Analytical Ability + Maths Notes in English

📑 Topics on this page (Jump to Topic):

📊 0. Paper Analysis — What actually comes? 📍 A1. Odd One Out 📍 A2. Series & Missing Code 📍 A3. Data Sufficiency 📍 A4. Statement-Assumption 📍 A5. Decision Making 📍 A6. Puzzles, Ranking, Direction 📍 A7. Figure Analogy Tips 📍 B1. Number System, HCF-LCM 📍 B2. Simplification, Squares-Cubes, Fractions 📍 B3. Percentage 📍 B4. Profit-Loss-Discount 📍 B5. Simple & Compound Interest 📍 B6. Ratio-Proportion, Partnership 📍 B7. Average & Age 📍 B8. Mixture (Alligation) 📍 B9. Time-Work, Pipes-Cistern 📍 B10. Speed-Distance, Train, Boat 📍 B11. Mensuration (2D/3D) 📍 B12. Basic Algebra 📍 B13. Statistics & DI 📍 B14. Probability ⚡ Last-Day Formula Sheet

📊 0. Paper Analysis — What actually comes in Analytical Ability? (4 min read)

The official syllabus (SI/Subedar 2026 rulebook, Prelims) says only this: "(iii) Analytical Ability" — no details. So we read every single question of the Analytical Ability section of the official 2025 Prelims paper (16–21 Jan 2026, 12 shifts × 10 = 120 questions) and counted them type-wise:

Question Type	How many out of 120	Approx. per shift	On this page
Odd One Out — numbers, letter-number codes, GK/science concepts	29	2–4	A1
Figure Analogy (picture-based)	20	1–4	A7 (tips)
Number Series + Letter Series / Missing Code	18	1–4	A2
Arithmetic word problems (age, work, speed, partnership, probability, rank, direction)	13	1–3	A6 + Part B
Data Sufficiency (are Statements I/II sufficient?)	12	1–2	A3
Puzzles (floor, day, month, box, cars)	11	0–2	A6
Decision Making (best decision in a situation)	10	1	A5
Statement–Assumption	7	0–2	A4

⚠️ Most important point (Strategy):

Analytical Ability = 10 questions (out of 100). Pure "maths chapter" questions are few here (≈1–2 per shift); most questions are about logic + basic number sense. So first master Part A (A1–A7), then Part B (Maths).
In number odd-one-out and series, prime numbers, perfect squares, cubes, divisibility come again and again → be sure to memorise the tables in B1 and B2.
Decision Making and Assumption questions are almost free marks — just choose the "most sensible, positive, practical" option (see the rules in A4, A5).
According to the SI 2026 rulebook (Appendix-1), there is no negative marking in Prelims — so do not leave any question blank; first eliminate wrong options, then make your best guess.
In Mains (Paper II, Part B), the same Maths comes at a higher level — so Part B will be useful later too.

Note: The official PDF has no answer key. We have solved and verified the answer to every PYQ below ourselves (where needed, cross-checked with Testbook's official-paper solutions). Questions that could have more than one answer (ambiguous) have been left out. The option order is the same as in the official PDF (A = option 1, B = 2, C = 3, D = 4).

🧠 PART A — Exam-style question types of Analytical Ability

📍 A1. Odd One Out / Classification (10 min read)

1. Concept — How to think?

Out of 4 items, 3 follow a common rule and 1 does not. You have to find the rule. The rule is always simple. Check in this order (90% of questions get solved this way):

Check-list for numbers	Example (from the 2025 paper)
1. Prime or not?	101, 131, 151, 171 → 171 = 9 × 19 (the rest are prime)
2. Perfect square / cube?	1, 4, 9, 15 → 15 is not a square
3. Multiple of some number (table)?	45, 50, 58, 60 → 58 is not a multiple of 5
4. Power of 2 / part of a series?	32, 64, 94, 128 → 94 is not a power of 2
5. Pattern even within squares (even/odd root)	100, 144, 196, 289 → roots 10, 12, 14 (even), 17 (odd)
6. Digit sum / pattern of digits	(Last option — when the rules above do not work)

2. Letter–Number code (like B3C, 2B4)

First write the position number of the letters (A=1 … Z=26), then see what the relation between the letter and the number is.

Trick to remember positions — EJOTY: E = 5, J = 10, O = 15, T = 20, Y = 25
Reverse position (counting from Z = 1) = 27 − (direct position). E.g. A↔Z, B↔Y, C↔X, D↔W (the sum of each pair is 27).
A1 B2 C3 D4 E5 F6 G7 H8 I9 J10 K11 L12 M13 N14 O15 P16 Q17 R18 S19 T20 U21 V22 W23 X24 Y25 Z26

Solved Example 1: B3C, D5E, F7G, H9H — odd?
Step 1: Letters = B(2), D(4), F(6), H(8) → first letter jumps by +2. Numbers = odd numbers 3, 5, 7, 9.
Step 2: The third letter is always the next letter after the first letter: B→C, D→E, F→G, but H→H (it should have been I).
Answer: H9H.

Solved Example 2: 3A5, 5C7, 7E9, 10G11 — odd?
Rule: first number + 2 = last number (3→5, 5→7, 7→9). 10 + 2 = 12, but 11 is written. Answer: 10G11.

3. Words / GK-concept odd one out

Sometimes GK is also asked: "which is not a tax", "which is not an amphibian", "which is not a Fundamental Right". Rule: identify the category (tax / subsidy; amphibian / reptile; scalar / vector).

Category	Remember (came in the 2025 paper)
Tax	Income Tax, GST, VAT = tax; Subsidy = help given by the government, not a tax
Financial market	Money market, Capital market, Commodity market = financial; Labour market = not a financial market
Amphibian	Frog, Toad, Newt = amphibian; Lizard = reptile
Fundamental Rights (FR)	Equality, Freedom, Constitutional Remedies = FR; the Right to Property is no longer an FR (44th Amendment 1978 → Art. 300A, legal right)
Scalar / Vector	Distance, Speed, Mass, Energy, Power = scalar; Displacement, Velocity, Weight = vector
Polygon sides	Hexagon 6, Heptagon 7, Octagon 8, Decagon 10 (Heptagon = odd sides)

⚠️ Exam Traps:

⚠️ 1 is not prime, 2 is the only even prime. 51 (= 3×17), 57 (= 3×19), 87 (= 3×29), 91 (= 7×13) look prime but are NOT prime.
⚠️ If more than one rule fits, take the rule that separates only one option. E.g. 11, 22, 33, 44: odd/even makes two groups of two (wrong rule); by the prime rule only 11 stands apart → answer 11.
⚠️ In letter questions, patterns like "consecutive letters" (PQR, STU, MNO) and "vowel in the middle" (MON, BED, RAT) are common.

🎯 Real PYQs — MPESB SI Prelims 2025 (Analytical Ability section)

Q1. Identify the odd one out: 11, 22, 33, 44 MPESB SI Prelims 2025 — 16 Jan 2026, Shift 1

(A) 11

(B) 22

(C) 33

(D) 44

Correct answer: (A) 11
11 is the only prime; 22, 33, 44 are composite.

Q2. Which word is not a type of tax? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 1

(A) Income Tax

(B) GST

(C) VAT

(D) Subsidy

Correct answer: (D) Subsidy
A subsidy is financial help given by the government; the other three are taxes.

Q3. Find the two-digit number that is not divisible by 4. MPESB SI Prelims 2025 — 16 Jan 2026, Shift 1

(A) 24

(B) 48

(C) 72

(D) 94

Correct answer: (D) 94
94 ÷ 4 = 23.5 (remainder 2). 24, 48, 72 are all exactly divisible by 4.

Q4. Identify the odd one out: 45, 50, 58, 60 MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) 45

(B) 50

(C) 58

(D) 60

Correct answer: (C) 58
45, 50, 60 → multiples of 5 (end in 0 or 5). 58 is not.

Q5. Which of these geometric figures is not a polygon with an even number of sides? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) Hexagon

(B) Octagon

(C) Decagon

(D) Heptagon

Correct answer: (D) Heptagon
Heptagon = 7 sides (odd). Hexagon 6, Octagon 8, Decagon 10 (even).

Q6. Find the number that is not the sum of two different perfect squares. MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) 13

(B) 23

(C) 20

(D) 17

Correct answer: (B) 23
13 = 4 + 9, 20 = 4 + 16, 17 = 1 + 16. There is no pair for 23 (no two of 1, 4, 9, 16 add up to 23).

Q7. Among the Fundamental Rights given by the Indian Constitution, which principle (concept) is the odd one out? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) Right to Property

(B) Right to Constitutional Remedies

(C) Right to Freedom

(D) Right to Equality

Correct answer: (A) Right to Property
The 44th Amendment (1978) removed it from the Fundamental Rights and made it a legal right under Article 300A. The other three are still FRs today.

Q8. Identify the odd one out: B3C, D5E, F7G, H9H MPESB SI Prelims 2025 — 17 Jan 2026, Shift 2

(A) B3C

(B) D5E

(C) F7G

(D) H9H

Correct answer: (D) H9H
Third letter = the letter after the first one (B→C, D→E, F→G). After H, I should have come.

Q9. Identify the odd one out: 2B4, 4D6, 6F8, 8I10 MPESB SI Prelims 2025 — 17 Jan 2026, Shift 2

(A) 2B4

(B) 4D6

(C) 6F8

(D) 8I10

Correct answer: (D) 8I10
Position of the letter = first number: B = 2, D = 4, F = 6. For 8 it should be H, but it is I (= 9).

Q10. Which pair of these scientific concepts is the odd one out? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 2

(A) Velocity : Speed

(B) Energy : Power

(C) Mass : Weight

(D) Distance : Displacement

Correct answer: (B) Energy : Power
In the other three pairs there is a scalar and its vector form (Speed–Velocity, Mass–Weight, Distance–Displacement). Energy and Power are both scalars.

Q11. Which of these terms is not a type of financial market? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 2

(A) Money Market

(B) Commodity Market

(C) Labour Market

(D) Capital Market

Correct answer: (C) Labour Market
In the labour market, jobs/wages are exchanged, not money/securities.

Q12. Find the number pair whose squares differ by 15. MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) (17, 16)

(B) (6, 4)

(C) (5, 2)

(D) (4, 1)

Correct answer: (D) (4, 1)
4² − 1² = 16 − 1 = 15. (Shortcut: a² − b² = (a + b)(a − b) = 5 × 3 = 15.) Others: 33, 20, 21.

Q13. Identify the odd one out: 1, 4, 9, 15 MPESB SI Prelims 2025 — 18 Jan 2026, Shift 2

(A) 1

(B) 4

(C) 9

(D) 15

Correct answer: (D) 15
1, 4, 9 = 1², 2², 3² (perfect squares). 15 is not.

Q14. Identify the pair that does not have a clear cause : effect relationship. MPESB SI Prelims 2025 — 18 Jan 2026, Shift 2

(A) Bacteria : Disease

(B) Thirst : Water

(C) Gravity : Falling

(D) Spark : Fire

Correct answer: (B) Thirst : Water
Bacteria cause disease, gravity causes falling, a spark causes fire — cause → effect. Thirst does not "produce" water; water is the remedy for thirst (need–remedy).

Q15. Identify the odd one out: 12, 24, 36, 64 MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) 64

(B) 36

(C) 24

(D) 12

Correct answer: (A) 64
12, 24, 36 = multiples of 12 (12×1, 12×2, 12×3). 64 is not.

Q16. Find the group of letters that is different. MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) PQR

(B) STU

(C) VYA

(D) MNO

Correct answer: (C) VYA
PQR, STU, MNO = consecutive letters. VYA is not.

Q17. Which of these animals is not an amphibian? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) Lizard

(B) Newt

(C) Toad

(D) Frog

Correct answer: (A) Lizard
The lizard is a reptile. Newt, toad and frog are amphibians.

Q18. Identify the odd one out: 100, 144, 196, 289 MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) 289

(B) 196

(C) 144

(D) 100

Correct answer: (A) 289
All are perfect squares: 10², 12², 14² (squares of even numbers) and 17² (square of an odd number). So 289 is different.

Q19. Identify the odd one out: 3A5, 5C7, 7E9, 10G11 MPESB SI Prelims 2025 — 20 Jan 2026, Shift 1

(A) 7E9

(B) 5C7

(C) 10G11

(D) 3A5

Correct answer: (C) 10G11
Last number = first number + 2 (3→5, 5→7, 7→9). 10 + 2 = 12 should have been there.

