Unit 1: Numbers, Quantification & Numerical Applications — Class 11 Applied Maths | Free MCQs & Solved Examples | Boundless Maths
Unit 1 of 7 10 Marks CBSE 2026–27

Unit 1: Numbers, Quantification
& Numerical Applications

CBSE Class 11 Applied Mathematics · Unit 1 · Free MCQs, Solved Examples & Case Studies

Unit 1 carries 10 marks in the CBSE Class 11 annual exam. Complete free resources: 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.

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Class 11 Applied Maths Unit 1 — Numbers, Quantification & Numerical Applications

This page covers all topics in Unit 1 of CBSE Class 11 Applied Mathematics — carrying 10 marks in the CBSE Class 11 annual exam. You'll find 26 MCQs with answers, 20 solved examples with step-by-step solutions, and 2 case studies based on real-world contexts. Sub-Unit A (Numbers & Quantification) covers Binary Numbers, Indices & Laws of Indices, Logarithms & Antilogarithms, and Laws of Logarithms. Sub-Unit B (Numerical Applications) covers Clock, Calendar, Time-Work-Distance (including Pipes & Cisterns and Speed-Distance-Time), and Seating Arrangements. All content aligned to the CBSE 2026-27 syllabus.

Binary NumbersIndices & Laws LogarithmsAntilogarithms Clock Angle FormulaCalendar Odd Days Time & WorkSeating Arrangements
Unit 1 · 10 Marks

Topics & Key Formulas

Two sub-units, eight topics. Memorise the formulas below — they appear in MCQs and short answers every year.

Sub-Unit A: Numbers & Quantification

1. Binary Numbers

Decimal ↔ binary conversion using repeated division; binary addition and subtraction

2. Indices & Laws of Indices

Product, quotient & power rules; zero, negative & fractional exponents: a⁰ = 1, a−n = 1/aⁿ, am/n = (ⁿ√a)ᵐ

3. Logarithms & Antilogarithms

Definition: logb x = y ⟺ by = x. Characteristic (integer part) & mantissa (decimal part). Log and antilog tables.

4. Laws of Logarithms

Product: log mn = log m + log n. Quotient: log(m/n) = log m − log n. Power: log mⁿ = n log m. Number of digits in N = ⌊log N⌋ + 1.

Sub-Unit B: Numerical Applications

5. Clock

Angle between hands = |30H − 5.5M|°. Minute hand gains 5.5° per minute over the hour hand. Hands coincide 22 times in 24 hours.

6. Calendar

Odd days: non-leap year = 1, leap year = 2. Century year is leap only if divisible by 400. Day shifts by odd days each year.

7. Time, Work & Distance

Combined rate = 1/x + 1/y. Time together = xy/(x+y). Pipes: filling = positive rate, draining = negative rate. Speed = Distance/Time.

8. Seating Arrangements

Position from right = (n+1) − position from left. Total = (rank from top + rank from bottom) − 1. Exchange problems: use new position to find total.

📐
All formulas for all 7 units of Class 11 Applied Maths — in one print-ready PDF

Binary steps, log laws, clock & calendar tricks, index rules — organised topic-wise, CBSE aligned

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Practice Questions

Practice MCQs — Unit 1

26 questions across all 8 topics. Click Show Answer to reveal the correct option and step-by-step explanation.

