Sunday, August 17, 2025

Answer Key with Short Reasoning (for Chapter 3 – Number Play)

 

Answer Key with Short Reasoning (for Chapter 3 – Number Play)

1. MCQs (20)

  1. (a) 121 – Reads same forward and backward (palindrome).

  2. (a) 6174 – Kaprekar’s constant.

  3. (b) 324 – Sum of digits = 9, divisible by 9.

  4. (a) 18 – 1+0+8+9 = 18.

  5. (c) 625 – 25×25 = 625.

  6. (d) Both (a) & (c) – 121 and 484 are palindrome squares.

  7. (b) 6174 – Result of Kaprekar’s process.

  8. (d) 50 – Not a cube number.

  9. (b) 1 – Collatz Conjecture ends at 1.

  10. (c) 366 days – Leap year.

  11. (a) 18 – 1+8=9, 18 divisible by 9.

  12. (d) 1234 – Not a palindrome.

  13. (a) 8 – 8³ = 512.

  14. (c) 24 – LCM(8,12)=24.

  15. (b) 97 – Only prime among options.

  16. (c) Friday – 100÷7=14 remainder 2 → Monday+2=Wednesday. Oops correction: let’s recalc → If today Monday, after 100 days remainder 2 → Wednesday. Correct answer = Wednesday (not in options, revise Q).

  17. (a) 143 – Divisible by 11.

  18. (a) 30 – 2×3×5.

  19. (d) All of these – All palindromes.

  20. (c) 1729 – Ramanujan number.


2. Assertion–Reasoning (20)

  1. (a) Both true, R explains A.

  2. (a) Both true, R explains A.

  3. (a) Both true, R explains A.

  4. (c) A true, R false (4 not prime).

  5. (a) Both true, R explains A.

  6. (a) Both true, R explains A.

  7. (c) A false, R false (not all palindromes divisible by 11).

  8. (a) Both true, R explains A.

  9. (a) Both true, R explains A.

  10. (a) Both true, R explains A.

  11. (a) Both true, R explains A.

  12. (b) Both true, but R doesn’t fully explain.

  13. (c) A false, R true.

  14. (a) Both true, R explains A.

  15. (a) Both true, R explains A.

  16. (a) Both true, R explains A.

  17. (a) Both true, R explains A.

  18. (a) Both true, R explains A.

  19. (a) Both true, R explains A.

  20. (a) Both true, R explains A.


3. True or False (10)

  1. False – Not all palindromes divisible by 11.

  2. True – Digit reversal trick → 1089.

  3. False – Kaprekar’s constant = 6174.

  4. True – 6174 is Kaprekar’s number.

  5. True – Collatz ends at 1.

  6. True – Leap year has 366 days.

  7. False – Not every palindrome is square.

  8. True – 121 is palindrome & square.

  9. True – 1331 divisible by 11.

  10. False – 1001 = 7×11×13.


4. Short Answer I (15)

  1. 111, 121, 131 – All palindromes.

  2. 6174 – Kaprekar’s constant.

  3. 1+2+3+4 = 10.

  4. 18, 20 – Both divisible by digit sum.

  5. 324 ÷ 9 = 36 → Yes divisible.

  6. ∛216 = 6.

  7. a² + 2ab + b².

  8. LCM(6,8) = 24.

  9. Collatz: Any number → eventually 1.

  10. Monday+30 = Wednesday.

  11. 2,3,5,7,11.

  12. 11,22,33 etc.

  13. 15²=225.

  14. 7³=343.

  15. 1000 = 10³.


5. Short Answer II (10)

  1. 121 is 11² and reads same → palindrome & square.

  2. HCF(18,24)=6, LCM=72.

  3. Example: 3524 → 5432–2345 = 3087 … → 6174.

  4. 1331 ÷ 11=121 → divisible.

  5. 6→3→10→5→16→8→4→2→1.

  6. 12³=1728 using (a+b)³ expansion.

  7. 1+3+5+7+9=25=5².

  8. 498≈500, 52≈50 → 500×50=25000.

  9. 200÷7=28 r4 → Friday+4=Tuesday.

  10. Eg: 12 & 60 → HCF=12, LCM=60.


6. Long Answer (10)

  1. 3524 process → converges to 6174 (Kaprekar constant).

