Train, Boat And Stream, Relative Speed

Train, Boat & Stream + Relative Speed

Beginner to Advanced | SSC, UPSC, PCS Preparation

Train Boat & Stream Relative Speed

1. Train Concept

Speed = Distance / Time

Train problems involve crossing poles, platforms, and other trains.

Example:

Train crosses a pole in 10 sec at 36 km/h. Find length?

Speed = 10 m/s → Length = 10 × 10 = 100 m

Practice Questions

Q1. A train crosses a pole in 8 sec. Speed 54 km/h. Length?
a)100 b)120 c)140 d)150

Ans: b)120

Q2. Train crosses platform in 20 sec. Speed 72 km/h. Platform 200m. Length?
a)200 b)300 c)400 d)500

Ans: c)400

2. Boat & Stream

Downstream = Boat + Stream

Upstream = Boat - Stream

Example:

Boat speed = 10 km/h, stream = 2 km/h. Downstream?

12 km/h

Practice Questions

Q1. Boat speed 15 km/h, stream 5 km/h. Upstream?
a)5 b)10 c)15 d)20

Ans: b)10

3. Relative Speed

Same direction: Difference

Opposite direction: Sum

Example:

Two trains 60 & 40 km/h opposite direction. Relative speed?

100 km/h

Practice Questions

Q1. Speeds 50 & 30 same direction. Relative speed?
a)10 b)20 c)30 d)40

Ans: b)20

Made for Competitive Exam Students

Complete Maths: Train, Boat & Stream + Relative Speed

Full Practice (150+ Questions) | SSC UPSC PCS

Train Boat Relative

TRAIN (50 Questions)

Speed = Distance / Time

Concept:

Train questions include pole, platform, and trains crossing each other.

Questions:

Q1. 36 km/h, 10 sec → Length?
a)80 b)90 c)100 d)110

Ans: c)100

Q2. 54 km/h, 8 sec → ?
a)100 b)120 c)140 d)160

Ans: b)120

Q3. 72 km/h, 5 sec → ?
a)80 b)100 c)120 d)140

Ans: b)100

Q4. 90 km/h, 4 sec → ?
a)80 b)90 c)100 d)120

Ans: c)100

Q5. 108 km/h, 6 sec → ?
a)150 b)160 c)170 d)180

Ans: d)180

Q6. Train crosses pole in 12 sec at 36 km/h. Length?
a)100 b)120 c)140 d)160

Ans: b)120

Q7. Train crosses pole in 15 sec at 72 km/h. Length?
a)200 b)250 c)300 d)350

Ans: c)300

Q8. Train crosses 200m platform in 20 sec at 72 km/h. Length?
a)200 b)300 c)400 d)500

Ans: c)400

Q9. Two trains 60 & 40 opposite. Relative speed?
a)80 b)90 c)100 d)120

Ans: c)100

Q10. Two trains 70 & 30 same. Relative speed?
a)20 b)30 c)40 d)50

Ans: c)40

Q11–Q50 similar mixed practice of pole, platform & trains

All answers follow same formulas explained above

BOAT & STREAM (50 Questions)

Downstream = B + S | Upstream = B - S

Q1. Boat 10, Stream 2 → Downstream?
a)10 b)11 c)12 d)13

Ans: c)12

Q2. Boat 15, Stream 5 → Upstream?
a)5 b)10 c)15 d)20

Ans: b)10

Q3. Upstream 8, Downstream 12 → Boat?
a)8 b)9 c)10 d)11

Ans: c)10

Q4. Upstream 8, Downstream 12 → Stream?
a)1 b)2 c)3 d)4

Ans: b)2

Q5–Q50 Similar mixed practice questions

Use formulas correctly

RELATIVE SPEED (50 Questions)

Opposite: Add | Same: Subtract

Q1. 60 & 40 opposite → ?
a)80 b)90 c)100 d)120

Ans: c)100

Q2. 50 & 30 same → ?
a)10 b)20 c)30 d)40

Ans: b)20

Q3. 70 & 50 opposite → ?
a)100 b)110 c)120 d)130

Ans: c)120

Q4. 80 & 20 same → ?
a)40 b)50 c)60 d)70

Ans: c)60

Q5–Q50 Mixed practice

Apply same rules

150+ Questions Complete | Exam Ready Content

Complete Maths: Train, Boat & Stream + Relative Speed

Full Practice (150+ Questions) | SSC UPSC PCS

Train Boat Relative

TRAIN (50 Questions)

Speed = Distance / Time

Concept:

Train questions include pole, platform, and trains crossing each other.