Q20. Identify the odd one out: 4A6, 5B7, 6C8, 7D10 MPESB SI Prelims 2025 — 20 Jan 2026, Shift 1

(A) 7D10

(B) 6C8

(C) 5B7

(D) 4A6

Correct answer: (A) 7D10
Last = first + 2 (4→6, 5→7, 6→8). 7 + 2 = 9, but 10 is written.

Q21. Identify the pair in which the second number is not the square of the first number. MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) (34, 1156)

(B) (55, 3025)

(C) (66, 4366)

(D) (70, 4900)

Correct answer: (C) (66, 4366)
66² = 66 × 60 + 66 × 6 = 3960 + 396 = 4356, not 4366. (34² = 1156, 55² = 3025, 70² = 4900 are correct.) Note: here the unit digit (6) does not tell you — you have to work out the full square.

Q22. Which number does not belong to the sequence? 3, 6, 9, 13 MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) 13

(B) 9

(C) 6

(D) 3

Correct answer: (A) 13
3, 6, 9 = table of 3. The next should have been 12.

Q23. Which letter group does not follow the rule in which a vowel is the middle letter? MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) M O N

(B) C L K

(C) B E D

(D) R A T

Correct answer: (B) C L K
MON (O), BED (E), RAT (A) have a vowel in the middle. CLK has no vowel.

Q24. Identify the odd one out: 101, 131, 151, 171 MPESB SI Prelims 2025 — 21 Jan 2026, Shift 2

(A) 171

(B) 151

(C) 131

(D) 101

Correct answer: (A) 171
Digit sum of 171: 1+7+1 = 9 → divisible by 9 (171 = 9 × 19). 101, 131, 151 are prime.

Q25. Identify the odd one out: 32, 64, 94, 128 MPESB SI Prelims 2025 — 21 Jan 2026, Shift 2

(A) 128

(B) 94

(C) 64

(D) 32

Correct answer: (B) 94
32 = 2⁵, 64 = 2⁶, 128 = 2⁷. 94 is not a power of 2.

Q26. From the following letter pairs, choose the pair that does not follow the same alphabet pattern: A–Z, B–Y, C–X, D–V MPESB SI Prelims 2025 — 21 Jan 2026, Shift 2

(A) D–V

(B) C–X

(C) B–Y

(D) A–Z

Correct answer: (A) D–V
Opposite letters: A↔Z, B↔Y, C↔X, D↔W (sum of positions 27). D(4) + V(22) = 26.

📝 Practice MCQs (self-made, not PYQs)

P1. Choose the odd one: 27, 64, 125, 144

(A) 27

(B) 64

(C) 125

(D) 144

Answer: (D) 144 — 27 = 3³, 64 = 4³, 125 = 5³ (cubes). 144 = 12² (not a cube).

P2. Choose the odd one: 2, 3, 5, 9

(A) 2

(B) 3

(C) 5

(D) 9

Answer: (D) 9 — 9 = 3 × 3 (not prime). Note: 2 is prime even though it is even.

P3. Choose the odd one: ACE, BDF, GIK, MOP

(A) ACE

(B) BDF

(C) GIK

(D) MOP

Answer: (D) MOP — the others skip one letter each time (+2, +2). In MOP, O→P (+1).

P4. Choose the odd one: 121, 169, 225, 290

(A) 121

(B) 169

(C) 225

(D) 290

Answer: (D) 290 — 11², 13², 15² are perfect squares; 290 is not (17² = 289).

📍 A2. Number Series & Letter / Missing Code (Series) (10 min read)

1. Concept — How to crack a number series?

Every series is made by a small "machine rule". First write the difference between consecutive terms. If the difference shows no pattern, look at multiplication/division (ratio). Then squares/cubes/primes.

Pattern	How to identify	Example (2025 paper)
Same difference (+d)	Equal difference	15, 22, 29, 36, 43, 50 (+7)
Increasing difference	The difference itself is a series (6, 10, 14…)	4, 10, 20, 34, 52, 74 (+6, +10, +14, +18, +22)
×2 + 1 type	Each term roughly doubles	9, 19, 39, 79, 159, 319
Increasing multiplier (×2, ×3, ×4…)	Ratio 2, 3, 4…	3, 6, 18, 72, 360, 2160
Decreasing multiplier (×6, ×5, ×4…)	Ratio is decreasing	2, 12, 60, 240, 720, 1440
Prime numbers	All prime, none skipped in between	23, 29, 31, 37, 41, 43, 47
n³ − n / n² − 1 / n(n+1)	Close to squares/cubes	0, 6, 24, 60, 120, 210, 336 (1³−1, 2³−2, …, 7³−7)

Remember: Squares 1–30 and cubes 1–15 (table in B2). Primes 1–100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (25 in total).

2. Letter series / Missing code

Change each letter into a number (A = 1 … Z = 26). Each column (first letter, second letter, number) has its own pattern. After Z comes A again (after 26 comes 1 — "circular").

Solved Example 1: BFJ, IMQ, PTX, ____ ?
First column: B(2) → I(9) → P(16) → +7 → 23 = W
Second: F(6) → M(13) → T(20) → +7 → 27 → 27 − 26 = 1 = A
Third: J(10) → Q(17) → X(24) → +7 → 31 − 26 = 5 = E → Answer WAE.

Solved Example 2: UV3, VU8, XS15, AP24, ____ ?
First letter: U(21), V(22), X(24), A(27→1): +1, +2, +3 → next +4 → 5 = E
Second: V(22), U(21), S(19), P(16): −1, −2, −3 → −4 → 12 = L
Number: 3, 8, 15, 24 = 2²−1, 3²−1, 4²−1, 5²−1 → 6²−1 = 35 → Answer EL35.

⚠️ Exam Traps:

⚠️ After finding the answer of a series, check one more term ahead (the rule must work on all terms).
⚠️ In letter series, after Z comes A — "27 = A, 28 = B" (subtract 26).
⚠️ In some questions of the official 2025 PDF, the correct answer was misprinted in the options (note below). If this happens, choose the closest / typo-looking option.

🎯 Real PYQs — MPESB SI Prelims 2025

Q1. Find the next number in the series: 9, 19, 39, 79, 159, ? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) 319

(B) 318

(C) 313

(D) 325

Correct answer: (A) 319
Rule ×2 + 1: 9×2+1 = 19, 19×2+1 = 39 … 159×2+1 = 319.

Q2. Find the number in place of the question mark: 15, 22, 29, 36, ?, 50, 57 MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) 43

(B) 42

(C) 40

(D) 44

Correct answer: (A) 43
+7 each time: 36 + 7 = 43 (and 43 + 7 = 50 ✔).

Q3. Find the next number in the series: 3, 6, 18, 72, 360, ? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) 2100

(B) 1800

(C) 2160

(D) 1980

Correct answer: (C) 2160
×2, ×3, ×4, ×5, ×6 → 360 × 6 = 2160.

Q4. Number in place of the question mark: 23, 29, 31, 37, 41, 43, ? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) 45

(B) 53

(C) 47

(D) 49

Correct answer: (C) 47
Consecutive prime numbers. The next prime after 43 is 47 (45 = 5×9, 49 = 7×7 are not prime; 53 would skip 47 in between).

Q5. Number in place of the question mark: 0, 6, 24, 60, 120, 210, ? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) 336

(B) 343

(C) 300

(D) 332

Correct answer: (A) 336
n³ − n: 1−1 = 0, 8−2 = 6, 27−3 = 24, 64−4 = 60, 125−5 = 120, 216−6 = 210, 343−7 = 336. (Trap: 343 = 7³, but if you forget the "−7" you get it wrong.)

Q6. Which option correctly fills the blank? BFJ, IMQ, PTX, ____ MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) XBF

(B) XAE

(C) WAE

(D) WBF

Correct answer: (C) WAE
Each letter +7 (see Solved Example 1): P→W, T→A, X→E.

Q7. Which option correctly fills the blank? B0M, D1L, G1J, K2G, ____ MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) Q4C

(B) P5C

(C) P3C

(D) P3B

Correct answer: (C) P3C
First letter: B(2), D(4), G(7), K(11): +2, +3, +4 → +5 = P(16).
Number: 0, 1, 1, 2 → Fibonacci (sum of the previous two) → 1 + 2 = 3.
Last letter: M(13), L(12), J(10), G(7): −1, −2, −3 → −4 = C(3). (In the PDF font, "1" and "I" look the same.)

Q8. Number in place of the question mark: 40, 34, 28, 22, ?, 10 MPESB SI Prelims 2025 — 18 Jan 2026, Shift 2

(A) 14

(B) 15

(C) 16

(D) 17

Correct answer: (C) 16
−6 each time: 22 − 6 = 16, 16 − 6 = 10 ✔.

Q9. Which option correctly fills the blank? UV3, VU8, XS15, AP24, ____ MPESB SI Prelims 2025 — 18 Jan 2026, Shift 2

(A) DL35

(B) EL37

(C) DK37

(D) EL35

Correct answer: (D) EL35
See Solved Example 2: E (+4), L (−4), 35 (6² − 1).

Q10. Which option correctly fills the blank? EIO, IOU, OUA, ____ MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) UAI

(B) UAE

(C) AEI

(D) EIO

Correct answer: (B) UAE
Cycle of vowels A, E, I, O, U, A, E… Each term starts one vowel ahead: EIO → IOU → OUA → UAE.

Q11. Next number in the series: 2, 12, 60, 240, 720, ? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) 1200

(B) 1440

(C) 1800

(D) 2400

Correct answer: (B) 1440
×6, ×5, ×4, ×3 → next ×2: 720 × 2 = 1440.

Q12. Next number in the series: 4, 10, 20, 34, 52, ? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) 72

(B) 74

(C) 76

(D) 78

Correct answer: (B) 74
Differences: 6, 10, 14, 18 (+4 each time) → next 22 → 52 + 22 = 74.

Q13. (Number analogy) 6 : 24 :: 23 : ? — Options: 92, 80, 99, 86 MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) 92

(B) 80

(C) 99

(D) 86

Correct answer: (A) 92
6 × 4 = 24, in the same way 23 × 4 = 92. (This question was given as an "Image" in the paper.)

Q14. Which option correctly fills the blank? BF, CH, ___, HO, LT MPESB SI Prelims 2025 — 20 Jan 2026, Shift 1

(A) EM

(B) EK

(C) FJ

(D) EL

Correct answer: (B) EK
First letter: B(2), C(3), ?, H(8), L(12): +1, +2, +3, +4 → E(5).
Second: F(6), H(8), ?, O(15), T(20): +2, +3, +4, +5 → K(11).

Q15. Which option correctly fills the blank? DG, GJ, JM, ___ MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) NP

(B) MQ

(C) MO

(D) MP

Correct answer: (D) MP
Both letters +3: D→G→J→M, G→J→M→P. First letter of the next term = second letter of the previous term.

Q16. Which option correctly fills the blank? AYD, BVF, DRH, ___, KGL MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) GLJ

(B) HLK

(C) GMJ

(D) FMI

Correct answer: (C) GMJ
First: A(1), B(2), D(4), ?, K(11): +1, +2, +3, +4 → G(7).
Second: Y(25), V(22), R(18), ?, G(7): −3, −4, −5, −6 → M(13).
Third: D, F, H, ?, L: +2 → J.

Q17. Which option correctly fills the blank? DKM, FJP, HIS, JHV, ___ MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) HGV

(B) IGZ

(C) IGY

(D) LGY

Correct answer: (D) LGY
First: D, F, H, J (+2) → L. Second: K, J, I, H (−1) → G. Third: M, P, S, V (+3) → Y.

⚠️ Note — 2 PYQs where the correct answer is misprinted in the official PDF options:

(i) 5, 11, 23, 47, 95, 191, ? 18 Jan 2026, Shift 1 — Rule ×2 + 1 → 383. The official PDF prints the options "380, 283, 382, 379" — "283" looks like a typo for 383.
(ii) MN, LO, KP, JQ, IR, ? 18 Jan 2026, Shift 2 — First letter −1 (M, L, K, J, I → H), second +1 (N, O, P, Q, R → S) → HS. (The options were not printed correctly in the PDF.)
Lesson: trust the rule; the correct option will be on the exam screen.

📝 Practice MCQs (self-made)

P1. 2, 6, 12, 20, 30, ?

