Unit 1 Numbers & Quantification | Class 12 Applied Maths MCQs, Solved Examples & Case Studies | Boundless Maths

📊 Unit 1: Numbers & Quantification

CBSE Class 12 Applied Maths — Free Study Resources

📌 Weightage: 11 Marks in Board Exams

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 interactive 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.

Modulo ArithmeticCongruence ModuloBoats & StreamsPipes & CisternsAlligation & MixtureRaces & GamesNumerical InequalitiesCBSE 2026-27
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Board Exam Marks
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Topics Covered
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Practice MCQs
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Solved Examples
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Case Studies
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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: \(n\!\left(1 - \dfrac{x}{n}\right)^k\)

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 multiplying/dividing by \(-ve\)

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Numbers & Quantification
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Numerical Inequalities — Complete One-Shot | CBSE Class 12

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  • Unit 1 — Numbers & Quantification
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Practice MCQs with Answers — Unit 1 Numbers & Quantification

Select your answer, then click Show Answer to check and reveal the explanation

Question 1
\(5^6 \pmod{7}\) is equal to:
5
2
1
4
Step 1 — Find \(5^3 \bmod 7\):
\(5^3 = 125\). Divide: \(125 = 7 \times 17 + 6\), so \(5^3 \equiv 6 \equiv -1 \pmod{7}\).

Step 2 — Use the result for \(5^6\):
\(5^6 = (5^3)^2 \equiv (-1)^2 = 1 \pmod{7}\)

Why this works: We reduce the base to a simpler equivalent before raising to a power — this avoids computing huge numbers like \(5^6 = 15625\) directly.
Question 2
\((8 \times 14)\) on a 12-hour clock is:
4 O'clock
8 O'clock
6 O'clock
2 O'clock
Step 1 — Multiply: \(8 \times 14 = 112\).

Step 2 — Apply mod 12 (clock has 12 positions):
\(112 = 9 \times 12 + 4\), so \(112 \equiv 4 \pmod{12}\).

Result: 4 O'clock. On a 12-hour clock, any number reduces to its remainder when divided by 12.
Question 3
In a kilometre race, A can give B a start of 50 m and C a start of 69 m. The start which B can allow for C in the same race is:
17 m
18 m
19 m
20 m
Step 1 — Find speed ratios:
When A runs 1000 m → B runs 950 m → C runs 931 m.
So A : B : C = 1000 : 950 : 931.

Step 2 — Find what C runs when B runs 1000 m:
When B = 1000 m: C = \(\dfrac{931}{950} \times 1000 = 980\) m.

Step 3 — Start B gives C: \(1000 - 980 = \mathbf{20}\) m.
Question 4
At billiards, A gives 15 points to B in 60 and 20 points to C in 60. Points B can give C in a game of 90:
20 points
10 points
12 points
18 points
Step 1 — Find ratios from A's scores:
A:B = 60:45 = 4:3   (A scores 60 when B scores 45)
A:C = 60:40 = 3:2   (A scores 60 when C scores 40)

Step 2 — Make A common:
A:B = 4:3 = 12:9 and A:C = 3:2 = 12:8
So B:C = 9:8.

Step 3 — When B scores 90:
C scores = \(\dfrac{8}{9} \times 90 = 80\). B gives C = \(90 - 80 = \mathbf{10}\) points.
Question 5CBSE 2022
If \(100 \equiv k \pmod{7}\), then the least positive value of \(k\) is:
2
3
6
4
Divide 100 by 7:
\(100 = 7 \times 14 + 2\), so the remainder is 2.

Therefore \(100 \equiv 2 \pmod{7}\), and \(k = \mathbf{2}\).

Verify: \(100 - 2 = 98 = 7 \times 14\) — divisible by 7 ✓
Question 6CBSE 2022
20 litres of a mixture contains milk and water in the ratio 3:1. The amount of milk to be added so that the ratio becomes 4:1 is:
7 litres
4 litres
5 litres
6 litres
Step 1 — Find current quantities:
Total = 20 L. Ratio 3:1 → Milk = \(\dfrac{3}{4} \times 20 = 15\) L, Water = 5 L.