Binary Numbers
Q1 Binary Numbers
The binary equivalent of decimal 13 is:
(A) 1101
(B) 1011
(C) 1110
(D) 1001
✓ Correct Answer: (A) 1101
Divide repeatedly by 2 and read remainders bottom-up:
13 ÷ 2 = 6 R1 · 6 ÷ 2 = 3 R0 · 3 ÷ 2 = 1 R1 · 1 ÷ 2 = 0 R1
Reading bottom-up: (13)10 = (1101)2
Q2 Binary Numbers
The decimal value of (10110)2 is:
(A) 20
(B) 22
(C) 24
(D) 26
✓ Correct Answer: (B) 22
1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰ = 16 + 0 + 4 + 2 + 0 = 22
Q3 Binary Numbers
The binary sum of (1011)2 + (1101)2 is:
(A) 10111
(B) 11000
(C) 11001
(D) 10110
✓ Correct Answer: (B) 11000
(1011)2 = 11 and (1101)2 = 13. Sum = 24 = 2⁴ + 2³ = (11000)2 ✓
Indices & Laws of Indices
Q4 Indices
The value of 2³ × 2−5 × 2⁴ is:
(A) 4
(B) 8
(C) 2
(D) 16
✓ Correct Answer: (A) 4
23+(−5)+4 = 2² = 4
Q5 Indices
Simplify: (x³)² ÷ x⁴
(A) x⁸
(B) x⁵
(C) x²
(D) x10
✓ Correct Answer: (C) x²
(x³)² = x⁶. Then x⁶ ÷ x⁴ = x6−4 = x²
Q6 Indices
The value of (27)2/3 is:
(A) 3
(B) 6
(C) 9
(D) 18
✓ Correct Answer: (C) 9
(27)2/3 = (³√27)² = 3² = 9
Logarithms & Antilogarithms
Q7 Logarithms
The value of log10 0.001 is:
(A) 3
(B) −1
(C) −2
(D) −3
✓ Correct Answer: (D) −3
0.001 = 10−3, so log10(10−3) = −3
Q8 Logarithms
If log10 x = 3.5724, the characteristic and mantissa are respectively:
(A) 5 and 0.3724
(B) 3 and 0.5724
(C) 0 and 3.5724
(D) 4 and 0.5724
✓ Correct Answer: (B) 3 and 0.5724
Characteristic = integer part = 3. Mantissa = decimal part = 0.5724.
Q9 Antilogarithm
If log x = 2, the value of x is:
(A) 2
(B) 20
(C) 100
(D) 200
✓ Correct Answer: (C) 100
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:
(A) 0.7781
(B) 0.1761
(C) 0.8451
(D) 0.6021
✓ Correct Answer: (A) 0.7781
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:
(A) 1
(B) 2
(C) 3
(D) 100
✓ Correct Answer: (B) 2
log 500 − log 5 = log(500/5) = log 100 = 2
Q12 Laws of Logarithms
The value of log2 32 + log2 4 is:
(A) 5
(B) 6
(C) 7
(D) 8
✓ Correct Answer: (C) 7
log2(32 × 4) = log2(128) = log2(2⁷) = 7
Clock
📌 Key Formula — Clock Angle

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

Q13 Clock
The angle between the hands of a clock at 3:30 is:
(A) 60°
(B) 70°
(C) 75°
(D) 90°
✓ Correct Answer: (C) 75°
Angle = |30 × 3 − 5.5 × 30| = |90 − 165| = 75°
Q14 Clock
How many times do the hands of a clock coincide in 24 hours?
(A) 20
(B) 22
(C) 24
(D) 44
✓ Correct Answer: (B) 22
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?
(A) 5°
(B) 5.5°
(C) 6°
(D) 6.5°
✓ Correct Answer: (B) 5.5°
Minute hand: 6°/min; hour hand: 0.5°/min. Relative gain = 6 − 0.5 = 5.5° per minute. This is why the angle formula uses 5.5M.
Calendar
📌 Key Rules — Calendar

Odd days: non-leap year = 1 odd day (365 = 52 weeks + 1); leap year = 2 odd days.
Leap year test: divisible by 4, except century years which must be divisible by 400.
Centuries: 100 years = 5 odd days; 200 years = 3; 300 years = 1; 400 years = 0.

