Unit 1: Numbers, Quantification & Numerical Applications — Class 11 Applied Maths Free Study Resources | Boundless Maths

Class 11 Applied Maths · Unit 1 · 9 Marks

Numbers, Quantification &
Numerical Applications

Complete free study resources for CBSE Class 11 Applied Maths Unit 1 — 26 MCQs with step-by-step answers, 20 solved examples, and 2 case studies. Covers Binary Numbers, Indices, Logarithms, Clock, Calendar, Time-Work-Distance, and Seating Arrangements — fully aligned to the CBSE 2026-27 syllabus.

📋 9 Marks ❓ 26 MCQs ✍️ 18 Solved Examples 📊 2 Case Studies 🎯 CBSE 2026-27
❓MCQs (26) ✍️Short Answers (20) 📊Case Studies (2)

Sub-Unit A: Numbers & Quantification

1. Binary Numbers

Decimal ↔ binary conversion, binary addition & subtraction

2. Indices & Laws of Indices

Product, quotient & power rules; zero, negative & fractional exponents

3. Logarithms & Antilogarithms

Definition, common log (base 10), characteristic, mantissa, log & antilog tables

4. Laws of Logarithms

Product, quotient & power rules; change of base formula, applications

Sub-Unit B: Numerical Applications

5. Clock

Clock angle formula, coincidence of hands, hands at right angles, time gain/loss

6. Calendar

Odd days, leap year rules, day-of-the-week problems, date calculations

7. Time, Work & Distance

Work-rate problems, pipes & cisterns, speed-distance-time, relative speed

8. Seating Arrangements

Linear & circular arrangements, position from left/right, rank-based reasoning

❓

Practice MCQs with Answers

26 questions across all topics — click "Show Answer" to reveal step-by-step explanations

Binary Numbers
Q1 Binary Numbers
The binary equivalent of decimal 13 is:
  • A1101
  • B1011
  • C1110
  • D1001
✓ Correct Answer: A — 1101

Explanation: 13÷2=6 R1 · 6÷2=3 R0 · 3÷2=1 R1 · 1÷2=0 R1
Reading remainders bottom-up: 1101

Q2 Binary Numbers
The decimal value of (10110)2 is:
  • A20
  • B22
  • C24
  • D26
✓ Correct Answer: B — 22

Explanation: 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22

Q3 Binary Numbers
The binary sum of 1011 + 1101 is:
  • A10111
  • B11000
  • C11001
  • D10110
✓ Correct Answer: B — 11000

Explanation: 1011 = 11 (decimal), 1101 = 13 (decimal). Sum = 24 = 2⁴ = 11000 in binary ✓

Indices & Laws of Indices
Q4 Indices
The value of 2³ × 2⁻⁵ × 2⁴ is:
  • A4
  • B8
  • C2
  • D16
✓ Correct Answer: A — 4

Explanation: 2^(3−5+4) = 2² = 4

Q5 Indices
Simplify: (x³)² ÷ x⁴
  • Ax⁸
  • Bx⁵
  • Cx²
  • Dx¹⁰
✓ Correct Answer: C — x²

Explanation: (x³)² = x⁶ → x⁶ ÷ x⁴ = x^(6−4) = x²

Q6 Indices
The value of (27)^(2/3) is:
  • A3
  • B6
  • C9
  • D18
✓ Correct Answer: C — 9

Explanation: (27)^(2/3) = (³√27)² = 3² = 9

Logarithms & Antilogarithms
Q7 Logarithms
The value of log10 0.001 is:
  • A3
  • B−1
  • C−2
  • D−3
✓ Correct Answer: D — −3

Explanation: 0.001 = 10⁻³ → log10(10⁻³) = −3

Q8 Logarithms
If log10 x = 3.5724, the characteristic and mantissa are respectively:
  • A5 and 0.3724
  • B3 and 0.5724
  • C0 and 3.5724
  • D4 and 0.5724
✓ Correct Answer: B — 3 and 0.5724

Explanation: Characteristic = integer part = 3. Mantissa = decimal part = 0.5724.

