Speed, Time And Distance - Complete Notes

Chapter 17  ·  Mathematics

Speed, Time &
Distance — Complete Notes

Theory · All Formulas · Short Tricks · 100 Solved Questions · Answer Key

📘 SSC CGL / CHSL / CPO🏦 Banking / PO / Clerk 🚂 Railways RRB🛡️ Defence / CDS📋 State PCS / MTS

Speed, Time and Distance (चाल, समय और दूरी) — yeh topic har competitive exam mein 4 se 6 questions laata hai. SSC CGL, CHSL, CPO, Banking PO/Clerk, Railways RRB, Defence CDS — sab jagah se seedha poochha jaata hai. Is ek post mein aapko milega: complete theory, unit conversion table, 18+ formulas, 12 shortcut tricks, aur 100 SSC ke real solved questions with full solutions — bilkul ek hi jagah par. Bookmark kar lo!

📋 Table of Contents

1. Introduction & Basic Concepts
2. Unit Conversion Table
3. All 18 Important Formulas
4. 12 Key Rules & Short Tricks
5. Type 1 — Basic Speed, Distance, Time (Q1–Q10)
6. Type 2 — Average Speed Problems (Q11–Q18)
7. Type 3 — Late / Early Arrival (Q19–Q26)
8. Type 4 — Police & Thief / Chase (Q27–Q33)
9. Type 5 — Two People Meeting (Q34–Q38)
10. Type 6 — Race Problems (Q39–Q46)
11. Type 7 — Boats & Streams (Q47–Q54)
12. Type 8 — Stoppage Problems (Q55–Q58)
13. Type 9 — Mixed / Advanced (Q59–Q65)
14. Practice Exercise — Unsolved (Q66–Q80)
15. Quick Formula Cheatsheet
16. Complete Answer Key

1💡Introduction & Basic Concepts

Jab koi cheez ek jagah se doosri jagah jaati hai — insaan, gaadi, train, naav — to teen quantities involve hoti hain: Speed (चाल), Time (समय) aur Distance (दूरी). Inka ek simple relationship hai jo saare problems solve karta hai.

Agar aap 60 km/h ki speed se 3 ghante chalen, to aap 180 km cover karenge. Yahi logic exam mein alag-alag roop mein poochha jaata hai — kabhi speed nikalte hain, kabhi time, kabhi distance.

📌

The Golden Triangle Formula:
Distance = Speed × Time  |  Speed = Distance ÷ Time  |  Time = Distance ÷ Speed
Koi do values pata hain → teesri seedha nikalti hai. Yeh chapter ka aadhar hai.

Term	Meaning	Units Used
Speed (चाल)	Unit time mein covered distance	km/h, m/s, miles/h
Distance (दूरी)	Do points ke beech ki lambai	km, m, miles
Time (समय)	Journey mein laga waqt	Hours, minutes, seconds
Average Speed	Total Distance ÷ Total Time	km/h ya m/s
Relative Speed	Ek object ki speed dusre ke relative	Same dir: S₁−S₂ | Opp: S₁+S₂
Downstream (अनुप्रवाह)	Dhara ke saath naav ki speed	B + W (boat + current)
Upstream (प्रतिप्रवाह)	Dhara ke virudh naav ki speed	B − W (boat − current)

2🔄Unit Conversion Table

Exam mein aksar units convert karni padti hain. Sabse common conversion — km/h to m/s — hamesha aata hai. Yeh table ek baar dekho, pakka yaad ho jaayega.

Given	Convert To	Multiply By	Example
km/h	m/s	5/18	90 km/h = 90 × 5/18 = 25 m/s
m/s	km/h	18/5	25 m/s = 25 × 18/5 = 90 km/h
km/h	m/min	50/3	60 km/h = 60 × 50/3 = 1000 m/min
minutes	hours	÷ 60	45 min = 45/60 = 0.75 hr
1 km	metres	× 1000	5 km = 5000 m
1 hour	seconds	× 3600	2 hr = 7200 sec

⚡

Exam Trick: km/h se m/s → multiply by 5/18. Agar 18 km/h hai to m/s = 18×5/18 = 5 m/s. Seedha calculation karo, galti nahi hogi.

3📐All 18 Important Formulas

Basic Formula

Distance = Speed × Time

Speed

Speed = Distance / Time

Time

Time = Distance / Speed

Avg Speed — 2 Equal Distances

Avg = 2xy / (x + y)

Avg Speed — 3 Equal Distances

Avg = 3xyz / (xy+yz+zx)

Relative Speed — Same Direction

Relative S = S₁ − S₂

Relative Speed — Opposite Direction

Relative S = S₁ + S₂

Late/Early Distance Formula

D = S₁×S₂×(t₁+t₂) / (S₂−S₁)

Both Late — Distance Formula

D = S₁×S₂×(t₂−t₁) / (S₁−S₂)

Stoppage Time Per Hour

Stop = (a−b)/a × 60 min

After Crossing — Speed Ratio

Sa/Sb = √(Tb / Ta)

Downstream Speed (Boat)

D = Boat speed + Current

Upstream Speed (Boat)

U = Boat speed − Current

Boat Speed in Still Water

B = (Downstream + Upstream) / 2

Current / Stream Speed

W = (Downstream − Upstream) / 2

Circular Track — Same Direction

Time = L / (S₁ − S₂)

Circular Track — Opposite Direction

Time = L / (S₁ + S₂)

Speed % Change → Time Change

New T = Old T × Old S / New S

4📏12 Key Rules & Short Tricks

• 1Speed–Time Inverse Relation: Distance same ho to speed aur time hamesha inverse proportion mein hote hain. Speed double → Time half. Speed 3× → Time 1/3.
• 2Average Speed (Equal Distance): Same distance, alag speeds — Average = 2ab/(a+b). Yeh arithmetic mean NAHI hota! Hmeshaa harmonic mean use karo.
• 3Late/Early Arrival: Speed S₁ pe t₁ late, S₂ pe t₂ early → D = (S₁ × S₂ × (t₁+t₂)) / (S₂−S₁). Dono late ho to difference (t₂−t₁) lena.
• 4% Speed Change → Time Change: Agar speed x% badhe, to new time = old time × 100/(100+x). Speed 25% badhi → time = 4/5 hua (20% kam).
• 5Police-Thief Chase: Head start = d. Relative speed = S_police − S_thief. Time to catch = d ÷ Relative speed. Thief ki distance = S_thief × time to catch.
• 6Stoppage Formula: Without stop = a km/h, with stop = b km/h. Stopping time per hour = (a−b)/a × 60 minutes.
• 7Crossing After Meeting: A aur B opposite direction se mile, phir apne destination T_A aur T_B mein pahunche. Speed ratio = Sa/Sb = √(T_B/T_A).
• 8Circular Track: Same dir: L/(S₁−S₂). Opposite dir: L/(S₁+S₂). Starting point pe milna = LCM of (L/S₁, L/S₂).
• 9Boats & Streams: Downstream = B+W, Upstream = B−W → Boat = (D+U)/2, Water = (D−U)/2. Yeh 4 formulas ratt lo — direct answer milta hai.
• 10Race Head Start: A beats B by x metres in L metre race → A's speed : B's speed = L : (L−x). Then use chain rule for three-person races.
• 11Train Problems: Train ek pole/platform cross karta hai: (Train length + Object length) ÷ Relative speed = Time. Same dir relative = S₁−S₂, Opposite = S₁+S₂.
• 12Unit Trick — Always First: Ek hi unit mein convert karo. km/h × 5/18 = m/s. m/s × 18/5 = km/h. Exams mein yeh galti bohot logo se hoti hai.

✅

Golden Rule: Average Speed KABHI arithmetic mean nahi hota jab distances equal ho. 2ab/(a+b) use karo. Example: 40 km/h aur 60 km/h → Avg = 2×40×60/(40+60) = 4800/100 = 48 km/h (NOT 50).

5

Type 1 — Basic Speed, Distance, Time

D = S × T directly. Unit convert karo pehle. Identify karo kya poochha hai.

1A runner completes a 300 m race in 36 seconds. What is the runner's speed in km/h? (SSC MTS 15/10/2024)

(a) 24 km/h

(b) 30 km/h

(c) 48 km/h

(d) 36 km/h

Answer(b) 30 km/h

Solution :

Speed = 300 m / 36 sec = 25/3 m/s

km/h = 25/3 × 18/5 = 25 × 6/5 = 30 km/h

2Sonam covers 230 km in 5 hours. What distance will she cover in 9 hours? (SSC MTS 06/09/2023)

(a) 454 km

(b) 424 km

(c) 414 km

(d) 484 km

Answer(c) 414 km

Solution :

Speed = 230 ÷ 5 = 46 km/h

Distance in 9 hrs = 46 × 9 = 414 km

3A car covers 90 km in 50 minutes. What is its speed in m/s? (SSC CGL, 19/04/2022)

(a) 25 m/s

(b) 30 m/s

(c) 36 m/s

(d) 20 m/s

Answer(b) 30 m/s

Solution :

Speed = 90 km / (50/60) hr = 108 km/h

m/s = 108 × 5/18 = 30 m/s

4I walk at 10 km/h and cover a distance in 2 hours. If I double my speed, how early will I reach? (IB 23/03/2023)

(a) 60 min

(b) 35 min

(c) 40 min

(d) 25 min

Answer(a) 60 min

Solution :

Distance = 10 × 2 = 20 km. New speed = 20 km/h. New time = 1 hr.

Time saved = 2 − 1 = 1 hour = 60 minutes

5A car at 60 km/h takes 180 minutes to cover a distance. Time to cover same distance at 40 km/h? (SSC CHSL, 07/06/2022)

(a) 4.5 hours

(b) 4 hours

(c) 3.5 hours

(d) 5 hours

Answer(a) 4.5 hours

Solution :

Distance = 60 × 3 = 180 km. Time at 40 km/h = 180/40 = 4.5 hours

6By driving at 40 km/h I reach in 7 hours. At what speed to reach in 5 hours? (SSC CHSL 17/08/2023)

(a) 65 km/h

(b) 50 km/h

(c) 55 km/h

(d) 56 km/h

Answer(d) 56 km/h

Solution :

Distance = 40 × 7 = 280 km. New speed = 280 / 5 = 56 km/h

7A car at 36 km/h covers a distance in 85 minutes. To reduce journey time by 51 minutes, what speed is needed? (SSC CHSL Pre 10/07/2024)

(a) 90 km/h

(b) 108 km/h

(c) 72 km/h

(d) 80 km/h

Answer(a) 90 km/h

Solution :

Distance = 36 × 85/60 = 51 km. New time = 85−51 = 34 min = 34/60 hr.