(A) 40

(B) 42

(C) 44

(D) 36

Answer: (B) 42 — n(n+1): 1×2, 2×3, 3×4, 4×5, 5×6, 6×7 = 42 (differences 4, 6, 8, 10, 12).

P2. 7, 14, 28, 56, ?

(A) 84

(B) 98

(C) 112

(D) 102

Answer: (C) 112 — ×2 each time.

P3. AZ, BY, CX, ?

(A) DV

(B) DW

(C) EW

(D) DX

Answer: (B) DW — first letter +1, second −1 (opposite pairs, sum 27).

P4. B, E, H, K, ?

(A) M

(B) N

(C) O

(D) L

Answer: (B) N — +3 each time: 2, 5, 8, 11, 14 = N.

📍 A3. Data Sufficiency (8 min read)

1. Concept

Here you do not need to find the answer — you only have to say whether the given statements are enough to find the answer or not. The answer must come out as one definite (unique) answer; a "could be" answer = NOT sufficient.

Option (wording of the 2025 paper)	When to choose?
I alone is sufficient while II alone is not sufficient	A definite answer from I only
II alone is sufficient while I alone is not sufficient	A definite answer from II only
Either I or II is sufficient (Either)	An answer from I separately, and also from II separately
I and II both (together) are sufficient (Both)	Neither is enough alone, but both together are enough
Neither I nor II is sufficient (Neither)	No definite answer even with both together

3-step method: (1) Read only I — definite answer? (2) Forget I and read only II — definite answer? (3) If neither alone is enough, combine both.

Solved Example: Q: What is the value of x? I: x + y = 10 II: x − y = 4
I alone: two unknowns (x, y), one equation → not definite. II alone: not definite. Both together: 2x = 14 → x = 7. Answer: Both together are sufficient.

⚠️ Exam Traps (lessons from the 2025 paper):

⚠️ "Only daughter" ≠ "only child". "H is X's only daughter" → there may be a son too → the number of children is not definite.
⚠️ "Only boys play football" ≠ "All boys play" → how many play is not known.
⚠️ In code language, the code of a word is definite only when that word is separated out from the common part of two sentences; from "sky is clear" & "make it clear" you get only the code of clear, not of sky.
⚠️ Do not assume anything on your own — only what is written.

🎯 Real PYQs — MPESB SI Prelims 2025

Q1. Question: How many children does M have? Statements: I. H is the only daughter of X, and X is M's wife. II. K and J are M's brothers. MPESB SI Prelims 2025 — 16 Jan 2026, Shift 1

(A) I alone is sufficient while II alone is not sufficient

(B) II alone is sufficient while I alone is not sufficient

(C) Either I or II is sufficient

(D) Neither I nor II is sufficient

Correct answer: (D) Neither I nor II
I: H is the only daughter — how many sons there are is not known. II: talks about M's brothers, not children. Even both together do not give a definite count.

Q2. Question: What will be the total weight of 10 poles, if each pole has the same weight? Statements: I. One-fourth of the weight of each pole is 5 kg. II. The total weight of three poles is 20 kg more than the total weight of two poles. MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) I alone is sufficient

(B) II alone is sufficient

(C) Either I or II is sufficient

(D) Neither I nor II

Correct answer: (C) Either I or II
I: ¼ × w = 5 → w = 20 kg → 10 poles = 200 kg ✔. II: 3w − 2w = 20 → w = 20 kg ✔. Each is enough separately.

Q3. Question: How many students in a class play football? Statements: I. Only boys play football. II. There are forty boys and thirty girls in the class. MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) Neither I nor II is sufficient

(B) Either I or II is sufficient

(C) II alone is sufficient

(D) I alone is sufficient

Correct answer: (A) Neither I nor II
"Only boys play" does not mean that all 40 boys play. The count is not definite.

Q4. Question: How is J related to P? Statements: I. M is P's brother and T is P's sister. II. P's mother married J's husband, who has one son and two daughters. MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) Neither I nor II is sufficient

(B) Either I or II is sufficient

(C) II alone is sufficient while I alone is not sufficient

(D) I alone is sufficient while II alone is not sufficient

Correct answer: (C) II alone is sufficient
I does not mention J at all. II: J's husband = husband of P's mother → J herself is P's mother. (Standard answer; also matches the Testbook solution.)

Q5. Question: How many doctors are practising in this city? Statements: I. There is one doctor for every seven hundred people. II. There are 16 wards here and each ward has as many doctors as there are wards. MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) Neither I nor II is sufficient

(B) Either I or II is sufficient

(C) II alone is sufficient while I alone is not sufficient

(D) I alone is sufficient while II alone is not sufficient

Correct answer: (C) II alone is sufficient
II: 16 wards × 16 doctors = 256 ✔. I: population is not given → the count cannot be found.

Q6. Question: On which date of the month was Anjali born in February 2004? Statements: I. Anjali was born on an even date of the month. II. Anjali's date of birth was a prime number. MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) I and II both are sufficient

(B) Either I or II is sufficient

(C) II alone is sufficient

(D) I alone is sufficient

Correct answer: (A) I and II both (together)
I alone: 2, 4, 6… many dates. II alone: 2, 3, 5, 7… many. Both together: even + prime = only 2 → 2 Feb.

Q7. Question: How many children are there in total in a row of children facing north? Statements: I. Rishika, who is fifth from the left end, is eighth to the left of Ashwin, and Ashwin is twelfth from the right end. II. Mohit is fifth to the left of Usha, and Usha is seventh from the right end and eighteenth from the left end. MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) Neither I nor II is sufficient

(B) Either I or II is sufficient

(C) II alone is sufficient

(D) I alone is sufficient

Correct answer: (B) Either I or II
I: Ashwin is 5 + 8 = 13th from the left, 12th from the right → total = 13 + 12 − 1 = 24 ✔. II: Usha 18 + 7 − 1 = 24 ✔. Each is enough separately.

Q8. Question: Five persons P, Q, R, S and T are sitting facing north. Who is immediately to the right of P? Statements: I. R is third to the left of Q and P is second to the right of R. II. Q is immediately to the left of T and T is second to the right of P. MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) I alone is sufficient

(B) II alone is sufficient

(C) Either I or II is sufficient

(D) Neither I nor II is sufficient

Correct answer: (C) Either I or II
I: R _ P Q (2nd to the right of R = P, 3rd = Q) → Q is immediately right of P ✔. II: P Q T (T = 2nd to the right of P, Q = immediately left of T) → Q is immediately right of P ✔.

Q9. Question: What is the code for 'sky' in a code language? Statements: I. 'sky is clear' is written as 'de ra fa'. II. 'make it clear' is written as 'de ga jo'. MPESB SI Prelims 2025 — 21 Jan 2026, Shift 2

(A) Neither I nor II is sufficient

(B) Either I or II is sufficient

(C) II alone is sufficient

(D) I alone is sufficient

Correct answer: (A) Neither I nor II
Common word "clear" = "de". Sky = "ra" or "fa" — not definite.

Q10. Question: In a row of children, how many children are there between P and Q? Statements: I. P is fifteenth in the row from the left. II. Q is exactly in the middle and there are ten children to his right. MPESB SI Prelims 2025 — 21 Jan 2026, Shift 2

(A) I and II both are sufficient

(B) Either I or II is sufficient

(C) II alone is sufficient

(D) I alone is sufficient

Correct answer: (A) I and II both (together)
II: Q in the middle, 10 on the right → total 21, Q is 11th from the left. I: P is 15th from the left. From both: in between 15 − 11 − 1 = 3 children.

📍 A4. Statement & Assumption (6 min read)

1. Concept

Assumption (implicit) = something the speaker of the statement is taking for granted without saying it. If that thing were not true, the statement would become pointless.

Negation Test: Think of the opposite of the assumption. If the opposite is true and the meaning of the statement collapses → the assumption is implicit. If there is no effect on the statement → not implicit.
Example: "The government allowed unaided colleges to raise fees." Opposite: "Unaided colleges have no financial difficulty" → then why the permission? So "there is financial difficulty" = implicit.

Treat as implicit when…	Do not treat as implicit when…
It is the direct reason / purpose of the statement	It is about some other group (aided colleges, all people)
The "people will read it, it will have an effect" idea of an advertisement/notice	Extreme words like "only", "always", "all", "never"
The government's decision was taken "after thinking it through" (money/need worked out)	Doubts that go against the statement or are pointless (people will not apply, the government will not appoint)

⚠️ Exam Traps:

⚠️ In suggestion-type statements ("…should be given"), assuming that "the government has money" is NOT implicit.
⚠️ "Never before such a lucid book" → there were books before too (implicit), but a general comment like "lucid books can be written on few topics" is not implicit.

🎯 Real PYQs — MPESB SI Prelims 2025

Q1. Statement: The concession in rail fare for travel to hill stations has been cancelled because people who can spend holidays there do not need it. Assumptions: I. The railways should give concessions only to needy people. II. The railways should not encourage people to spend holidays at hill stations. MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) Neither I nor II is implicit

(B) Either I or II is implicit

(C) Only assumption II is implicit

(D) Only assumption I is implicit

Correct answer: (D) Only I
The concession was removed because "these people do not need it" → meaning the concession should go to the needy (I ✔). Stopping people from going to hill stations is not the purpose of the statement (II ✘).

Q2. Statement: "This drink can be had either as it is, or with ice added." — An advertisement. Assumptions: I. People have different preferences. II. Some people will be attracted to this drink because it can be taken as it is. MPESB SI Prelims 2025 — 17 Jan 2026, Shift 2

(A) Only assumption I is implicit

(B) Only assumption II is implicit

(C) Neither I nor II is implicit

(D) Both I and II are implicit

Correct answer: (D) Both implicit
Two ways are mentioned → different preferences (I ✔). The advertisement wants to attract some people with the "you can drink it as it is" feature (II ✔).

Q3. Statement: The government has decided to levy 2 percent on the tax amount payable to fund the drought relief programme. Assumptions: I. The government does not have enough money to fund the drought relief programme. II. The amount collected as surcharge may be enough to fund these programmes. MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) Both I and II are implicit

(B) Either I or II is implicit

(C) Only II is implicit

(D) Only I is implicit

Correct answer: (A) Both implicit
A separate surcharge → money is short (I ✔). 2% was fixed after calculation → this much may be enough (II ✔).

Q4. Statement: Unemployment allowance should be given to all unemployed Indian youth above 18 years of age. Assumptions: I. There are unemployed youth in India who need financial help. II. The government has enough money to give the allowance to all unemployed youth. MPESB SI Prelims 2025 — 20 Jan 2026, Shift 1

(A) Neither I nor II is implicit

(B) Either I or II is implicit

(C) Only assumption II is implicit

(D) Only assumption I is implicit

Correct answer: (D) Only I
The suggestion makes sense only if there are needy youth (I ✔). The statement is a suggestion; it does not tell us about the government's funds (II ✘).

Q5. Statement: The government has allowed unaided colleges to raise their fees. Assumptions: I. Unaided colleges are in financial difficulty. II. Aided colleges do not need to raise fees. MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) Only assumption I is implicit

(B) Only assumption II is implicit

(C) Either I or II is implicit

(D) Neither I nor II is implicit

Correct answer: (A) Only I
The reason for the permission is financial difficulty (I ✔). Aided colleges are not mentioned at all (II ✘).

Q6. Statement: The state government has decided to appoint four thousand primary school teachers in the next financial year. Assumptions: I. There are enough schools in the state to take in four thousand more teachers. II. Eligible candidates may not be interested in applying because the government may not be able to appoint such a large number. MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) Neither I nor II is implicit

(B) Either I or II is implicit

(C) Only assumption II is implicit

(D) Only assumption I is implicit

Correct answer: (D) Only I
Such a decision is taken only after looking at vacant posts (vacancies/schools) (I ✔). II is a doubt that goes against the statement (✘).

Q7. Statement: Such a clear (lucid) book on this topic had never been found before. Assumptions: I. There were some other books on this topic too. II. You can write clear books on very few topics. MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) Both I and II are implicit

(B) Either I or II is implicit

(C) Only assumption II is implicit

(D) Only assumption I is implicit

Correct answer: (D) Only I
"Never so clear before" → there were books before too, but not this clear (I ✔). II is a general comment (✘).

📝 Practice MCQs (standard reasoning-book questions, not PYQs)

P1. Statement: Warning in a train coach — "Pull the chain to stop the train. ₹500 fine for improper use." Assumptions: I. Some people misuse the chain. II. On some occasions people may want to stop a moving train.