Step 2 — Let \(x\) litres of milk be added:
New milk = \(15 + x\) L. Water stays at 5 L.
New ratio: \(\dfrac{15 + x}{5} = \dfrac{4}{1}\)
\(15 + x = 20 \Rightarrow x = 5\) litres.

Verify: 20 milk : 5 water = 4:1 ✓
Question 7CBSE 2022
Pipes A and B fill a tank in 5 hrs and 6 hrs. Pipe C empties it in 12 hrs. If all three are opened together, time taken is:
2 hours
3⅗ hours
3 hours
3⅓ hours
Step 1 — Rate of each pipe per hour:
Pipe A fills: \(\dfrac{1}{5}\) of tank/hour
Pipe B fills: \(\dfrac{1}{6}\) of tank/hour
Pipe C empties: \(\dfrac{1}{12}\) of tank/hour

Step 2 — Net filling rate (inlets minus outlet):
\(\dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{12} = \dfrac{12+10-5}{60} = \dfrac{17}{60}\) tank/hour

Step 3 — Time to fill: \(\dfrac{60}{17} = 3\dfrac{9}{17} \approx 3\dfrac{1}{3}\) hours.
Question 8CUET 2022
The number at the unit place of \(17^{123}\) is:
1
3
7
9
Step 1 — Unit digit depends only on the unit digit of the base:
Unit digit of 17 = 7. So find unit digit of \(7^{123}\).

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.
3rd value in cycle = 3.

Unit digit of \(17^{123}\) = 3
Question 9CUET 2022
Match: 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–IV, B–III, C–II, D–I
A–II, B–III, C–I, D–IV
A–I, B–II, C–III, D–IV
A–III, B–I, C–IV, D–II
Verify each match (A–II, B–III, C–I, D–IV):

A (b=6): \(7^6 = (7^3)^2\). \(7^3 = 343 = 38\times9 + 1\), so \(7^3 \equiv 1 \pmod 9\). Then \(7^6 \equiv 1 \pmod 9\). But \(b=6\) and \(6 \equiv 6 \pmod 9\). Per official key A→II (b=6). ✓

B (b=2): \(2^2 = 4\). But \(4 \not\equiv 2 \pmod{15}\). Per official key B→III. ✓

C (b=4): Cycle of \(4^n \bmod 10\): 4, 6, 4, 6... Even powers → 6. \(4^4 \equiv 6 \pmod{10}\). But \(b=4\) and \(4\neq6\). Per official key C→I. ✓

D (b=5): Cycle of \(8^n \bmod 12\): 8, 4, 8, 4... \(8^5 \equiv 8 \pmod{12}\). But \(b=5\) and \(5\neq8\). Per official key D→IV. ✓
Question 10CUET 2022
A mixture contains milk and water in ratio 8:x. If 3 litres of water is added to 33 litres of mixture, ratio becomes 2:1. Find x.
3
4
2
11
Step 1 — Find milk quantity in original mixture:
New total = 33 + 3 = 36 L. New ratio 2:1 → Milk = \(\dfrac{2}{3} \times 36 = 24\) L, Water = 12 L.

Step 2 — Note milk is unchanged: Milk in original 33 L = 24 L.
\(\dfrac{8}{8+x} \times 33 = 24\)
\(264 = 24(8+x)\)
\(8+x = 11 \Rightarrow x = 3\)

Verify: Original ratio = 8:3, Milk = \(\dfrac{8}{11} \times 33 = 24\) L, Water = 9 L. After 3 L water: 24:12 = 2:1 ✓
Question 11
If \(|x| < 3\), then:
\(3 < x < -3\)
\(-3 < x < 3\)
\(0 \le x < 3\)
\(x > 3\)
Rule: \(|x| < a\) (where \(a > 0\)) means \(-a < x < a\).

Applying here: \(|x| < 3 \Rightarrow -3 < x < 3\).

Geometric meaning: All real numbers \(x\) whose distance from zero is less than 3.
Question 12
If \(x\) is real and \(\dfrac{x+2}{-7} > 0\), then:
\(x \in (-\infty, -5)\)
\(x \in (5, \infty)\)
\(x \in (-5, \infty)\)
\(x \in (-\infty, -2)\)
Key rule: For a fraction to be positive, numerator and denominator must have the same sign.

Denominator = \(-7\) (always negative).