Q9 Antilogarithm
If log x = 2, the value of x is:
  • A2
  • B20
  • C100
  • D200
✓ Correct Answer: C — 100

Explanation: log x = 2 → x = antilog(2) = 10² = 100

Laws of Logarithms
Q10 Laws of Logarithms
If log 2 = 0.3010 and log 3 = 0.4771, the value of log 6 is:
  • A0.7781
  • B0.1761
  • C0.8451
  • D0.6021
✓ Correct Answer: A — 0.7781

Explanation: log 6 = log(2×3) = log 2 + log 3 = 0.3010 + 0.4771 = 0.7781

Q11 Laws of Logarithms
The value of log 500 − log 5 is:
  • A1
  • B2
  • C3
  • D100
✓ Correct Answer: B — 2

Explanation: log 500 − log 5 = log(500/5) = log 100 = 2

Q12 Laws of Logarithms
The value of log2 32 + log2 4 is:
  • A5
  • B6
  • C7
  • D8
✓ Correct Answer: C — 7

Explanation: log2(32×4) = log2(128) = log2(2⁷) = 7

Clock
Q13 Clock
The angle between the hands of a clock at 3:30 is:
  • A60°
  • B70°
  • C75°
  • D90°
✓ Correct Answer: C — 75°

Explanation: Angle = |30H − 5.5M| = |30×3 − 5.5×30| = |90 − 165| = 75°

Q14 Clock
How many times do the hands of a clock coincide in 24 hours?
  • A20
  • B22
  • C24
  • D44
✓ Correct Answer: B — 22

Explanation: Hands coincide 11 times every 12 hours. In 24 hours: 11 × 2 = 22 times.

Q15 Clock
The minute hand gains how many degrees over the hour hand in one minute?
  • A5°
  • B5.5°
  • C6°
  • D6.5°
✓ Correct Answer: B — 5.5°

Explanation: Minute hand: 6°/min; hour hand: 0.5°/min. Relative gain = 6 − 0.5 = 5.5° per minute. This is why the angle formula is |30H − 5.5M|.

Calendar
Q16 Calendar
Which of the following is NOT a leap year?
  • A2000
  • B1900
  • C2024
  • D1600
✓ Correct Answer: B — 1900

Explanation: A century year is a leap year only if divisible by 400. 1900 is not divisible by 400 → NOT a leap year. 2000 and 1600 are divisible by 400.

Q17 Calendar
If 1st January 2024 is a Monday, what day of the week is 1st January 2025?
  • AMonday
  • BTuesday
  • CWednesday
  • DThursday
✓ Correct Answer: C — Wednesday

Explanation: 2024 is a leap year → 366 days = 52 weeks + 2 odd days. Monday + 2 = Wednesday.

Q18 Calendar
A non-leap year has how many odd days?
  • A0
  • B1
  • C2
  • D3
✓ Correct Answer: B — 1

Explanation: A non-leap year has 365 days = 52 weeks + 1 odd day. A leap year has 366 days = 52 weeks + 2 odd days.

Time, Work & Distance
Q19 Time & Work
A can do a piece of work in 12 days and B in 18 days. Together they finish it in:
  • A6 days
  • B7.2 days
  • C8 days
  • D9 days
✓ Correct Answer: B — 7.2 days

Explanation: Combined rate = 1/12 + 1/18 = 3/36 + 2/36 = 5/36 → Time = 36/5 = 7.2 days

Q20 Pipes & Cisterns
A pipe fills a tank in 4 hours; another empties it in 12 hours. When both are open together, the tank fills in:
  • A4 hours
  • B6 hours
  • C8 hours
  • D12 hours
✓ Correct Answer: B — 6 hours

Explanation: Net rate = 1/4 − 1/12 = 3/12 − 1/12 = 2/12 = 1/6 → Time = 6 hours

Q21 Speed & Distance
A train 150 m long passes a pole in 15 seconds. Its speed in km/h is:
  • A30 km/h
  • B36 km/h
  • C40 km/h
  • D54 km/h
✓ Correct Answer: B — 36 km/h

Explanation: Speed = 150/15 = 10 m/s → 10 × (18/5) = 36 km/h

Q22 Speed & Distance
Two persons walk towards each other from points 18 km apart at 4 km/h and 5 km/h. After how many hours do they meet?
  • A1 hour
  • B1.5 hours
  • C2 hours
  • D2.5 hours
✓ Correct Answer: C — 2 hours

Explanation: Relative speed = 4 + 5 = 9 km/h → Time = 18 / 9 = 2 hours

Seating Arrangements
Q23 Seating Arrangement
In a row of 20 students, Ravi is 8th from the left. His position from the right is:
  • A11th
  • B12th
  • C13th
  • D14th
✓ Correct Answer: C — 13th

Explanation: Position from right = (Total + 1) − Position from left = 21 − 8 = 13th

Q24 Seating Arrangement
P is 7th from the left. Q is 9th from the right. They exchange positions; now P is 11th from the left. How many students are in the row?
  • A17
  • B18
  • C19
  • D20
✓ Correct Answer: C — 19