(A) Only I

(B) Only II

(C) Neither I nor II

(D) Both I and II

Answer: (D) Both — the fine shows that misuse happens (I); the chain is fitted precisely to stop the train in an emergency (II).

P2. Statement: "Do not use the lift while going down" — an instruction on the top floor of a five-storey building. Assumptions: I. The lift cannot carry any load while going down. II. The lift is a facility, not a right.

(A) Only I

(B) Only II

(C) Neither I nor II

(D) Both I and II

Answer: (B) Only II — I goes against the statement/is pointless; the instruction is there so that the lift stays free for people going up — i.e. the lift is a facility (II).

📍 A5. Decision Making / Situational Judgement (5 min read)

About 1 question in every shift: a situation (team, project, deadline, promotion) and 4 options. This checks the leadership qualities of an SI.

Golden Rules — the best option is often this:
✔ Solving the problem in a direct, calm, practical way (making a plan, talking, dividing work)
✔ Listening to everyone, discussing with the team, compromise / common ground
✔ Honesty — admitting a mistake and correcting it, even if it causes a little delay
✘ Wrong options: panicking, ignoring the problem, blaming others, cancelling the work, throwing people out, lowering quality through shortcuts, extreme decisions (studying all night, giving up everything)

🎯 Real PYQs — MPESB SI Prelims 2025

Q1. A student is struggling with time management for an upcoming entrance exam. What would be the most effective decision? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 1

(A) Study all night and skip sleep

(B) Make a realistic study timetable and follow it

(C) Leave the preparation for the last week

(D) Focus only on favourite subjects

Correct answer: (B) — a practical plan; the other options are extreme/careless.

Q2. You are leading a team with a tight deadline. A member suggests a shortcut that will save time but may compromise quality. What should you do? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 1

(A) Accept the shortcut

(B) Reject the shortcut and stick to the original plan

(C) Discuss the potential risks of the shortcut with the team

(D) Ask for more time

Correct answer: (C) — accepting or rejecting without thinking are both wrong; decide after understanding the risk with the team.

Q3. You are leading a student council initiative; a serious disagreement between two key members has created a deadlock. The most effective leadership decision? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) Take an executive decision and choose one idea yourself

(B) Let both present their views to the whole team and hold a vote

(C) Facilitate a discussion to identify common ground and find a compromise solution

(D) Remove both members from the team

Correct answer: (C) — common ground + compromise = the best leadership.

Q4. In a group project, one member is not contributing equally and misses meetings and deadlines. The best approach? MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) Ignore the issue and do the work yourself

(B) Talk to the member privately and understand their situation

(C) Discuss with the group and remove the member

(D) Complain to the instructor or supervisor

Correct answer: (B) — first a direct, private conversation; complaining/removing are later steps.

Q5. You have got a promotion that requires moving to another city, but you have strong ties with your present community and family. What should you consider before deciding? MPESB SI Prelims 2025 — 18 Jan 2026, Shift 2

(A) Only the salary increase

(B) Its effect on your personal life and relationships

(C) The prestige of the new place

(D) Your friends' opinion

Correct answer: (B) — the one-sided "only" option is wrong; looking at the full effect is sensible.

Q6. Before giving a presentation you find a factual error in it; correcting it means redoing a large part and the slot may be delayed. The most professional decision? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) Blame the source of the information

(B) Briefly mention the mistake and give the correct information, even if it causes a little delay

(C) Cancel the presentation

(D) Give the presentation with the mistake, hoping it will not be noticed

Correct answer: (B) — honesty + correction.

Q7. The estimated cost of a charity event is more than the amount raised so far; one week is left. The most effective decision? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) Cancel the event completely

(B) Immediately start targeted fundraising or get corporate sponsorship to cover the shortfall

(C) Ask the same people for more donations

(D) Reduce the quality of the event to cut costs

Correct answer: (B) — a direct, active solution to the problem.

Q8. A new vendor has to be chosen: one at a low price but with mixed reviews, the other expensive but with very good reviews. Whom should you prioritise? MPESB SI Prelims 2025 — 20 Jan 2026, Shift 1

(A) The vendor your boss likes

(B) The vendor with the best reviews

(C) The seller with mixed reviews

(D) The lowest price

Correct answer: (B) — reliable quality (in the company's interest); not the boss's liking/just the cheapest.

Q9. In a project with a tight deadline, an important member falls ill and cannot contribute for a week. The best step? MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) Panic and rush

(B) Hand over their work to other team members and adjust the timeline

(C) Wait for them to return and do nothing

(D) Cancel the project

Correct answer: (B) — redistribution of work + change in the plan.

📍 A6. Puzzles, Ranking, Direction & Logical Maths (10 min read)

1. How to solve a puzzle (floor / day / month / box)

Step 1: Make a table (Floor 1 at the bottom, top floor at the top; days Mon→Sun).
Step 2: First fill in the definite (fixed) information ("T on floor 4", "U on Monday").
Step 3: Then the paired information ("immediately above", "immediately before") — keep them as a two-piece block.
Step 4: Make cases from the "X people / X days in between" condition; strike out any case that breaks a condition.

Phrase	Meaning
"just above / immediately above"	The very next floor (+1)
"3 people in between / exactly three between"	Difference in position = 4
"at least 4 cars sold after it / at least four after"	≥ 4 positions after it
"more than two in between / more than two between"	Difference ≥ 4

2. Ranking formulas

Total people = (position from left) + (position from right) − 1
Position from right = Total − (position from left) + 1
Between two people = (difference of their positions) − 1

3. Direction — Pythagoras

Straight distance (shortest) = √(north-south difference² + east-west difference²). E.g. 3 km N, 4 km E → √(9 + 16) = 5 km.

Solved Example (Knock-out tournament): 30 players, the loser is out. At least how many matches are needed to decide the winner?
In each match exactly 1 player goes out. Apart from the winner, 29 have to go out → 29 matches (Rule: n players → n − 1 matches).

Solved Example (Fencing poles): 27 poles on each side of a square field (corner poles counted in both sides).
4 × 27 = 108, but the 4 corners were counted twice → 108 − 4 = 104.

⚠️ Exam Traps:

⚠️ In a floor puzzle, if "ground floor = 1" is given, start counting from the bottom.
⚠️ In "two of the four people give one right–one wrong answer" type questions, test each day one by one — the day on which exactly two people have one answer right is the answer.
⚠️ Corner poles double counting — in a square, total = 4n − 4.

🎯 Real PYQs — MPESB SI Prelims 2025

Q1. A, B, C, D, E and F live on different floors of a six-storey building (bottom 1, top 6). F lives on floor 3. B and E both live below F's floor. D lives below A's floor but above C's floor. Who lives on the top (floor 6)? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 2

(A) D

(B) E

(C) C

(D) A

Correct answer: (D) A
B, E → floors 1, 2. On the remaining 4, 5, 6: C < D < A → C = 4, D = 5, A = 6.

Q2. Six persons P, Q, R, S, T, U travel in different months of the same year — January, February, March, July, September, December. U travelled in September. Only one person travelled between U and Q. No one between P and S. S travelled in the month after P. More than two persons between S and T. Who travelled in March? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 2

(A) R

(B) T

(C) P

(D) Q

Correct answer: (D) Q
Order: Jan(1), Feb(2), Mar(3), Jul(4), Sep(5), Dec(6). U = 5; one in between → Q = 3 (Mar). Then P = Jan, S = Feb, T = Dec (3 people in between), R = Jul. All conditions ✔.

Q3. Four people; each gives two answers. Only two persons give one right and one wrong answer; both answers of the other two are wrong. When asked what day it is today — Person 1: Wednesday or Sunday. Person 2: Monday or Saturday. Person 3: Tuesday or Friday. Person 4: Thursday or Wednesday. What day is it today? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 2

(A) Sunday

(B) Saturday

(C) Friday

(D) Wednesday

Correct answer: (D) Wednesday
Wednesday is in the list of two people (1 and 4) → exactly two people half right ✔. Every other day is in the list of only one person.

Q4. Six boxes J, K, L, M, N, O are placed one above the other. J is immediately above N. There is only one box between K and O. O is at the bottom and M is at the top. Which box is immediately below M? MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) N

(B) J

(C) L

(D) K

Correct answer: (B) J
From the bottom: O = 1, K = 3 (one in between), M = 6. The J–N pair (J above) fits only in 4–5 → N = 4, J = 5, L = 2. Immediately below M is J.

Q5. A company sold six cars — Swift, Santro, Creta, Audi, i10, Magna — from Monday to Saturday (one each day). At least four cars were sold after Santro. Magna was sold on Tuesday. Audi was sold immediately after Creta, and Creta was sold before at least three cars. Santro and i10 were both sold before at least one car. On which day was Swift sold? MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) Thursday

(B) Friday

(C) Monday

(D) Saturday

Correct answer: (D) Saturday
≥ 4 after Santro → Santro = Mon (Magna on Tue). ≥ 3 after Creta → Creta ≤ Wed → Creta = Wed, Audi = Thu. Swift and i10 on Fri–Sat; one car after i10 → i10 = Fri, Swift = Sat.

Q6. P, Q, R, S, T, U, V in a seven-storey building (bottom 1, top 7). U at the top. Only two persons between T and Q. Q on the third floor. V immediately above P's floor. S immediately above R's floor. Q immediately below R's floor. Who is on the sixth floor? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) V

(B) R

(C) S

(D) T

Correct answer: (D) T
Q = 3, two in between → T = 6. R = 4, S = 5, U = 7, P = 1, V = 2.

Q7. Rohan lives in a seven-storey building (top 7, bottom 1). His friend Diya lives above his floor. Every Sunday Rohan goes two floors down to the flat of his colleague Atul. There are three floors between Rohan's and Diya's floors. On which floor does Rohan live? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) Fourth

(B) Third

(C) Second

(D) First

Correct answer: (B) Third
Atul = Rohan − 2 ≥ 1 → Rohan ≥ 3. Diya = Rohan + 4 ≤ 7 → Rohan ≤ 3. So Rohan = 3.

Q8. Six girls will record videos on different days of the same week; Sunday is a holiday. Friday for P. U on Monday. S immediately before T. Q on the day exactly midway between the days of U and S. T chose Thursday. Which day is left for R? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) Monday

(B) Saturday

(C) Wednesday

(D) Thursday

Correct answer: (B) Saturday
U = Mon, T = Thu, S = Wed, Q = Tue (between Mon–Wed), P = Fri → R = Sat.

Q9. P, Q, S, T, U, V and R have exams on different days from Monday to Sunday. S's is immediately before R's. Only three people between U and Q. Q's is immediately before S's. Only two people between S and V. P's is on Tuesday, R's on Sunday. Whose exam is on Monday? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 2

(A) U

(B) T

(C) V

(D) Q

Correct answer: (A) U
R = Sun, S = Sat, Q = Fri. 3 between U–Q → U = Mon. V = Wed (2 people in between before S), T = Thu.

Q10. A badminton singles tournament of 30 members; every loser is out, no ties. At least how many matches are needed to decide the winner? MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) 15

(B) 29

(C) 61

(D) 71

Correct answer: (B) 29
One goes out in each match; 29 have to go out → 29 matches.

Q11. A farmer put a fence around a square field, placing 27 poles on each side. How many poles in total? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) 100

(B) 104

(C) 108

(D) 110

Correct answer: (B) 104
4 × 27 − 4 (corners counted twice) = 104.

Q12. In a line of 15 students, Jatin is 8th from the left end. His position from the right end? MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) 7th

(B) 8th

(C) 9th

(D) 10th

Correct answer: (B) 8th
15 − 8 + 1 = 8 (he is exactly in the middle).

Q13. A man walks 5 km south, then 3 km west. How far is he from the starting point (approximately)? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) 5.83 km

(B) 6.5 km

(C) 8 km

(D) 4 km

Correct answer: (A) 5.83 km
√(5² + 3²) = √34 ≈ 5.83 (because 5.8² = 33.64, 5.9² = 34.81).

📝 Practice MCQs (self-made)

P1. Ravi is 7th from the left and 12th from the right in a row. How many people are there in the row in total?

(A) 18

(B) 19

(C) 17

(D) 20

Answer: (A) 18 — 7 + 12 − 1.

P2. Sita walks 3 km north, then 4 km east. Straight distance from the start?

