🔔 BPSC TRE 4.0 ONLINE FORM 2026 – APPLY NOW AT bpsc.bihar.gov.in 🔔

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🏛️ Bihar Public Service Commission

BPSC School Teacher TRE 4.0 Online Form 2026

Apply online for 45,198 Teacher Vacancies across Primary, Middle, Secondary & Senior Secondary schools in Bihar

📋 45,198 Vacancies

📅 Last Date: March 2026

✍️ Exam: Sep 22–27, 2026

🌐 bpsc.bihar.gov.in

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📌 Recruitment Overview

Organization

Bihar Public Service Commission (BPSC)

Exam Name

Teacher Recruitment Exam (TRE 4.0)

Total Vacancies

45,198 Posts

Application Mode

Online Only

Exam Date

22–27 September 2026

Official Website

bpsc.bihar.gov.in

#BPSCTRE4 #BiharTeacherBharti #SarkariNaukri2026 #Bihar Government Jobs #TRE4.0 #बिहार शिक्षक भर्ती

📅 Important Dates

• 🔔Short Notification Released11 February 2026
• 📝OTR Registration StartFebruary 2026
• 🖊️Online Application Start DateFebruary / March 2026
• ⏰Last Date to ApplyMarch 2026
• 📄Detailed Notification ExpectedMarch 2026
• 🎟️Admit Card ReleaseBefore Exam (To Be Announced)
• ✍️Written Examination Dates22–27 September 2026
• 🏆Result DeclarationNovember 2026 (Expected)

📊 Vacancy Details – Post Wise

Post Name	Class Level	Post Type	Qualification	Pay Level
Primary Teacher (PRT)	Class 1–5	Primary	12th + D.El.Ed / CTET Paper 1	Level 6
Middle School Teacher	Class 6–8	Upper Primary	Graduation + B.Ed / STET Paper 1	Level 7
Secondary Teacher (TGT)	Class 9–10	Secondary	Graduation + B.Ed / STET Paper 2	Level 8
Sr. Secondary Teacher (PGT)	Class 11–12	Senior Secondary	Post-Graduation + B.Ed / STET	Level 9

🎂 Age Limit (Category-Wise)

Age is calculated as on 01 August 2025. Minimum age: 18 years (Primary) / 21 years (Secondary & Senior Secondary).

Category	Max Age – Primary (1–5)	Max Age – Middle (6–8)	Max Age – Secondary/Sr. Sec.
General (Male)	37 Years	37 Years	37 Years
BC / OBC / EBC	40 Years	40 Years	40 Years
Women (General)	40 Years	40 Years	40 Years
SC / ST	42 Years	42 Years	42 Years
PwD (Divyang)	As per rules	As per rules	As per rules

🎓 Educational Eligibility

🏫 Primary Teacher (Class 1–5)

• 12th with 50% marks + D.El.Ed (2-year)
• 12th with 45% marks + D.El.Ed (relaxed)
• 12th with 50% + 4-year B.El.Ed
• Graduation 50% + D.El.Ed / B.Ed
• CTET Paper 1 or Bihar STET Paper 1 required

📚 Middle School (Class 6–8)

• Graduation + B.Ed from recognised university
• 50% marks in graduation preferred
• Bihar STET Paper 1 or CTET Paper 2 required
• Subject specialisation as per post

🔬 Secondary TGT (Class 9–10)

• Graduation in relevant subject + B.Ed
• Minimum 50% marks in graduation
• Bihar STET Paper 2 qualification mandatory
• Subject-specific paper in exam

🎓 Sr. Secondary PGT (Class 11–12)

• Post-Graduation in relevant subject
• B.Ed from NCTE recognised institution
• 55% marks in Post-Graduation required
• Bihar STET qualification required

💳 Application Fee

General / EWS / OBC

₹750

+ Additional ₹200 if no Aadhaar

BC / EBC (Bihar)

₹200

SC / ST

₹200

PwD / Female Bihar

₹200

⚠️ Payment must be made online only via Credit Card / Debit Card / Net Banking. Candidates without Aadhaar must pay an additional ₹200 as biometric fee.

🖊️ How to Apply Online

⚠️ Important: OTR Registration is Mandatory First!

Before filling the application form, candidates must complete One-Time Registration (OTR) on the BPSC portal and also register on DigiLocker to upload/fetch academic documents automatically.

1

Complete OTR (One-Time Registration)

Visit bpsc.bihar.gov.in → Go to Online Application section → Complete OTR with your basic personal details and mobile number / email ID.