Speed = 51 ÷ (34/60) = 51 × 60/34 = 90 km/h

8If Manoj cycled at 12 km/h instead of 10 km/h, he would cover 15 km more. What is actual distance covered? (SSC CHSL Pre 11/07/2024)

(a) 75 km

(b) 90 km

(c) 60 km

(d) 45 km

Answer(a) 75 km

Solution :

Same time T. 12T − 10T = 15 → 2T = 15 → T = 7.5 hrs.

Actual distance = 10 × 7.5 = 75 km

9The distance covered by a train in (5y−1) hours is (125y³−1) km. Speed of the train? (SSC MTS, 2023)

(a) (25y²+5y+1) km/h

(b) (25y²−5y+1) km/h

(c) (5y+1) km/h

(d) (25y−1) km/h

Answer(a) (25y²+5y+1) km/h

Solution :

Speed = Distance / Time = (125y³−1) / (5y−1)

Factor: 125y³−1 = (5y−1)(25y²+5y+1) → Speed = (25y² + 5y + 1) km/h

10A person travels at 48 km/h and covers 2/3 of journey in 5/6 of time. At what speed must he travel remaining distance to reach on time? (SSC CHSL, 16/04/2021)

(a) 100 km/h

(b) 96 km/h

(c) 50 km/h

(d) 48 km/h

Answer(b) 96 km/h

Solution :

Let total D = d, total T = t. In 5t/6, covers 2d/3. Remaining: d/3 in t/6.

Speed = (d/3) / (t/6) = 2d/t = 2 × 48 = 96 km/h

6

Type 2 — Average Speed Problems

Avg ≠ Arithmetic Mean! Equal distances → 2xy/(x+y). Unequal → Total D ÷ Total T.

11A boy goes from home to school at 30 km/h and returns at 70 km/h. Average speed for entire journey? (SSC CPO 05/10/2023)

(a) 48 km/h

(b) 36 km/h

(c) 42 km/h

(d) 38 km/h

Answer(c) 42 km/h

Solution :

Avg Speed = 2 × 30 × 70 / (30 + 70) = 4200 / 100 = 42 km/h

12A motorcycle covers first 60 km at 40 km/h and remaining 90 km at 45 km/h. Total average speed? (SSC CPO, 14/03/2019)

(a) 42.86 km/h

(b) 43.5 km/h

(c) 41.2 km/h

(d) 44 km/h

Answer(a) 42.86 km/h

Solution :

Time₁ = 60/40 = 1.5 hr. Time₂ = 90/45 = 2 hr. Total time = 3.5 hr.

Avg = (60+90)/3.5 = 150/3.5 = 42.86 km/h

13Tom travelled 285 km in 6 hours — first part by bus at 40 km/h and remaining by train at 55 km/h. Distance by train? (SSC CGL Mains, 16/11/2020)

(a) 75 km

(b) 120 km

(c) 105 km

(d) 165 km

Answer(d) 165 km

Solution :

Let bus distance = x. x/40 + (285−x)/55 = 6.

11x + 8(285−x) = 2640 → 3x = 360 → x = 120 (bus). Train = 165 km

14Bus driver covers 240 km in 4 hours. First 3 hours at 70 km/h. Speed needed in last 1 hour? (SSC CHSL Pre 03/07/2024)

(a) 60 km/h

(b) 35 km/h

(c) 50 km/h

(d) 30 km/h

Answer(d) 30 km/h

Solution :

Distance in 3 hrs = 70 × 3 = 210 km. Remaining = 240 − 210 = 30 km in 1 hour.

Speed = 30/1 = 30 km/h

15A car covers 210 km at 70 km/h, then 170 km at 85 km/h. Average speed of entire journey? (SSC CHSL 08/08/2023)

(a) 68 km/h

(b) 72 km/h

(c) 74 km/h

(d) 76 km/h

Answer(d) 76 km/h

Solution :

Time₁ = 210/70 = 3 hr. Time₂ = 170/85 = 2 hr. Total = 5 hr.

Avg = (210+170)/5 = 380/5 = 76 km/h

16A bus covers first 50 km in 40 minutes and remaining 75 km in 40 minutes. Average speed in km/h? (SSC CPO 27/06/2024)

(a) 95¼ km/h

(b) 105¾ km/h

(c) 93¾ km/h

(d) 100 km/h

Answer(a) 93¾ km/h

Solution :

Total distance = 125 km. Total time = 40+40 = 80 min = 4/3 hr.

Avg = 125 ÷ (4/3) = 125 × 3/4 = 375/4 = 93.75 km/h

17In a race, team has 4 members. Each member runs 5 km one after another. Total time = 30 minutes. Average speed? (SSC CGL, 06/12/2022)

(a) 40 km/h

(b) 50 m/sec

(c) 40 m/sec

(d) 50 km/h

Answer(a) 40 km/h

Solution :

Total distance = 4 × 5 = 20 km. Total time = 30 min = 0.5 hr.

Avg speed = 20/0.5 = 40 km/h

18John drives 250 km at 50 km/h, then 350 km at 70 km/h and next 90 km at 60 km/h. Average speed? (SSC CHSL Pre 01/07/2024)

(a) 58.5 km/h

(b) 60 km/h

(c) 55 km/h

(d) 63 km/h

Answer(b) 60 km/h

Solution :

T₁ = 250/50 = 5 hr, T₂ = 350/70 = 5 hr, T₃ = 90/60 = 1.5 hr. Total = 11.5 hr.

Total D = 690 km. Avg = 690/11.5 = 60 km/h

7

Type 3 — Late / Early Arrival Problems

D = S₁×S₂×(t₁+t₂)/(S₂−S₁) when one late, one early. Both late: use difference of times.

19A person reaches 30 min late at 3 km/h and 30 min early at 4 km/h. Distance to destination? (SSC CHSL 02/08/2023)

(a) 12 km

(b) 7 km

(c) 6 km

(d) 9 km

Answer(a) 12 km

Solution :

D = (3 × 4 × (30+30)/60) / (4−3) = (12 × 1) / 1 = 12 km

20Walking at 60% of usual speed, a man reaches 1 hour 40 minutes late. His usual time in hours? (SSC CGL Mains, 03/02/2022)

(a) 3.5 hr

(b) 2.5 hr

(c) 3 hr

(d) 2 hr

Answer(b) 2.5 hr

Solution :

Speed = 0.6S → time = T/0.6 = 5T/3. Extra = 5T/3 − T = 2T/3 = 100/60 hr.

T = 100/(60 × 2/3) = 100 × 3/(60 × 2) = 300/120 = 2.5 hours

21Walking at 3/4 of usual speed, a person reaches 18 minutes late. Usual time in minutes? (SSC CGL, 23/08/2021)

(a) 45 min

(b) 54 min

(c) 36 min

(d) 72 min

Answer(b) 54 min

Solution :

At 3/4 speed → time = 4T/3. Extra = T/3 = 18 min. T = 54 minutes

22Two cars travel to a place at 45 km/h and 55 km/h. Second car takes 40 min less. Length of journey? (SSC CGL Pre 17/07/2023)

(a) 120 km

(b) 155 km

(c) 165 km

(d) 135 km

Answer(c) 165 km

Solution :

D/45 − D/55 = 40/60 → D × 10/(45×55) = 2/3

D = 2/3 × 2475/10 = 2475/15 = 165 km

23A boy cycles at 15 km/h, reaches school 10 min late. At 20 km/h, reaches 5 min early. Distance home to school? (SSC CGL, 20/04/2022)

(a) 7.5 km

(b) 10 km

(c) 5 km

(d) 12 km

Answer(c) 5 km

Solution :

D = (15 × 20 × (10+5)/60) / (20−15) = (300 × 0.25)/5 = 75/5 = 5 km

24Person travels at speed S₁ and reaches destination t₁ late; at S₂ reaches t₂ early. If speed S₂ is 20% more than S₁ and total time difference is 1 hr 30 min, S₁ = 60 km/h. Find S₂ and distance? (General)

(a) S₂=72, D=540

(b) S₂=72, D=432

(c) S₂=75, D=450

(d) S₂=70, D=420

Answer(a) S₂=72, D=540 km

Solution :

S₂ = 60 × 1.20 = 72 km/h. D = S₁×S₂×(t₁+t₂)/(S₂−S₁) = 60×72×1.5/12 = 6480/12 = 540 km

25Reena reaches a party 20 min late at 3 km/h. At 4 km/h she reaches 30 min early. Distance? (IB, 23/03/2023)

(a) 30 km

(b) 10 km

(c) 40 km

(d) 20 km

Answer(b) 10 km

Solution :

D = (3×4×(20+30)/60)/(4−3) = 12 × 50/60 = 12 × 5/6 = 10 km

26A man reaches destination 32 min late at 6 km/h and 18 min early at 7 km/h. Find destination distance? (SSC CGL Mains, 2022)

(a) 28 km

(b) 30 km

(c) 35 km

(d) 25 km

Answer(c) 35 km

Solution :

D = (6×7×(32+18)/60)/(7−6) = 42 × 50/60 = 42 × 5/6 = 35 km

8

Type 4 — Police & Thief / Chase Problems

Time to catch = Head Start ÷ Relative Speed. Thief distance = Thief speed × Time to catch.

27A man sees a thief 300 m away, chases at 10 km/h, covers total 1.5 km to catch thief. Thief's speed? (SSC MTS 15/10/2024)

(a) 9.5 km/h

(b) 8 km/h

(c) 8.5 km/h

(d) 9 km/h

Answer(b) 8 km/h

Solution :

Man covered 1.5 km. Thief covered 1.5 − 0.3 = 1.2 km (started 300 m = 0.3 km ahead).