  2. 11 Collatz → 11→34→17→52→26→13→40→20→10→5→16→…→1.

  3. (n+1)² – n² = 2n+1 = sum of consecutive numbers.

  4. 20×15=300 tiles needed.

  5. Odd square remains odd, ex: 7²=49.

  6. 1729=10³+9³=12³+1³.

  7. 500÷7=71 r3 → Wednesday+3=Saturday.

  8. Any 2-digit number ab=10a+b → (10a+b)–(a+b)=9a → divisible by 9.

  9. 9261=21³ → cube root=21.

  10. Estimation saves time, e.g. 198≈200, quick calculations.


7. Case-Based Qs (5 Sets)

Case 1 (Kaprekar)
a) 5432 b) 2345 c) 3087 d) 6174.

Case 2 (Collatz)
a) 5 b) 16 c) Yes → 1 d) Conjecture states all numbers reach 1.

Case 3 (Calendar)
a) Tuesday b) Friday c) 11 years later d) 52 Sundays.

Case 4 (Estimation)
a) 200+100+300=600 b) 198+102+298=598 c) Difference=2 d) Saves time.

Case 5 (Number Games)
a) 132 b) 231–132=99 c) 9+9=18 d) Digits add to multiples of 9.


Saturday, August 16, 2025

Chapter 3: Number Play – Study Material cum Worksheet


Chapter 3: Number Play – Study Material cum Worksheet


Chapter Subsections Covered

3.1 Numbers can Tell us Things

3.2 Supercells

3.3 Patterns of Numbers on the Number Line

3.4 Playing with Digits

3.5 Pretty Palindromic Patterns

3.6 The Magic Number of Kaprekar

3.7 Clock and Calendar Numbers

3.8 Mental Math

3.9 Playing with Number Patterns

3.10 Collatz Conjecture

3.11 Simple Estimation

3.12 Games and Winning Strategies



1. Multiple Choice Questions (MCQs)

Which of the following is a palindromic number?

 a) 121 b) 132 c) 145 d) 234 (Recognising palindromes)

The Kaprekar constant is:

 a) 6174 b) 1729 c) 1089 d) 1001 (Number pattern recognition

Which of the following numbers is divisible by 9?

 a) 234 b) 324 c) 245 d) 561 (Divisibility rule for 9)

The sum of digits of 1089 is:

 a) 18 b) 10 c) 9 d) 19 (Playing with digits)

The square of 25 is:

 a) 525 b) 650 c) 625 d) 600 (Square evaluation)

Which number is a palindrome and also a perfect square?

 a) 121 b) 144 c) 484 d) Both (a) and (c ) (Pattern identification)

In Kaprekar’s process, repeatedly subtracting the largest and smallest 4-digit numbers formed by digits leads to:

 a) 1729 b) 6174 c) 1000 d) 9999 (Kaprekar’s constant)

Which of the following is not a perfect cube?

 a) 27 b) 64 c) 125 d) 50 (Identifying cubes)

The Collatz Conjecture states that any positive integer will eventually reach:

 a) 0 b) 1 c) 2 d) 4 (Understanding conjectures)

A leap year has:

 a) 364 days b) 365 days c) 366 days d) 367 days (Calendar numbers)

Which number is a Harshad number (divisible by the sum of its digits)?

 a) 18 b) 19 c) 20 d) 21 (Divisibility test)

Which of the following is not a palindrome?

 a) 2002 b) 2332 c) 3223 d) 1234 (Palindrome test)

The cube root of 512 is:

 a) 8 b) 9 c) 7 d) 6 (Cube roots)

The LCM of 8 and 12 is:

 a) 48 b) 36 c) 24 d) 16 (LCM computation)

Which of the following is a prime number?

 a) 91 b) 97 c) 87 d) 77 (Prime identification)

If today is Monday, what day will it be after 100 days?

 a) Tuesday b) Thursday c) Friday d) Sunday (Modular arithmetic with calendars)

Which of the following is divisible by 11?

 a) 143 b) 234 c) 345 d) 456 (Divisibility by 11)

The product of first 3 prime numbers is:

 a) 30 b) 36 c) 42 d) 60 (Prime multiplication)

Which number is a palindrome?

 a) 1001 b) 1111 c) 1221 d) All of these (Palindromes)

Which one is known as Ramanujan’s number (sum of two cubes in two ways)?

 a) 1089 b) 6174 c) 1729 d) 1001 (Famous numbers)



2. Assertion and Reasoning Questions

Directions:

 a) A and R are true, R explains A.

 b) A and R are true, but R does not explain A.

 c) A is true, R is false.

 d) A is false, R is true.

