 days, B 1 day, repeat): cycle = 3 days, work = 2A-days + 1B-day।

4. किसी cycle में कोई worker absent हो - उसकी contribution निकाल दें और नया cycle-work निकालें।

Trick to Always Use

  • Step 1 → Times को daily rates (1/day) में बदलें।
  • Step 2 → Full alternation cycle का work निकालें।
  • Step 3 → Integer cycles से ज़्यादातर काम पूरा करें और remaining fraction निकालें।
  • Step 4 → Remaining काम को correct turn order में day-by-day पूरा करें।

Summary

Alternate Working Days problems में:

  • हमेशा एक cycle बनाएं (A/B के लिए 2-day cycle) और उसका work निकालें।
  • Integer cycles से काम reduce करें, फिर बचे हुए days को क्रम के अनुसार solve करें।
  • अगर next worker की full-day output से कम work बाकी है, तो काम उसी day के हिस्से में पूरा हो जाएगा - fractional day निकालें।
  • Quick check: cycles + leftover contributions को जोड़कर total = 1 आना चाहिए।

Practice

(1/5)
1. A can do a job in 10 days and B in 15 days. They work on alternate days starting with A (A on day 1, B on day 2, ...). In how many days will the job be finished?
easy
A. 12 days
B. 11 days
C. 10 days
D. 13 days

Solution

  1. Step 1: Identify daily rates:

    A = 1/10 per day; B = 1/15 per day.

  2. Step 2: Work in one 2-day cycle (A then B):

    Cycle work = 1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6.

  3. Step 3: Use full cycles:

    After 5 full cycles (10 days) work done = 5×(1/6) = 5/6. Remaining = 1 - 5/6 = 1/6.

  4. Step 4: Finish remaining day-by-day:

    Day 11 (A's turn) does 1/10 < 1/6, remaining becomes 1/6 - 1/10 = 1/15. Day 12 (B's turn) does 1/15 = remaining → finishes at end of Day 12.

  5. Final Answer:

    12 days → Option A.

  6. Quick Check:

    6 full 2-day cycles = 12 days → 6×(1/6)=1 (matches) ✅
Hint: Compute cycle work (A+B), use integer cycles, then finish leftover day-by-day.
Common Mistakes: Using time ratios instead of daily work rates; forgetting turn order for remainder.
2. A can finish a task in 8 days and B in 12 days. They work on alternate days starting with B (B on day 1, A on day 2, ...). How long will the task take?
easy
A. 9 days 2 hours
B. 9 2/3 days
C. 10 days
D. 8 2/3 days

Solution

  1. Step 1: Identify daily rates:

    A = 1/8 per day; B = 1/12 per day.

  2. Step 2: Work in one 2-day cycle (B then A):

    Cycle work = 1/12 + 1/8 = (2+3)/24 = 5/24.

  3. Step 3: Use full cycles:

    After 4 full cycles (8 days) work done = 4×(5/24) = 20/24 = 5/6. Remaining = 1/6.

  4. Step 4: Finish remaining day-by-day (next is day 9: B):

    Day 9 (B) does 1/12 < 1/6 → remaining after day 9 = 1/6 - 1/12 = 1/12. Day 10 (A) will finish; fraction of A's day needed = (1/12) ÷ (1/8) = 2/3 day. So additional time after day 8 = 1 + 2/3 = 1 2/3 days → total = 8 + 1 2/3 = 9 2/3 days.

  5. Final Answer:

    9 2/3 days → Option B.

  6. Quick Check:

    8 days → 5/6 done; day9 (B) adds 1/12 → 5/6+1/12=(10+1)/12=11/12; remaining 1/12 done by 2/3 of A's day ✅
Hint: If a worker's full day exceeds remaining, compute fractional day needed = remaining ÷ their rate.
Common Mistakes: Forgetting who works next after cycles (start matters).
3. A can do a job in 9 days and B in 6 days. They work alternately starting with A. How many days to finish the job?
easy
A. 8 days
B. 7 days
C. 7 1/3 days
D. 6 2/3 days

Solution

  1. Step 1: Identify daily rates:

    A = 1/9 per day; B = 1/6 per day.