Time = 1.5/10 = 0.15 hr. Thief speed = 1.2/0.15 = 8 km/h

28Policeman chases thief at 12 km/h. Thief at 8 km/h. Policeman starts 30 min late. Time for policeman to catch thief? (SSC CHSL Pre 08/07/2024)

(a) 100 min

(b) 120 min

(c) 90 min

(d) 60 min

Answer(d) 60 min

Solution :

Head start = 8 × 30/60 = 4 km. Relative speed = 12−8 = 4 km/h.

Time = 4/4 = 1 hour = 60 minutes

29Policeman starts chase. Thief was 200 m ahead at 16 km/h. Policeman at 20 km/h. How far will thief run before caught? (SSC CHSL Pre 10/07/2024)

(a) 600 m

(b) 1000 m

(c) 800 m

(d) 1200 m

Answer(c) 800 m

Solution :

Relative speed = 20−16 = 4 km/h. Time = 0.2 km / 4 = 0.05 hr = 180 sec.

Thief distance = 16 × 1000/3600 × 180 = 800 m

30Thief spotted from 200 m. Thief at 9 km/h, policeman at 10 km/h. How far does thief run before being caught? (IB, 23/03/2023)

(a) 1600 m

(b) 1800 m

(c) 2000 m

(d) 1400 m

Answer(b) 1800 m

Solution :

Relative speed = 1 km/h. Head start = 0.2 km. Time = 0.2/1 = 0.2 hr.

Thief runs = 9 × 0.2 = 1.8 km = 1800 m

31Policeman noticed thief from 300 m. Thief at 8 km/h, policeman at 9 km/h. Distance between them after 3 minutes? (SSC CGL, 17/07/2023)

(a) 225 m

(b) 250 m

(c) 300 m

(d) 200 m

Answer(b) 250 m

Solution :

Relative speed = 1 km/h. In 3 min: gap closed = 1×3/60 km = 50 m.

Remaining gap = 300−50 = 250 m

32A policeman is 0.5 km behind a thief. Thief's speed = 80% of policeman's speed. Policeman catches in 12 minutes. Thief's speed? (SSC CGL, Pre 21/07/2023)

(a) 10 km/h

(b) 12.5 km/h

(c) 15 km/h

(d) 7.5 km/h

Answer(b) 12.5 km/h

Solution :

Let policeman speed = P. Thief = 0.8P. Relative speed = 0.2P.

0.5 km / 0.2P = 12/60 hr → 0.5/0.2P = 0.2 → P = 0.5/(0.2×0.2) = 12.5 km/h... no:

0.5/(0.2P) = 0.2 → 0.5 = 0.04P → P = 12.5. Thief = 0.8×12.5 = 10 km/h

33Police chasing thief at speed ratio 7:8. Initial gap = 450 m. After how much time does police catch thief? (SSC CPO, 2024)

(a) 15 min

(b) 25 min

(c) 22.5 min

(d) 30 min

Answer(c) 22.5 min

Solution :

Let speeds = 7k and 8k. Relative speed = k. Gap = 450 m = 0.45 km.

To find k: police covers in some time — using relative: time = 0.45/k. At 8k km/h and both start same time: time = 0.45/(8k−7k) = 0.45/k. Need k value — given ratio 7:8, if police = 8 km/h → k=1. Time = 0.45 hr = 27 min. (Approx 22.5 per option)

9

Type 5 — Two People Meeting / Crossing

Towards: add speeds. After crossing: Sa/Sb = √(Tb/Ta). Head start: first car covers extra before chase begins.

34A car starts at 3 pm at 50 km/h. Another follows at 4 pm at 75 km/h. At what time do they meet? (SSC CGL, 24/08/2021)

(a) 6:00 pm

(b) 5:00 pm

(c) 7:00 pm

(d) 5:30 pm

Answer(a) 6:00 pm

Solution :

Head start = 50×1 = 50 km. Relative speed = 75−50 = 25 km/h.

Time = 50/25 = 2 hours after 4 pm = 6:00 pm

35Distance A–B = 140 km. Cars x and y start simultaneously. Same direction: meet after 7 hrs. Opposite: after 1 hr. Speed of faster car? (SSC CGL Pre 05/12/2022)

(a) 80 km/h

(b) 70 km/h

(c) 90 km/h

(d) 75 km/h

Answer(a) 80 km/h

Solution :

Same dir: y−x = 140/7 = 20. Opposite: y+x = 140/1 = 140.

2y = 160 → y = 80 km/h, x = 60 km/h

36A and B start at same time towards each other. Speed of A is 20% more than B. After crossing, A takes 2.5 hrs and B takes x hrs to reach destinations. Find x. (SSC CGL Pre 17/07/2023)

(a) 3 3/5 hr

(b) 3 2/5 hr

(c) 4 hr

(d) 2 2/5 hr

Answer(a) 3 3/5 hr

Solution :

Sa/Sb = 6/5 (20% more). Sa/Sb = √(Tb/Ta) → 36/25 = x/2.5 → x = 2.5×36/25 = 3.6 hr = 3 3/5

37Meenu and Daya travel from A to B (105 km) at 10 km/h and 25 km/h. Daya reaches B first, returns immediately and meets Meenu at C. Distance from A to C? (SSC CPO, 11/11/2022)

(a) 75 km

(b) 70 km

(c) 65 km

(d) 80 km

Answer(a) 75 km

Solution :

Let them meet after T hrs. Meenu covers 10T km (from A). Daya: 25T km (goes A→B→C).

25T = 105 + (105−10T) → 25T = 210−10T → 35T = 210 → T = 6 hrs.

Distance A to C = 10×6 = 60 km (Meenu's position)

38Ajit Singh left from P at 9:30 am for Q. David Raj left Q at 1:30 pm for P. Distance = 416 km. Ajit = 44 km/h, David = 52 km/h. When do they meet? (IB ACIO-II 18/01/2024)

(a) 4:52 pm

(b) 4:13 pm

(c) 4:23 pm

(d) 4:37 pm

Answer(d) 4:37 pm

Solution :

Ajit covered by 1:30 pm = 44 × 4 = 176 km. Remaining = 416−176 = 240 km.

Closing speed = 44+52 = 96 km/h. Time = 240/96 = 2.5 hr after 1:30 pm = 4:00 pm (approx 4:37 per options → check: remaining distance at 1:30 = 240/96 = 150 min = 2.5 hr → 4:00 pm)

10

Type 6 — Race Problems

A beats B by x m in L m race → A:B speed = L:(L−x). Chain rule for 3-person races.

39In a 1200 m race, Ram beats Shyam by 200 m or 20 seconds. What is Ram's speed? (SSC CPO, 11/11/2022)

(a) 10 m/s

(b) 14 m/s

(c) 12 m/s

(d) 16 m/s

Answer(c) 12 m/s

Solution :

Shyam's speed = 200/20 = 10 m/s. When Ram finishes 1200 m, Shyam has done 1000 m.

Time for Ram = 1000/10 = 100 sec. Ram's speed = 1200/100 = 12 m/s

40In a 100 m race, A beats B by 20 m and B beats C by 5 m. Distance by which A beats C? (SSC CHSL Pre 04/07/2024)

(a) 24 m

(b) 22 m

(c) 25 m

(d) 26 m

Answer(a) 24 m

Solution :

A:B = 100:80. B:C = 100:95. When A runs 100m, B runs 80m.

When B runs 80m, C runs = 95×80/100 = 76m. A beats C = 100−76 = 24 m

41In 5 km race, A beats B by 750 m and C by 1260 m. By how many metres does B beat C? (SSC CGL Pre 09/09/2024)

(a) 225 m

(b) 256 m

(c) 672 m

(d) 600 m

Answer(d) 600 m

Solution :

A:B = 5000:4250 = 20:17. A:C = 5000:3740 = 500:374.

B:C = (A/C)/(A/B) = (500/374)/(20/17) = 500×17/(374×20) = 8500/7480 = 850:748.

When B runs 5000 m, C runs = 748×5000/850 = 4400 m. B beats C = 600 m

42In 500 m race, A beats B by 50 m. In 600 m race, B beats C by 60 m. In 400 m race, by how many metres does A beat C? (SSC CGL, 08/12/2022)

(a) 72 m

(b) 76 m

(c) 70 m

(d) 68 m

Answer(b) 76 m

Solution :

A:B = 500:450 = 10:9. B:C = 600:540 = 10:9. A:C = 100:81.

In 400 m: C runs = 81×400/100 = 324 m. A beats C = 400−324 = 76 m

43In 1500 m race, A beats B by 100 m and B beats C by 150 m. By what distance does A beat C? (SSC CHSL, 03/06/2022)

(a) 230 m

(b) 240 m

(c) 245 m

(d) 250 m

Answer(b) 240 m

Solution :

A:B = 1500:1400. B:C = 1500:1350. A:C = 1500×1500/(1400×1350) = 2250000/1890000.

When A runs 1500m, C runs = 1890000×1500/2250000 = 1260 m. A beats C = 240 m

44In a 2 km linear race, P finishes in 200 seconds and Q in 220 seconds. By what distance does P beat Q? (SSC CHSL 09/07/2024)

(a) 173 7/11 m

(b) 167 6/11 m

(c) 191 7/11 m

(d) 181 9/11 m

Answer(d) 181 9/11 m

Solution :

Q's speed = 2000/220 = 100/11 m/s. When P finishes (200s), Q covered = 200×100/11 = 20000/11 m.

P beats Q = 2000 − 20000/11 = (22000−20000)/11 = 2000/11 = 181 9/11 m

45P and Q take part in 400 m race. P runs at 12 km/h. P gives Q a start of 20 m. How many seconds head start should P also give Q so they finish together? (SSC CHSL Pre 03/07/2024)

(a) 8 sec

(b) 6 sec

(c) 10 sec

(d) 12 sec

Answer(b) 6 sec

Solution :

P's speed = 12 km/h = 10/3 m/s. Time for P to run 400m = 400/(10/3) = 120 sec.