A: 121 is a palindrome.

 R: A number that reads the same backward and forward is a palindrome. (Palindrome recognition)

A: 1729 is called Ramanujan’s number.

 R: 1729 can be expressed as the sum of two cubes in two ways. (Number properties)

A: 6174 is Kaprekar’s constant.

 R: Repeated subtraction of the largest and smallest numbers formed by digits leads to 6174. (Kaprekar’s process)

A: Every even number greater than 2 is a prime.

 R: 4 is an even prime. (Prime understanding)

A: A number divisible by 2 and 5 is divisible by 10.

R: LCM of 2 and 5 is 10. (Divisibility rule)

A: 1089 always appears in a digit reversal trick.

 R: Subtracting a 3-digit number and its reverse, then reversing again, gives 1089. (Digit manipulation patterns)

A: All palindromic numbers are divisible by 11.

 R: Alternating sum of digits in palindromes is always divisible by 11. (Divisibility by 11)

A: Collatz Conjecture ends with 1 for any positive integer.

 R: Even numbers are halved, odd numbers multiplied by 3 and increased by 1. (Collatz conjecture)

A: 1331 is divisible by 11.

 R: The alternating sum of digits is divisible by 11. (Divisibility rule)

A: 25² = 625.

 R: Square of 25 means 25 × 25. (Square calculation)

A: Every cube of a number ends in 0, 1, 4, 5, 6, or 9.

 R: Cube endings follow fixed patterns for digits. (Cube pattern recognition)

A: Calendar years repeat after 4 years.

 R: Leap years add an extra day. (Calendar arithmetic)

A: Estimation is always more accurate than actual calculation.

 R: Estimation reduces time and is close to the actual answer. (Estimation)

A: 49 is a perfect square.

 R: 7² = 49. (Square numbers)

A: 27 is a perfect cube.

 R: 3³ = 27. (Cube numbers)

A: Kaprekar’s constant 6174 is reached in at most 7 steps for any 4-digit number (not all digits same).

 R: The process converges by repeated subtraction. (Kaprekar’s process)

A: Collatz sequence starting at 6 ends at 1.

 R: The sequence goes 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1. (Collatz conjecture)

A: Estimation is useful in shopping.

 R: It helps in quickly checking if money is sufficient. (Estimation in daily life)

A: The sum of first n odd numbers is n².

 R: 1 + 3 + 5 + … + (2n–1) = n². (Number patterns)

A: Any year divisible by 400 is a leap year.

 R: This is an exception to the century year rule. (Leap year rule)

3. True or False – 10 Statements

All palindromes are divisible by 11. (False) (Palindrome test)

1089 is obtained in a digit reversal trick. (True) (Playing with digits)

Kaprekar’s constant is 1729. (False) (Kaprekar’s process)

6174 is known as Kaprekar’s number. (True) (Number property)

Collatz Conjecture ends at 1. (True) (Conjecture understanding)

A leap year has 366 days. (True) (Calendar numbers)

Every palindrome is a square number. (False) (Palindrome recognition)

121 is both a palindrome and a perfect square. (True) (Pattern recognition)

1331 is divisible by 11. (True) (Divisibility by 11)

1001 is a prime number. (False) (Prime recognition)

4. Short Answer I (2 Marks) – 15 Questions

Write any three palindromic numbers between 100 and 200. (Palindrome identification)

State Kaprekar’s constant. (Kaprekar’s process)

Find the sum of digits of 1234. (Playing with digits)

Write two Harshad numbers. (Divisibility test)

Find whether 324 is divisible by 9. (Divisibility rule)

Write the cube root of 216. (Cube roots)

Expand (a+b)². (Algebraic identity)

Find the LCM of 6 and 8. (LCM)

State Collatz Conjecture in your own words. (Understanding conjectures)

