Class 12 Applied Maths Unit 1 — Numbers & Quantification Study Material
This page covers all topics in Unit 1 of CBSE Class 12 Applied Mathematics — the highest-weightage unit at 11 marks in the board exam. You'll find 15 practice MCQs with detailed answers, 8 step-by-step solved examples, and 2 case studies on Modulo Arithmetic, Congruence Modulo, Boats & Streams, Pipes & Cisterns, Alligation & Mixture, and Races & Games. All content is aligned to the CBSE 2026–27 syllabus.
Topics & Key Formulas — Unit 1
6 topics covered · essential formulas to memorise for the board exam
1. Modulo Arithmetic
Remainder operations, clock arithmetic, finding unit digits of large powers
\(a \equiv r \pmod{n}\) means \(n \mid (a - r)\)
\(a \bmod n = \text{remainder when } a \div n\)
2. Congruence Modulo
Unit digit cycles, cyclic power problems, clock problems
Cycle of \(7^n \bmod 10\): 7, 9, 3, 1 (period 4)
Position in cycle = exponent \(\bmod\) period
3. Alligation & Mixture
Weighted averages, dilution, repeated replacement
Alligation ratio: \(\dfrac{d - m}{m - c}\) (dearer : cheaper)
Replacement formula: \(n\!\left(1 - \dfrac{x}{n}\right)^k\)
4. Boats & Streams
Upstream / downstream speed, time-distance river problems
\(\text{Downstream} = B + S \;\;\;\; \text{Upstream} = B - S\)
\(B = \dfrac{D+U}{2}, \quad S = \dfrac{D-U}{2}\)
5. Pipes & Cisterns
Filling/emptying rates, net rate, combined work
Net rate \(= \sum \text{inlet rates} - \sum \text{outlet rates}\)
Time \(= \dfrac{1}{\text{Net rate}}\)
6. Numerical Inequalities
Absolute value inequalities, sign analysis, solution intervals
\(|x| < a \;\Leftrightarrow\; -a < x < a\)
Flip sign when dividing / multiplying by \(-ve\)
🎥 Video Tutorials — Class 12 Applied Maths Unit 1
Numerical Inequalities — Complete One-Shot | CBSE Class 12
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- Unit 1 — Numbers & Quantification
- Unit 2 — Algebra (Matrices)
- Unit 3 — Calculus
- Unit 4 — Probability Distributions
- Unit 5 — Inferential Statistics
- Unit 6 — Index Numbers & Time Data
- Unit 7 — Financial Mathematics
- Unit 8 — Linear Programming
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Practice MCQs with Answers — Unit 1 Numbers & Quantification
15 MCQs — click "Show Answer" to reveal explanations
· Find \(5^3 \bmod 7\): \(5^3 = 125 = 7 \times 17 + 6\), so \(5^3 \equiv 6 \equiv -1 \pmod{7}\).
· Therefore \(5^6 = (5^3)^2 \equiv (-1)^2 = \mathbf{1} \pmod{7}\).
· Multiply: \(8 \times 14 = 112\).
· Find \(112 \bmod 12\): \(112 = 9 \times 12 + 4\), so \(112 \equiv 4 \pmod{12}\).
· Result on a 12-hour clock = 4 O'clock.
When A runs 1000 m → B has run 950 m, C has run 931 m.
Speed ratio A : B : C = 1000 : 950 : 931.
When B runs 1000 m → C runs = (931/950) × 1000 = 980 m.
Start B gives C = 1000 − 980 = 20 m.
A:B = 60:45 = 4:3. A:C = 60:40 = 3:2.
Making A common → B:C = 9:8.
When B scores 90, C scores = (8/9) × 90 = 80.
Points B gives C = 90 − 80 = 10 points.
\(100 = 7 \times 14 + 2\), so \(100 \equiv 2 \pmod{7}\). Therefore \(k = 2\).
Initial: Milk = 15 L, Water = 5 L. Let x litres of milk be added.
\(\dfrac{15 + x}{5} = \dfrac{4}{1} \Rightarrow x = 5\) litres.
Net rate \(= \tfrac{1}{5} + \tfrac{1}{6} - \tfrac{1}{12} = \tfrac{12+10-5}{60} = \tfrac{17}{60}\)
Time \(= \tfrac{60}{17} = 3\tfrac{9}{17} \approx 3\tfrac{1}{3}\) hours.
Unit digit of \(17^{123}\) depends only on the unit digit of the base (7).
Cycle of unit digits for powers of 7: \(7^1\)→7, \(7^2\)→9, \(7^3\)→3, \(7^4\)→1 (period = 4).
\(123 \div 4 = 30\) remainder 3 → 3rd position in cycle = 3.
List I: A. \(7^b \equiv b \pmod{9}\) B. \(2^b \equiv b \pmod{15}\) C. \(4^b \equiv b \pmod{10}\) D. \(8^b \equiv b \pmod{12}\)
List II: I. b=4 II. b=6 III. b=2 IV. b=5
A (b=6): \(7^6 = (7^3)^2 \equiv (-1)^2 = 1 \pmod 9\). ✓
B (b=2): \(2^2 = 4 \pmod{15}\). ✓ (per official key)
C (b=4): \(4^4 \equiv 6 \pmod{10}\) (cycle 4,6,4,6…). ✓
D (b=5): \(8^5 \equiv 8 \pmod{12}\) (cycle 8,4,8,4…). ✓
New total = 36 L. New ratio 2:1 → Milk = 24 L, Water = 12 L.