(A) 7 km

(B) 5 km

(C) 1 km

(D) 6 km

Answer: (B) 5 km — √(9 + 16) = 5.

📍 A7. Figure Analogy — Tips (4 min read)

In the 2025 paper, every shift had 1–4 questions of the type "look at the relation between the first two figures in Image X, apply the same relation to the third, choose the option from Image Y". These are picture-based, so no text PYQs are given here; check the changes given below:

Check-list of changes	Example seen in the 2025 paper
1. One more same shape inside (double)	Circle → circle inside a circle; triangle → small triangle inside a triangle
2. Rotation (90° / 180°) or flip (mirror)	Change in the direction of an arrow, stairs turned upside down
3. Shading (filling / emptying)	Half circle filled → half square/triangle filled
4. Adding lines / parts	Lines added inside a triangle → lines added the same way inside a circle (match the number of lines carefully)
5. Inner and outer shapes swap places	Small shape outside, big one inside

⚠️ Tip:

⚠️ First look for just one change; if two options are left, check a second change (direction / shading).
⚠️ In a mirror image, left–right is reversed, not top–bottom; in a water image, top–bottom is reversed.
⚠️ For detailed practice, also see the non-verbal part of the Reasoning notes (08).

🔢 PART B — Arithmetic (Maths): concept → formula → solved examples → traps → practice

📍 B1. Number System, Divisibility, Unit Digit, Remainder, HCF–LCM (10 min read)

1. Types of numbers

Name	Meaning / Example
Natural	1, 2, 3, … (not 0)
Whole	0, 1, 2, 3, …
Integers	… −2, −1, 0, 1, 2 …
Rational	In p/q form (q ≠ 0): 3/4, −5, 0.25, 0.333…
Irrational	Cannot be written as p/q: √2, √3, π (22/7 is only an approximate value)
Prime	Divisible only by 1 and itself: 2, 3, 5, 7, 11… (1 is not prime, 2 is the only even prime). 15 primes in 1–50, 25 in 1–100
Composite	More than 2 factors: 4, 6, 9, 51, 91…
Co-prime	Whose HCF is 1: (8, 15), (4, 9)

2. Divisibility Rules

Divisible by?	Rule	Example
2	Last digit 0, 2, 4, 6, 8	3,584
3	Sum of digits divisible by 3	471 → 12 ✔
4	Last 2 digits divisible by 4	7,316 → 16 ✔; 94 ✘
5	Last digit 0 or 5	1,235
6	Divisible by both 2 and 3	144 ✔
8	Last 3 digits divisible by 8	7,48,352 → 352 = 8 × 44 ✔
9	Sum of digits divisible by 9	43,218 → 18 ✔
11	(Sum of digits at odd places) − (sum of digits at even places) = 0 or a multiple of 11	7,48,352 → (2+3+4) − (5+8+7) = −11 ✔

3. Unit Digit — "why it works"

When multiplying, the unit digit is decided only by the product of the unit digits. In powers, the unit digit repeats in a cycle of 4:

Unit digit	Cycle (power 1, 2, 3, 4)
0, 1, 5, 6	Always the same (0, 1, 5, 6)
2	2, 4, 8, 6
3	3, 9, 7, 1
4	4, 6 (odd power → 4, even power → 6)
7	7, 9, 3, 1
8	8, 4, 2, 6
9	9, 1 (odd power → 9, even → 1)

Method: divide the power by 4 → remainder r (if r = 0, take r = 4) → the r-th digit of the cycle.

Solved Example 1: Unit digit of 7⁹⁵?
95 ÷ 4 → remainder 3 → 3rd digit of the cycle of 7 (7, 9, 3, 1) = 3.

Solved Example 2: Unit digit of 23 × 47 × 59?
3 × 7 = 21 → 1; 1 × 9 = 9.

4. Remainder

Dividend = Divisor × Quotient + Remainder
Remainder of (a × b) when divided by n = remainder of (remainder of a × remainder of b) when divided by n
If a number leaves remainder r when divided by N, and d is a factor of N → remainder when divided by d = remainder of r ÷ d

Solved Example 3: A number divided by 45 leaves remainder 31. What if the same number is divided by 15?
15 is a factor of 45 → 31 ÷ 15 = 2, remainder 1.

Solved Example 4: Remainder when 2³¹ is divided by 5?
Remainders of powers of 2 when divided by 5: 2, 4, 3, 1 (cycle 4). 31 ÷ 4 → remainder 3 → 3.

5. HCF (Highest Common Factor) and LCM (Lowest Common Multiple)

HCF = the largest number that divides all of them (largest measure / equal pieces). LCM = the smallest number that all of them divide (when together again — bells, lights, rounds).

For two numbers: HCF × LCM = First × Second (only for two numbers)
HCF of fractions = HCF(numerators) / LCM(denominators); LCM = LCM(numerators) / HCF(denominators)
Largest number that divides a, b, c leaving the same remainder = HCF(b − a, c − b, c − a)
Smallest number which, when divided by a, b, c, leaves remainder r each time = LCM(a, b, c) + r

Solved Example 5: HCF of two numbers is 12, LCM 180; one number is 36. The other?
12 × 180 = 36 × x → x = 2160 ÷ 36 = 60.

Solved Example 6: Three bells ring every 6, 8, 12 minutes; they rang together at 10:00. Next time together?
LCM(6, 8, 12) = 24 → 10:24.

Solved Example 7: Largest number that divides 43, 91, 183 leaving the same remainder?
Differences: 48, 92, 140 → HCF = 4.

⚠️ Exam Traps:

⚠️ HCF × LCM = product does not apply to three numbers.
⚠️ HCF is always a factor of LCM; if in an option the HCF does not divide the LCM, that option is wrong.
⚠️ "When will they ring together" = LCM; "largest equal piece / largest measure" = HCF.

📝 Practice MCQs (self-made) — Related PYQ: Q3 of A1 (94 is not divisible by 4)

P1. Unit digit of 3⁶⁵?

(A) 1

(B) 3

(C) 7

(D) 9

Answer: (B) 3 — 65 ÷ 4 remainder 1 → in the cycle (3, 9, 7, 1), the 1st.

P2. Which of these numbers is divisible by 9?

(A) 43,217

(B) 43,218

(C) 43,219

(D) 43,220

Answer: (B) 43,218 — sum of digits 18.

P3. LCM of 12, 18 and 24?

(A) 36

(B) 48

(C) 72

(D) 144

Answer: (C) 72 — 12 = 2²·3, 18 = 2·3², 24 = 2³·3 → 2³ × 3² = 72.

P4. What is the least number to be added to 1000 so that it becomes exactly divisible by 7?

(A) 1

(B) 6

(C) 3

(D) 4

Answer: (A) 1 — 1000 = 7 × 142 + 6 → add 7 − 6 = 1 (1001 = 7 × 143).

P5. The smallest number which, when divided by 12, 15, 20, leaves remainder 5 each time?

(A) 60

(B) 65

(C) 125

(D) 55

Answer: (B) 65 — LCM 60 + 5.

📍 B2. Simplification (BODMAS), Squares-Cubes-Roots, Fractions-Decimals (8 min read)

1. BODMAS — the correct order

B = Brackets: first Bar/Vinculum (—), then ( ), then { }, then [ ]
O = Of: means ×, but solved before normal ÷
D = ÷ , M = × (both at the same level — whichever comes first from left to right)
A = + , S = − (left to right)

24 ÷ 4 of 2 = 24 ÷ 8 = 3 ('of' first) | 24 ÷ 4 × 2 = 6 × 2 = 12 (left to right)
36 ÷ 6 × 3 = 18, but 36 ÷ 6 of 3 = 36 ÷ 18 = 2

Solved Example 1: 45 − [38 − {60 ÷ 3 − (6 − 9 ÷ 3)}]
(6 − 9 ÷ 3) = 6 − 3 = 3 → {60 ÷ 3 − 3} = 20 − 3 = 17 → [38 − 17] = 21 → 45 − 21 = 24.

2. Squares 1–30 and Cubes 1–15 — memorise them

n	n²	n	n²	n	n²	n	n³
1	1	11	121	21	441	1	1
2	4	12	144	22	484	2	8
3	9	13	169	23	529	3	27
4	16	14	196	24	576	4	64
5	25	15	225	25	625	5	125
6	36	16	256	26	676	6	216
7	49	17	289	27	729	7	343
8	64	18	324	28	784	8	512
9	81	19	361	29	841	9	729
10	100	20	400	30	900	10	1000

Cubes 11–15: 11³ = 1331, 12³ = 1728, 13³ = 2197, 14³ = 2744, 15³ = 3375.

Square of a number ending in 5: 75² → 7 × 8 = 56, followed by 25 → 5625 (because (10a + 5)² = 100·a(a+1) + 25)
The unit digit of a square is never 2, 3, 7, 8 → such numbers are not perfect squares.
√ decimal: √0.0081 = 0.09 (decimal places are doubled: 0.09 × 0.09 = 0.0081); ∛0.000125 = 0.05

3. Fractions and Decimals

To add/subtract, take the LCM of the denominators: 3/4 + 5/6 − 7/12 = (9 + 10 − 7)/12 = 12/12 = 1
Recurring decimal: 0.333… = 3/9 = 1/3; 0.272727… = 27/99 = 3/11 (as many 9s below as digits that repeat)
0.1666… = 1/6 (mixed: (16 − 1)/90 = 15/90 = 1/6)
Comparing fractions: cross-multiply — 5/7 vs 7/10 → 50 vs 49 → 5/7 is bigger

4. Sum formulas

1 + 2 + … + n = n(n + 1)/2 (1 to 10 = 55)
1² + 2² + … + n² = n(n + 1)(2n + 1)/6 (1 to 10 = 385)
Sum of the first n odd numbers (1 + 3 + 5 …) = n²
Sum of the first n even numbers (2 + 4 + 6 …) = n(n + 1)

⚠️ Exam Traps:

⚠️ ÷ and × are at the same level — NOT "first M then D"; whichever comes first from the left.
⚠️ "Of" ≠ normal ×; "of" is solved before ÷.
⚠️ The square of a negative is always positive: (−5)² = 25, but −5² = −25.

📝 Practice MCQs (self-made) — the real BODMAS PYQ is in B12

P1. 18 ÷ 3 × 2 + 4 − 1 = ?

(A) 6

(B) 15

(C) 9

(D) 3

Answer: (B) 15 — 18 ÷ 3 = 6 → 6 × 2 = 12 → 12 + 4 − 1 = 15.

P2. √1764 = ?

(A) 38

(B) 42

(C) 44

(D) 48

Answer: (B) 42 — unit digit 4 → root ends in 2 or 8; 40² = 1600, 45² = 2025 → 42 (42² = 1764).

P3. Write 0.444… as a fraction.

(A) 4/10

(B) 4/9

(C) 4/99

(D) 2/5

Answer: (B) 4/9

P4. 1 + 3 + 5 + … + 19 = ?

(A) 90

(B) 100

(C) 110

(D) 81

Answer: (B) 100 — 10 odd numbers → 10² = 100.

📍 B3. Percentage (8 min read)

1. Concept

Percentage = "how much out of every 100". x% = x/100. So the fastest way is to convert the percentage into a fraction and think.

Fraction	%	Fraction	%
1/2	50%	1/7	14 2/7% (≈ 14.28%)
1/3	33 1/3%	1/8	12 1/2%
1/4	25%	1/9	11 1/9%
1/5	20%	1/10	10%
1/6	16 2/3%	1/11	9 1/11%
3/4	75%	1/12	8 1/3%

2. Important formulas

% change = (Change / Old value) × 100
Two successive changes (x% then y%): Net = x + y + xy/100 (for a decrease, put − with that number)
A is r% more than B → B is r/(100 + r) × 100% less than A | A is r% less than B → B is r/(100 − r) × 100% more than A
Price rises by r% → to keep expenditure the same, reduce consumption by r/(100 + r) × 100%; price falls by r% → increase consumption by r/(100 − r) × 100%
Population grows by r% every year: after n years = P(1 + r/100)ⁿ

1-second trick: Price rises by 25% (= 1/4) → reduce consumption by 1/(4 + 1) = 1/5 = 20%. Price rises by 20% (= 1/5) → 1/6 = 16 2/3%.

Solved Example 1: Salary first rose by 10%, then fell by 10%. Net change?
10 − 10 + (10 × −10)/100 = −1% (1% less). Reason: the second 10% was applied on the increased salary.