2

Register on DigiLocker

Create a DigiLocker account and upload your 10th, 12th, Graduation, and B.Ed/D.El.Ed certificates. Link your Aadhaar with active mobile number for OTP verification.

3

Login & Open Application Form

Login using your OTR registration credentials → Click on BPSC TRE 4.0 Application Form link → Start filling in all required details.

4

Fill Form & Upload Documents

Enter personal details, educational qualifications, and category details. Upload scanned passport photo, signature, and required certificates.

5

Pay Application Fee Online

Pay the fee as per your category via Credit/Debit Card or Net Banking. Keep the payment receipt safe for future reference.

6

Submit & Take Printout

Review all filled details carefully → Submit the form → Download and take a printout of the final submitted application form for your records.

🏆 Selection Process

✍️

Stage 1

Written Examination

Offline OMR-based objective type exam. Held from 22–27 September 2026.

📂

Stage 2

Document Verification (DV)

Shortlisted candidates verify educational, eligibility, and category certificates.

🏅

Stage 3

Merit List

Final merit list based on written exam marks. Category-wise cut-off published separately.

💰 Salary & Pay Scale

Post	Pay Level	Basic Pay (Approx.)	Gross Salary
Primary Teacher (PRT)	Level 6	₹35,400 – ₹1,12,400	₹45,000+ (Est.)
Middle School Teacher	Level 7	₹44,900 – ₹1,42,400	₹55,000+ (Est.)
Secondary Teacher (TGT)	Level 8	₹47,600 – ₹1,51,100	₹60,000+ (Est.)
Sr. Secondary Teacher (PGT)	Level 9	₹53,100 – ₹1,67,800	₹68,000+ (Est.)

*Salary figures include DA, HRA as per Bihar government norms. Exact figures will be confirmed in the official notification.

🔗 Important Links

🖊️ Apply Online (Official) 📄 Download Notification PDF 📋 OTR Registration 🗄️ DigiLocker Registration 🌐 Official BPSC Website 📑 Bihar STET Certificate

⚠️ All links lead to official government portals. Always apply through official website only. Beware of fraud websites.

📌 Important Notices

• Candidates must read the official detailed notification carefully before applying.
• Ensure Aadhaar number is linked with active mobile number for OTR verification.
• Name in DigiLocker must exactly match the BPSC application details.
• Application must be submitted before the deadline. No offline mode available.
• Maximum 3 attempts allowed for the examination (as per BPSC rules).
• Age calculated as on 01 August 2025. Check eligibility carefully.
• This blog is for informational purposes only. Always verify from official BPSC website.

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BPSC TRE 4.0 | Bihar Teacher Recruitment 2026

Official Website: bpsc.bihar.gov.in  |  Contact BPSC: 0612-2237999  |  Email: bpscpat-bih@nic.in

Disclaimer: This blog post is created for informational purposes only. All details are sourced from official BPSC announcements. For the most accurate and legally valid information, always refer to bpsc.bihar.gov.in.

Hindu UNESCO World Heritage Sites in India

A Comprehensive Study Guide for UPSC, SSC, PCS & State Competitive Examinations

1. Sun Temple, Konark

• 📍 Location: Puri District, Odisha
• 📅 Year of Inscription: 1984
• 👑 Built By: King Narasimhadeva I (Eastern Ganga Dynasty, 13th Century)

Exam Highlights: Designed in the shape of a colossal chariot for the Sun God (Surya), complete with 24 elaborately carved stone wheels pulled by seven horses. It is a masterpiece of Kalinga architecture and is also historically referred to as the "Black Pagoda" by European sailors.

2. Group of Monuments at Hampi

• 📍 Location: Vijayanagara District, Karnataka
• 📅 Year of Inscription: 1986
• 👑 Built By: Rulers of the Vijayanagara Empire (14th-16th Century)

Exam Highlights: Situated on the banks of the Tungabhadra River, it was the prosperous capital of the Vijayanagara Empire. Key Hindu structures include the Virupaksha Temple (an active place of worship) and the Vittala Temple complex, famous for its iconic stone chariot and musical pillars. It represents the pinnacle of Dravidian architecture.

3. Khajuraho Group of Monuments

• 📍 Location: Chhatarpur District, Madhya Pradesh
• 📅 Year of Inscription: 1986
• 👑 Built By: Chandela Dynasty (10th-12th Century)

Exam Highlights: Famous for their Nagara-style architectural symbolism and intricate erotic sculptures. Originally a group of 85 temples, only about 25 remain today. The Kandariya Mahadeva temple is the largest and most famous. The Matangeshvara Temple is still an active site of Hindu worship.