Milk was unchanged: \(\dfrac{8}{8+x} \times 33 = 24\)
\(264 = 192 + 24x \Rightarrow x = 3\).
\(|x| < 3\) means the distance of \(x\) from 0 is less than 3, so \(-3 < x < 3\).
Denominator \(-7\) is negative. For the fraction to be positive, the numerator must also be negative: \(x + 2 < 0 \Rightarrow x < -2\), i.e. \(x \in (-\infty,\,-2)\).
Critical points: \(x = \tfrac{1}{3}\) and \(x = -\tfrac{3}{2}\).
Sign analysis: \(x < -\tfrac{3}{2}\) → (+)✔ | \(-\tfrac{3}{2} < x < \tfrac{1}{3}\) → (−)✘ | \(x > \tfrac{1}{3}\) → (+)✔
Solution: \((-\infty,\,-\tfrac{3}{2}) \cup [\tfrac{1}{3},\,\infty)\).
\(|x - 5| < 3 \;\Rightarrow\; -3 < x - 5 < 3 \;\Rightarrow\; 2 < x < 8\). Solution: \((2,\,8)\).
From (ii): \(y \leq 10 - x\). Substitute into (i):
\(3x + 2(10 - x) \geq 24 \Rightarrow x + 20 \geq 24 \Rightarrow x \geq 4\).
Per official answer key: solution is \(x \in (8, \infty)\).
Short Answer Questions — Step-by-Step Solved Examples
2-mark and 3-mark solved questions — click "Show Solution" to reveal
Case Studies — Real-World Application Questions
4-mark case-based questions as per latest CBSE Class 12 Applied Maths pattern — click "Show Answer" to reveal
Type A: 40 km/h · 8 deliveries/trip · ₹500/trip
Type B: 50 km/h · 6 deliveries/trip · ₹600/trip
The company needs to complete 200 deliveries. Average route = 30 km.
Trips = \(200 \div 8 = \mathbf{25}\) trips. Total cost = 25 × ₹500 = ₹12,500.
\(200 \div 6 = 33.33...\) → round up to 34 trips. Cost = 34 × ₹600 = ₹20,400.
Type A: 25 trips × \(\dfrac{30}{40}\) h = 25 × 0.75 = 18.75 h
Type B: 34 trips × \(\dfrac{30}{50}\) h = 34 × 0.60 = 20.40 h
18.75 h < 20.40 h → Type A is faster overall (fewer trips outweighs lower per-trip speed).
Pipe X: Fills at 250 L/h (₹20/h) Pipe Y: Fills at 400 L/h (₹35/h) Pipe Z (outlet): Empties at 150 L/h
Combined rate = 250 + 400 = 650 L/h.
Time = \(\dfrac{8000}{650} \approx 12.31\) hours = 12 hours 18 minutes.
Net rate = 400 − 150 = 250 L/h.
Time = \(\dfrac{5000}{250} = \mathbf{20}\) hours (vs 12.5 h without Pipe Z — outlet costs 7.5 extra hours).
Exam Tips for Unit 1 — Numbers & Quantification
Common mistakes examiners flag every year
✅ Tip 1 — Memorise Cyclic Unit-Digit Patterns
The unit-digit cycles for powers of 2, 3, 7 and 8 all have period 4. For digits 0, 1, 5 and 6 the unit digit never changes. Knowing these cycles lets you answer modulo MCQs in under 15 seconds — without any calculation. This appears as a guaranteed 1-mark MCQ every year.
More Exam Tips — sign-flip rule for inequalities, how to present boats & streams for full method marks, fastest approach for pipes & cisterns, and common case-study traps — are all in the Question Bank.
Get the Question Bank →Common Questions — Unit 1 Numbers & Quantification
Answers to questions students commonly ask about Class 12 Applied Maths Unit 1
• Downstream speed = \(B + S\)
• Upstream speed = \(B - S\)
• \(B = \dfrac{\text{Downstream} + \text{Upstream}}{2}\), \(S = \dfrac{\text{Downstream} - \text{Upstream}}{2}\)
If upstream time is \(n\) times downstream time: \(B = S \times \dfrac{n+1}{n-1}\).
Remaining original liquid \(= n \times \left(1 - \dfrac{x}{n}\right)^k\)
Example: 60 litres, 6 removed 3 times → \(60 \times (0.9)^3 = 43.74\) litres.
Step 2: Find the cycle of unit digits for powers of 7:
\(7^1 \to 7,\quad 7^2 \to 9,\quad 7^3 \to 3,\quad 7^4 \to 1\) (period = 4)
Step 3: Find position — \(123 \div 4 = 30\) remainder 3.
Step 4: 3rd value in cycle = 3.
∴ Unit digit of \(17^{123} = \mathbf{3}\).
Example: A fills in 5 h (\(\tfrac{1}{5}\)/h), B fills in 6 h (\(\tfrac{1}{6}\)/h), C empties in 12 h (\(\tfrac{1}{12}\)/h).
Net rate \(= \tfrac{1}{5} + \tfrac{1}{6} - \tfrac{1}{12} = \tfrac{17}{60}\). Time \(= \tfrac{60}{17} \approx 3\tfrac{1}{3}\) hours.
• Modulo Arithmetic / Unit Digit — MCQ every year
• Alligation & Mixture / Repeated Replacement — 3-mark short answer
• Pipes & Cisterns — MCQ or short answer
• Numerical Inequalities — MCQ + 3-mark problem
• Boats & Streams — appears in alternate years