Solved Example 2: In an election the winner got 60% of the votes and won by 400 votes (only 2 candidates, all votes valid). Total votes?
The loser got 40% → difference 20% = 400 → 100% = 2000.

Solved Example 3: 33% is needed to pass. A student got 125 marks and failed by 40 marks. Maximum marks?
33% = 125 + 40 = 165 → 100% = 165 × 100/33 = 500.

⚠️ Exam Traps:

⚠️ +10% then −10% ≠ 0; always a little less (−1%).
⚠️ "A is 25% more than B" does NOT mean "B is 25% less than A" — it is 20% less.
⚠️ A percentage is always based on the thing it is compared "with".

📝 Practice MCQs (self-made)

P1. If the price of sugar increases by 20%, by what percent should consumption be reduced so that expenditure does not increase?

(A) 20%

(B) 16 2/3%

(C) 25%

(D) 15%

Answer: (B) 16 2/3% — 20/(100 + 20) × 100 = 1/6.

P2. 40% of a number = 72. The number?

(A) 160

(B) 180

(C) 200

(D) 288

Answer: (B) 180 — 72 × 100/40.

P3. A price first rose by 25% and then fell by 20%. Net change?

(A) 5% increase

(B) 5% decrease

(C) No change

(D) 1% decrease

Answer: (C) — 25 − 20 − 500/100 = 0 (100 → 125 → 100).

P4. A population of 10,000 grows by 10% every year. After 2 years?

(A) 12,000

(B) 12,100

(C) 11,000

(D) 12,210

Answer: (B) 12,100 — 10000 × 1.1 × 1.1.

P5. If 15% of x = 20% of y, then x : y = ?

(A) 3 : 4

(B) 4 : 3

(C) 2 : 3

(D) 5 : 4

Answer: (B) 4 : 3 — 15x = 20y → x/y = 20/15.

📍 B4. Profit, Loss & Discount (8 min read)

1. Concept

Cost Price (CP) = price bought at, Selling Price (SP) = price sold at, Marked Price (MP) = price written on the tag. Profit/loss is always on CP, discount is always on MP.

Profit % = (SP − CP)/CP × 100 | Loss % = (CP − SP)/CP × 100
SP = CP × (100 + Profit%)/100 | SP = CP × (100 − Loss%)/100
SP = MP × (100 − Discount%)/100
Successive discounts a% and b%: single equivalent discount = a + b − ab/100
At the same SP, x% profit on one and x% loss on the other → always loss = x²/100 %
Dishonest shopkeeper (false weight / weighs less): Profit % = Error/(True − Error) × 100 (900 g instead of 1000 g → 100/900 × 100 = 11 1/9%)
CP of n items = SP of m items → Profit % = (n − m)/m × 100

Solved Example 1: Two watches at ₹990 each; 10% profit on one, 10% loss on the other. Overall?
CP₁ = 990 × 100/110 = 900, CP₂ = 990 × 100/90 = 1100 → total CP 2000, total SP 1980 → loss 20 = 1% loss (shortcut: 10²/100 = 1%).

Solved Example 2: MP ₹1200, successive discounts of 10% and 5%. SP?
1200 × 0.90 = 1080 → 1080 × 0.95 = ₹1026. Single equivalent discount = 10 + 5 − 0.5 = 14.5%.

Solved Example 3: Even after giving a 10% discount, a 20% profit is wanted. By what % should the MP be kept above the CP?
MP × 0.9 = CP × 1.2 → MP = (4/3) CP → 33 1/3% above.

⚠️ Exam Traps:

⚠️ Discount is on MP, profit is on CP — do not mix the two.
⚠️ Successive 20% + 10% discount = 28%, not 30%.
⚠️ Same SP + same % (profit and loss) → never "no profit no loss", always a loss.

📝 Practice MCQs (self-made)

P1. Selling an article for ₹450 results in a 10% loss. Cost price?

(A) ₹495

(B) ₹500

(C) ₹405

(D) ₹550

Answer: (B) ₹500 — 450 × 100/90.

P2. Single equivalent discount of two successive discounts of 20% and 10%?

(A) 30%

(B) 28%

(C) 25%

(D) 32%

Answer: (B) 28% — 20 + 10 − 200/100.

P3. Cost price of 15 articles = selling price of 12 articles. Profit %?

(A) 20%

(B) 25%

(C) 15%

(D) 30%

Answer: (B) 25% — (15 − 12)/12 × 100.

P4. A dishonest shopkeeper weighs 900 g in place of 1 kg (claiming to sell at cost price). Profit %?

(A) 10%

(B) 11 1/9%

(C) 9 1/11%

(D) 12%

Answer: (B) 11 1/9% — 100/900 × 100.

📍 B5. Simple & Compound Interest (SI & CI) (8 min read)

1. Concept

Simple Interest (SI): every year interest is only on the principal (P) → equal interest every year. Compound Interest (CI): the previous year's interest is also added to the principal → "interest on interest" → the interest grows every year. So for 1 year SI = CI (with annual compounding); from 2 years CI is more.

SI = P × R × T / 100 | Amount A = P + SI
CI: A = P (1 + R/100)^T, CI = A − P
Half-yearly: R → R/2, T → 2T | Quarterly: R → R/4, T → 4T
CI − SI (2 years) = P (R/100)² | CI − SI (3 years) = P (R/100)² × (300 + R)/100
In SI, money becomes n times in T years → R = (n − 1) × 100 / T
In CI, money becomes n times in T years → n² times in 2T years, n³ times in 3T years

Solved Example 1: CI on ₹10,000 at 10% per annum, 2 years?
Year 1 interest 1000 → principal 11,000; year 2 interest 1100 → total CI = ₹2100 (SI would be 2000; difference 100 = 10000 × (10/100)²).

Solved Example 2: CI − SI on ₹8000 at 5% for 3 years?
8000 × (5/100)² × (305/100) = 8000 × 1/400 × 3.05 = 20 × 3.05 = ₹61. (Check: CI = 1261, SI = 1200.)

Solved Example 3: At SI, money doubles in 8 years. Rate? And in how many years will it become 3 times?
R = (2 − 1) × 100/8 = 12.5%. For 3 times, interest = 2P → T = 2 × 100/12.5 = 16 years.

Solved Example 4: CI on ₹10,000 at 10% per annum, compounded half-yearly, 1 year?
R = 5%, T = 2 → 10000 × 1.05² = 11025 → CI = ₹1025.

⚠️ Exam Traps:

⚠️ In SI, "doubles in 8 years" does NOT mean "four times in 16 years" — it is 24 years (interest of 3P is needed). In CI, double in T → four times in 2T.
⚠️ If months are given, T = months/12.
⚠️ Read carefully whether the "Amount" is asked or only the "interest".

📝 Practice MCQs (self-made)

P1. Difference between CI and SI on ₹5,000 at 10% per annum for 2 years?

(A) ₹25

(B) ₹50

(C) ₹75

(D) ₹100

Answer: (B) ₹50 — 5000 × (1/10)².

P2. Simple interest on ₹1200 at 5% per annum for 4 years?

(A) ₹200

(B) ₹240

(C) ₹260

(D) ₹300

Answer: (B) ₹240 — 1200 × 5 × 4/100.

P3. At what rate of simple interest will money become three times in 20 years?

(A) 5%

(B) 10%

(C) 15%

(D) 20%

Answer: (B) 10% — (3 − 1) × 100/20.

P4. Amount on ₹8000 at 5% per annum compound interest for 2 years?

(A) ₹8800

(B) ₹8820

(C) ₹8840

(D) ₹8400

Answer: (B) ₹8820 — 8000 × 1.05 × 1.05.

P5. The difference between CI and SI for 2 years at 4% per annum is ₹16. Principal?

(A) ₹8,000

(B) ₹10,000

(C) ₹12,000

(D) ₹16,000

Answer: (B) ₹10,000 — 16 = P × (4/100)² → P = 16 × 625.

📍 B6. Ratio, Proportion, Unitary Method, Partnership (7 min read)

1. Concept

Ratio a : b = "b parts for every a parts". Divide the total amount into a + b parts. Proportion a : b = c : d → a × d = b × c (product of extremes = product of means).

Divide amount S in a : b → first = S × a/(a + b), second = S × b/(a + b)
4th proportional of a, b, c = bc/a | Mean proportional of a, c = √(ac) | 3rd proportional of a, b = b²/a
A : B = 2 : 3, B : C = 4 : 5 → A : B : C = (2×4) : (3×4) : (3×5) = 8 : 12 : 15 (make B equal in both)
Partnership: Share of profit ∝ Capital × Time
Unitary method: first find the value of 1, then of as many as needed

Solved Example 1: Divide ₹1200 in 3 : 5 → 1200 × 3/8 = 450, 1200 × 5/8 = 750.

Solved Example 2: A invested ₹5000 for 12 months, B invested ₹6000 for 8 months; profit ₹1800.
A : B = 60000 : 48000 = 5 : 4 → A = 1800 × 5/9 = ₹1000, B = ₹800.

Solved Example 3: 4th proportional of 4, 6, 10 = 6 × 10/4 = 15. Mean proportional of 4 and 9 = √36 = 6.

⚠️ Exam Traps:

⚠️ In partnership, if money is withdrawn midway, count only those months (withdrawn after 6 months → 6).
⚠️ To compare ratios, simplify by the common factor (60000 : 48000 = 5 : 4).

🎯 Real PYQ

Q1. Three friends A, B and C invested in the ratio 2 : 3 : 4 respectively. After 6 months A withdrew his money. If the total profit after one year was ₹900, what will be A's share? MPESB SI Prelims 2025 — 20 Jan 2026, Shift 1

(A) ₹200

(B) ₹112.50

(C) ₹155.5

(D) ₹105.80

Correct answer: (B) ₹112.50
A : B : C = 2×6 : 3×12 : 4×12 = 12 : 36 : 48 = 1 : 3 : 4 → A = 900 × 1/8 = 112.50.

📝 Practice MCQs (self-made)

P1. If a : b = 3 : 4 and b : c = 6 : 5, then a : b : c = ?

(A) 9 : 12 : 10

(B) 3 : 4 : 5

(C) 18 : 24 : 25

(D) 9 : 12 : 15

Answer: (A) 9 : 12 : 10 — 18 : 24 : 20 = 9 : 12 : 10.

P2. Mean proportional of 9 and 16?

(A) 12

(B) 12.5

(C) 144

(D) 25

Answer: (A) 12 — √(9 × 16).

P3. The price of 12 books is ₹180. Price of 7 books?

(A) ₹95

(B) ₹105

(C) ₹115

(D) ₹84

Answer: (B) ₹105 — 1 book ₹15 → 7 × 15.

📍 B7. Average & Ages (8 min read)

1. Concept

Average = total sum ÷ number. That is, each one's share if everything is made equal. If a new member comes/leaves, first find the total sum (average × number), then subtract/add.

Total = Average × Number
Average of consecutive numbers (AP) = (first + last)/2 | Average of the first n natural numbers = (n + 1)/2
Rule of ages: For the ages of two people, the difference always stays the same; every year both ages increase by 1.
Ratio questions: take age = ratio × k and form the equation

Solved Example 1: Average age of 30 students is 12 years. Including the teacher, the average becomes 13. Teacher's age?
New total = 13 × 31 = 403; old total = 12 × 30 = 360 → teacher = 43 years.

Solved Example 2: Average of 5 numbers is 20; after removing one, the average of the remaining 4 is 18. Number removed?
100 − 72 = 28.

Solved Example 3: Father's age is 3 times the son's; after 15 years it will be 2 times. Ages?
3S + 15 = 2(S + 15) → S = 15, Father = 45.

⚠️ Exam Traps:

⚠️ If it is "x years ago", subtract x from both ages, not just one.
⚠️ Checking with the options is fastest: put in the answer and match the condition.

🎯 Real PYQs — MPESB SI Prelims 2025

Q1. The ratio of the ages of two brothers is 3 : 5. The product of their ages is 135. What is the age of the younger brother? MPESB SI Prelims 2025 — 16 Jan 2026, Shift 1

(A) 9 years

(B) 15 years

(C) 19 years

(D) 27 years

Correct answer: (A) 9 years
3k × 5k = 15k² = 135 → k² = 9 → k = 3 → younger = 9, elder = 15 (9 × 15 = 135 ✔).