  2. Step 2: Work in one 2-day cycle (A then B):

    Cycle work = 1/9 + 1/6 = (2+3)/18 = 5/18.

  3. Step 3: Use full cycles:

    After 3 cycles (6 days) work done = 3×(5/18) = 15/18 = 5/6. Remaining = 1/6.

  4. Step 4: Finish remaining day-by-day:

    Day 7 (A) does 1/9 < 1/6 → remaining becomes 1/6 - 1/9 = 1/18. Day 8 (B) will finish; fraction of B's day needed = (1/18) ÷ (1/6) = 1/3 day. Total = 7 + 1/3 = 7 1/3 days.

  5. Final Answer:

    7 1/3 days → Option C.

  6. Quick Check:

    6 days = 5/6; A adds 1/9 → 5/6+1/9=(15+2)/18=17/18; B’s 1/3 day gives (1/3)×(1/6)=1/18 → total 1 ✅
Hint: If a full day overshoots remaining, compute fractional day = remaining ÷ rate of that worker.
Common Mistakes: Stopping at cycles without accounting for which worker comes next.
4. A can complete a work in 7 days and B in 11 days. They work alternately starting with A. After 8 days how much of the work will be done and how long more (fractional day) will A need when his next turn comes to finish it?
medium
A. After 8 days: 72/77 done; A needs 5/11 of a day
B. After 8 days: 70/77 done; A needs 3/11 of a day
C. After 8 days: 72/77 done; A needs 5/77 of a day
D. After 8 days: 70/77 done; A needs 5/11 of a day

Solution

  1. Step 1: Daily rates:

    A = 1/7, B = 1/11.

  2. Step 2: Cycle work (A then B):

    1/7 + 1/11 = (11 + 7)/77 = 18/77 per 2-day cycle.

  3. Step 3: Work after 4 full cycles (8 days):

    4×(18/77) = 72/77 done; remaining = 1 - 72/77 = 5/77.

  4. Step 4: Next turn is A (day 9). Fraction of A's day needed = (remaining) ÷ (A's rate) = (5/77) ÷ (1/7) = (5/77)×7 = 35/77 = 5/11 day.

  5. Final Answer:

    After 8 days 72/77 done; A needs 5/11 of a day on his next turn → Option A.

  6. Quick Check:

    A’s partial day contribution = (5/11)×(1/7)=5/77 → 72/77+5/77=1 ✅
Hint: Compute cycle total, multiply by cycles, then convert remainder into fractional day using next worker's rate.
Common Mistakes: Converting remainder incorrectly (mixing denominators) or using wrong next-worker turn.
5. B can do a job in 6 days and A in 4 days. They work alternately starting with B. How many days will it take to finish the job?
medium
A. 4 1/2 days
B. 5 1/3 days
C. 4 2/3 days
D. 5 days

Solution

  1. Step 1: Daily rates:

    B = 1/6 per day; A = 1/4 per day.

  2. Step 2: Cycle work (B then A):

    1/6 + 1/4 = (2 + 3)/12 = 5/12 per 2-day cycle.

  3. Step 3: Use full cycles:

    Two full cycles (4 days) give 2×(5/12)=10/12=5/6. Remaining = 1/6.

  4. Step 4: Next is day 5 (B's turn):

    B does 1/6 which exactly equals remaining → job finishes at end of day 5.

  5. Final Answer:

    5 days → Option D.

  6. Quick Check:

    4 days → 5/6 done; day5 (B) adds 1/6 → total 1 ✅
Hint: Check if the next worker's full day exactly matches the remainder to finish cleanly.
Common Mistakes: Assuming equal-day contributions or forgetting start order.