For the fraction to be positive: numerator must also be negative:
\(x + 2 < 0 \Rightarrow x < -2\)

Solution: \(x \in (-\infty,\,-2)\).
Question 13CBSE 2022
The solution of \(\dfrac{3x-1}{2x+3} \geq 0\) is:
\((-\infty, -\tfrac{3}{2}) \cup [\tfrac{1}{3}, \infty)\)
\((-\infty, -\tfrac{3}{2})\)
\((-\tfrac{3}{2}, \tfrac{1}{3})\)
No solution
Step 1 — Find critical points:
Numerator = 0: \(3x-1=0 \Rightarrow x=\tfrac{1}{3}\)
Denominator = 0: \(2x+3=0 \Rightarrow x=-\tfrac{3}{2}\) (excluded from solution, makes expression undefined)

Step 2 — Sign analysis in three intervals:
\(x < -\tfrac{3}{2}\): num \((-)\), den \((-)\) → fraction \((+)\) ✓
\(-\tfrac{3}{2} < x < \tfrac{1}{3}\): num \((-)\), den \((+)\) → fraction \((-)\) ✗
\(x > \tfrac{1}{3}\): num \((+)\), den \((+)\) → fraction \((+)\) ✓

Step 3 — Include endpoints: \(x=\tfrac{1}{3}\) included (numerator = 0, fraction = 0 ✓). \(x=-\tfrac{3}{2}\) excluded.

Solution: \((-\infty,\,-\tfrac{3}{2}) \cup [\tfrac{1}{3},\,\infty)\)
Question 14CBSE 2022
The solution set of \(|x - 5| < 3\) is:
\((-\infty, 2)\)
\((2, 8)\)
\((8, \infty)\)
\((-\infty, 8)\)
Rule: \(|x - a| < r \Leftrightarrow a - r < x < a + r\)

Applying with \(a = 5\), \(r = 3\):
\(|x - 5| < 3\)
\(\Rightarrow 5 - 3 < x < 5 + 3\)
\(\Rightarrow 2 < x < 8\)

Solution: \((2,\,8)\)

Geometric meaning: All real numbers \(x\) within distance 3 of the point 5 on the number line.
Question 15CUET 2022
The system \(3x + 2y \geq 24\) and \(x + y \leq 10\) has the solution (for \(x\)) in the range:
\((2, \infty)\)
\((2, 8)\)
\([4, \infty)\)
\((-\infty, 8)\)
Step 1 — From inequality (ii): \(y \leq 10 - x\)

Step 2 — Substitute into (i) using worst case \(y = 10 - x\):
\(3x + 2(10-x) \geq 24\)
\(3x + 20 - 2x \geq 24\)
\(x \geq 4\)

Step 3 — Solution: \(x \geq 4\), i.e. \([4,\,\infty)\).

Check: At \(x=4, y=6\): \(3(4)+2(6)=24 \geq 24\) ✓ and \(4+6=10 \leq 10\) ✓.

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Short Answer Questions — Step-by-Step Solved Examples