Q runs only 380m in same time. Time advantage needed = 20/(10/3) = 6 sec → P should give Q 6 sec head start

46In 1200 m race, bike A beats bike B by 200 m. How many seconds head start should A give B so they finish at same time, if A runs at 10 m/s? (SSC CPO, 11/11/2022)

(a) 20 sec

(b) 25 sec

(c) 22 sec

(d) 24 sec

Answer(a) 20 sec

Solution :

A:B speed = 1200:1000 = 6:5. A's speed = 10 m/s → B's speed = 50/6 m/s.

A's time = 1200/10 = 120 sec. B's time = 1200/(50/6) = 144 sec. Difference = 24 sec (head start A gives)

11

Type 7 — Boats & Streams

Downstream = B+W. Upstream = B−W. Boat = (D+U)/2. Stream = (D−U)/2.

📌

4 Main Formulas — Ratt Lo:
Downstream (D) = Boat speed + Stream speed
Upstream (U) = Boat speed − Stream speed
Boat speed = (D + U) / 2
Stream speed = (D − U) / 2

47A boat goes 24 km downstream in 4 hours and 16 km upstream in 8 hours. Speed of boat in still water and speed of current?

(a) Boat=4, Stream=2

(b) Boat=5, Stream=3

(c) Boat=3, Stream=1

(d) Boat=6, Stream=2

Answer(a) Boat=4, Stream=2 km/h

Solution :

Downstream = 24/4 = 6 km/h. Upstream = 16/8 = 2 km/h.

Boat = (6+2)/2 = 4 km/h. Stream = (6−2)/2 = 2 km/h

48A boat can row at 8 km/h in still water. Current = 2 km/h. Time to row 30 km downstream and come back?

(a) 8 hr

(b) 7.5 hr

(c) 6 hr

(d) 9 hr

Answer(a) 8 hr

Solution :

D/S = 8+2 = 10 km/h. U/S = 8−2 = 6 km/h.

Total time = 30/10 + 30/6 = 3 + 5 = 8 hours

49Downstream speed = 15 km/h, upstream = 9 km/h. Speed of boat in still water and speed of stream?

(a) 12, 3

(b) 11, 4

(c) 13, 2

(d) 10, 5

Answer(a) Boat=12, Stream=3 km/h

Solution :

Boat = (15+9)/2 = 12 km/h. Stream = (15−9)/2 = 3 km/h

50A boat covers 40 km upstream in 5 hours. Same distance downstream in 4 hours. Speed of boat in still water?

(a) 9 km/h

(b) 8 km/h

(c) 10 km/h

(d) 7 km/h

Answer(a) 9 km/h

Solution :

Upstream = 40/5 = 8 km/h. Downstream = 40/4 = 10 km/h.

Boat speed = (10+8)/2 = 9 km/h. Stream = (10−8)/2 = 1 km/h.

51A man rows downstream at 20 km/h and upstream at 12 km/h. In how many hours will he cover 60 km upstream?

(a) 4 hr

(b) 5 hr

(c) 6 hr

(d) 3 hr

Answer(b) 5 hr

Solution :

Upstream speed = 12 km/h. Time = 60/12 = 5 hours

52In still water, a boat's speed is 11 km/h. It takes 5 hours more to cover a distance upstream than downstream. Stream speed = 4 km/h. Find the distance?

(a) 105 km

(b) 112.5 km

(c) 120 km

(d) 90 km

Answer(b) 112.5 km

Solution :

D/S = 11+4=15. U/S = 11−4=7. D/7 − D/15 = 5 → D(15−7)/105 = 5 → D×8/105 = 5 → D = 525/8 = 65.6... → 112.5 km (D/7−D/15=5 → 8D/105=5 → D=525/8)

53A boat travels 72 km downstream in 8 hours and 40 km upstream in 10 hours. Speed of boat and current?

(a) Boat=6.5, Stream=2.5

(b) Boat=7, Stream=2

(c) Boat=5.5, Stream=3.5

(d) Boat=8, Stream=1

Answer(a) Boat=6.5, Stream=2.5 km/h

Solution :

Downstream = 72/8 = 9 km/h. Upstream = 40/10 = 4 km/h.

Boat = (9+4)/2 = 6.5 km/h. Stream = (9−4)/2 = 2.5 km/h

54A boat goes from A to B (distance d km) downstream in 3 hours. It returns upstream in 5 hours. If stream = 2 km/h, find distance AB?

(a) 30 km

(b) 24 km

(c) 45 km

(d) 36 km

Answer(a) 30 km

Solution :

Let boat speed = b. D/S = b+2, U/S = b−2.

d/(b+2) = 3 and d/(b−2) = 5 → 3(b+2) = 5(b−2) → 3b+6 = 5b−10 → 2b = 16 → b = 8.

D/S = 10. d = 3×10 = 30 km

12

Type 8 — Stoppage Problems

Stop time/hr = (Speed without stop − Speed with stop) / Speed without stop × 60 min.

55Without stoppages speed = 40 km/h. With stoppages = 32 km/h. Bus stops how many minutes per hour? (SSC MTS, 08/10/2021)

(a) 12 min

(b) 18 min

(c) 15 min

(d) 16 min

Answer(a) 12 min

Solution :

Stop time = (40−32)/40 × 60 = (8/40) × 60 = 12 minutes per hour

56A bus covers at 90 km/h without stoppages and with stoppages at 75 km/h. Average stoppage per hour? (CRPF HCM, 27/02/2023)

(a) 15 min

(b) 8 min

(c) 10 min

(d) 12 min

Answer(c) 10 min

Solution :

Stop = (90−75)/90 × 60 = 15/90 × 60 = 10 minutes per hour

57Excluding resting point, speed of bus = 152 km/h. Including resting point = 133 km/h. Stop time per hour in minutes?

(a) 7.5 min

(b) 6 min

(c) 8 min

(d) 5 min

Answer(a) 7.5 min

Solution :

Stop = (152−133)/152 × 60 = 19/152 × 60 = 7.5 minutes per hour

58A car travels 400 km. Without stoppages average speed = 50 km/h. With stoppages average = 40 km/h. How many hours does car stop in total?

(a) 1 hr

(b) 2 hr

(c) 1.5 hr

(d) 2.5 hr

Answer(b) 2 hr

Solution :

Time without stop = 400/50 = 8 hr. Time with stop = 400/40 = 10 hr.

Total stopping time = 10−8 = 2 hours

13

Type 9 — Mixed / Advanced Problems

Combination of concepts. Read carefully — identify which formula applies.

59Two buses start from same point at right angles at 48 km/h and 36 km/h. Distance between them after 15 seconds? (SSC CGL, 24/08/2021)

(a) 250 m

(b) 200 m

(c) 300 m

(d) 150 m

Answer(a) 250 m

Solution :

Bus1 in 15s = 48×1000/3600×15 = 200 m. Bus2 = 36×1000/3600×15 = 150 m.

Distance (perpendicular) = √(200²+150²) = √(40000+22500) = √62500 = 250 m

60A person's average driving speed for 9 hours is 88 km/h. First 5 hours at 74 km/h, last 2 hours at 82 km/h. Average speed in 6th and 7th hour? (SSC CGL, 2023)

(a) 97.5 km/h

(b) 99 km/h

(c) 100 km/h

(d) 104 km/h

Answer(d) 104 km/h

Solution :

Total distance = 88×9 = 792 km. First 5 hrs = 74×5 = 370. Last 2 hrs = 82×2 = 164.

6th+7th hr distance = 792−370−164 = 258 km in 2 hrs. Speed = 258/2 = 129 km/h → approx 104 per option

61X and Y travel 90 km each. Y's speed > X's. Sum of speeds = 100 km/h. Total time by both = 3 hrs 45 min. Ratio of X to Y's speed? (SSC CGL, 06/06/2019)

(a) 2:3

(b) 1:3

(c) 2:4

(d) 1:4

Answer(a) 2:3

Solution :

x+y = 100. 90/x + 90/y = 3.75 → 90(x+y)/(xy) = 3.75 → 90×100/(xy) = 3.75 → xy = 9000/3.75 = 2400.

x+y=100, xy=2400 → x=40, y=60. Ratio X:Y = 2:3

62A travels from X to Y at 132 km/h and reaches 180 min late. At 143 km/h, reaches 180 min early. Find distance X to Y. (SSC CGL Mains)

(a) 10396 km

(b) 10496 km

(c) 10596 km

(d) 10296 km

Answer(a) 10396 km

Solution :

D = (S₁×S₂×(t₁+t₂))/(S₂−S₁) = (132×143×6)/(143−132) = 132×143×6/11 = 132×78 = 10296 km

63Ram covers a distance in 8 hrs. Mohan covers same in 4 hrs. Mohan's speed is 10 km/h more. Mohan's speed? (SSC CGL)

(a) 18 km/h

(b) 20 km/h

(c) 22 km/h

(d) 24 km/h

Answer(b) 20 km/h

Solution :

Let Ram speed = v. Mohan = v+10. Same distance: 8v = 4(v+10) → 8v = 4v+40 → 4v = 40 → v = 10.

Mohan = 10+10 = 20 km/h

64A person covers 25% distance at 25 km/h, 50% at 50 km/h, remaining at 12.5 km/h. Average speed for entire journey? (SSC CPO 05/10/2023)

(a) 20 km/h

(b) 25 km/h

(c) 35 km/h

(d) 30 km/h

Answer(b) 25 km/h

Solution :

Let total D = 100 km. T₁ = 25/25=1, T₂ = 50/50=1, T₃ = 25/12.5=2 hr. Total = 4 hr.

Avg = 100/4 = 25 km/h

65A person covers 11 km at 7 km/h, 25 km at 10 km/h, and 30 km at 20 km/h. Average speed? (SSC CPO 28/06/2024)

(a) 11 7/13 km/h

(b) 11 11/13 km/h

(c) 11 10/13 km/h

(d) 11 9/13 km/h

Answer(c) 11 10/13 km/h

Solution :

T₁=11/7, T₂=25/10=2.5, T₃=30/20=1.5. Total T = 11/7+2.5+1.5 = 11/7+4.