Q16 Calendar
Which of the following is NOT a leap year?
(A) 2000
(B) 1900
(C) 2024
(D) 1600
✓ Correct Answer: (B) 1900
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 → leap years.
Q17 Calendar
If 1st January 2024 is a Monday, what day of the week is 1st January 2025?
(A) Monday
(B) Tuesday
(C) Wednesday
(D) Thursday
✓ Correct Answer: (C) Wednesday
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?
(A) 0
(B) 1
(C) 2
(D) 3
✓ Correct Answer: (B) 1
A non-leap year has 365 days = 52 × 7 + 1, so 1 odd day. A leap year has 366 days = 52 × 7 + 2, so 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:
(A) 6 days
(B) 7.2 days
(C) 8 days
(D) 9 days
✓ Correct Answer: (B) 7.2 days
Combined rate = 1/12 + 1/18 = (3+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:
(A) 4 hours
(B) 6 hours
(C) 8 hours
(D) 12 hours
✓ Correct Answer: (B) 6 hours
Net rate = 1/4 − 1/12 = (3−1)/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:
(A) 30 km/h
(B) 36 km/h
(C) 40 km/h
(D) 54 km/h
✓ Correct Answer: (B) 36 km/h
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?
(A) 1 hour
(B) 1.5 hours
(C) 2 hours
(D) 2.5 hours
✓ Correct Answer: (C) 2 hours
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:
(A) 11th
(B) 12th
(C) 13th
(D) 14th
✓ Correct Answer: (C) 13th
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?
(A) 17
(B) 18
(C) 19
(D) 20
✓ Correct Answer: (C) 19
After exchange, P is at Q's original position (9th from right) and now 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?
(A) 37
(B) 38
(C) 39
(D) 40
✓ Correct Answer: (C) 39
Total = (Rank from top + Rank from bottom) − 1 = (8 + 32) − 1 = 39
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?
(A) A
(B) B
(C) C
(D) D
✓ Correct Answer: (B) B
D is left of A; A is left of B; C is right of B; E must be at one end.
Order: D — A — B — C — E → B is 3rd of 5 = middle

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  • Binary conversion steps, clock angle formula, log and index laws
  • Calendar tricks and work-rate formulas
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Step-by-Step Solutions

Solved Examples — All 8 Topics

20 questions with detailed solutions. Click Show Solution to reveal each answer.