Which day of the week is it 30 days from Monday? (Calendar use)

Write first 5 prime numbers. (Prime sequence)

Write a palindrome less than 100. (Palindrome recognition)

Find the square of 15. (Square computation)

Find the cube of 7. (Cube computation)

Express 1000 as cube of a number. (Perfect cube)



5. Short Answer II (3 Marks) – 10 Questions

Show that 121 is both a palindrome and a square. (Application of patterns)

Find the LCM and HCF of 18 and 24. (LCM/HCF)

Show with an example how Kaprekar’s process works. (Kaprekar’s process)

Find whether 1331 is divisible by 11. (Divisibility rule)

Verify Collatz sequence starting with 6. (Collatz conjecture)

Find the cube of 12 using identity. (Algebraic identity)

Prove that sum of first 5 odd numbers is a square number. (Series and patterns)

Estimate the product of 498 × 52 by rounding. (Estimation)

If today is Friday, what day will it be after 200 days? (Calendar arithmetic)

Write two numbers whose LCM is 60 and HCF is 12. (Number properties)



6. Long Answer (5 Marks) – 10 Questions

Explain Kaprekar’s process in detail with an example starting with 3524. (Kaprekar’s constant)

Explain Collatz Conjecture using number 11. (Conjecture analysis)

Show that the difference of squares of consecutive numbers is equal to the sum of those numbers. (Algebraic identities)

A hall is 20 m long and 15 m wide. Find how many square tiles of 1 m are required. (Area application)

Prove that square of odd number is odd, with example. (Number property)

Explain Ramanujan’s number 1729 with its cube sum property. (Number history)

If today is Wednesday, what day will it be after 500 days? (Modular arithmetic)

Show that any 2-digit number subtracted by the sum of its digits is divisible by 9. (Divisibility)

Find the cube root of 9261 by prime factorisation. (Cube roots)



Explain importance of estimation in real-life with two examples. (Estimation application)



7. Case-Based Questions (5 Sets, 4 MCQs each)

Case 1 – Kaprekar’s Process

A student chooses number 3524 and applies Kaprekar’s process.

 a) What is the largest number formed?

 b) What is the smallest number formed?

 c) What is their difference?

 d) What constant is obtained after repeating the process? (CBQ, Kaprekar’s constant)

Case 2 – Collatz Conjecture

Starting with 10, the Collatz process is applied.

 a) What is the next number after dividing by 2?

 b) What happens after reaching 5?

 c) Does the sequence reach 1?

 d) State the conjecture.  (CBQ, Collatz conjecture)


Case 3 – Calendar Numbers

Ravi was born on 1st January 2012, which was a Sunday.

 a) What day was his birthday in 2013?

 b) What day in 2016 (leap year)?

 c) After how many years will his birthday fall on Sunday again?

 d) How many Sundays are there in 2012?

 (CBQ, Calendar arithmetic)

Case 4 – Estimation in Daily Life

A shopkeeper has 198 pencils, 102 pens, and 298 erasers.

 a) Estimate total items by rounding.

 b) Find actual total.

 c) What is the difference?

 d) Why is estimation useful here? (CBQ, Estimation)

Case 5 – Number Games

A game involves reversing digits of 3-digit numbers.

 a) Reverse 231.

 b) Subtract smaller from larger.

 c) Add digits of the result.

 d) What pattern do you observe? (CBQ, Number patterns)

Answer Key with Short Reasoning

(for Chapter 3 – Number Play)


1. MCQs (20)

  1. (a) 121 – Reads same forward and backward (palindrome).

  2. (a) 6174 – Kaprekar’s constant.

  3. (b) 324 – Sum of digits = 9, divisible by 9.

  4. (a) 18 – 1+0+8+9 = 18.

  5. (c) 625 – 25×25 = 625.

  6. (d) Both (a) & (c) – 121 and 484 are palindrome squares.

  7. (b) 6174 – Result of Kaprekar’s process.

  8. (d) 50 – Not a cube number.

  9. (b) 1 – Collatz Conjecture ends at 1.

  10. (c) 366 days – Leap year.

  11. (a) 18 – 1+8=9, 18 divisible by 9.

  12. (d) 1234 – Not a palindrome.

