Vice Principal & Principal 2026 — 5,847 explained questions, 18 chapters and 13 full papers for the DoE Delhi CRT.Start reading
Principal Sarathi VP & Principal 2026

Reasoning & Quantitative Aptitude

Syllabus head (iii) — Reasoning ability and Quantitative Aptitude | Chapter 12 of 18

50questions, every one explained
19items in the 2022 paper drew on this chapter

Why this chapter matters

The 2022 CRT paper for Vice Principal (Directorate of Education, GNCTD) carried 120 questions for 300 marks — 2.5 marks per question, with one-third negative marking. Around 20 questions (about 50 marks) came from this syllabus head. That is more weight than most single education-theory chapters in this book.

Two features of these 20 questions matter for your strategy:

Score arithmetic: getting 16 of these 20 right adds about 40 marks; careless errors on 6 of them can swing your total by over 20 marks once negative marking is counted. No other 20 questions in the paper are this controllable.

Three questions from this chapter, solved

Every one of the 50 items is written this way — the answer, and why each wrong option is wrong.

Republic Day (26 January) 2026 falls on a Monday. On which day will 26 January 2027 fall? (No 29 February lies in this period — 2026 is not a leap year.)

  1. (a) Monday
  2. (b) Wednesday
  3. (c) Sunday
  4. (d) Tuesday
Answer (d).

Tuesday — From 26-01-2026 to 26-01-2027 is 365 days (no 29 February in between): 365 ÷ 7 leaves 1 odd day. Monday + 1 = Tuesday. Trap: (b) Wednesday adds 2 odd days as if a leap day intervened. Answer grid for quick marking: | Q | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | 1–10 | b | c | b | c | d | c | b | a | a | a | | 11–20 | b | c | a | d | c | b | a | d | b | c | | 21–30 | a | d | d | b | c | a | c | d | a | c | | 31–40 | d | b | c | d | b | d | a | c | b | c | | 41–50 | a | a | c | c | a | c | d | b | c | d |

Directions for 36–37: Five students — J, K, L, M, N — sit in a row facing north. L sits exactly in the middle. K sits second to the left of L. N is an immediate neighbour of L but not of K. J sits somewhere to the right of M.

Who sits second to the right of M?

  1. (a) N
  2. (b) L
  3. (c) J
  4. (d) K
Answer (a).

N — Second to the right of M(2) is seat 4 = N. Trap: (b) L is the IMMEDIATE right of M — "second to the right" means two seats along. Set-up for 38–39. Seats 1–6 clockwise, all facing the centre; put U at 1. Opposite: X = 4. Facing the centre, left = clockwise, so V (second to U's left) = 3. W neighbours X (3 or 5); 3 is taken, so W = 5 (and 5 is not adjacent to 3 — consistent). Y neighbours V (2 or 4); 4 is taken, so Y = 2, leaving Z = 6. Clockwise: U, Y, V, X, W, Z.

The librarian walks 12 m west, then 5 m south, then 12 m east. Where is she relative to her starting point?

  1. (a) 5 m north
  2. (b) 5 m south
  3. (c) 29 m south
  4. (d) 12 m west
Answer (b).

5 m south — East–west: −12 + 12 = 0; north–south: 5 m south. She stands exactly 5 m south of the start. Trap: (c) 29 m is the distance WALKED (12 + 5 + 12), not the displacement. Set-up for 36–37. Seats 1–5 left to right, all facing north. L = 3 (middle). K is second to L's left: K = 1. N neighbours L (2 or 4) but not K, so N ≠ 2, N = 4. Remaining seats 2 and 5 for M and J with J to the right of M: M = 2, J = 5. Row: K, M, L, N, J.

THE COMPLETE METHOD TOOLKIT (TOPICS A TO Q)

Seventeen topics cover the entire question spread. The route map below shows what each topic trains and its typical share of the 20 exam questions:

Topic What it trains Typical share
A Percentages, pass/fail overlap, successive change 2–3 questions
B Averages: add/remove/replace/transfer 2–3 questions
C Weighted (combined) mean across sections 1 question
D Ratio, fund division, ages 1–2 questions
E Linear equations from word problems; profit–loss 1–2 questions
F Marks-rescaling (mean and SD under scaling) 0–1 question
G Number, letter and alphanumeric series 2 questions
H Coding–decoding 1–2 questions
I Blood relations 1 question
J Direction sense 1 question
K Seating arrangements (linear and circular) 1–2 questions
L Syllogisms 1–2 questions
M Coded inequalities 0–1 question
N Data sufficiency 0–1 question
O Permutation–combination, invigilator derangements 0–1 question
P Clocks and calendars 0–1 question
Q Data interpretation (tables) 1–2 questions

Work through them in order: A–F build the arithmetic engine, G–Q are the reasoning battery.