Binary Numbers
Example 1
Convert (45)10 to binary and verify by converting back to decimal.
Divide repeatedly by 2 and record remainders:
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
Example 2
Subtract (1010)2 from (11001)2. Verify using decimal.
(11001)2 = 16+8+1 = 25   and   (1010)2 = 8+2 = 10
25 − 10 = 15 in decimal
Convert 15 to binary: 15÷2=7R1, 7÷2=3R1, 3÷2=1R1, 1÷2=0R1 → (1111)2
(11001)2 − (1010)2 = (1111)2
Example 3
Add (10111)2 and (01101)2. Express the answer in both binary and decimal.
(10111)2 = 16+4+2+1 = 23   and   (01101)2 = 8+4+1 = 13
23 + 13 = 36 in decimal
Convert 36 to 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
Example 4
Simplify: (a²b³)² × (a³b)−1
(a²b³)² = a⁴b⁶
(a³b)−1 = a−3b−1
Product = a⁴b⁶ × a−3b−1 = a4−3·b6−1 = ab⁵
(a²b³)² × (a³b)−1 = ab⁵
Example 5
Find the value of: 43/2 + 82/3 − 251/2
43/2 = (√4)³ = 2³ = 8
82/3 = (³√8)² = 2² = 4
251/2 = √25 = 5
= 8 + 4 − 5 = 7
43/2 + 82/3 − 251/2 = 7
Logarithms & Antilogarithms
Example 6
Given log 2 = 0.3010 and log 3 = 0.4771, find the value of log 72.
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
Example 7
Find x if log x = 2.8451. (Given: antilog 0.8451 = 7.000)
Characteristic = 2, Mantissa = 0.8451
antilog(0.8451) = 7.000 (given)
Characteristic = 2 → the number has 2+1 = 3 digits before the decimal point
x = 7.000 × 10² = 700
x = 700
Laws of Logarithms
Example 8
Prove that: log 2 + log 3 + log 4 + log 5 = log 120
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
Example 9
If log 2 = 0.3010, find the number of digits in 220.
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
Example 10
Find the angle between the hands of a clock at 4 hours 20 minutes.
Formula: 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°
Example 11
Between 3 and 4 o'clock, at what time are the hands of a clock at right angles (90°)?
Set |30H − 5.5M| = 90° with H = 3:   |90 − 5.5M| = 90
Case 1: 90 − 5.5M = 90 → M = 0 (gives 3:00, boundary — before the interval)
Case 2: 5.5M − 90 = 90 → 5.5M = 180 → M = 180/5.5 = 360/11 ≈ 32 8/11 min
Between 3 and 4, hands are at 90° at 3 h 32 8/11 min (≈ 3:32:43)
Example 12
At what time between 6 and 7 o'clock do the minute and hour hands coincide?
At 6:00, the hour hand is at the 30-min 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 min after 6
Hands coincide at 6 h 32 8/11 min (≈ 6:32:43)
Calendar
Example 13
What day of the week was 15th August 1947? (Given: 1st January 1947 was a Wednesday.)
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 = 226 days
226 ÷ 7 = 32 weeks + 2 odd days → Wednesday + 2 = Friday
15th August 1947 was a Friday
Example 14
If today is Thursday 20th March 2025, what day of the week will it be 100 days later?
100 ÷ 7 = 14 weeks + 2 odd days
Thursday + 2 = Saturday
100 days from Thursday will be a Saturday
Example 15
In which year will the calendar of 2025 repeat again?
A calendar repeats when the total odd days accumulated = a multiple of 7 AND the leap-year pattern matches.
2025 is a non-leap year (1 Jan 2025 = Wednesday). Counting odd days from 2025:
2025: +1 (total 1) · 2026: +1 (2) · 2027: +1 (3) · 2028 (leap): +2 (5) · 2029: +1 (6) · 2030: +1 → total 7 ≡ 0 (mod 7)
Total odd days = 7 at the start of 2031. Also verify: 2031 is a non-leap year, same as 2025 ✓
The calendar of 2025 will repeat in 2031
Example 16
On which days of the week will 1st January fall between 2001 and 2020?
1st January 2001 = Monday. Each non-leap year adds 1 odd day; each leap year (2004, 2008, 2012, 2016, 2020) adds 2 odd days.
2001 Mon · 2002 Tue · 2003 Wed · 2004 Thu · 2005 Sat · 2006 Sun · 2007 Mon · 2008 Tue · 2009 Thu · 2010 Fri · 2011 Sat · 2012 Sun · 2013 Tue · 2014 Wed · 2015 Thu · 2016 Fri · 2017 Sun · 2018 Mon · 2019 Tue · 2020 Wed
1st January falls on every day of the week between 2001–2020. 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
Example 17
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?
Tap's filling rate = 1/6 tank/hr  |  Net rate with leak = 1/8 tank/hr
Leak rate = 1/6 − 1/8 = (4−3)/24 = 1/24 tank/hr
The leak alone empties the full tank in 24 hours
Example 18
Two trains start simultaneously from stations A and B (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?
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
They meet after 2.5 hours, 150 km from A
Seating Arrangements
Example 19
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?
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
Example 20
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?
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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4-Mark Questions

Case Study Questions

Board-pattern case studies. Read the context carefully, then click Show Answers for each case.

💡 About Case Study Questions

Case studies carry 4 marks each in the CBSE Class 11 annual exam (Section E). Each has a real-world scenario followed by 3–4 sub-questions. Practice reading the scenario carefully and identifying which topic and 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.
(i)

Convert the ASCII code of 'A' (decimal 65) into binary. Show all steps. (2 marks)

(ii)

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

(iii)

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

(i)(65)10 = (1000001)2. 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
(ii)2⁸ = 256 characters. Verification: log2(256) = log2(2⁸) = 8 bits ✓
(iii)Sum = (10000000)2 = 128. (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.
(i)

If Priya and Arjun work together every day, in how many days will they complete the project? (1 mark)

(ii)

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

(iii)

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

(i)6 days. Combined rate = 1/15 + 1/10 = (2+3)/30 = 5/30 = 1/6. Time = 6 days.
(ii)Saturday, 8th March 2025. Day 1: Mon 3 Mar → Day 2: Tue 4 Mar → Day 3: Wed 5 Mar → Day 4: Thu 6 Mar → Day 5: Fri 7 Mar → Day 6: Sat 8 Mar 2025
(iii)9:15 AM → 172.5° | 3:45 PM → 157.5°. At 9:15 AM: |30×9 − 5.5×15| = |270 − 82.5| = 187.5°. Since this exceeds 180°, the actual (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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Score Full Marks

Exam Tips — Unit 1

Common mistakes students make in the exam — and how to avoid them.