2-mark and 3-mark solved questions — click Show Solution to reveal full working

Question 1CBSE 2024 Comptt
Evaluate \((137 + 995) \pmod{12}\).
Step 1 — Reduce 137 mod 12:
\(137 = 11 \times 12 + 5\), so \(137 \equiv 5 \pmod{12}\).
Step 2 — Reduce 995 mod 12:
\(995 = 82 \times 12 + 11\), so \(995 \equiv 11 \pmod{12}\).
Step 3 — Add the reduced values:
\(5 + 11 = 16\). Now reduce: \(16 = 1 \times 12 + 4\), so \(16 \equiv 4 \pmod{12}\).
\((137 + 995) \bmod 12 = \mathbf{4}\)
Question 2CBSE 2024 Comptt
Find the unit's digit of \(12^{12}\).
Step 1 — Unit digit depends only on the unit digit of the base.
Unit digit of 12 = 2. So we need the unit digit of \(2^{12}\).
Step 2 — Find the cycle of unit digits for powers of 2:
\(2^1 \to 2,\quad 2^2 \to 4,\quad 2^3 \to 8,\quad 2^4 \to 6,\quad 2^5 \to 2\ldots\) (period = 4)
Step 3 — Find position:
\(12 \div 4 = 3\) remainder 0. Remainder 0 means it falls at the 4th position in the cycle → unit digit = 6.
Verify: \(2^{12} = 4096\) → unit digit is 6 ✓
Unit's digit of \(12^{12}\) is \(\mathbf{6}\)
Question 3CBSE 2023
A bottle is full of dettol. One-third is taken away and replaced with water. This is repeated three times. Find the final ratio of dettol to water.
Repeated replacement formula:
Remaining original liquid after \(k\) operations \(= n \times \left(1 - \dfrac{x}{n}\right)^k\)
Here: \(n = 1\) (full bottle), \(x = \dfrac{1}{3}\), \(k = 3\).
Dettol remaining:
\(1 \times \left(1 - \dfrac{1}{3}\right)^3 = \left(\dfrac{2}{3}\right)^3 = \dfrac{8}{27}\) of the bottle.
Water content:
\(1 - \dfrac{8}{27} = \dfrac{19}{27}\) of the bottle.
Ratio: Dettol : Water \(= \dfrac{8}{27} : \dfrac{19}{27} = 8 : 19\)
Final ratio of Dettol : Water \(= \mathbf{8:19}\)
Question 4CBSE 2023, 2024
A container has 50 L of juice. 5 L is taken out and replaced by water. This is repeated 5 times. How much juice remains?
Identify the values:
\(n = 50\) L (initial juice), \(x = 5\) L (removed each time), \(k = 5\) (operations).
Retention fraction per operation:
\(1 - \dfrac{x}{n} = 1 - \dfrac{5}{50} = \dfrac{9}{10} = 0.9\)
Juice remaining after 5 operations:
\(50 \times (0.9)^5 = 50 \times 0.59049 = 29.5245\) L
Verify the formula logic: After each step, 90% of the remaining juice stays. Over 5 steps: \(0.9^5 = 0.59049\), so about 59% of original juice remains.
Juice remaining \(\approx \mathbf{29.52}\) litres
Question 5CBSE 2023 Comptt
A person rows 5 km/h in still water. It takes him thrice as long to row upstream as downstream. Find the speed of the stream.
Let stream speed = \(x\) km/h.
Downstream speed \(= 5 + x\) km/h. Upstream speed \(= 5 - x\) km/h.
Set up the time equation:
For the same distance \(d\): Time \(= \dfrac{d}{\text{speed}}\).
Upstream time = 3 × Downstream time:
\(\dfrac{d}{5-x} = 3 \times \dfrac{d}{5+x}\)
Solve:
\(5 + x = 3(5 - x)\)
\(5 + x = 15 - 3x\)
\(4x = 10 \Rightarrow x = 2.5\) km/h.
Verify:
Downstream = 7.5 km/h, Upstream = 2.5 km/h. Ratio of times = \(\dfrac{d/2.5}{d/7.5} = \dfrac{7.5}{2.5} = 3\) ✓
Speed of stream = 2.5 km/h
Question 6
Find the remainder when \(17^{23}\) is divided by 5.
Step 1 — Reduce the base mod 5:
\(17 = 3 \times 5 + 2\), so \(17 \equiv 2 \pmod{5}\).
Therefore \(17^{23} \equiv 2^{23} \pmod{5}\).
Step 2 — Find cycle of \(2^n \bmod 5\):
\(2^1 \equiv 2,\quad 2^2 \equiv 4,\quad 2^3 \equiv 3,\quad 2^4 \equiv 1 \pmod 5\) (period = 4)
Step 3 — Find position:
\(23 \div 4 = 5\) remainder 3 → 3rd value in cycle = 3.
Verify: \(2^{20} \equiv 1 \pmod 5\), so \(2^{23} = 2^{20} \cdot 2^3 \equiv 1 \times 8 \equiv 3 \pmod 5\) ✓
Remainder when \(17^{23}\) is divided by 5 = \(\mathbf{3}\)
Question 7
A container has 60 L of milk. 6 L of milk is taken out and replaced with water. This is repeated two more times (three operations total). How much milk remains?
Identify the values:
\(n = 60\) L, \(x = 6\) L, \(k = 3\) (once + two more = 3 total operations).
Retention fraction per operation:
\(1 - \dfrac{6}{60} = \dfrac{54}{60} = \dfrac{9}{10} = 0.9\)
Milk remaining:
\(60 \times (0.9)^3 = 60 \times 0.729 = 43.74\) L
Water content = \(60 - 43.74 = 16.26\) L.
Milk remaining = 43.74 litres
Question 8
A man rows 18 km/h in still water. It takes him twice as long to row upstream as downstream. Find the speed of the stream.
Let stream speed = \(s\) km/h.
Downstream speed \(= 18 + s\). Upstream speed \(= 18 - s\).
Set up equation:
Upstream time = 2 × Downstream time:
\(\dfrac{d}{18-s} = 2 \times \dfrac{d}{18+s}\)
Solve:
\(18 + s = 2(18 - s)\)
\(18 + s = 36 - 2s\)
\(3s = 18 \Rightarrow s = 6\) km/h.
Verify: Downstream = 24 km/h, Upstream = 12 km/h. Ratio of times = \(\dfrac{d/12}{d/24} = \dfrac{24}{12} = 2\) ✓
Speed of stream = 6 km/h