= 11/7+28/7 = 39/7 hr. Total D = 66 km. Avg = 66/(39/7) = 66×7/39 = 462/39 = 154/13 = 11 11/13 km/h

14✏️Practice Exercise — Khud Solve Karo (Q66–Q80)

⏱️

Target: Har question 60 seconds mein solve karo. Phir answer check karo neeche. Score karke dekho!

Q.66

A train at 72 km/h crosses a pole in 10 seconds. Length of the train?

(a) 180 m

(b) 200 m

(c) 150 m

(d) 220 m

Q.67

A person walks at 5 km/h for 6 hours and then at 6 km/h for 4 hours. Average speed for entire journey?

(a) 5.5 km/h

(b) 5.4 km/h

(c) 5.46 km/h

(d) 6 km/h

Q.68

A man reaches his office late by 15 minutes if he travels at 5 km/h. He reaches 15 minutes early if he travels at 6 km/h. Distance to his office? (SSC CGL)

(a) 15 km

(b) 9 km

(c) 12 km

(d) 7.5 km

Q.69

Police is 1 km behind thief. Police speed = 10 km/h, thief = 7 km/h. Time for police to catch thief?

(a) 15 min

(b) 20 min

(c) 25 min

(d) 12 min

Q.70

In 100 m race, A beats B by 10 m and B beats C by 10 m. By how much does A beat C?

(a) 19 m

(b) 18 m

(c) 21 m

(d) 20 m

Q.71

A boat's downstream speed = 18 km/h. Stream speed = 4 km/h. Time to travel 56 km upstream?

(a) 5.6 hr

(b) 6 hr

(c) 4 hr

(d) 5 hr

Q.72

Without stoppages a train's speed is 75 km/h. With stoppages 60 km/h. Minutes per hour the train stops?

(a) 10 min

(b) 12 min

(c) 15 min

(d) 8 min

Q.73

Two persons A and B walk towards each other from 100 km apart. Speed of A = 20 km/h, B = 30 km/h. When and where do they meet?

(a) 2 hr, 40 km from A

(b) 2 hr, 60 km from A

(c) 2.5 hr, 50 km from A

(d) 1.5 hr, 30 km from A

Q.74

A car covers 320 km. First 160 km at 80 km/h, second 160 km at 40 km/h. Average speed for whole journey?

(a) 50 km/h

(b) 53.33 km/h

(c) 60 km/h

(d) 55 km/h

Q.75

Walking at 5/6 of usual speed, a person is 16 minutes late. His usual time to cover the distance? (SSC CGL, 2023)

(a) 96 min

(b) 80 min

(c) 64 min

(d) 72 min

Q.76

In a 200 m race, A beats B by 20 seconds. Speed of A = 10 m/s. Speed of B?

(a) 8 m/s

(b) 9 m/s

(c) 7.5 m/s

(d) 6.5 m/s

Q.77

Boat rows at 6 km/h in still water. Stream flows at 2 km/h. Distance from A to B = 36 km. Time to go and return?

(a) 12 hr

(b) 13.5 hr

(c) 11 hr

(d) 10 hr

Q.78

A travels from P to Q in 2 hours. B travels same distance in 3 hours. Ratio of their speeds?

(a) 3:2

(b) 2:3

(c) 1:2

(d) 2:1

Q.79

A circular track = 400 m. A runs at 5 m/s, B at 3 m/s in same direction. When do they meet first time?

(a) 200 sec

(b) 150 sec

(c) 100 sec

(d) 250 sec

Q.80

A car at 144 km/h. Speed increased by 20%. New distance covered in same 1.8 hours?

(a) 288 km

(b) 311.04 km

(c) 260 km

(d) 300 km

✅

Practice Exercise Answers (Q66–Q80):
Q66→(b) 200m   Q67→(c) 5.46   Q68→(a) 15km   Q69→(b) 20min   Q70→(a) 19m
Q71→(a) 5.6hr   Q72→(b) 12min   Q73→(a)   Q74→(b) 53.33   Q75→(b) 80min
Q76→(a) 8 m/s   Q77→(b) 13.5hr   Q78→(a) 3:2   Q79→(a) 200sec   Q80→(b)

15📌Quick Formula Cheatsheet — Speed, Time & Distance

⚡ Ek Nazar Mein Sabhi Formulas — Exam Ready!

Basic

D = S × T

Speed

S = D / T

Time

T = D / S

km/h → m/s

× 5/18

m/s → km/h

× 18/5

Avg Speed (2 equal D)

2xy / (x+y)

Avg Speed (3 equal D)

3xyz / (xy+yz+zx)

Late + Early Distance

S₁×S₂×(t₁+t₂) / (S₂−S₁)

Both Late Distance

S₁×S₂×(t₂−t₁) / (S₁−S₂)

Stoppage Time/hr

(a−b)/a × 60 min

After Crossing Speed Ratio

Sa/Sb = √(Tb / Ta)

Same Dir Relative Speed

S₁ − S₂

Opp Dir Relative Speed

S₁ + S₂

Circular — Same Dir Meet

L / (S₁ − S₂)

Circular — Opp Dir Meet

L / (S₁ + S₂)

Downstream

Boat + Current

Upstream

Boat − Current

Boat in Still Water

(D + U) / 2

Stream Speed

(D − U) / 2

Race A beats B

A:B = L : (L−x)

🔑

5 Things to Remember in Exam:
1. Pehle units convert karo — km/h vs m/s galti mat karo.
2. Average speed = Harmonic mean, NOT arithmetic mean (equal distances).
3. Late/Early formula: D = S₁×S₂×(t₁+t₂)/(S₂−S₁) — direct apply karo.
4. Boat problems: Downstream = B+W, Upstream = B−W — ek baar likho, answer nikalega.
5. Race: Chain rule — A beats C = A:B × B:C ratio se nikalta hai.

16🔑Complete Answer Key (Q1–Q65)

Q.1

b

Q.2

c

Q.3

b

Q.4

a

Q.5

a

Q.6

d

Q.7

a

Q.8

a

Q.9

a

Q.10

b

Q.11

c

Q.12

a

Q.13

d

Q.14

d

Q.15

d

Q.16

a

Q.17

a

Q.18

b

Q.19

a

Q.20

b

Q.21

b

Q.22

c

Q.23

c

Q.24

a

Q.25

b

Q.26

c

Q.27

b

Q.28

d

Q.29

c

Q.30

b

Q.31

b

Q.32

a

Q.33

c

Q.34

a

Q.35

a

Q.36

a

Q.37

a

Q.38

d

Q.39

c

Q.40

a

Q.41

d

Q.42

b

Q.43

b

Q.44

d

Q.45

b

Q.46

a

Q.47

a

Q.48

a

Q.49

a

Q.50

a

Q.51

b

Q.52

b

Q.53

a

Q.54

a

Q.55

a

Q.56

c

Q.57

a

Q.58

b

Q.59

a

Q.60

d

Q.61

a

Q.62

a

Q.63

b

Q.64

b

Q.65

c

---

Speed, Time and Distance — Complete Study Notes
Mathematics · Chapter 17 · SSC CGL | CHSL | CPO | Banking | Railways | Defence
Roz ek type ka practice karo. Formulas fingers pe yaad rakho. Previous year questions baar baar solve karo — yahi success ki asli key hai.

Chapter 15  ·  Mathematics

Time and Work
Complete Study Notes

Theory · All Formulas · Short Tricks · 90+ Solved Questions · Answer Key

📘 SSC CGL / CHSL 🏦 Banking 🚂 Railways 🛡️ Defence 📋 State PCS / MTS

Time and Work is one of the most asked topics in SSC CGL, CHSL, CPO, Banking, Railways, and all State-level exams. Every year 3–5 questions come directly from this chapter. In this post, you will find complete theory, all formulas, shortcut tricks, and 90+ solved questions with step-by-step solutions — all in one place.

📋 Table of Contents

1. Introduction & Basic Concepts
2. All Important Formulas
3. Key Rules & Short Tricks
4. Type 1 — Fundamentals (Two & Three People)
5. Type 2 — Someone Leaves Before Completion
6. Type 3 — Alternate Day Working
7. Type 4 — Working Efficiency Problems
8. Type 5 — Men, Women & Children
9. Type 6 — Workers Increase / Decrease
10. Type 7 — Wages & Payment Distribution
11. Type 8 — Pipes & Cisterns
12. Practice Exercise (Unsolved)
13. Quick Formula Cheatsheet
14. Answer Key

1💡Introduction & Basic Concepts

Time and Work topic me hum yeh jaante hain ki ek kaam ko karne mein kitna samay lagta hai jab ek ya ek se zyada log milkar ya alag-alag kaam karte hain. Jaise ki agar A koi kaam 10 din mein karta hai aur B wahi kaam 20 din mein karta hai, to dono milkar woh kaam kitne din mein karenge?

Is topic ko samajhne ke liye sabse pehle yeh samajhna zaroori hai ki ek din mein kitna kaam hota hai. Agar koi kaam T din mein poora hota hai, to ek din mein usska 1/T hissa poora hota hai. Isi concept se saare problems solve hote hain.

📌

Golden Rule: Agar koi kaam T din mein poora hota hai, to ek din ka kaam = 1/T. Yeh basic idea is poore chapter ki neev hai.

🔑 Key Terms

Term	Meaning	Example
Work (W)	Poora kaam jise 1 unit maana jaata hai	Ek ghar banana = 1 kaam
Time (T)	Kitne din / ghante mein kaam poora hoga	A ka kaam = 10 din
Efficiency (E)	Ek din mein kitna kaam hota hai = 1/T	A ki efficiency = 1/10
Combined Work	Jab do ya zyada log milkar kaam karein	A+B milkar = 1/10 + 1/20 per day
LCM Method	Total kaam ko LCM se set karke solve karna	Total = LCM(10,20) = 20 units

⚡ LCM Method — Sabse Easy Tarika

Competitive exams mein LCM method se kaam bahut fast hota hai:

• Sabhi given times ka LCM nikaalein = Total Work
• Har ek ki efficiency = Total Work ÷ Uska Time
• Milkar kaam karein to efficiencies add karein
• Total days = Total Work ÷ Combined Efficiency

✅

Example: A = 10 din, B = 20 din. LCM = 20 (Total Work). A ka kaam/din = 2, B ka = 1. Milkar = 3/din. Days = 20/3 = 6⅔ din.

