Video summary

Time and Work - Shortcuts & Tricks for Placement Tests, Job Interviews & Exams

Main summary

Key takeaways

Educational

Main ideas / lessons

  • Core shortcut: Inversion (reciprocal)

    • If total work is done in N days, then work done in 1 day = 1/N.
    • If 1 day work is 1/N, then total time to finish = N.
    • This “invert” idea is repeatedly used to switch between:
      • days ↔ one-day work (rate)

  • Core proportional reasoning

    • If the number of workers increases, then time/days decreases (because total work per day increases).
    • A faster worker means less time; a slower worker means more time.

  • Standard workflow for time & work problems

    • Usually do:
      1. Find (1) one-day work for each person,
      2. (2) add when working together,
      3. (3) invert to get total days.
    • For cases where workers leave/join, use a line diagram / timeline approach.

Methodologies / step-by-step instructions (as taught)

Inversion-based method (general)

  1. Convert “total days” into “one-day work” using inversion:
    • Given: A can finish work in N days
    • Then: A’s work in 1 day = 1/N
  2. When multiple people work together:
    • Total one-day work = sum of individual one-day work rates
  3. Final step:
    • If total one-day work = x, then total days = 1/x (invert again)

Problem-wise concepts (as conveyed)

Q1. A vs B speed + difference in time

Given

  • A is 5 times faster than B
  • A takes 60 days less than B

Approach (rate/time relation)

  • If B takes N days, then A takes N/5 days
  • Also A = B − 60 ⇒ N/5 = N − 60
  • Solve for N, then compute each time.

Result

  • A = 15 days, B = 75 days

Q2. Changing number of workers (men)

Common mistake highlighted

  • Direct cross-multiplication can be wrong because more men ⇒ fewer days, and time does not scale linearly.

Correct taught method (invert via one-day work)

  • 24 men finish in 10 days ⇒ work in 1 day = 1/10
  • Find 1-day work for 30 men:
    • work_rate ∝ number_of_workers
    • (30/24) × (1/10) = 1/8
  • Invert:
    • If work per day = 1/8, then total days = 8

Result

  • 30 men = 8 days

Q3. Three workers together (find time)

Given

  • A in 3 days, B in 6 days, C in 7 days

Method

  • One-day work = 1/3 + 1/6 + 1/7
  • LCM (3,6,7) effectively gives 42
  • Total one-day work = 9/14
  • Total days = 1 ÷ (9/14) = 14/9

Result

  • 14/9 days

Q4. Three persons with pairwise completion times (PQ, QR, RP)

Given

  • (P+Q) in 12 days ⇒ one-day work = 1/12
  • (Q+R) in 16 days ⇒ one-day work = 1/16
  • (R+P) in 24 days ⇒ one-day work = 1/24

Method

  • Let P, Q, R be one-day rates.
  • Add equations carefully:
    • (P+Q) + (Q+R) + (R+P) = 2(P+Q+R)
  • Solve for total one-day work of (P+Q+R), then invert.

Result

  • Total time = 32/3 days

Q5. Efficiency increase percentage

Given

  • P completes in 30 days
  • Q is 25% more efficient than P

Method

  • P one-day work = 1/30
  • Q rate = 125% of P ⇒ one-day work = (1.25) × (1/30) = 1/24
  • Invert ⇒ days = 24

Result

  • Q = 24 days

Q6. Men vs boys equivalence

Given

  • 3 men in 2 days
  • 4 boys in 6 days

Method taught

  • Use time ratio to infer rate ratio:
    • Boys take 3× more time ⇒ men work 3× more per person (as derived)
  • Establish equivalence:
    • 1 man = 4 boys
  • Convert workers into one unit (boys):
    • 8 men = 32 boys, plus 8 boys ⇒ 40 boys
  • 4 boys in 6 days ⇒ 4-boy one-day work = 1/6
  • 40-boy one-day work = (40/4) × (1/6) = 10/6 (as used in the steps), then invert following the video’s timeline conclusion.

Result (as stated)

  • They complete in effectively “6 × 10 days”, and the final numeric conclusion given is: 60 days.

Q7. Someone leaves after some time (timeline/line diagram)

Given

  • Sita completes in 20 days
  • Gita completes in 25 days
  • Together, then Sita leaves; Gita finishes remaining work in 10 days

Method (timeline/line diagram)

  1. One-day work:
    • Sita: 1/20, Gita: 1/25
  2. Work done by Gita in last 10 days:
    • 10 × (1/25) = 2/5
  3. Remaining work:
    • 1 − 2/5 = 3/5
  4. Combined one-day work before leaving:
    • 1/20 + 1/25 = 9/100
  5. Time for remaining:
    • (3/5) ÷ (9/100) = (3/5)×(100/9) = 20/3 days

Result

  • Sita leaves after 20/3 days

Q8. “P alone is X more than (P+Q)” and “Q alone is Y more”

Given

  • P alone takes 25 days more than (P+Q) together
  • Q alone takes 9 days more than (P+Q) together

Shortcut formula

  • Let time for (P+Q) together = n
  • Then n = √(25 × 9)

Result

  • n = √225 = 15 days

Q9. Daily hours differ (convert days to hours)

Given

  • A: 12 days, works 8 hours/day
  • B: 8 days, works 10 hours/day

Method

  • Convert to total “work-time units”:
    • A: 12×8 = 96
    • B: 8×10 = 80
  • One-hour rates:
    • 1/96 and 1/80
  • Together per hour: add rates → invert to get total hours
  • Convert hours to days by dividing by 8 hours/day

Result (as stated)

  • Days = 60/11 days

Q10. Alternate-day working (Raj and Surj)

Given

  • Raj alone: 16 days
  • Surj alone: 12 days
  • They work on alternate days, Raj starts

Method (pattern repetition with line/blocks)

  1. One-day work amounts:
    • Raj: 1/16, Surj: 1/12
  2. In a 2-day block (Raj+Surj):
    • work = 1/16 + 1/12
  3. Repeat 2-day block until remaining work is less than 1 and next person/day fits.
  4. Handle leftover by comparing which worker’s one-day capacity can finish remaining work.

Result (as concluded in the video)

  • 55/4 days (shown as 13 + 3/4 days = 55/4 days)

Speakers / sources featured

  • No specific named speaker is explicitly identified in the subtitles.
  • Source referenced: career-right.com (mentioned as the platform where practice questions are available).

Original video