✓ Tip 1 — Always Read Remainders Bottom-Up in Binary Conversion

The most common error in binary conversion is reading remainders top-down. After dividing repeatedly by 2, write out the chain of remainders and then read them from bottom to top. A safe habit: circle the last remainder first and arrow upward.

✓ Tip 2 — Write the Clock Angle Formula Before Substituting

Always write Angle = |30H − 5.5M| before plugging in numbers. Examiners award 1 mark for writing the correct formula even if you make an arithmetic error. Also: if your answer exceeds 180°, subtract from 360° to get the smaller angle — and note this step explicitly.

✓ Tip 3 — State Characteristic and Mantissa Separately in Log Questions

For antilog questions, always write: (1) identify the characteristic, (2) identify the mantissa, (3) look up antilog of the mantissa, (4) place the decimal using the characteristic. Missing any step costs marks. The rule: characteristic = c means c+1 digits before the decimal point for positive c.

❌ Common Mistakes to Avoid in Unit 1

  • Reading binary conversion remainders top-down instead of bottom-up
  • Forgetting that ⌊−2.4⌋ = −3, not −2 — same logic applies to negative characteristics in log
  • Using the clock angle formula with hours past midnight instead of the 12-hour clock value (use H = 3 for 3 PM, not 15)
  • Treating 1900 as a leap year — century years must be divisible by 400
  • In work problems: forgetting to subtract the draining pipe rate (it is 1/4 − 1/12, not 1/4 + 1/12)
  • Seating: using Total instead of (Total + 1) when computing rank from the other end
  • Indices: writing (x³)² = x⁵ instead of x⁶ — multiply exponents, do not add
Common Questions

Frequently Asked Questions

Key questions students ask about Unit 1 — formulas, concepts, and exam strategy.

Unit 1 — Numbers, Quantification & Numerical Applications carries 10 marks in the CBSE Class 11 Applied Mathematics school final exam (80-mark theory paper).
Sub-Unit A (Numbers & Quantification): Binary Numbers, Indices & Laws of Indices, Logarithms & Antilogarithms, Laws of Logarithms.
Sub-Unit B (Numerical Applications): Clock, Calendar, Time-Work-Distance (including Pipes & Cisterns and Speed-Distance-Time), and Seating Arrangements.
Repeatedly divide the number by 2 and record each remainder. Read the remainders from bottom to top. Example: 13 ÷ 2 = 6 R1 · 6 ÷ 2 = 3 R0 · 3 ÷ 2 = 1 R1 · 1 ÷ 2 = 0 R1 → reading bottom-up: (13)10 = (1101)2.
Angle between clock hands = |30H − 5.5M|°, where H = hour (12-hour clock) and M = minutes past the hour. If the result exceeds 180°, subtract from 360° to get the smaller angle. The minute hand gains 5.5° per minute over the hour hand.
A year is a leap year if divisible by 4, except century years (ending in 00) which must be divisible by 400. So 2000 ✓ leap, 1900 ✗ not leap, 2024 ✓ leap. Leap year = 366 days = 2 odd days. Non-leap year = 365 days = 1 odd day.
If A finishes in x days and B in y days, their combined rate = 1/x + 1/y and time together = xy/(x+y) days. For pipes & cisterns: filling pipe = positive rate, draining pipe = negative rate. Net rate = fill rate − drain rate.
Position from right = (n+1) − position from left, where n is the total number of people. For rank problems: Total = rank from top + rank from bottom − 1. For exchange problems: use the new position to determine the total.
Number of digits in a positive integer N = ⌊log10 N⌋ + 1 = characteristic + 1. Example: log(220) = 20 × 0.3010 = 6.020. Characteristic = 6, so 220 has 6 + 1 = 7 digits.
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