Explanation: After exchange, P is at Q's original position (9th from right) and is 11th from left. Total = (11 + 9) − 1 = 19 students

Q25 Seating Arrangement
Amit is ranked 8th from the top and 32nd from the bottom in a class. How many students are in the class?
  • A37
  • B38
  • C39
  • D40
✓ Correct Answer: C — 39

Explanation: Total = (Rank from top + Rank from bottom) − 1 = (8 + 32) − 1 = 39 students

Q26 Seating Arrangement
5 people A, B, C, D, E sit in a row. A is to the immediate left of B, C is to the immediate right of B, and D is to the immediate left of A. Who is in the middle?
  • AA
  • BB
  • CC
  • DD
✓ Correct Answer: B — B

Explanation: D left of A; A left of B; C right of B; E is rightmost.
Order: D — A — B — C — E → B is in the middle (3rd of 5).

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Short Answer Questions with Step-by-Step Solutions

20 questions covering all 8 topics — detailed solutions for 2-mark and 3-mark board patterns

Binary Numbers
Q1 Binary Numbers
Convert (45)10 to binary and verify by converting back to decimal.
Solution
45÷2=22 R1 · 22÷2=11 R0 · 11÷2=5 R1 · 5÷2=2 R1 · 2÷2=1 R0 · 1÷2=0 R1
Reading remainders bottom-up: (45)10 = (101101)2
Verification: 1×32 + 0×16 + 1×8 + 1×4 + 0×2 + 1×1 = 32+8+4+1 = 45 ✓
✓ (45)10 = (101101)2
Q2 Binary Numbers
Subtract (1010)2 from (11001)2. Verify using decimal.
Solution
(11001)2 = 16+8+1 = 25    (1010)2 = 8+2 = 10
25 − 10 = 15 (decimal)
15 in binary: 15÷2=7R1, 7÷2=3R1, 3÷2=1R1, 1÷2=0R1 → (1111)2
✓ (11001)2 − (1010)2 = (1111)2
Q3 Binary Numbers
Add the binary numbers (10111)2 and (01101)2. Express the answer in both binary and decimal.
Solution
(10111)2 = 16+4+2+1 = 23    (01101)2 = 8+4+1 = 13
23 + 13 = 36 (decimal)
36 in binary: 36÷2=18R0, 18÷2=9R0, 9÷2=4R1, 4÷2=2R0, 2÷2=1R0, 1÷2=0R1 → (100100)2
✓ (10111)2 + (01101)2 = (100100)2 = 36 in decimal
Indices & Laws of Indices
Q4 Indices
Simplify: (a²b³)² × (a³b)⁻¹
Solution
(a²b³)² = a⁴b⁶
(a³b)⁻¹ = a⁻³b⁻¹
Product = a⁴b⁶ × a⁻³b⁻¹ = a^(4−3) × b^(6−1) = ab⁵
✓ Final Answer: ab⁵
Q5 Indices
Find the value of: 4^(3/2) + 8^(2/3) − 25^(1/2)
Solution
4^(3/2) = (√4)³ = 2³ = 8
8^(2/3) = (³√8)² = 2² = 4
25^(1/2) = √25 = 5
= 8 + 4 − 5 = 7
✓ Final Answer: 7
Logarithms & Antilogarithms
Q6 Logarithms
Given log 2 = 0.3010 and log 3 = 0.4771, find the value of log 72.
Solution
72 = 2³ × 3²
log 72 = 3 log 2 + 2 log 3 = 3(0.3010) + 2(0.4771)
= 0.9030 + 0.9542 = 1.8572
✓ log 72 = 1.8572
Q7 Antilogarithm
Find x if log x = 2.8451. (Given: antilog 0.8451 = 7.000)
Solution
Characteristic = 2, Mantissa = 0.8451
Antilog(0.8451) = 7.000 (given)
Characteristic = 2 → number has (2+1) = 3 digits before decimal: x = 7.000 × 10² = 700
✓ x = 700
Laws of Logarithms
Q8 Laws of Logarithms
Prove that: log 2 + log 3 + log 4 + log 5 = log 120
Solution
LHS = log 2 + log 3 + log 4 + log 5
Using product rule: = log(2 × 3 × 4 × 5) = log(120) = RHS ✓
✓ Hence proved: log 2 + log 3 + log 4 + log 5 = log 120
Q9 Laws of Logarithms
If log 2 = 0.3010, find the number of digits in 2²⁰.
Solution
log(2²⁰) = 20 × log 2 = 20 × 0.3010 = 6.020
Characteristic = 6
Number of digits = characteristic + 1 = 6 + 1 = 7
✓ 2²⁰ has 7 digits
Clock
📌 Key Formula — Clock Angle

Angle between clock hands = |30H − 5.5M| degrees, where H = hour, M = minutes past the hour. The minute hand gains 5.5° per minute over the hour hand.