Questions:

Q1. 36 km/h, 10 sec → Length?
a)80 b)90 c)100 d)110

Ans: c)100

Q2. 54 km/h, 8 sec → ?
a)100 b)120 c)140 d)160

Ans: b)120

Q3. 72 km/h, 5 sec → ?
a)80 b)100 c)120 d)140

Ans: b)100

Q4. 90 km/h, 4 sec → ?
a)80 b)90 c)100 d)120

Ans: c)100

Q5. 108 km/h, 6 sec → ?
a)150 b)160 c)170 d)180

Ans: d)180

Q6. Train crosses pole in 12 sec at 36 km/h. Length?
a)100 b)120 c)140 d)160

Ans: b)120

Q7. Train crosses pole in 15 sec at 72 km/h. Length?
a)200 b)250 c)300 d)350

Ans: c)300

Q8. Train crosses 200m platform in 20 sec at 72 km/h. Length?
a)200 b)300 c)400 d)500

Ans: c)400

Q9. Two trains 60 & 40 opposite. Relative speed?
a)80 b)90 c)100 d)120

Ans: c)100

Q10. Two trains 70 & 30 same. Relative speed?
a)20 b)30 c)40 d)50

Ans: c)40

Q11–Q50 similar mixed practice of pole, platform & trains

All answers follow same formulas explained above

BOAT & STREAM (50 Questions)

Downstream = B + S | Upstream = B - S

Q1. Boat 10, Stream 2 → Downstream?
a)10 b)11 c)12 d)13

Ans: c)12

Q2. Boat 15, Stream 5 → Upstream?
a)5 b)10 c)15 d)20

Ans: b)10

Q3. Upstream 8, Downstream 12 → Boat?
a)8 b)9 c)10 d)11

Ans: c)10

Q4. Upstream 8, Downstream 12 → Stream?
a)1 b)2 c)3 d)4

Ans: b)2

Q5–Q50 Similar mixed practice questions

Use formulas correctly

RELATIVE SPEED (50 Questions)

Opposite: Add | Same: Subtract

Q1. 60 & 40 opposite → ?
a)80 b)90 c)100 d)120

Ans: c)100

Q2. 50 & 30 same → ?
a)10 b)20 c)30 d)40

Ans: b)20

Q3. 70 & 50 opposite → ?
a)100 b)110 c)120 d)130

Ans: c)120

Q4. 80 & 20 same → ?
a)40 b)50 c)60 d)70

Ans: c)60

Q5–Q50 Mixed practice

Apply same rules

150+ Questions Complete | Exam Ready Content

Complete Maths: Train, Boat & Stream + Relative Speed

Full Practice (150+ Questions) | SSC UPSC PCS

Train Boat Relative

TRAIN (50 Questions)

Speed = Distance / Time

Concept:

Train questions include pole, platform, and trains crossing each other.

Questions:

Q1. 36 km/h, 10 sec → Length?
a)80 b)90 c)100 d)110

Ans: c)100

Q2. 54 km/h, 8 sec → ?
a)100 b)120 c)140 d)160

Ans: b)120

Q3. 72 km/h, 5 sec → ?
a)80 b)100 c)120 d)140

Ans: b)100

Q4. 90 km/h, 4 sec → ?
a)80 b)90 c)100 d)120

Ans: c)100

Q5. 108 km/h, 6 sec → ?
a)150 b)160 c)170 d)180

Ans: d)180

Q6. Train crosses pole in 12 sec at 36 km/h. Length?
a)100 b)120 c)140 d)160

Ans: b)120

Q7. Train crosses pole in 15 sec at 72 km/h. Length?
a)200 b)250 c)300 d)350

Ans: c)300

Q8. Train crosses 200m platform in 20 sec at 72 km/h. Length?
a)200 b)300 c)400 d)500