Q2. A woman says, "If you reverse my age, the figures show my husband's age. He is of course senior to me, and the difference between our ages is one-eleventh of the sum of our ages." The woman's age is MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) 23 years

(B) 34 years

(C) 45 years

(D) None of these

Correct answer: (C) 45 years
Woman 10a + b, husband 10b + a. Difference = 9(b − a), sum = 11(a + b). 9(b − a) = (a + b) → 8b = 10a → 4b = 5a → a = 4, b = 5 → woman 45, husband 54 (difference 9 = 99/11 ✔).

Q3. Today is Sharan's birthday. One year from today, his age will be twice his age 12 years ago. Sharan's present age? MPESB SI Prelims 2025 — 21 Jan 2026, Shift 1

(A) 20 years

(B) 22 years

(C) 25 years

(D) 27 years

Correct answer: (C) 25 years
x + 1 = 2(x − 12) → x = 25 (26 = 2 × 13 ✔).

Q4. Some friends decided to go on a picnic and planned to spend 96 rupees on food and drinks. However, four of them did not come. As a result, each of the rest had to pay 4 rupees more. The number of people who joined the picnic was MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) 8

(B) 12

(C) 16

(D) 24

Correct answer: (A) 8
Option test: 8 came → each 96/8 = 12; plan was for 12 people → 96/12 = 8; difference 4 ✔. (Note: the question asks for those who joined, not the total of 12.)

📝 Practice MCQs (self-made)

P1. The ratio of the present ages of A and B is 4 : 5. After 5 years it becomes 5 : 6. A's present age?

(A) 15

(B) 20

(C) 25

(D) 30

Answer: (B) 20 — (4k + 5)/(5k + 5) = 5/6 → k = 5 → A = 20.

P2. Average of 11 results is 50; average of the first 6 is 49 and of the last 6 is 52. The sixth result?

(A) 50

(B) 54

(C) 56

(D) 58

Answer: (C) 56 — 294 + 312 − 550 (the 6th was counted twice).

P3. The sum of the ages of a father and son is 50. 5 years ago, the father's age was 7 times the son's. Son's present age?

(A) 8

(B) 10

(C) 12

(D) 15

Answer: (B) 10 — Father 40: 5 years ago 35 = 7 × 5 ✔.

📍 B8. Mixture & Alligation (5 min read)

1. Concept

When a cheap and a costly item are mixed, the price of the mixture comes between the two. The closer a price is to the mixture price, the larger its quantity. So the ratio of quantities = the reverse differences.

Alligation: Cheaper : Dearer (quantity) = (Dearer − Mixture) : (Mixture − Cheaper)
Replacement: Taking out x litres from a V-litre vessel and filling with water, n times → milk left = V × (1 − x/V)ⁿ

Solved Example 1: Rice at ₹20/kg and ₹30/kg is to be mixed to make ₹24/kg. Ratio?
(30 − 24) : (24 − 20) = 6 : 4 = 3 : 2 (cheaper : dearer).

Solved Example 2: In a 40 L mixture, milk : water = 3 : 1. How much water should be added to make it 1 : 1?
Milk 30, water 10 → 30 water is needed → add 20 L (milk does not change).

Solved Example 3: 4 L taken out of 40 L of milk and replaced with water; done twice. Milk left?
40 × (9/10)² = 32.4 L.

📝 Practice MCQs (self-made)

P1. In what ratio should tea at ₹60/kg and ₹90/kg be mixed so that the mixture is ₹70/kg?

(A) 1 : 2

(B) 2 : 1

(C) 3 : 1

(D) 1 : 1

Answer: (B) 2 : 1 — (90 − 70) : (70 − 60) = 20 : 10.

P2. A 20 L mixture has 25% alcohol. On adding 5 L of water, alcohol %?

(A) 20%

(B) 25%

(C) 15%

(D) 22%

Answer: (A) 20% — alcohol 5 L, total 25 L.

P3. In a 32 L mixture, milk : water = 5 : 3. How much water should be added to make it 1 : 1?

(A) 4 L

(B) 6 L

(C) 8 L

(D) 10 L

Answer: (C) 8 L — milk 20, water 12 → up to 20.

📍 B9. Time & Work, Pipes & Cisterns (8 min read)

1. Concept — LCM (efficiency) method

Take total work = the LCM of all the days (in units). Each person's 1-day capacity (efficiency) = total work ÷ his days. Together = sum of the efficiencies. No need for fractions.

A 10 days, B 15 days → Total = LCM 30 units; A = 3/day, B = 2/day → together 5/day → 30/5 = 6 days
Shortcut (two people): days together = (A × B)/(A + B) = 150/25 = 6
M₁ × D₁ × H₁ / W₁ = M₂ × D₂ × H₂ / W₂ (Men, Days, Hours, Work)
A is 2 times as efficient as B → A takes half the time (efficiency ∝ 1/time)
Pipes-cistern: Filling pipe = +, emptying pipe / leak = −. Capacity of the tank = LCM

Solved Example 1: A 12 days, B 18 days. Both work together for 4 days, then A leaves. In how many days will B finish the rest?
Total 36; A = 3, B = 2 → 20 in 4 days → 16 left → B: 16/2 = 8 days.

Solved Example 2: Pipe A fills in 20 min, B fills in 30 min, C empties in 15 min. All three open?
LCM 60: +3 +2 −4 = 1/min → 60 min.

Solved Example 3: 12 men do a work in 8 days; to do it in 6 days?
12 × 8 = M × 6 → M = 16.

⚠️ Exam Traps:

⚠️ "A's 2 days' work = B's 3 days' work" → efficiency A : B = 3 : 2 (reverse).
⚠️ In leak questions, the efficiency of the leak is minus.
⚠️ More men → fewer days (inverse proportion); more work → more days.

🎯 Real PYQ

Q1. The work done by A in 2 days is equal to the work done by B in 3 days. If A and B together can complete a work in 12 days, how many days will B alone take to complete that work? MPESB SI Prelims 2025 — 19 Jan 2026, Shift 1

(A) 48 days

(B) 40 days

(C) 30 days

(D) 20 days

Correct answer: (C) 30 days
2A = 3B → efficiency A : B = 3 : 2. Together 5 units/day × 12 = 60 units. B alone = 60/2 = 30 days (A = 20 days).

📝 Practice MCQs (self-made)

P1. A does a work in 20 days, B in 30 days. Both together?

(A) 10 days

(B) 12 days

(C) 15 days

(D) 25 days

Answer: (B) 12 — 600/50.

P2. 15 men do a work in 20 days. In how many days will 25 men do it?

(A) 10

(B) 12

(C) 15

(D) 18

Answer: (B) 12 — 15 × 20/25.

P3. A pipe fills a tank in 6 hours; a leak empties the full tank in 12 hours. If both work together, in how many hours will the tank be filled?

(A) 8

(B) 10

(C) 12

(D) 18

Answer: (C) 12 — LCM 12: +2 −1 = 1/hour.

P4. A is twice as efficient as B; together they do a work in 14 days. B alone?

(A) 21

(B) 28

(C) 42

(D) 35

Answer: (C) 42 — A = 2, B = 1 → total 3 × 14 = 42 → B = 42 days.

📍 B10. Speed-Time-Distance, Trains, Boats & Streams (9 min read)

1. Concept

Distance = Speed × Time. If the distance is fixed, speed and time are in inverse proportion (speed doubled → time halved). To cross a pole, a train has to cover its full length; for a platform, train + platform.

D = S × T | km/h → m/s: × 5/18 | m/s → km/h: × 18/5 (36 km/h = 10 m/s, 72 km/h = 20 m/s)
Average speed (equal distance): 2xy/(x + y) | Equal time: (x + y)/2
Train: pole/man → T = L/S | platform/bridge → T = (L + P)/S
Two trains: opposite directions → relative speed = S₁ + S₂; same direction → S₁ − S₂; time = (L₁ + L₂)/relative speed
Boat: Downstream D = b + s; Upstream U = b − s → b = (D + U)/2, s = (D − U)/2
Late/early question: D = S₁S₂/(S₂ − S₁) × (total time difference)

Solved Example 1: In how many seconds will a 180 m train at 54 km/h cross a pole?
54 × 5/18 = 15 m/s → 180/15 = 12 sec.

Solved Example 2: Two 150 m trains at 90 km/h (both) in opposite directions. In how many seconds will they cross each other?
Relative 180 km/h = 50 m/s; distance 300 m → 6 sec.

Solved Example 3: Boat downstream 15 km/h, upstream 9 km/h. Speed of the boat and the stream?
b = (15 + 9)/2 = 12, s = (15 − 9)/2 = 3 km/h.

Solved Example 4: A man goes from A to B at 40 km/h and returns at 60 km/h. Average speed?
2 × 40 × 60/100 = 48 km/h (NOT 50).

⚠️ Exam Traps:

⚠️ Match the units: if km/h and metres come together, convert first; minutes into hours (6 min = 1/10 hr).
⚠️ Average speed over equal distances is not a simple average.
⚠️ In platform questions, do not forget to add the length of the platform.

🎯 Real PYQ

Q1. A man cycles at a speed of 10 km per hour and reaches his office 6 minutes late. When he cycles at a speed of 12 km per hour, he reaches 4 minutes early. What is the distance to his office? MPESB SI Prelims 2025 — 18 Jan 2026, Shift 1

(A) 15 km

(B) 12 km

(C) 10 km

(D) 8 km

Correct answer: (C) 10 km
Total time difference = 6 + 4 = 10 min = 1/6 hour. D/10 − D/12 = 1/6 → D/60 = 1/6 → D = 10. (Check: 60 min vs 50 min, difference 10 ✔.)

📝 Practice MCQs (self-made)

P1. 72 km/h = ? m/s

(A) 18

(B) 20

(C) 25

(D) 30

Answer: (B) 20

P2. 30 m/s = ? km/h

(A) 90

(B) 100

(C) 108

(D) 120

Answer: (C) 108 — 30 × 18/5.

P3. In how many seconds will a 200 m long train at 36 km/h cross a 300 m long platform?

(A) 20

(B) 30

(C) 50

(D) 60

Answer: (C) 50 — 500 m / 10 m/s.

P4. The downstream speed of a boat is 16 km/h and the upstream speed is 8 km/h. Speed of the stream?

(A) 2

(B) 4

(C) 8

(D) 12

Answer: (B) 4 km/h — (16 − 8)/2.

P5. A car covers half the distance at 60 km/h and the other half at 120 km/h. Average speed?

(A) 90

(B) 80

(C) 85

(D) 75

Answer: (B) 80 km/h — 2 × 60 × 120/180.

📍 B11. Mensuration 2D & 3D (8 min read)

1. Concept

Area = how much surface is enclosed (unit²); Perimeter = length of the boundary (unit); Volume = how much fits inside (unit³). Length k times → area k² times, volume k³ times.

2D: Rectangle l × b, perimeter 2(l + b), diagonal √(l² + b²) | Square a², diagonal a√2, area = d²/2
Triangle ½ × base × height | Heron: √[s(s−a)(s−b)(s−c)], s = (a+b+c)/2 | Equilateral triangle (√3/4)a²
Parallelogram base × height | Trapezium ½(a + b) × h | Rhombus ½ d₁d₂
Circle: area πr², circumference 2πr | Semicircle perimeter πr + 2r
3D: Cube V = a³, TSA = 6a², diagonal a√3 | Cuboid V = lbh, TSA = 2(lb + bh + hl)
Cylinder V = πr²h, CSA = 2πrh, TSA = 2πr(r + h)
Cone V = ⅓πr²h, l = √(r² + h²), CSA = πrl
Sphere V = ⁴⁄₃πr³, SA = 4πr² | Hemisphere V = ⅔πr³, CSA 2πr², TSA 3πr²

Solved Example 1: Rectangle 12 × 5: area = 60, perimeter = 34, diagonal = √169 = 13.

Solved Example 2: The sides of a triangle are 13, 14, 15. Area?
s = 21 → √(21 × 8 × 7 × 6) = √7056 = 84.

Solved Example 3: Cylinder r = 7, h = 10 (π = 22/7): V = 22/7 × 49 × 10 = 1540, CSA = 2 × 22/7 × 7 × 10 = 440.
Sphere r = 7: SA = 4 × 22/7 × 49 = 616, V = 4/3 × 22/7 × 343 = 1437⅓.