Q10 Clock
Find the angle between the hands of a clock at 4 hours 20 minutes.
Solution
Angle = |30H − 5.5M|, where H = 4, M = 20
= |30×4 − 5.5×20| = |120 − 110| = 10°
✓ The angle between the hands at 4:20 is 10°
Q11 Clock
Between 3 and 4 o'clock, at what time are the hands of a clock at right angles (90°)?
Solution
|30H − 5.5M| = 90° with H = 3 → |90 − 5.5M| = 90
Case 1: 90 − 5.5M = 90 → M = 0 (boundary; gives 3:00, before the interval)
Case 2: 5.5M − 90 = 90 → 5.5M = 180 → M = 180/5.5 ≈ 32 min 43 sec
✓ Between 3 and 4, hands are at 90° at 3 hrs 32 min 43 sec
Q12 Clock
At what time between 6 and 7 o'clock do the minute and hour hands coincide?
Solution
At 6:00, the hour hand is at the 30-minute mark (180° from 12). The minute hand is at 0°. Gap = 30 minute-spaces.
Time to close the gap = H × (60/11) = 6 × (60/11) = 360/11 = 32 8/11 minutes after 6
✓ Hands coincide at 6 hrs 32 8/11 minutes (≈ 6:32:43)
Calendar
Q13 Calendar
What day of the week was 15th August 1947? (Given: 1st January 1947 was a Wednesday.)
Solution
Days from 1 Jan to 15 Aug: Jan(31)+Feb(28)+Mar(31)+Apr(30)+May(31)+Jun(30)+Jul(31)+Aug 1–15(15) = 227 days
Days after 1 Jan = 227 − 1 = 226 days
226 ÷ 7 = 32 weeks + 2 odd days → Wednesday + 2 = Friday
✓ 15th August 1947 was a Friday
Q14 Calendar
If today is Thursday 20th March 2025, what day of the week will it be 100 days later?
Solution
100 ÷ 7 = 14 weeks + 2 odd days
Thursday + 2 = Saturday
✓ 100 days from Thursday will be a Saturday
Q15 Calendar
In which year will the calendar of 2025 repeat again?
Solution
A calendar repeats when the total odd days accumulated equal a multiple of 7 AND the leap year pattern is the same.
2025 is a non-leap year starting on Wednesday (1 Jan 2025 = Wednesday).
Count odd days year by year from 2025:
2025 (non-leap) = 1 odd day → running total: 1
2026 (non-leap) = 1 → total: 2
2027 (non-leap) = 1 → total: 3
2028 (leap) = 2 → total: 5
2029 (non-leap) = 1 → total: 6
2030 (non-leap) = 1 → total: 7 ≡ 0 (mod 7)
Total odd days = 7 (i.e., 0 mod 7) reached at the start of 2031. Also verify: 2031 is a non-leap year, same as 2025 — the calendar structure matches.
✓ The calendar of 2025 will repeat in 2031
Q16 Calendar
On which days of the week will the 1st of January fall between 2001 and 2020?
Solution
1st January 2001 = Monday (given / reference point).
Each non-leap year adds 1 odd day; each leap year adds 2 odd days. Leap years in this range: 2004, 2008, 2012, 2016, 2020.
Year-by-year 1st January day:
2001 — Monday
2002 — Tuesday (2001: non-leap, +1)
2003 — Wednesday (+1)
2004 — Thursday (+1)
2005 — Saturday (2004: leap, +2)
2006 — Sunday (+1)
2007 — Monday (+1)
2008 — Tuesday (+1)
2009 — Thursday (2008: leap, +2)
2010 — Friday (+1)
2011 — Saturday (+1)
2012 — Sunday (+1)
2013 — Tuesday (2012: leap, +2)
2014 — Wednesday (+1)
2015 — Thursday (+1)
2016 — Friday (+1)
2017 — Sunday (2016: leap, +2)
2018 — Monday (+1)
2019 — Tuesday (+1)
2020 — Wednesday (+1)
✓ Between 2001–2020, 1st January falls on every day of the week. Monday: 2001, 2007, 2018 · Tuesday: 2002, 2008, 2019 · Wednesday: 2003, 2014, 2020 · Thursday: 2004, 2009, 2015 · Friday: 2010, 2016 · Saturday: 2005, 2011 · Sunday: 2006, 2012, 2017
Time, Work & Distance
Q17 Time & Work
A tap fills a tank in 6 hours. Due to a leak, it takes 8 hours to fill. In how many hours will the leak alone empty the full tank?
Solution
Tap's filling rate = 1/6 tank/hr    Net rate with leak = 1/8 tank/hr
Leak rate = 1/6 − 1/8 = 4/24 − 3/24 = 1/24 tank/hr
✓ The leak alone empties the full tank in 24 hours
Q18 Speed & Distance
Two trains start simultaneously from stations A and B which are 250 km apart, moving towards each other at 60 km/h and 40 km/h. After how many hours do they meet, and how far from A?
Solution
Relative speed = 60 + 40 = 100 km/h
Time to meet = 250 ÷ 100 = 2.5 hours
Distance covered by train from A = 60 × 2.5 = 150 km from A
✓ They meet after 2.5 hours, 150 km from A
Seating Arrangements
Q19 Seating Arrangement
In a class of 35 students, Meena ranks 10th from the top. Priya is 5 ranks below Meena. What is Priya's rank from the bottom?
Solution
Priya's rank from top = 10 + 5 = 15th
Priya's rank from bottom = (Total + 1) − Rank from top = 36 − 15 = 21st
✓ Priya's rank from the bottom is 21st
Q20 Seating Arrangement
In a row, Anil is 12th from the left and 8th from the right. Suresh is 3 places to the right of Anil. How many people are in the row, and what is Suresh's position from the right?
Solution
Total = (12 + 8) − 1 = 19 people
Suresh's position from left = 12 + 3 = 15th
Suresh's position from right = (19 + 1) − 15 = 5th
✓ Total = 19 people · Suresh is 5th from the right