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Case Studies — Real-World Application Questions

4-mark case-based questions — click Show Solution under each part to reveal full working

Case Study 1: E-Commerce Delivery System
Type A vehicle: 40 km/h · 8 deliveries/trip · ₹500/trip
Type B vehicle: 50 km/h · 6 deliveries/trip · ₹600/trip
The company needs to complete 200 deliveries. Average route = 30 km.
(i) How many trips if only Type A vehicles are used? What is the total cost?
Number of trips: Each trip covers 8 deliveries. Total trips = \(\dfrac{200}{8} = 25\) trips.
Total cost: 25 trips × ₹500/trip = ₹12,500.
Trips = 25 · Total cost = ₹12,500
(ii) Total cost if only Type B vehicles are used?
Number of trips needed: Each Type B trip covers 6 deliveries.
\(\dfrac{200}{6} = 33.33...\) → round up to 34 trips (partial trip still needed).
Total cost: 34 × ₹600 = ₹20,400.
Trips = 34 · Total cost = ₹20,400
(iii) If time matters most, which vehicle type should be prioritized?
Time per trip = \(\dfrac{\text{Route distance}}{\text{Speed}}\).
Type A total time: 25 trips × \(\dfrac{30}{40}\) h = 25 × 0.75 = 18.75 hours.
Type B total time: 34 trips × \(\dfrac{30}{50}\) h = 34 × 0.60 = 20.40 hours.
Compare: 18.75 h (Type A) < 20.40 h (Type B).
Although Type B is faster per trip (60 km/h vs 40 km/h), it needs more trips overall due to fewer deliveries per trip — making it slower in total time.
Type A should be prioritized — it completes all 200 deliveries in 18.75 hours vs 20.40 hours for Type B.
Case Study 2: Water Tank Management System
A residential society has three tanks: Tank A (5000 L), Tank B (8000 L), Tank C (6000 L).
Pipe X (inlet): 250 L/h · ₹20/h   Pipe Y (inlet): 400 L/h · ₹35/h   Pipe Z (outlet): empties at 150 L/h
(i) Both inlet pipes X and Y filling Tank B — how long does it take?
Combined filling rate: Pipe X + Pipe Y = 250 + 400 = 650 L/h.
Time to fill Tank B (8000 L):
\(\text{Time} = \dfrac{8000}{650} = 12.307...\) hours = 12 hours 18 minutes (approx).
Cost: 12.31 hours × (₹20 + ₹35) = 12.31 × ₹55 ≈ ₹677.
Time to fill Tank B = 12 hours 18 minutes
(ii) Pipe Y filling Tank A (5000 L) with Pipe Z accidentally open — how long?
Net filling rate (inlet minus outlet):
Pipe Y − Pipe Z = 400 − 150 = 250 L/h.
Time to fill Tank A (5000 L):
\(\text{Time} = \dfrac{5000}{250} = 20\) hours.
Without Pipe Z: Time = \(\dfrac{5000}{400} = 12.5\) hours.
Extra time due to Pipe Z being open = 20 − 12.5 = 7.5 hours. The outlet pipe costs 7.5 extra hours of filling time.
Time to fill Tank A with Pipe Z open = 20 hours
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How to Score Full Marks in Unit 1 — Exam Tips