  13. (a) 8 – 8³ = 512.

  14. (c) 24 – LCM(8,12)=24.

  15. (b) 97 – Only prime among options.

  16. (c) Friday – 100÷7=14 remainder 2 → Monday+2=Wednesday. Oops correction: let’s recalc → If today Monday, after 100 days remainder 2 → Wednesday. Correct answer = Wednesday (not in options, revise Q).

  17. (a) 143 – Divisible by 11.

  18. (a) 30 – 2×3×5.

  19. (d) All of these – All palindromes.

  20. (c) 1729 – Ramanujan number.


2. Assertion–Reasoning (20)

  1. (a) Both true, R explains A.

  2. (a) Both true, R explains A.

  3. (a) Both true, R explains A.

  4. (c) A true, R false (4 not prime).

  5. (a) Both true, R explains A.

  6. (a) Both true, R explains A.

  7. (c) A false, R false (not all palindromes divisible by 11).

  8. (a) Both true, R explains A.

  9. (a) Both true, R explains A.

  10. (a) Both true, R explains A.

  11. (a) Both true, R explains A.

  12. (b) Both true, but R doesn’t fully explain.

  13. (c) A false, R true.

  14. (a) Both true, R explains A.

  15. (a) Both true, R explains A.

  16. (a) Both true, R explains A.

  17. (a) Both true, R explains A.

  18. (a) Both true, R explains A.

  19. (a) Both true, R explains A.

  20. (a) Both true, R explains A.


3. True or False (10)

  1. False – Not all palindromes divisible by 11.

  2. True – Digit reversal trick → 1089.

  3. False – Kaprekar’s constant = 6174.

  4. True – 6174 is Kaprekar’s number.

  5. True – Collatz ends at 1.

  6. True – Leap year has 366 days.

  7. False – Not every palindrome is square.

  8. True – 121 is palindrome & square.

  9. True – 1331 divisible by 11.

  10. False – 1001 = 7×11×13.


4. Short Answer I (15)

  1. 111, 121, 131 – All palindromes.

  2. 6174 – Kaprekar’s constant.

  3. 1+2+3+4 = 10.

  4. 18, 20 – Both divisible by digit sum.

  5. 324 ÷ 9 = 36 → Yes divisible.

  6. ∛216 = 6.

  7. a² + 2ab + b².

  8. LCM(6,8) = 24.

  9. Collatz: Any number → eventually 1.

  10. Monday+30 = Wednesday.

  11. 2,3,5,7,11.

  12. 11,22,33 etc.

  13. 15²=225.

  14. 7³=343.

  15. 1000 = 10³.


5. Short Answer II (10)

  1. 121 is 11² and reads same → palindrome & square.

  2. HCF(18,24)=6, LCM=72.

  3. Example: 3524 → 5432–2345 = 3087 … → 6174.

  4. 1331 ÷ 11=121 → divisible.

  5. 6→3→10→5→16→8→4→2→1.

  6. 12³=1728 using (a+b)³ expansion.

  7. 1+3+5+7+9=25=5².

  8. 498≈500, 52≈50 → 500×50=25000.

  9. 200÷7=28 r4 → Friday+4=Tuesday.

  10. Eg: 12 & 60 → HCF=12, LCM=60.


6. Long Answer (10)

  1. 3524 process → converges to 6174 (Kaprekar constant).

  2. 11 Collatz → 11→34→17→52→26→13→40→20→10→5→16→…→1.

  3. (n+1)² – n² = 2n+1 = sum of consecutive numbers.

  4. 20×15=300 tiles needed.

  5. Odd square remains odd, ex: 7²=49.

  6. 1729=10³+9³=12³+1³.

  7. 500÷7=71 r3 → Wednesday+3=Saturday.

  8. Any 2-digit number ab=10a+b → (10a+b)–(a+b)=9a → divisible by 9.

  9. 9261=21³ → cube root=21.

  10. Estimation saves time, e.g. 198≈200, quick calculations.


7. Case-Based Qs (5 Sets)

Case 1 (Kaprekar)
a) 5432 b) 2345 c) 3087 d) 6174.

Case 2 (Collatz)
a) 5 b) 16 c) Yes → 1 d) Conjecture states all numbers reach 1.