A. Percentages

A percentage is a fraction with denominator 100: "45% of the students" means 45 out of every 100. The fastest exam tool is converting between fractions and percentages both ways.

The reference table below must become memory. Blank line, then the table:

Fraction Percent Fraction Percent
1/2 50% 1/10 10%
1/3 33.33% 1/12 8.33%
1/4 25% 1/16 6.25%
1/5 20% 1/20 5%
1/6 16.67% 1/25 4%
1/8 12.5% 3/4 75%
2/5 40% 5/8 62.5%

Use it in reverse for speed: 12.5% of 640 is 640 × 1/8 = 80; 16.67% of 540 is 540 × 1/6 = 90.

METHOD — Percentage change and successive change

  1. Percentage change is always measured on the original (old) value: % change = (new − old) ÷ old × 100.
  2. For two successive changes of a% and b% (use minus signs for decreases): net % = a + b + (a × b)/100.
  3. Marks as percentage: % = marks obtained ÷ maximum marks × 100. Example: 342 out of 450 gives 342/450 × 100 = 76%.
  4. For pass/fail in two subjects, use Venn logic: % passing at least one = % passing subject 1 + % passing subject 2 − % passing both, and % failing both = 100 − % passing at least one.

Worked Example A1 (percentage change). A Directorate school's enrolment rose from 1,250 in 2024 to 1,450 in 2025. Find the percentage increase.

Solution. Increase = 1,450 − 1,250 = 200. Base is the old value 1,250. So % increase = 200/1250 × 100 = 16%. (Check with the table: 200/1250 = 4/25, and 1/25 = 4%, so 4/25 = 16%.)

Worked Example A2 (pass/fail with overlap). In a Class X exam, 70% of students passed Mathematics, 65% passed Science, and 55% passed both. If 20 students failed both subjects, how many students appeared?

Solution. Passed at least one subject = 70 + 65 − 55 = 80%. Failed both = 100 − 80 = 20%. So 20% of the total = 20 students, giving total = 20 ÷ 0.20 = 100 students.

Successive change in one line: a fee raised by 20% and later reduced by 10% changes by 20 − 10 + (20 × (−10))/100 = +8% net — never +10%.

Trap: The base of a percentage is whatever follows the word "of" or precedes "increased/decreased". Dividing the change by the new value is the most common wrong option built into these questions.

Speed drill A (do mentally; answers in brackets): 25% of 480 [120]; enrolment 640 → 800, per cent increase [25%]; 15% of ₹2,400 [₹360]; 30% of 60% of 1,500 [270 — multiply the fractions: 0.3 × 0.6 × 1500].

B. Averages

The average (mean) of n values is average = sum ÷ n. Every exam variation is bookkeeping on the two quantities sum and n.

METHOD — The five average moves

  1. Basic: sum = average × n. Always convert to sums first.
  2. Add a member: new average = (n × old average + new value) ÷ (n + 1).
  3. Remove a member: new average = (n × old average − removed value) ÷ (n − 1).
  4. Replace a member: the count stays n, so new member = old member + n × (change in average). A rise in average means the newcomer brought a surplus of n × shift.
  5. Transfer between two sections: the mover's value leaves one sum and joins the other, and both counts change. Recompute each section separately.
  6. First n natural numbers: average = (n + 1)/2. So roll numbers 1 to 20 average 21/2 = 10.5, and 1 to 25 average 13.

Worked Example B1 (adding a member). The average score of 24 students of X-A in a unit test is 56. A new student who scores 81 is admitted. Find the new class average.

Solution. Old sum = 24 × 56 = 1,344. New sum = 1,344 + 81 = 1,425. New count = 25. New average = 1425 ÷ 25 = 57. Shortcut: the newcomer is 25 above the old mean; spread over 25 students, the mean rises by exactly 1.

Worked Example B2 (replacement). When one student weighing 45 kg was replaced by a new student, the average weight of a class of 40 rose by 0.5 kg. Find the new student's weight.

Solution. Count is unchanged at 40, so the total rose by 40 × 0.5 = 20 kg. New student = 45 + 20 = 65 kg. No need to know the class average at all.

Worked Example B3 (removing a member). Thirty students of VII-A average 62 marks. The topper, who scored 91, is promoted to another section. Find the average of the remaining 29.

Solution. Old sum = 30 × 62 = 1,860. New sum = 1,860 − 91 = 1,769. New average = 1769 ÷ 29 = 61. Removing an above-average member pulls the average down; removing a below-average member pulls it up (see Practice 8).

Worked Example B4 (transfer between sections). Section A (30 students, average 60) and Section B (20 students, average 70). A student scoring 91 moves from B to A. Find both new averages.

The rest of this chapter

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