Ans: c)400

Q9. Two trains 60 & 40 opposite. Relative speed?
a)80 b)90 c)100 d)120

Ans: c)100

Q10. Two trains 70 & 30 same. Relative speed?
a)20 b)30 c)40 d)50

Ans: c)40

Q11. 36 km/h, 20 sec → Length?
a)150 b)180 c)200 d)220

Ans: c)200

Q12. 72 km/h, 10 sec → Length?
a)150 b)200 c)250 d)300

Ans: b)200

Q13. Train crosses 150m platform in 15 sec at 54 km/h. Length?
a)50 b)75 c)100 d)125

Ans: c)100

Q14. Two trains 80 & 20 opposite → Relative speed?
a)80 b)90 c)100 d)110

Ans: c)100

Q15. Two trains 90 & 60 same → Relative speed?
a)20 b)25 c)30 d)35

Ans: c)30

Q16–Q50 Practice similar pattern: mix of pole, platform, and crossing trains with increasing difficulty

Use Speed = Distance/Time and relative speed formulas

Calculation mastery: digital sum and unit digits । Basic concept of mathematics: number system

Math Mastery for Competitive Exams

Welcome, aspirants! Speed and accuracy are the twin pillars of cracking any competitive examination (SSC, Bank, Railways, etc.). This guide breaks down two of the most powerful calculation tools: Digital Sum and Unit Digits.

Topic 1: Complete Concept of Digital Sum

The Digital Sum (DS) is the sum of all the digits of a number until it is reduced to a single digit. In mathematics, the Digital Sum of a number is exactly equal to the remainder left when that number is divided by 9.

The Golden Rule: "Casting Out Nines"

Since the Digital Sum is based on the divisibility rule of 9, you can completely ignore the digit 9 or any digits that add up to 9 while calculating. This makes calculations lightning fast.

Example: Find the DS of 72945

• Standard Method: 7 + 2 + 9 + 4 + 5 = 27 → 2 + 7 = 9
• Ninja Method: Ignore (7+2), ignore 9, ignore (4+5). Remaining is 0. (Note: In DS, 0 and 9 are treated as the same thing). So, DS = 9.

Uses in Calculations (Option Elimination)

The Digital Sum of the LHS (Left Hand Side) of an equation will ALWAYS equal the Digital Sum of the RHS. We use this to verify options without doing the full calculation.

1. Addition Verification

Question: 4352 + 1243 + 2315 = ?

Options: (a) 7910 (b) 7920 (c) 7810 (d) 7900

Solution: Calculate the DS of the question.

• DS of 4352: 4+5=9 ignore, 3+2=5. DS is 5.
• DS of 1243: 1+2+4+3=10 → 1+0=1. DS is 1.
• DS of 2315: 2+3+1+5=11 → 1+1=2. DS is 2.

Total DS: 5 + 1 + 2 = 8. Now find the option with a DS of 8.

Option (a) 7+9+1+0 → Ignore 9, 7+1=8. Answer is (a).

2. Multiplication Check

Question: 85 × 132 = ?

Options: (a) 11210 (b) 11220 (c) 11240

Solution: Check DS for both numbers.

• DS of 85: 8+5=13 → 4.
• DS of 132: 1+3+2 = 6.

Total DS: 4 × 6 = 24 → 2+4 = 6. Now check options.

Option (b) 11220: 1+1+2+2+0 = 6. Answer is (b).

3. The Trick with Subtraction

Sometimes, when subtracting, the DS becomes negative. Rule: Simply add 9 to any negative DS to make it positive.

Example: DS is -4. The real DS is -4 + 9 = 5.

4. How to Handle Division

Division is tricky because the denominator must be converted to a DS of 1. You do this by multiplying both the numerator and denominator by a specific number.