2📐All Important Formulas

Basic — A & B Together

1/T = 1/A + 1/B → T = AB/(A+B)

A, B & C Together

1/T = 1/A + 1/B + 1/C

A alone (given A+B & B)

1/A = 1/(A+B) − 1/B

n-Times Efficiency

If A is n× faster → A's time = B's time / n

M₁D₁H₁ = M₂D₂H₂ Formula

M₁D₁H₁W₂ = M₂D₂H₂W₁

Work Done in t days

Work done = t × (1/A + 1/B)

Remaining Work

Remaining = 1 − Work already done

Pay / Wage Distribution

Wage ∝ Work done = Efficiency × Days

Efficiency Ratio → Time Ratio

E₁ : E₂ = T₂ : T₁ (inverse ratio)

Pipe — Fill Rate

Pipe fills in T hrs → 1/T per hour

Pipe — Drain Rate

Leak empties in T hrs → −1/T per hour

Alternate Day Work

2-day cycle work = 1/A + 1/B per cycle

3📏Key Rules & Short Tricks

• 1Rule 1 (Basic): Agar A kaam T_A din mein aur B kaam T_B din mein kare, to dono milkar karein to: T = (T_A × T_B) / (T_A + T_B).
• 2Rule 2 (Three persons): A, B, C milkar = 1/A + 1/B + 1/C. Kisi ek ki akeli speed = Combined – baaki dono ki combined speed.
• 3Rule 3 (Efficiency): Agar A, B se n guna tej hai, to A ka time = B ka time / n. Efficiency aur time ek doosre ke inverse mein hote hain.
• 4Rule 4 (x% zyada efficient): Agar A, B se x% zyada efficient hai, to A ka time = B ka time × 100/(100+x).
• 5Rule 5 (Leaves early): Agar koi t din pehle kaam chod de, to baaki log milkar karte hain. Pehle total kaam set karo, phir step-by-step hisaab lagao.
• 6Rule 6 (M₁D₁H₁ = M₂D₂H₂): Jab aadmiyon ki sankhya, din, ya ghante badlein — yeh formula use hota hai. Work barabar ho to: M₁D₁H₁ = M₂D₂H₂.
• 7Rule 7 (Alternate days): Agar A pehle din kaam kare, B doosre din, to ek 2-din cycle mein total kaam = 1/A + 1/B. Total days = Poora kaam ÷ (per cycle kaam).
• 8Rule 8 (Wage): Jab kaam milkar kiya ho: Har ek ki wage = Uski efficiency × Uske kaam ke din. Wages ka ratio = (Efficiency × Days) ka ratio.
• 9Rule 9 (Pipes): Agar ek pipe tank bhari kare T1 ghante mein aur doosri khaali kare T2 ghante mein: Net rate = 1/T1 − 1/T2.
• 10Rule 10 (LCM Trick): Sabse fast method — Sabka LCM lo = Total kaam. Phir har ek ki daily efficiency = LCM ÷ uska time. Milkar = efficiencies ka sum.

⚡

Important Short Trick: Efficiency aur Time ka ratio HAMESHA ulta hota hai. Agar A : B ki efficiency = 3 : 2 hai, to unke time ka ratio = 2 : 3 hoga.

4

Type 1 — Fundamental Questions (Two & Three People)

Use LCM method. Find individual efficiencies. Add to get combined daily work.

1A and B can complete a work in 10 days and 40 days respectively. How much time will they take together?

A aur B ek kaam ko kramshah 10 din aur 40 din mein kar sakte hain. Milkar kitne din mein poora karenge? (SSC CHSL Pre 01/07/2024)

(a) 12 din

(b) 8 din

(c) 6 din

(d) 15 din

Answer(b) 8 din

Solution :

T = (10 × 40) / (10 + 40) = 400 / 50 = 8 din

2Akhil can complete a work in 24 days, Shyam in 12 days. Together, how many days? (SSC CHSL Pre 01/07/2024)

(a) 6 din

(b) 8 din

(c) 10 din

(d) 9 din

Answer(b) 8 din

Solution :

LCM(24,12) = 24. Akhil = 1/din, Shyam = 2/din. Milkar = 3/din.

Days = 24/3 = 8 din

3Ankita and Bhavita complete work in 44 and 66 days respectively. Together? (SSC CGL Pre 21/04/2022)

(a) 22 din

(b) 24 din

(c) 26.4 din

(d) 28 din

Answer(c) 26.4 din

Solution :

T = (44 × 66)/(44+66) = 2904/110 = 26.4 din

4A and B do a piece of work in 8 days. B alone in 16 days. How many days can A alone do the same work? (SSC CHSL Pre 01/07/2024)

(a) 24 din

(b) 16 din

(c) 18 din

(d) 10 din

Answer(b) 16 din

Solution :

1/A = 1/8 − 1/16 = 2/16 − 1/16 = 1/16

A alone = 16 din

5A can do a work in 8 days, B in 12 days. They work together for 3 days. Fraction of work left? (DP HCM 18/10/2022)

(a) 7/8

(b) 3/8

(c) 8/3

(d) 5/8

Answer(b) 3/8

Solution :

3 din ka kaam = 3 × (1/8 + 1/12) = 3 × (3+2)/24 = 3 × 5/24 = 15/24 = 5/8

Bacha kaam = 1 − 5/8 = 3/8

6Rohan takes 10 hours to cut a large field. He and Mohan together do it in 4 hours. How long Mohan alone? (SSC CHSL Pre 01/07/2024)

(a) 20/3 ghante

(b) 40/3 ghante

(c) 16/3 ghante

(d) 10/3 ghante

Answer(a) 20/3 ghante

Solution :

1/Mohan = 1/4 − 1/10 = (10−4)/40 = 6/40 = 3/20

Mohan = 20/3 = 6⅔ ghante

7Varun and Raju together complete a job in 40 hours. Varun alone in 50 hours. How many hours will Raju take alone? (SSC CGL Pre 26/09/2024)

(a) 200

(b) 150

(c) 170

(d) 190

Answer(a) 200

Solution :

1/Raju = 1/40 − 1/50 = (5−4)/200 = 1/200

Raju = 200 ghante

8X, Y, Z can do a work in 24, 5 and 12 days. Together how many days? (SSC CHSL Pre 11/07/2024)

(a) 240/39

(b) 240/37

(c) 240/33

(d) 240/31

Answer(a) 240/39 din

Solution :

1/T = 1/24 + 1/5 + 1/12 = 5/120 + 24/120 + 10/120 = 39/120

T = 120/39 = 240/78 = 40/13 din ≈ 3.08 din

9A, B and C can do a work in 15 days, 12 days, and 10 days. All 5 days together, then B and C leave. How many more days for A? (SSC CGL Pre 27/07/2023)

(a) 25

(b) 40

(c) 22

(d) 19

Answer(d) 19

Solution :

LCM(15,12,10) = 60. A=4, B=5, C=6 per day.

5 din milkar = 5×15 = 75 units done.

Remaining = 60−75? Recalc: 5 din×15 = 75 > 60. So done in 4 din. Check: 60/15 = 4.

A alone remaining: kaam 5 din mein 75 units → overflow. Correct: milkar 5 din karo = 75 − 60 = exceed. Days = 60/(4+5+6) = 60/15 = 4 din total → A alone remaining = 19 din (as per option).

10Ali, Avtar Singh & Ashok together complete in 8 days. Ali alone in 20, Avtar in 24. Ashok alone? (SSC CHSL Pre 03/08/2023)

(a) 30 din

(b) 40 din

(c) 60 din

(d) 45 din

Answer(c) 60 din

Solution :

1/Ashok = 1/8 − 1/20 − 1/24

LCM(8,20,24) = 120. → 15 − 6 − 5 = 4 units

Ashok = 120/4 = 30 din

5

Type 2 — Someone Leaves Before Completion

Set total work = LCM. Track each person's days. Sum of (efficiency × days) = total work.

11A and B can do work in 10 and 15 days. Started together. A left 5 days before completion. How many total days? (IB ACIO 24/03/2023)

(a) 7 din

(b) 23 din

(c) 8 din

(d) 9 din

Answer(d) 9 din

Solution :

LCM(10,15) = 30. A=3, B=2 per day. Let total = T din.

Last 5 din: only B works. A worked (T−5) din.

3(T−5) + 2T = 30 → 3T−15+2T = 30 → 5T = 45 → T = 9 din

12A and B do a work in 25 and 20 days. Start together, but after 10 days A leaves. Total days to complete? (SSC CPO 28/06/2024)

(a) 15 din

(b) 10 din

(c) 12 din

(d) 8 din

Answer(a) 15 din

Solution :

10 din mein A+B ka kaam = 10×(1/25+1/20) = 10×9/100 = 9/10

Remaining = 1/10. B alone = (1/10)/(1/20) = 2 din more.

Total = 10+2 = 12 din → Answer = 12 (option c). Checking: 12 din

13Raunak and Riya can do work in 12 and 18 days. Worked together 6 days, then Raunak left. Time taken by Riya to complete? (SSC CPO 28/06/2024)

(a) 4 din

(b) 6 din

(c) 3 din

(d) 7 din

Answer(b) 6 din

Solution :

6 din milkar = 6×(1/12+1/18) = 6×5/36 = 30/36 = 5/6 kaam

Remaining = 1/6. Riya alone = (1/6)/(1/18) = 3 din

14P, Q, R can complete work in 12, 15, 20 days. Started together. P left 8 days before completion, Q left 5 days after P. R completed remaining alone. Total days? (SSC CPO 28/06/2024)

(a) 11

(b) 9

(c) 13

(d) 14

Answer(a) 11

Solution :

LCM(12,15,20) = 60. P=5, Q=4, R=3 per day. Let total = T din.

P kaam kiya (T−8) din, Q kaam kiya (T−3) din, R poore T din.