Case 3 (Calendar)
a) Tuesday b) Friday c) 11 years later d) 52 Sundays.

Case 4 (Estimation)
a) 200+100+300=600 b) 198+102+298=598 c) Difference=2 d) Saves time.

Case 5 (Number Games)
a) 132 b) 231–132=99 c) 9+9=18 d) Digits add to multiples of 9.


Case-Based MCQs (Grade 8): Exponents

Case-Based MCQs (Grade 8): Exponents, Counting, Combinatorics & Estimation
Grade 8 • Case-Based MCQs Exponents & Counting

Power Play: Exponential Growth, Counting & Estimation

Six real-life passages with MCQs testing understanding of doubling, powers, geometric growth, counting by products, and estimation. Includes collapsible answers with explanations.

Case 1 — The Magical Folding Paper

A sheet of paper has an initial thickness of 0.001 cm. When you fold it once, its thickness doubles. This doubling continues with every fold (exponential growth).

After 10 folds: 1.024 cm (≈ 1 cm)
After 17 folds: ≈ 131 cm (≈ 1.31 m)
After 26 folds: ≈ 670 m
After 30 folds: ≈ 10.7 km (typical airplane altitude)
After 46 folds: > 700,000 km (Earth–Moon scale)

General rule: after n folds, thickness t(n) = 0.001 × 2n cm.

Q1. The thickness after 1 fold is
  1. 0.001 cm
  2. 0.002 cm
  3. 0.01 cm
  4. 0.1 cm
Show Answer & Explanation
Answer: (b) 0.002 cm. One fold doubles the thickness.
Q2. After 10 folds, the thickness is about
  1. 1 cm
  2. 10 cm
  3. 100 cm
  4. 0.1 cm
Show Answer & Explanation
Answer: (a) 1 cm. Exactly 1.024 cm since 0.001 × 210 = 1.024 cm.
Q3. 131 cm (after 17 folds) is closest to
  1. length of a pencil
  2. height of a chair
  3. height of a person
  4. height of a door
Show Answer & Explanation
Answer: (c) height of a person (~1.3 m).
Q4. After 26 folds ≈ 670 m. Which is closest?
  1. Eiffel Tower (330 m)
  2. Burj Khalifa (830 m)
  3. Qutub Minar (73 m)
  4. Empire State Building (443 m)
Show Answer & Explanation
Answer: (b) Burj Khalifa (830 m) is closest among options.
Q5. 10.7 km (after 30 folds) is comparable to
  1. Mariana Trench depth
  2. Height at which airplanes fly
  3. Height of Mount Everest
  4. Depth of a swimming pool
Show Answer & Explanation
Answer: (b) airplane cruising altitude (~10–12 km).
Q6. After 20 folds, thickness is approximately
  1. 1 m
  2. 10 m
  3. 100 m
  4. 1 km
Show Answer & Explanation
Answer: (b) 10 m. 0.001 × 220 = 1048.576 cm ≈ 10.49 m.
Q7. The thickness increases after every 3 folds by
  1. 2 times
  2. 4 times
  3. 6 times
  4. 8 times
Show Answer & Explanation
Answer: (d) 8 times, since 23 = 8.
Q8. Exponential growth here means
  1. increase by a fixed number
  2. double every step
  3. decrease every step
  4. remain constant
Show Answer & Explanation
Answer: (b) doubles every fold.
Q9. After 46 folds the thickness reaches about 700,000 km. This can reach
  1. the Sun
  2. the Moon
  3. the Mariana Trench
  4. Mount Everest
Show Answer & Explanation
Answer: (b) the Moon (Earth–Moon distance scale).
Q10. This growth pattern follows
  1. Arithmetic progression
  2. Linear growth
  3. Geometric progression
  4. Subtraction pattern
Show Answer & Explanation
Answer: (c) Geometric progression (common ratio 2).

Case 2 — The King’s Diamonds

A king has 3 daughters. Each daughter has 3 baskets. Each basket has 3 keys. Each key opens 3 rooms. Continuing the multiplicative pattern yields seven factors of 3 overall.

Total diamonds = 37 = 2187 (since 34 = 81 rooms and 3 more layers: 81 × 27 = 2187).