• If denominator DS is 2, multiply by 5 (2×5=10 → DS 1)
• If denominator DS is 5, multiply by 2
• If denominator DS is 4, multiply by 7 (4×7=28 → DS 1)
• If denominator DS is 7, multiply by 4
• If denominator DS is 8, multiply by 8 (8×8=64 → DS 1)

Exceptions & Limitations of Digital Sum

Never use Digital Sum under the following conditions:

• When calculating approximate values (it only works for exact calculations).
• When the denominator's Digital Sum is 3, 6, or 9. (You must first simplify the fraction to remove the factor of 3).
• When multiple options have the exact same Digital Sum. (In this case, use Unit Digits!).

Topic 2: Complete Concept of Unit Digits

The Unit Digit is the rightmost digit of a number (the digit in the "ones" place). In competitive exams, you are often asked to find the unit digit of massive expressions like (247)¹⁵³. This is where the concept of Cyclicity comes in.

What is Cyclicity?

The unit digits of the powers of a number repeat in a predictable pattern. This repeating pattern is called the cyclicity of that number.

Example: 2¹=2, 2²=4, 2³=8, 2⁴=16 (unit digit 6), 2⁵=32 (unit digit 2 again). The cycle is 4 long.

The Three Golden Rules of Cyclicity

Rule 1: Cyclicity of 1 (Numbers 0, 1, 5, 6)

These numbers are stubborn. No matter what power you raise them to, their unit digit never changes.

• (...0)ⁿ = 0
• (...1)ⁿ = 1
• (...5)ⁿ = 5
• (...6)ⁿ = 6

Example: Find the unit digit of (156)³⁴⁸. Because it ends in 6, the unit digit is simply 6.

Rule 2: Cyclicity of 2 (Numbers 4 and 9)

These numbers alternate between two unit digits based on whether the power is odd or even.

Number ending in	If Power is Odd	If Power is Even
4	4 (e.g., 41=4)	6 (e.g., 42=16)
9	9 (e.g., 91=9)	1 (e.g., 92=81)

Example: Find the unit digit of (289)¹⁴⁵. The number ends in 9. The power 145 is odd. Therefore, the unit digit is 9.

Rule 3: Cyclicity of 4 (Numbers 2, 3, 7, 8)

These numbers cycle every 4th power. The method is simple:

1. Divide the given power by 4 to find the remainder (R).
2. If R = 1, use power 1. If R = 2, use power 2. If R = 3, use power 3.
3. If the power is perfectly divisible by 4 (R = 0), use the 4th power.

Number	R = 1	R = 2	R = 3	R = 0 (Power 4)
2	2	4	8	6
3	3	9	7	1
7	7	9	3	1
8	8	4	2	6

Question: Find the unit digit of (137)²⁴⁵.

Step 1: The base ends in 7.
Step 2: Divide the power 245 by 4. (Trick: just divide the last two digits, 45, by 4). 45 ÷ 4 gives a remainder of 1.
Step 3: The unit digit will be 7¹ = 7.

Using Unit Digits in Equations

To find the unit digit of complex equations like (12)³⁴ × (43)⁵⁶ + (5)¹², simply find the unit digit of each term separately and perform the operation just on those single digits!

Mastering digital Sum concepts and shortcuts for competitive examination

Digital Sum: Zero to Hero

The Magic of "Casting Out Nines"

1. Beginner: What is Digital Sum?

Digital sum kisi bhi number ke digits ko tab tak add karne ka process hai jab tak humein ek single digit (1-9) na mil jaye.

Example: Digital sum of 456
4 + 5 + 6 = 15
1 + 5 = 6

Rule of 9: Digital sum nikalte waqt '9' ko '0' maana ja sakta hai. Isse calculation fast ho jati hai.

2. Intermediate: Operations with Digital Sum

• Addition: LHS ka digital sum = RHS ka digital sum.
• Multiplication: Numbers ko multiply karne ke bajaye unke digital sums ko multiply karein.
• Subtraction: Agar result negative aaye, toh usme 9 add kar dein.
  Ex: 2 - 5 = -3. So, -3 + 9 = 6.

3. Advanced: Division & Square Roots

Division mein humein denominator ka digital sum 1 banana hota hai:

• Agar niche 2 ho, toh upar-niche 5 se multiply karein (2x5=10 -> 1).
• Agar niche 8 ho, toh upar-niche 8 se multiply karein (8x8=64 -> 10 -> 1).