4. Great Living Chola Temples

• 📍 Location: Tamil Nadu
• 📅 Year of Inscription: 1987 (Expanded in 2004)
• 👑 Built By: Kings of the Chola Empire (11th-12th Century)

Exam Highlights: This site comprises three magnificent 11th and 12th-century temples: the Brihadeeswarar Temple at Thanjavur (built by Rajaraja I), the Brihadeeswarar Temple at Gangaikonda Cholapuram (built by Rajendra I), and the Airavatesvara Temple at Darasuram (built by Rajaraja II). They are outstanding examples of Dravidian architecture, bronze casting, and painting.

5. Group of Monuments at Mahabalipuram

• 📍 Location: Chengalpattu District, Tamil Nadu
• 📅 Year of Inscription: 1984
• 👑 Built By: Pallava Dynasty (7th-8th Century)

Exam Highlights: Founded by Pallava kings (notably Narasimhavarman I and II). The site is famous for its rock-cut caves and structural temples. Important monuments include the Shore Temple (dedicated to Shiva and Vishnu), the Pancha Rathas (monolithic rock-cut temples shaped like chariots), and the giant open-air rock relief known as the "Descent of the Ganges" or "Arjuna's Penance".

6. Group of Monuments at Pattadakal

• 📍 Location: Bagalkot District, Karnataka
• 📅 Year of Inscription: 1987
• 👑 Built By: Chalukya Dynasty (8th Century)

Exam Highlights: Pattadakal was the holy city for the royal coronation of the Chalukya kings. It represents a harmonious blend of architectural forms from northern (Rekha-Nagara) and southern (Dravida Vimana) India. The Virupaksha Temple, built by Queen Lokamahadevi to commemorate her husband's victory over the Pallavas, is the masterpiece of this group.

7. Ellora Caves (Hindu Caves)

• 📍 Location: Chhatrapati Sambhajinagar District, Maharashtra
• 📅 Year of Inscription: 1983
• 👑 Built By: Rashtrakuta and Yadava Dynasties (600-1000 CE)

Exam Highlights: While Ellora houses Buddhist, Hindu, and Jain caves (showcasing ancient India's religious tolerance), Caves 13 to 29 are Hindu. The crown jewel is Cave 16, the Kailasanatha Temple, built by King Krishna I of the Rashtrakuta dynasty. It is the largest single monolithic rock excavation in the world, carved top-down from a single volcanic basalt rock.

8. Elephanta Caves

• 📍 Location: Elephanta Island (Gharapuri), Mumbai, Maharashtra
• 📅 Year of Inscription: 1987
• 👑 Built By: Kalachuri and Rashtrakuta Dynasties (5th-7th Century)

Exam Highlights: A collection of rock-cut caves primarily dedicated to Lord Shiva. The main cave (Cave 1) contains magnificent rock-cut stone sculptures showing syncretic Hindu spiritual ideas, the most famous being the 20-foot high Trimurti (three-faced Shiva representing the creator, preserver, and destroyer), Gangadhara, and Ardhanarishvara.

9. Kakatiya Rudreshwara (Ramappa) Temple

• 📍 Location: Mulugu District, Telangana
• 📅 Year of Inscription: 2021
• 👑 Built By: Recherla Rudra (General of Kakatiya King Ganapati Deva, 1213 CE)

Exam Highlights: Dedicated to Lord Shiva, this is the only temple in India named after its sculptor (Ramappa). It is uniquely built using "sandbox technology" for the foundation to protect against earthquakes, and the roof is made of incredibly light, floating bricks. It is a stunning example of Kakatiya architecture.

10. Sacred Ensembles of the Hoysalas

• 📍 Location: Karnataka (Belur, Halebidu, Somanathapura)
• 📅 Year of Inscription: 2023
• 👑 Built By: Hoysala Empire (12th-13th Century)

Exam Highlights: The most recent addition to the list. It includes the Chennakeshava Temple (Belur), Hoysaleswara Temple (Halebidu), and Keshava Temple (Somanathapura). Known for their unique star-shaped ground plans (stellate), raised platforms (jagati), and hyper-detailed soapstone carvings that cover almost every inch of the temples.

Created for UPSC, SSC, and PCS Aspirants. Best of luck with your preparations!

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

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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.