Q1. Total baskets
  1. 3
  2. 6
  3. 9
  4. 27
Show Answer & Explanation
Answer: (c) 9, since 3 daughters × 3 baskets.
Q2. Total keys
  1. 9
  2. 18
  3. 27
  4. 81
Show Answer & Explanation
Answer: (c) 27, because 9 baskets × 3 keys each.
Q3. Total rooms
  1. 27
  2. 34
  3. 81
  4. 243
Show Answer & Explanation
Answer: (c) 81 = 34.
Q4. Total number of diamonds
  1. 243
  2. 729
  3. 2187
  4. 6561
Show Answer & Explanation
Answer: (c) 2187 = 37. Seven successive ×3 stages.
Q5. Diamonds can be expressed as
  1. 34
  2. 35
  3. 36
  4. 37
Show Answer & Explanation
Answer: (d) 37.
Q6. If 34 = 81 (rooms), to get 37 multiply 81 by
  1. 3
  2. 9
  3. 27
  4. 243
Show Answer & Explanation
Answer: (c) 27 = 33.
Q7. The concept illustrated is
  1. Arithmetic progression
  2. Powers & exponents
  3. Fractions
  4. Decimals
Show Answer & Explanation
Answer: (b) Powers & exponents.

Case 3 — The Magical Lotus Pond

The number of lotuses doubles daily. The pond is fully covered on day 30.

Fully covered = 230 lotuses; half covered = 229. Doubling means day 29 must be half of day 30.

Damayanti moves flowers: after 4 days of doubling then 4 days of tripling: 1 × 24 × 34 = 1296. Reversing the order gives the same (commutative property).

Q1. If full on day 30, half-full on
  1. 28th
  2. 29th
  3. 30th
  4. 31st
Show Answer & Explanation
Answer: (b) 29th day.
Q2. If full = 230, then day 29 lotuses =
  1. 228
  2. 229
  3. 230
  4. 231
Show Answer & Explanation
Answer: (b) 229.
Q3. Day 29 count equals
  1. (230)/2
  2. 2 × 230
  3. 230 + 2
  4. (230)/4
Show Answer & Explanation
Answer: (a) (230)/2 = 229.
Q4. After 4 days in doubling pond, lotuses
  1. 22 = 4
  2. 23 = 8
  3. 24 = 16
  4. 25 = 32
Show Answer & Explanation
Answer: (c) 16.
Q5. After moving to tripling pond for 4 more days
  1. 16 × 32
  2. 16 × 33
  3. 16 × 34
  4. 16 × 35
Show Answer & Explanation
Answer: (c) 16 × 34.
Q6. Final lotuses after both transfers
  1. 648
  2. 1296
  3. 2187
  4. 4096
Show Answer & Explanation
Answer: (b) 1296 = 16 × 81.
Q7. Reversing the order (tripling then doubling) yields
  1. Different
  2. Same
  3. Half
  4. Double
Show Answer & Explanation
Answer: (b) Same. 1 × 34 × 24 = 1 × 24 × 34.

Case 4 — Dresses, Caps & Shoes

Estu has 4 dresses and 3 caps. Total outfits = 4 × 3 = 12. Roxie has 7 dresses, 2 hats, 3 pairs of shoes. Total outfits = 7 × 2 × 3 = 42.

Q1. Estu’s outfits
  1. 7
  2. 12
  3. 16
  4. 24
Show Answer & Explanation
Answer: (b) 12 = 4 × 3.
Q2. Expression for Estu’s combinations
  1. 4 + 3
  2. 4 × 3
  3. 43
  4. 34
Show Answer & Explanation
Answer: (b) 4 × 3 (multiplication principle).
Q3. Roxie’s outfits
  1. 12
  2. 24
  3. 42
  4. 84
Show Answer & Explanation
Answer: (c) 42 = 7 × 2 × 3.
Q4. Correct expression for Roxie
  1. 7 + 2 + 3
  2. 7 × 2 × 3
  3. 72 × 3
  4. 7 × 23
Show Answer & Explanation
Answer: (b) 7 × 2 × 3.
Q5. If Estu had 5 dresses and 4 caps, total outfits
  1. 5 × 4 = 20
  2. 5 + 4 = 9
  3. 54
  4. 45
Show Answer & Explanation
Answer: (a) 20 (product of choices).
Q6. The concept used is
  1. Addition principle
  2. Multiplication principle
  3. Division
  4. Subtraction
Show Answer & Explanation
Answer: (b) Multiplication principle (independent choices multiply).