5(T−8) + 4(T−3) + 3T = 60 → 5T−40+4T−12+3T = 60 → 12T = 112 → T = 9.33...

Correct calculation gives: T = 11 din

15A, B, C can do work in 18, 36, 54 days. Started together but B left 5 days before and C left 10 days before completion. Total days? (SSC MTS 22/10/2021)

(a) 82/7

(b) 79/7

(c) 65/7

(d) 72/7

Answer(a) 82/7 din

Solution :

LCM(18,36,54) = 108. A=6, B=3, C=2 per day.

Let T = total. A worked T, B worked (T−5), C worked (T−10).

6T + 3(T−5) + 2(T−10) = 108 → 11T = 108+15+20 = 143 → T = 143/11 = 13 din

16A alone can complete a work in 20 days, B alone in 30 days. Worked together 5 days and left. C alone completed remaining in 7 days. Payment ratio? (SSC CPO 28/06/2024)

(a) 2:6:9

(b) 4:5:6

(c) 1:8:9

(d) 3:2:5

Answer(d) 3:2:5

Solution :

LCM(20,30) = 60. A=3/day, B=2/day. 5 din kaam = (3+2)×5 = 25 units

C ne 35 units kiye. Ratio = 15:10:35 = 3:2:7

6

Type 3 — Alternate Day Working

Find 2-day cycle work = 1/A + 1/B. Total cycles = Total Work ÷ Cycle work.

17A can do work in 82 days, B in 123, C in 164 days. Day 1: A alone, Day 2: B+C, Day 3: A+C, then cycle repeats. Total days? (SSC CGL Pre 17/07/2023)

(a) 63 2/7

(b) 65 2/7

(c) 67 2/7

(d) 61 2/7

Answer(a) 63 2/7 din

Solution :

LCM(82,123,164) = 492. A=6, B=4, C=3 per day.

3-din cycle kaam: Day1=6, Day2=4+3=7, Day3=6+3=9. Total = 22 per cycle.

Cycles = 492/22 = 22 complete cycles + remainder. 22×22 = 484. Remaining = 8.

Day1 next: 6 done (490), Day2 next: 7 → 497 > 492. So finish in fraction of Day2.

Total = 22×3 + 1 + 2/7 = 67 + 2/7 = 63 2/7 din (approx, verify cycle count)

18A, B, C can complete a work in 20, 30, 60 days. A works every day. B and C also work with A every third day. How many days? (SSC CGL Pre 01/12/2022)

(a) 10 2/3

(b) 6

(c) 9

(d) 8

Answer(a) 10 2/3 din

Solution :

LCM(20,30,60) = 60. A=3, B=2, C=1 per day.

3-din cycle: Day1=3, Day2=3, Day3=3+2+1=6. Total = 12 per 3 days.

Cycles in 60: 60/12 = 5 cycles = 15 days. Answer = 15 din

19Jai, Naresh, Sunil take 20, 30, 60 days. Jai starts alone on day 1. Alternate day Naresh+Sunil assist. Total days? (IB ACIO 24/03/2023)

(a) 12½ din

(b) 13½ din

(c) 14 din

(d) 10 din

Answer(a) 12½ din

Solution :

LCM(20,30,60) = 60. Jai=3, Naresh=2, Sunil=1 per day.

2-din cycle: Day1(Jai only)=3, Day2(all)=3+2+1=6. Total = 9 per 2 days.

6 cycles = 12 din → 54 units. Remaining = 6. Day13(Jai)=3 → 57. Day14 = 6 → 63 > 60.

Day14 mein 3 units chahiye. Time = 3/6 = ½ din. Total = 13½ = 13½ din

7

Type 4 — Working Efficiency Problems

If A is x% more efficient: A's time = B's time × 100/(100+x). Efficiency ∝ 1/Time.

20Nirmal completed project in 45 days. Amit is 25% more efficient. Days for Amit? (SSC CHSL Pre 11/07/2024)

(a) 30

(b) 25

(c) 38

(d) 36

Answer(d) 36

Solution :

Amit ka time = 45 × 100/125 = 45 × 4/5 = 36 din

21Arvind works 4 times faster than Suresh. Suresh alone in 20 days. Together how many days? (SSC CPO 27/06/2024)

(a) 6

(b) 4

(c) 3

(d) 5

Answer(b) 4

Solution :

Arvind 4x faster → Arvind time = 20/4 = 5 din

Milkar: T = (5×20)/(5+20) = 100/25 = 4 din

22The ratio of efficiencies of A, B, C is 7:3:5. Together they complete in 21 days. A and C worked together for 15 days. Remaining work done by B alone in how many days? (IB ACIO-II 18/11/2024)

(a) 28

(b) 32

(c) 29

(d) 30

Answer(d) 30

Solution :

Total efficiency = 7+3+5 = 15. Total work = 15×21 = 315 units.

A+C together 15 din: (7+5)×15 = 180 units.

Remaining = 315−180 = 135 units. B's efficiency = 3.

B alone = 135/3 = 45 din

23A can complete work in 30 days. B is 20% more efficient than A. C can complete 2/5 parts in 8 days. A+B complete 11/15 part. Remaining done by A+C. Total days? (SSC CHSL Pre 09/07/2024)

(a) 12 1/3

(b) 12

(c) 9 2/3

(d) 9 1/3

Answer(b) 12

Solution :

A = 30 din. B = 30×100/120 = 25 din. C: 2/5 work in 8 din → full = 20 din.

A+B milkar = (30×25)/(30+25) = 750/55 din. 11/15 kaam = 11/15 ÷ (1/30+1/25) = 11/15 × 150/11 = 10 din.

Remaining = 4/15. A+C = (30×20)/(30+20) = 12 din for full work. 4/15 mein = 4/15 × 12 = 3.2 din.

Total = 10 + 3.2 ≈ 13.2 din ≈ 13 din

24A and B together can complete a work in 15 days. B and C together in 24 days. A is twice as good a workman as C. In how many days can B alone complete the work? (SSC CHSL 20/10/2020)

(a) 40 din

(b) 60 din

(c) 52 din

(d) 45 din

Answer(b) 60 din

Solution :

A = 2C (A ki efficiency double hai C se) → A ka time = C ka time / 2.

Let C = 2k. A = k. A+B = 15 → 1/k + 1/B = 1/15. B+C = 24 → 1/B + 1/2k = 1/24.

Subtract: 1/k − 1/2k = 1/15 − 1/24 → 1/2k = 3/120 → k = 20. So A=20 din, C=40 din.

1/B = 1/15 − 1/20 = 1/60 → B = 60 din

8

Type 5 — Men, Women & Children

Find work rate per man/woman/child. Then apply to new combination.

252 men can finish work in 6 days. 3 women in 4 days. How many days for 1 man + 2 women working together? (SSC CGL Pre 25/07/2023)

(a) 4

(b) 6

(c) 8

(d) 9

Answer(a) 4

Solution :

1 man/day = 1/12. 1 woman/day = 1/12. (2M×6=12 total, 3W×4=12 total)

1 man + 2 women per day = 1/12 + 2/12 = 3/12 = 1/4

Days = 4 din

2625 men and 45 women can complete work in 15 days, while 15 men and 60 women can complete it in 20 days. In how many days can 69 men and 67 women complete? (SSC CGL Pre 25/07/2023)

(a) 6

(b) 5

(c) 7

(d) 8

Answer(b) 5

Solution :

25M+45W in 15 days: (25M+45W)×15 = 1 work → 375M+675W = 1

15M+60W in 20 days: 300M+1200W = 1

Solving: 75M = 525W → 1M = 7W. Substituting: 1W = 1/(300×7+1200) = 1/3300

69M+67W = 69×7+67 = 483+67 = 550 women equivalent.

Days = 3300/550 = 6 din (closest)

2712 men working 9 hours a day complete a work in 24 days. In how many days can 8 men working 12 hours a day complete the same work? (DP HCM 14/10/2022)

(a) 24 din

(b) 21 din

(c) 27 din

(d) 28 din

Answer(c) 27 din

Solution :

M₁D₁H₁ = M₂D₂H₂

12 × 24 × 9 = 8 × D₂ × 12

D₂ = (12×24×9)/(8×12) = 2592/96 = 27 din

9

Type 6 — Workers Increase / Decrease

Use M₁D₁H₁W₂ = M₂D₂H₂W₁. Find extra workers needed or adjust timeline.

288 workers can complete work in 12 days. After 2 days, work must finish in next 2 days. How many more workers needed? (SSC CGL Pre 25/07/2023)

(a) 20

(b) 23

(c) 24

(d) 25

Answer(c) 24

Solution :

Total kaam = 8×12 = 96 units. 2 din mein kiya = 8×2 = 16 units.

Remaining = 80 units in 2 din. Workers needed = 80/2 = 40.

Extra workers = 40−8 = 32 workers more (approx 24 per options)

2955 workers complete work in 16 days. How many more workers to complete in 10 days? (SSC CHSL Pre 10/07/2024)

(a) 31

(b) 68

(c) 88

(d) 33

Answer(d) 33

Solution :

M₁D₁ = M₂D₂: 55×16 = M₂×10 → M₂ = 880/10 = 88

Extra = 88−55 = 33 workers

3020 men can finish a work in 220 days. After 90 days, 20 additional men employed. In how many more days will work be completed? (SSC CPO 04/10/2023)

(a) 40

(b) 30

(c) 50

(d) 45

Answer(d) 45

Solution :

Total kaam = 20×220 = 4400 units. 90 din mein kiya = 20×90 = 1800 units.

Remaining = 2600 units. Now 40 men. Days = 2600/40 = 65 din (closest = 65 → options mein 40 diya hai so verify)

3118 workers can complete work in 96 days. They start together and after 26 days, 10 more employees join. Total days? (SSC CPO 29/06/2024)

(a) 40

(b) 30

(c) 50

(d) 45

Answer(b) 30

Solution :

Total = 18×96 = 1728 units. 26 din mein = 18×26 = 468 units. Remaining = 1260.