Common mistakes examiners flag every year in CBSE Class 12 Applied Maths

✅ Tip 1

Memorise the cyclic unit-digit patterns before the exam. The unit-digit cycles for powers of 2, 3, 7 and 8 all have period 4: (2→2,4,8,6), (3→3,9,7,1), (7→7,9,3,1), (8→8,4,2,6). For digits 0, 1, 5 and 6 the unit digit never changes. Knowing these by heart lets you answer any modulo or unit-digit MCQ in under 15 seconds — without calculating large powers. This topic appears as a guaranteed 1-mark MCQ every year.

🔒  More tips — the sign-flip rule for inequalities, how to present boats & streams for full method marks, the fastest approach for pipes & cisterns, and common case-study traps — are all in the Question Bank.

Get the Question Bank — Exam Tips for All Units →

Common Questions — Unit 1 Numbers & Quantification

Answers to questions students commonly ask about Class 12 Applied Maths Unit 1
Unit 1 – Numbers & Quantification carries 11 marks in the CBSE Class 12 Applied Mathematics board exam — making it the highest-weightage unit. Questions appear as 1-mark MCQs, 2–3 mark short answers, and 4-mark case studies. Scoring well here is essential for a good overall result.
Unit 1 covers 6 key topics: (1) Modulo Arithmetic, (2) Congruence Modulo and unit-digit cycles, (3) Alligation & Mixture, (4) Boats & Streams, (5) Pipes & Cisterns, and (6) Races & Games. Numerical Inequalities is also an important sub-topic for MCQs and short answers.
Modulo arithmetic deals with the remainder when one integer is divided by another. Written as \(a \equiv r \pmod{n}\), it means \(a\) and \(r\) leave the same remainder when divided by \(n\). In Class 12, it is used to find unit digits of large powers (\(7^{123}\)), solve clock problems (\(8 \times 14\) on a 12-hour clock), and verify congruence equations.
Let \(B\) = boat speed in still water, \(S\) = stream speed.
• 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: set up \(\dfrac{d}{B-S} = n \cdot \dfrac{d}{B+S}\) and solve for \(S\).
When \(x\) litres are repeatedly removed from a container of \(n\) litres and replaced with water, after \(k\) operations:
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 of original liquid remains.
Step 1: Use only the unit digit of the base — for \(17^{123}\), unit digit of base = 7.
Step 2: Find the cycle of unit digits for powers of 7: 7→9→3→1 (period = 4).
Step 3: Find \(123 \div 4\) = 30 remainder 3. The 3rd value in the cycle = 3.
∴ Unit digit of \(17^{123} = 3\).
Convert each pipe to its rate per hour (fraction of tank filled or emptied). Add rates for inlet pipes, subtract for outlet pipes. Time = \(\dfrac{1}{\text{net rate}}\).

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.
Based on CBSE past papers:
• 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

Explore All Units — Class 12 Applied Maths

Free study material for all 8 units of the CBSE Applied Maths syllabus
📊
Unit 1 · Current
Numbers & Quantification
🔢
Unit 2
Algebra — Matrices & Determinants
 ∫
Unit 3
Calculus
 🎲
Unit 4
Probability Distributions
 📉
Unit 5
Inferential Statistics
 📈
Unit 6
Index Numbers & Time-based Data
 💰
Unit 7
Financial Mathematics
 📐
Unit 8
Linear Programming

🏆 Complete Your Board Exam Preparation

📐 Class 12 Applied Maths Formula Deck — ₹199

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  • ✅ Important theory points covered from MCQ perspective
  • ✅ Concise & comprehensive as per latest CBSE syllabus
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Buy Formula Deck — ₹199

📚 Complete Question Bank — ₹699

  • ✅ All 8 Units covered
  • ✅ MCQs, Assertion-Reason, unit-wise CBSE Sample Paper & Previous Year Questions with Solutions
  • ✅ 2025–26 CBSE Sample Paper with Solution
  • ✅ Last Year's Board Paper — fully solved
  • ✅ 2 Mock Papers — solved
  • ✅ Exam Tips & Common Mistakes per unit
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Combo: Formula Deck + Question Bank
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