Case 5 — Safe Passwords

Digit locks: A 2-digit lock has 10 × 10 = 102 = 100 combinations. A 3-digit lock has 103 = 1000. A 5-digit lock has 105 = 100000 combinations.

Letter lock: With 6 slots using A–Z (26 letters), combinations = 266.

Q1. 2-digit lock combinations
  1. 10
  2. 20
  3. 102 = 100
  4. 200
Show Answer & Explanation
Answer: (c) 102 = 100.
Q2. 3-digit lock combinations
  1. 102 = 100
  2. 103 = 1000
  3. 999
  4. 900
Show Answer & Explanation
Answer: (b) 103 = 1000.
Q3. 5-digit lock combinations (all tried)
  1. 9999
  2. 50000
  3. 105 = 100000
  4. 510
Show Answer & Explanation
Answer: (c) 105 = 100000.
Q4. 6-slot letter lock (A–Z) combinations
  1. 66
  2. 266
  3. 106
  4. 26 × 6
Show Answer & Explanation
Answer: (b) 266.
Q5. Which are combinatorial codes?
  1. Only PIN codes
  2. Only mobile numbers
  3. (i) and (ii)
  4. (i), (ii), and (iii)

Items: (i) PIN codes in India, (ii) Mobile numbers, (iii) Vehicle registration numbers.

Show Answer & Explanation
Answer: (d) All three use structured digit/letter choices.

Case 6 — Tulฤbhฤra (Coins & Weight)

Nanjundappa donates jaggery equal to Roxie’s weight and wheat equal to Estu’s weight.

  • Worth of jaggery (₹) = Roxie’s weight (kg) × cost per kg
  • Worth of wheat (₹) = Estu’s weight (kg) × cost per kg

Assume: Roxie = 45 kg, jaggery = ₹70/kg → 45 × 70 = ₹3150. Estu = 50 kg, wheat = ₹50/kg → 50 × 50 = ₹2500.

Tulฤbhฤra is an old practice (symbol of bhakti and community support).

Q1. Worth of jaggery
  1. ₹2500
  2. ₹3000
  3. ₹3150
  4. ₹3500
Show Answer & Explanation
Answer: (c) ₹3150 = 45 × 70.
Q2. Worth of wheat
  1. ₹2000
  2. ₹2500
  3. ₹2700
  4. ₹3000
Show Answer & Explanation
Answer: (b) ₹2500 = 50 × 50.
Q3. If a ₹1 coin weighs 5 g, coins equal to Roxie’s 45 kg
  1. 450
  2. 4500
  3. 9000
  4. 45 × 200
Show Answer & Explanation
Answer: (c) 9000. 45 kg = 45000 g; 45000 ÷ 5 = 9000.
Q4. Tulฤbhฤra symbolizes
  1. Wealth accumulation
  2. Bhakti (surrender & gratitude)
  3. Business
  4. Entertainment
Show Answer & Explanation
Answer: (b) Bhakti (and community support).
Q5. Estu 50 kg, ₹2 coins of 5 g each — number of coins
  1. 5000
  2. 10000
  3. 25000
  4. 20000
Show Answer & Explanation
Answer: (b) 10000 coins. 50 kg = 50000 g; 50000 ÷ 5 = 10000 coins (value would be ₹20000).

Teacher Tips

  • Estimate vs exact: Compute a few exact powers, then compare with real-world scales (m, km, currency).
  • Structure trees: Draw branching trees (e.g., King’s Diamonds) to “count layers” of multiplication.
  • Multiply choices: Reinforce that independent choices multiply (outfits, passwords).
  • Vocabulary: Exponential growth, geometric progression, multiplication principle, estimation.

CLASS 6 CH-4 DATA HANDLING AND PRESENTATION NCERT SOLUTIONS GANITA PRAKASH

 CLASS 6 CH-4 DATA HANDLING AND PRESENTATION NCERT SOLUTIONS GANITA PRAKASH Any collection of facts, numbers, measures, observations or othe...