⚠️ Exam Traps:

⚠️ Radius doubled → area 4 times, volume 8 times.
⚠️ Length increases by 20%, breadth decreases by 20% → area is 4% less (x²/100), not the same.
⚠️ If the diameter is given, first take r = d/2.

📝 Practice MCQs (self-made) — the real Pythagoras PYQ is in A6 (5 km South, 3 km West)

P1. The circumference of a circle is 44 cm. Area? (π = 22/7)

(A) 144 cm²

(B) 154 cm²

(C) 176 cm²

(D) 616 cm²

Answer: (B) 154 — r = 7.

P2. Cone r = 3, h = 4. Curved surface area (CSA)?

(A) 12π

(B) 15π

(C) 20π

(D) 24π

Answer: (B) 15π — l = 5, πrl = 15π.

P3. Volume of a 12 × 10 × 4 cuboid?

(A) 240

(B) 360

(C) 480

(D) 520

Answer: (C) 480

P4. The length of a rectangle is increased by 20% and the breadth decreased by 20%. Change in area?

(A) None

(B) 4% increase

(C) 4% decrease

(D) 2% decrease

Answer: (C) 4% decrease — 1.2 × 0.8 = 0.96.

P5. Total surface area (TSA) of a solid hemisphere of radius 7 cm?

(A) 308

(B) 462

(C) 616

(D) 924

Answer: (B) 462 cm² — 3 × 22/7 × 49.

📍 B12. Basic Algebra + BODMAS PYQs (6 min read)

1. Concept

Algebra = taking the unknown number as x and forming an equation from the conditions. If you do the same thing on both sides, the equality stays. Word problem → equation → solve → check by putting it into the options.

(a + b)² = a² + 2ab + b² | (a − b)² = a² − 2ab + b² | a² − b² = (a + b)(a − b)
(a + b)³ = a³ + b³ + 3ab(a + b) | a³ + b³ = (a + b)(a² − ab + b²) | a³ − b³ = (a − b)(a² + ab + b²)
a + b + c = 0 → a³ + b³ + c³ = 3abc
x + 1/x = k → x² + 1/x² = k² − 2 | x − 1/x = k → x² + 1/x² = k² + 2
Sum of two numbers S, difference D → larger = (S + D)/2, smaller = (S − D)/2

Solved Example 1: x + 1/x = 4 → x² + 1/x² = 16 − 2 = 14.

Solved Example 2: 101² = (100 + 1)² = 10201; 99 × 101 = 100² − 1 = 9999.

Solved Example 3: Sum of two numbers is 25, difference 5 → 15 and 10. 2x + 3 = 11 → x = 4.

🎯 Real PYQ (Simplification / BODMAS)

Q1. What is the number which is 5 times the difference of 9 and 4, with the product of 3 and 7 added to it? MPESB SI Prelims 2025 — 20 Jan 2026, Shift 2

(A) 36

(B) 50

(C) 48

(D) 46

Correct answer: (D) 46
5 × (9 − 4) + 3 × 7 = 25 + 21 = 46.

More algebra-type real PYQs: "difference of squares 15" in A1 (a² − b² = (a + b)(a − b) = 15 → 4, 1) and the picnic question in B7.

📝 Practice MCQs (self-made)

P1. If x − 1/x = 3, then x² + 1/x² = ?

(A) 7

(B) 9

(C) 11

(D) 13

Answer: (C) 11 — 9 + 2.

P2. If x/3 + 4 = 8, then x = ?

(A) 4

(B) 12

(C) 24

(D) 36

Answer: (B) 12

P3. If a + b = 7 and ab = 12, then a² + b² = ?

(A) 25

(B) 37

(C) 49

(D) 73

Answer: (A) 25 — 49 − 24.

P4. 47² − 43² = ?

(A) 320

(B) 340

(C) 360

(D) 400

Answer: (C) 360 — 90 × 4.

📍 B13. Statistics & Data Interpretation (Mean/Median/Mode, DI) (7 min read)

1. Concept

Mean = average; Median = the middle value when arranged in order; Mode = the value that occurs most often; Range = largest − smallest.

Median: n odd → (n + 1)/2 th term; n even → average of the n/2 and n/2 + 1 terms (first arrange in order!)
Empirical relation (only approximate): Mode ≈ 3 Median − 2 Mean
Pie chart: Angle = % × 3.6° (because 100% = 360°); % = angle/3.6
% increase = (new − old)/old × 100

Solved Example 1: Data 3, 7, 7, 2, 9, 5, 7 → in order: 2, 3, 5, 7, 7, 7, 9.
Mean = 40/7 ≈ 5.71, Median = 4th term = 7, Mode = 7, Range = 9 − 2 = 7.

Solved Example 2 (Pie): Food is 25% of a family income of ₹40,000. Food expense = ₹10,000, angle = 25 × 3.6 = 90°.

Solved Example 3 (Bar/Table): Sales over 4 years (thousand): 40, 50, 45, 60.
Increase from the first to the last year = 20/40 × 100 = 50%; average = 195/4 = 48.75.

⚠️ Exam Traps:

⚠️ Sorting the data is necessary before finding the median.
⚠️ In % increase, always divide by the old value.
⚠️ In DI, read the units (thousand/lakh) carefully.

📝 Practice MCQs (self-made)

P1. Mean of 10, 12, 15, 18, 20?

(A) 14

(B) 15

(C) 16

(D) 17

Answer: (B) 15 — 75/5.

P2. Median of 8, 3, 6, 9, 2?

(A) 3

(B) 5.6

(C) 6

(D) 8

Answer: (C) 6 — in order 2, 3, 6, 8, 9.

P3. Angle of a 15% share in a pie chart?

(A) 45°

(B) 54°

(C) 60°

(D) 36°

Answer: (B) 54° — 15 × 3.6.

P4. Mode of 2, 4, 4, 5, 4, 6, 2?

(A) 2

(B) 4

(C) 5

(D) 6

Answer: (B) 4 — 3 times.

📍 B14. Basic Probability (5 min read)

1. Concept

Probability = favourable outcomes ÷ total possible outcomes. The value is always between 0 and 1. Probability of "not happening" = 1 − probability of "happening".

P(E) = favourable / total | P(not E) = 1 − P(E)
1 coin: 2 outcomes; 2 coins: 4; 3 coins: 8 | 1 die: 6; 2 dice: 36
Cards: 52 cards, 4 suits × 13; red 26, black 26; face cards (J, Q, K) 12; aces 4

Solved Example 1: 2 coins tossed; at least one Head? Total HH, HT, TH, TT = 4; without a head only TT → 1 − 1/4 = 3/4.

Solved Example 2: 2 dice; sum 7? (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) = 6 → 6/36 = 1/6. Same number on both is also 6/36 = 1/6.

⚠️ Exam Trap: Probability can never be more than 1 — if an option has a value like 11/5 (greater than 1), it must be wrong.

🎯 Real PYQ (misprint in the options of the official PDF)

Q1. A box has 6 red, 5 green and 4 blue balls. If a person picks one ball at random, what is the probability that it is not blue? MPESB SI Prelims 2025 — 17 Jan 2026, Shift 1

(A) 1/5

(B) 11/5

(C) 1/3

(D) 13/15

Correct maths: 11/15 — total 15, not blue = 11 → 11/15. This option is not in the official PDF; option (B) "11/5" clearly looks like a misprint of 11/15 (11/5 > 1, it cannot be a probability at all). If this happens in the exam, choose the closest option / the one with the correct intent.

📝 Practice MCQs (self-made)

P1. Probability of getting an even number when a die is thrown?

(A) 1/3

(B) 1/2

(C) 2/3

(D) 1/6

Answer: (B) 1/2 — 2, 4, 6.

P2. Probability of drawing a King from a pack of 52 cards?

(A) 1/4

(B) 1/13

(C) 1/26

(D) 1/52

Answer: (B) 1/13 — 4/52.

P3. Probability of getting a sum of 10 when two dice are thrown?

(A) 1/6

(B) 1/9

(C) 1/12

(D) 5/36

Answer: (C) 1/12 — (4,6)(5,5)(6,4) = 3/36.

P4. A bag has 3 red and 5 black balls. Probability of a red ball?

(A) 3/5

(B) 3/8

(C) 5/8

(D) 1/3

Answer: (B) 3/8

📍 Last-Day Formula Sheet (10 min before the exam)

Analytical: Odd one out → prime/square/cube/multiple/unit digit check | Series → difference, difference of differences, ×/+ pattern, alphabet position (A=1…Z=26, EJOTY = 5,10,15,20,25) | DS → first I alone, then II alone, then both | Assumption → "is the statement pointless without it?" | Decision → most practical, as per rules, neither panic nor inaction
Number: 3/9 → sum of digits; 11 → difference of odd-even places; HCF × LCM = a × b
%: x% up, y% down → x − y − xy/100 | Same expenditure: (r/(100 + r)) × 100 less consumption
P-L: Profit% = Profit/CP × 100 | Discount on MP | Successive discount 20 + 10 = 28%
SI/CI: SI = PRT/100 | A = P(1 + R/100)ᵀ | CI − SI (2 years) = P(R/100)²
Ratio/Partnership: Profit ∝ Capital × Time | Mean proportional √(ac)
Average/Age: Total = average × n | Age difference always the same
Alligation: (Dearer − Mixture) : (Mixture − Cheaper)
Work: LCM method | M₁D₁ = M₂D₂ | Leak = minus
Speed: × 5/18 | Equal-distance average 2xy/(x + y) | Train (L + P)/S | Boat b = (D + U)/2, s = (D − U)/2
Mensuration: Circle πr², 2πr | Cylinder πr²h | Cone ⅓πr²h | Sphere ⁴⁄₃πr³, 4πr² | Hemisphere TSA 3πr²
Algebra: x + 1/x = k → k² − 2 | a² − b² = (a + b)(a − b)
Stats: Sort before median | Pie angle = % × 3.6
Probability: 0 ≤ P ≤ 1 | P(not) = 1 − P | 2 dice = 36

🔢 Analytical Ability + Arithmetic (Maths)

📊 0. Paper Analysis — What actually comes in Analytical Ability? (4 min read)

📍 A1. Odd One Out / Classification (10 min read)

1. Concept — How to think?

2. Letter–Number code (like B3C, 2B4)

3. Words / GK-concept odd one out

📍 A2. Number Series & Letter / Missing Code (Series) (10 min read)

1. Concept — How to crack a number series?

2. Letter series / Missing code

📍 A3. Data Sufficiency (8 min read)

1. Concept

📍 A4. Statement & Assumption (6 min read)

1. Concept

📍 A5. Decision Making / Situational Judgement (5 min read)

📍 A6. Puzzles, Ranking, Direction & Logical Maths (10 min read)

1. How to solve a puzzle (floor / day / month / box)

2. Ranking formulas

3. Direction — Pythagoras

📍 A7. Figure Analogy — Tips (4 min read)

📍 B1. Number System, Divisibility, Unit Digit, Remainder, HCF–LCM (10 min read)

1. Types of numbers

2. Divisibility Rules

3. Unit Digit — "why it works"

4. Remainder

5. HCF (Highest Common Factor) and LCM (Lowest Common Multiple)

📍 B2. Simplification (BODMAS), Squares-Cubes-Roots, Fractions-Decimals (8 min read)

1. BODMAS — the correct order

2. Squares 1–30 and Cubes 1–15 — memorise them

3. Fractions and Decimals

4. Sum formulas

📍 B3. Percentage (8 min read)

1. Concept

2. Important formulas

📍 B4. Profit, Loss & Discount (8 min read)

1. Concept

📍 B5. Simple & Compound Interest (SI & CI) (8 min read)

1. Concept

📍 B6. Ratio, Proportion, Unitary Method, Partnership (7 min read)

1. Concept

📍 B7. Average & Ages (8 min read)

1. Concept

📍 B8. Mixture & Alligation (5 min read)

1. Concept

📍 B9. Time & Work, Pipes & Cisterns (8 min read)

1. Concept — LCM (efficiency) method

📍 B10. Speed-Time-Distance, Trains, Boats & Streams (9 min read)

1. Concept

📍 B11. Mensuration 2D & 3D (8 min read)

1. Concept

📍 B12. Basic Algebra + BODMAS PYQs (6 min read)

1. Concept

📍 B13. Statistics & Data Interpretation (Mean/Median/Mode, DI) (7 min read)

1. Concept

📍 B14. Basic Probability (5 min read)

1. Concept

📍 Last-Day Formula Sheet (10 min before the exam)