Chapter 17 · Mathematics
Speed, Time &
Distance — Complete Notes
Theory · All Formulas · Short Tricks · 100 Solved Questions · Answer Key
๐ SSC CGL / CHSL / CPO๐ฆ Banking / PO / Clerk
๐ Railways RRB๐ก️ Defence / CDS๐ State PCS / MTS
Speed, Time and Distance (เคाเคฒ, เคธเคฎเคฏ เคเคฐ เคฆूเคฐी) — 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!
1๐กIntroduction & Basic Concepts
Jab koi cheez ek jagah se doosri jagah jaati hai — insaan, gaadi, train, naav — to teen quantities involve hoti hain: Speed (เคाเคฒ), Time (เคธเคฎเคฏ) aur Distance (เคฆूเคฐी). 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.
๐The 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 (เคाเคฒ) | Unit time mein covered distance | km/h, m/s, miles/h |
| Distance (เคฆूเคฐी) | Do points ke beech ki lambai | km, m, miles |
| Time (เคธเคฎเคฏ) | 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 (เค
เคจुเคช्เคฐเคตाเคน) | Dhara ke saath naav ki speed | B + W (boat + current) |
| Upstream (เคช्เคฐเคคिเคช्เคฐเคตाเคน) | Dhara ke virudh naav ki speed | B − W (boat − current) |
2๐Unit 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.
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.
๐4 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)
15๐Quick Formula Cheatsheet — Speed, Time & Distance
⚡ Ek Nazar Mein Sabhi Formulas — Exam Ready!
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)
Boat in Still Water
(D + U) / 2
Race A beats B
A:B = L : (L−x)
๐
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.
No comments:
Post a Comment