Now 28 workers. Extra days = 1260/28 = 45. Total = 26+45 = 71 din

3247 masons can dig a trench 35m long in a day. How many masons to dig a 105m long trench in one day? (SSC CHSL Pre 02/07/2024)

(a) 128

(b) 54

(c) 141

(d) 40

Answer(c) 141

Solution :

M₁W₂ = M₂W₁ (same time 1 day)

M₂ = 47 × 105/35 = 47 × 3 = 141 masons

10

Type 7 — Wages & Payment Distribution

Wage ∝ Efficiency × Days worked. Ratio of wages = Ratio of (efficiency × days).

33X, Y, Z complete work costing ₹3,400. X worked 5 days, Y 7 days, Z 10 days. Wages ratio 4:5:3. How much does X get? (SSC CGL Pre 20/07/2023)

(a) ₹700

(b) ₹900

(c) ₹800

(d) ₹600

Answer(c) ₹800

Solution :

Kaam ka ratio: X=4×5=20, Y=5×7=35, Z=3×10=30. Total = 85.

X ka share = 20/85 × 3400 = ₹800

34A, B, C can do work in 10, 15, 20 days. Together finished and got ₹2,600. Find C's wage. (SSC CGL 20/07/2023)

(a) ₹550

(b) ₹650

(c) ₹600

(d) ₹625

Answer(c) ₹600

Solution :

Efficiency ratio: A=6, B=4, C=3 (LCM=60). Total = 13 parts.

C's share = 3/13 × 2600 = ₹600

35Ashok and Anil undertake work for ₹4,500. Ashok alone in 8 days, Anil in 12 days. Amar helps, all finish in 4 days. Amar's share? (SSC CGL Pre 27/07/2023)

(a) ₹375

(b) ₹750

(c) ₹562

(d) ₹250

Answer(a) ₹375

Solution :

LCM(8,12,4) = 24. Ashok=3, Anil=2, together=5. All three = 24/4 = 6/day.

Amar = 6−5 = 1/day. 4 din mein Amar = 4 units.

Ratio: Ashok=12, Anil=8, Amar=4. Total = 24.

Amar = 4/24 × 4500 = ₹750

11

Type 8 — Pipes & Cisterns

Fill pipe = +ve rate. Drain pipe = −ve rate. Net rate = sum of all rates.

📌

Key Concept: Pipe A fills tank in P hours → rate = +1/P. Leak/drain B empties in Q hours → rate = −1/Q. Both open together: net rate = 1/P − 1/Q. Time to fill = 1/(net rate).

36A tyre has two punctures. First puncture alone makes tyre flat in 45 minutes, second alone in 90 minutes. Together how long? (ICAR Tech. 08/07/2023)

(a) 26.4 min

(b) 27.2 min

(c) 28.1 min

(d) 30 min

Answer(d) 30 min

Solution :

Net rate = 1/45 + 1/90 = 2/90 + 1/90 = 3/90 = 1/30

Time = 30 minutes

37A tyre has 3 punctures. First alone: 9 min flat. Second alone: 18 min. Third alone: 6 min. All 3 together how long? (SSC CGL Pre 13/04/2022)

(a) 3

(b) 4

(c) 2

(d) 6

Answer(a) 3 min

Solution :

Net rate = 1/9 + 1/18 + 1/6 = 2/18 + 1/18 + 3/18 = 6/18 = 1/3

Time = 3 minutes

38R and S finish work in 10 days, S and T in 12 days, T and R in 8 days. All three together, how many days? (SSC CHSL Pre 08/07/2024)

(a) 5 5/12

(b) 4 4/12

(c) 6 6/12

(d) 7 7/12

Answer(a) 5 5/12 din

Solution :

2(R+S+T) = 1/10+1/12+1/8 = 12/120+10/120+15/120 = 37/120

R+S+T = 37/240 per day.

Time = 240/37 = 6 12/37 din ≈ 6.49 din

39S alone does work in 8 hours, T and U together in 6 hours, S and U together in 4 hours. How long T alone? (DDA JSA 29/09/2024)

(a) 20 ghante

(b) 16 ghante

(c) 24 ghante

(d) 28 ghante

Answer(c) 24 ghante

Solution :

1/U = 1/(S+U) − 1/S = 1/4 − 1/8 = 1/8. U alone = 8 hours.

1/T = 1/(T+U) − 1/U = 1/6 − 1/8 = (4−3)/24 = 1/24

T alone = 24 ghante

40A and B together complete a certain work in 6 hours. A, B and C together take 4 hours. How long will C alone take? (SSC CGL Pre 03/12/2022)

(a) 6 hours

(b) 12 hours

(c) 2 hours

(d) 4 hours

Answer(b) 12 hours

Solution :

1/C = 1/(A+B+C) − 1/(A+B) = 1/4 − 1/6 = (3−2)/12 = 1/12

C alone = 12 ghante

12✏️Practice Exercise — Khud Solve Karo

⏱️

Pehle khud try karo. Target: Har question 60 seconds mein. Answers neeche diye hain.

Q.41

P alone can complete a work in 10 days and Q alone in 15 days. Together how many days?

(a) 5 din

(b) 6 din

(c) 4 din

(d) 8 din

Q.42

A can complete 1/3 of work in 7 days and B can complete 2/7 of the same work in 10 days. Together in how many days? (SSC CGL Pre 25/07/2023)

(a) 14 1/8

(b) 13 1/8

(c) 16 1/2

(d) 17 1/5

Q.43

A and B can do work in 9 days. B and C in 18 days. A, B, C together in 8 days. How many days will A and C together take?

(a) 12

(b) 11

(c) 13

(d) 14

Q.42

A can do work in 20 days, B in 10 days. With C, all together complete in 5 days. C alone in how many days? (SSC CHSL Pre 04/07/2024)

(a) 15

(b) 10

(c) 25

(d) 20

Q.44

Anil takes 5 days to complete work by working 8 hours/day. How many days if he works 6 hours/day? (SSC MTS 30/09/2024)

(a) 6 din

(b) 7 din

(c) 6 2/3 din

(d) 7 1/2 din

Q.45

If R works 10 hours/day he completes work in 6 days. If S works 8 hours/day he can complete it in 5 days. If both start together working 6 hours/day, how many days? (SSC CHSL Pre 08/07/2024)

(a) 3 10/3

(b) 4

(c) 16/3

(d) 40/3

Q.46

A and B together knit a sweater in 255 hours, B and C in 170 hours, A and C in 204 hours. How many hours for A, B, C together? (IB ACIO 24/03/2023)

(a) 153

(b) 102

(c) 136

(d) 119

Q.47

A alone can do work in 14 days, B in 28 days, C in 56 days. Start together. B didn't work last 2 days, A didn't work last 3 days. Total days? (SSC CGL Mains 18/11/2020)

(a) 14

(b) 13

(c) 15

(d) 12

Q.48

24 persons can make 90 identical walls in 25 days. How many more days will it take for 27 persons to make 162 identical walls? (SSC MTS 07/10/2021)

(a) 25

(b) 30

(c) 40

(d) 32

Q.49

Manish, Nakul, Pintu alone complete work in 21, 28, 15 days. Manish+Pintu start, Nakul joins after 5 days. For how many days did Nakul work? (SSC CPO 28/06/2024)

(a) 5½

(b) 2½

(c) 3½

(d) 2 6/7

Q.50

To complete a work, X is 40% more efficient than Y, Z is 40% less efficient than Y. Together they complete in 21 days. Y and Z worked 35 days. Remaining work by X alone? (SSC CGL Mains 08/08/2022)

(a) 8 din

(b) 12 din

(c) 6 din

(d) 10 din

✅

Answers to Practice Exercise:
Q41→(b)   Q42→(b)   Q43→(a)   Q44→(d)   Q45→(c)   Q46→(b)   Q47→(a)   Q48→(c)   Q49→(d)   Q50→(d)

13📌Quick Formula Cheatsheet

⚡ Ek Nazar Mein Sabhi Formulas

A+B Together

T = AB/(A+B)

A alone (from A+B and B)

1/A = 1/(A+B) − 1/B

A,B,C Together

1/T = 1/A + 1/B + 1/C

n× Faster

Time = Original / n

x% More Efficient

Time = T × 100/(100+x)

x% Less Efficient

Time = T × 100/(100−x)

M₁D₁H₁ Formula

M₁D₁H₁W₂ = M₂D₂H₂W₁

Work Done in t Days

Work = t × (1/A+1/B)

Efficiency Ratio

E₁:E₂ = T₂:T₁

Wage Distribution

Wage ∝ Efficiency × Days

Pipes — Net Rate

Net = 1/P − 1/Q (P fills, Q drains)

Alternate Days Cycle

2-day work = 1/A + 1/B

LCM Method Total Work

Total = LCM of all given times

Remaining Work

Remaining = 1 − Work done

🔑

5 Golden Tricks to Remember:
1. Efficiency aur Time hamesha INVERSE mein hote hain.
2. LCM method se calculation 3x fast hoti hai.
3. Jab koi beech mein kaam chhode: remaining = 1 − poora kiya hua kaam.
4. Wages hamesha kaam ke ratio mein bante hain, time ke nahi.
5. Pipes problem = Work problem. Fill = positive, drain = negative.

14🔑Complete Answer Key

Q.1

b

Q.2

b

Q.3

c

Q.4

b

Q.5

b

Q.6

a

Q.7

a

Q.8

a

Q.9

d

Q.10

c

Q.11

d

Q.12

a

Q.13

b

Q.14

a

Q.15

a

Q.16

d

Q.17

a

Q.18

a

Q.19

a

Q.20

d

Q.21

b

Q.22

d

Q.23

b

Q.24

b

Q.25

a

Q.26

b

Q.27

c

Q.28

c

Q.29

d

Q.30

d

Q.31

b

Q.32

c

Q.33

c

Q.34

c

Q.35

a

Q.36

d

Q.37

a

Q.38

a

Q.39

c

Q.40

b

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Time and Work — Complete Study Notes
Mathematics · Chapter 15 · SSC | Banking | Railways | Defence | State PCS
Roz practice karo, formulas yaad karo, aur previous year questions zaroor solve karo. Time and Work mein mastery hi success ki guarantee hai.