Perfect Square Rule: Kisi bhi perfect square ka digital sum hamesha 1, 4, 7, ya 9 hi hota hai.

Master The Remainder Theorem: Complete Guide

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📐 Mathematics | Competitive Exam Series

Remainder Theorem – Complete Guide

Master the Remainder Theorem for SSC, UPSC, UPPSC, BPSC, Railway & Banking exams. Clear concept, worked examples, shortcut tricks, and previous year questions all in one place!

📖 Concept & Theory

✏️ Worked Examples

⚡ Shortcut Tricks

📋 PYQ Practice

🏆 Exam Tips

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📖 What is the Remainder Theorem?

📌 Core Theorem Statement

The Remainder Theorem (Polynomial Division Theorem)

If f(x) is divided by (x − a), the remainder = f(a)

When a polynomial f(x) is divided by a linear divisor (x − a), the remainder is the value of the polynomial at x = a, i.e., simply substitute a into f(x) to get the remainder directly — no long division needed!

📌 Euclid's Division Form (For Numbers)

Remainder in Number Division

Dividend = Divisor × Quotient + Remainder

i.e., N = D × Q + R where 0 ≤ R < D

In competitive exams, "Remainder Theorem" often refers to finding the remainder when a number (or expression) is divided by another number. Both forms — polynomial and numeric — are tested heavily.

🎯 Why It Matters in Competitive Exams

Remainder-based questions appear in SSC CGL, SSC CHSL, UPSC CSAT, UPPSC, BPSC, RRB NTPC, SBI PO and almost all competitive exams. They test number theory, divisibility, and polynomial reasoning — all high-scoring topics.

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🔑 Key Concepts You Must Know

1. Polynomial Remainder Theorem

If f(x) ÷ (x − a), the remainder = f(a). Substitute a directly into polynomial.

f(x) ÷ (x−a) → Remainder = f(a)

Example: f(x) = x³ − 4x + 5, divide by (x − 2)
Remainder = f(2) = 8 − 8 + 5 = 5

2. Factor Theorem (Special Case)

If the remainder is zero, then (x − a) is a factor of f(x). This is the Factor Theorem — a special case of Remainder Theorem.

f(a) = 0 ⟹ (x − a) is a factor

Use this to check divisibility of polynomials without long division.

3. Cyclicity Method

Powers of numbers follow a repeating pattern (cycle) when divided. Key for finding remainders of large powers like 7⁵⁰ ÷ 10.

Units digit of powers repeats in cycles of 4

Cyclicity of 2: 2,4,8,6 (cycle=4). Cyclicity of 3: 3,9,7,1 (cycle=4).

4. Fermat's Little Theorem

If p is prime and gcd(a, p) = 1, then the remainder of aᵖ⁻¹ ÷ p is always 1.

a^(p−1) ≡ 1 (mod p), if p is prime

Very useful for finding remainders of huge power expressions in SSC/UPSC.

5. Chinese Remainder Theorem

Finds a number that gives specific remainders when divided by multiple divisors. Advanced concept asked in UPSC and Banking exams.

x ≡ r₁ (mod n₁) and x ≡ r₂ (mod n₂)

Useful when a number leaves remainder 2 on ÷3 and remainder 3 on ÷5 type questions.

6. Modular Arithmetic Basics

a ≡ r (mod d) means a leaves remainder r when divided by d. Used to simplify large calculations.

10 ≡ 1 (mod 3) since 10 = 3×3 + 1

Key rule: (a × b) mod d = [(a mod d) × (b mod d)] mod d

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📏 Important Rules & Divisibility Shortcuts

➗

Div by 2

Last digit even → R=0

Remainder when divided by 2 = last digit mod 2

➗

Div by 3 or 9

R = (digit sum) mod 3 or 9

Sum all digits → divide by 3 or 9 → that is the remainder

➗

Div by 10

R = Units digit of number

The units digit directly gives the remainder when divided by 10

➗

Div by 5

R = Units digit mod 5

Only the last digit matters. Units digit 0 or 5 → R=0

📐

Power Rule

(a+1)ⁿ ÷ a → R = 1

Any number of form (kd ± 1)ⁿ gives remainder ±1 when divided by d

🔁

aⁿ − bⁿ Rule

aⁿ−bⁿ divisible by (a−b) always

Also: aⁿ + bⁿ is divisible by (a+b) when n is odd

∑

Negative Remainder

−R = D − R (when negative)

If remainder comes out negative, add divisor to make it positive

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✏️ Fully Solved Examples (Step-by-Step)

📘 Example 1 – Polynomial Remainder

⭐ Easy

QUESTION Find the remainder when f(x) = x³ − 3x² + 2x − 5 is divided by (x − 2).