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Case Studies

Real-world application questions — CBSE board exam pattern

💡 About Case Study Questions

Case studies carry 4 marks each in the CBSE board exam (Section E). Each has a real-world scenario followed by 3–4 sub-questions. Practice reading the scenario carefully and identifying which topic/formula applies before solving.

Case Study 1
📡 Digital Data & Binary Communication
A computer science teacher explains that computers store all data in binary (base 2) using only 0s and 1s. She tells the class: "The number of unique values storable in n bits is 2n. Conversely, the number of bits required = log2(total values)."

She gives the class tasks using the ASCII code for the letter 'A', which equals 65 in decimal.
Question 1 (2 marks)
Convert the ASCII code of 'A' (decimal 65) into binary. Show all steps.
✓ Answer

65÷2=32 R1 · 32÷2=16 R0 · 16÷2=8 R0 · 8÷2=4 R0 · 4÷2=2 R0 · 2÷2=1 R0 · 1÷2=0 R1
Reading bottom-up: (65)10 = (1000001)2

Question 2 (1 mark)
8-bit ASCII encoding can represent 2⁸ characters. How many is that? Verify using log2.
✓ Answer

2⁸ = 256 characters.
Verification: log2(256) = log2(2⁸) = 8 bits ✓

Question 3 (1 mark)
Add the binary numbers (01001011)2 and (00110101)2. Convert both to decimal, add, then express the result in binary.
✓ Answer

(01001011)2 = 64+8+2+1 = 75
(00110101)2 = 32+16+4+1 = 53
75 + 53 = 128 = 2⁷ = (10000000)2 ✓

Case Study 2
🏢 Office Scheduling — Clock, Calendar & Work Efficiency
Priya joins a new office on Monday, 3rd March 2025. She is assigned to a project team:

Priya can complete the project alone in 15 days. Arjun can complete the same project alone in 10 days.

The team has a daily stand-up meeting at 9:15 AM and a review meeting at 3:45 PM.
Question 1 (1 mark)
If Priya and Arjun work together every day, in how many days will they complete the project?
✓ Answer

Combined rate = 1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6
They finish in 6 days.

Question 2 (1 mark)
Starting Monday 3rd March 2025, on what date and day of the week is the project completed?
✓ Answer

Day 1: Mon 3 Mar → Day 6: Saturday, 8th March 2025

Question 3 (2 marks)
Find the angle between the clock hands at the 9:15 AM stand-up meeting and at the 3:45 PM review meeting.
✓ Answer

At 9:15 AM: |30×9 − 5.5×15| = |270 − 82.5| = 187.5° → smaller angle = 360 − 187.5 = 172.5°

At 3:45 PM: |30×3 − 5.5×45| = |90 − 247.5| = 157.5°

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Binary steps, log laws, clock & calendar tricks, index rules — every formula for every unit, in one organised printable PDF.

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