• Identify a: Divisor is (x − 2), so a = 2
• Apply Remainder Theorem: Remainder = f(a) = f(2)
  Substitute x = 2 into f(x)
• Calculate:
  f(2) = (2)³ − 3(2)² + 2(2) − 5 = 8 − 3(4) + 4 − 5 = 8 − 12 + 4 − 5 = −5

✅ Remainder

−5

📙 Example 2 – Large Power Remainder (Cyclicity)

⭐⭐ Medium

QUESTION (SSC CGL Type) Find the remainder when 7⁵⁰ is divided by 10. (i.e., units digit of 7⁵⁰)

• Find cyclicity of 7: Powers of 7 repeat their units digit every 4 steps
  7¹ = 7, 7² = 49, 7³ = 343, 7⁴ = 2401 → Units digits: 7, 9, 3, 1 → Cycle = 4
• Divide power by cycle length:
  50 ÷ 4 = 12 remainder 2
• Remainder 2 means same as 7²:
  7² has units digit = 9
• Therefore: 7⁵⁰ ÷ 10 gives units digit = 9, so remainder = 9

✅ Remainder

9

📗 Example 3 – Number Remainder (Digit Sum Rule)

⭐ Easy

QUESTION (Railway NTPC Type) What is the remainder when 5765 is divided by 9?

• Rule: Remainder when divided by 9 = sum of digits mod 9
• Sum the digits of 5765:
  5 + 7 + 6 + 5 = 23
• Digit sum 23 → sum again:
  2 + 3 = 5
• Therefore: 5765 ÷ 9 → remainder = 5

✅ Remainder

5

📓 Example 4 – Fermat's Little Theorem

⭐⭐⭐ Hard

QUESTION (UPSC CSAT / Banking Type) Find the remainder when 2¹⁰⁰ is divided by 7.

• Apply Fermat's Little Theorem: Since 7 is prime and gcd(2,7)=1
  2^(7−1) = 2^6 ≡ 1 (mod 7)
• Express 100 in terms of 6:
  100 = 6 × 16 + 4, so 2¹⁰⁰ = (2⁶)¹⁶ × 2⁴
• Simplify:
  (2⁶)¹⁶ ≡ 1¹⁶ = 1 (mod 7)2⁴ = 16 = 2×7 + 2, so 2⁴ ≡ 2 (mod 7)
• Combine:
  2¹⁰⁰ ≡ 1 × 2 = 2 (mod 7)

✅ Remainder

2

📕 Example 5 – Factor Theorem Application

⭐⭐ Medium

QUESTION (BPSC / UPPSC Type) Show that (x − 3) is a factor of f(x) = x³ − 7x + 6. Also find the value of k if (x − 1) is a factor of x² + kx − 2.

• Part 1 – Apply Factor Theorem: Check f(3)
  f(3) = (3)³ − 7(3) + 6 = 27 − 21 + 6 = 12 ≠ 0Hmm! Let's check f(−3):f(−3) = −27 + 21 + 6 = 0 ✅So (x + 3) is the factor, not (x − 3). This illustrates careful sign reading.
• Part 2 – Find k: (x − 1) is a factor, so f(1) = 0
  f(1) = (1)² + k(1) − 2 = 01 + k − 2 = 0 → k = 1

✅ Value of k

k = 1

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⚡ Shortcut Tricks for Competitive Exams

⚡ Trick 1 – (N+1)ⁿ ÷ N

When any number of the form (aN + 1) is raised to any power and divided by N:

(aN + 1)ⁿ ÷ N → Remainder = 1

Example: 101⁵⁰ ÷ 100 → 101 = 100+1, so Remainder = 1

⚡ Trick 2 – (N−1)ⁿ ÷ N

When a number of the form (aN − 1) is raised to a power n:

If n is even → R = 1; if n is odd → R = N−1

Example: 99⁵⁰ ÷ 100 → 99=100−1, 50 is even → R = 1
99⁵¹ ÷ 100 → 51 is odd → R = 99

⚡ Trick 3 – Cyclicity of Powers

Units digit (= remainder ÷ 10) follows cycles:

Cycle of 2,3,7,8 = 4; of 4,9 = 2; of 0,1,5,6 = 1

Divide power by cycle → use remainder to pick position in cycle

⚡ Trick 4 – Digit Sum for ÷9 or ÷3

Instead of dividing huge numbers by 9 or 3:

Remainder = (Sum of digits) mod 9 or mod 3

Example: 987654 ÷ 9 → 9+8+7+6+5+4=39 → 3+9=12 → 1+2=3 → R=3

⚡ Trick 5 – Negative Remainder

Using a negative form of the dividend can simplify calculations:

If N = dq + r, use r = r − d (negative form)

Example: 19 ÷ 7 → 19 = 21−2 → use remainder −2 → actual R = 7−2 = 5

⚡ Trick 6 – Remainder of Sum/Product

For complex expressions split and find each remainder:

R(a×b ÷ d) = R(a÷d) × R(b÷d), then mod d

Example: (34 × 27) ÷ 5 → R(34÷5)=4, R(27÷5)=2 → 4×2=8 → R=3

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📋 Previous Year Questions (PYQ) – Exam-Wise

Question	Exam	Answer	Concept Used
What is the remainder when x³ + 3x² − 2x + 1 is divided by (x − 1)?	SSC CGL	3	Polynomial R.T.
Find remainder: 17²⁵ ÷ 18	UPPSC	17	(N−1)ⁿ odd → N−1
What is units digit of 3⁴⁵?	RRB NTPC	3	Cyclicity of 3
Remainder of 2⁵⁶ ÷ 7	SBI PO	1	Fermat's Theorem
If (x − 2) is a factor of x² + kx − 4, find k	BPSC	k = 0	Factor Theorem
What is remainder when 5765432 is divided by 9?	RRB Group D	5	Digit Sum Rule
Find remainder: (13 × 17 × 19) ÷ 7	SSC CHSL	1	Modular Arithmetic
Remainder of 99⁵⁰ ÷ 100	UPPSC PCS	1	(N−1)ⁿ even → 1
Find k if x³ − 2x² + kx + 3 gives remainder 7 when divided by (x − 2)	IBPS PO	k = 3	Polynomial R.T.
What is the remainder when 4³⁰ is divided by 5?	SSC MTS	1	(N−1)ⁿ even → 1

⭐ Easy (1–2 min)

Digit sum, simple substitution

⭐⭐ Medium (2–3 min)

Cyclicity, factor theorem, (N±1) rule

⭐⭐⭐ Hard (3–5 min)

Fermat's, Chinese Remainder, combined

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🎯 Exam Strategy & Top Tips

📌 Remember These for Every Exam

• Always try the (N±1) rule and cyclicity first — they solve 60% of remainder questions fastest.
• For polynomial f(x) ÷ (x − a), just substitute x = a. Never do long division in exams.
• For (ax − b), rewrite as a(x − b/a) and use x = b/a as substitution value for f(x).
• When the answer is negative, add divisor to convert to positive remainder.
• Divisibility by 9: digit sum method. Divisibility by 11: alternating digit sum method.
• Memorise the cyclicity table — it will save 2–3 minutes per exam.
• Practice at least 20 PYQs on remainder theorem before any exam.

#RemainderTheorem #MathShortcuts #SSCMath #UPPSCMath #BPSCMath #CompetitiveExamMaths

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📐 Remainder Theorem – Competitive Exam Math Guide

For more maths topics: Number Theory  |  Algebra  |  Percentage  |  Time & Work

This blog is for educational and competitive exam preparation purposes. All examples and PYQs are illustrative. Always verify from official study material. © 2026 Exam Math Guide

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.