REAL NUMBER - CHAPTER QUESTION BANK 1
SECTION A
Q1. The HCF of 48 and 72 is
Correct Answer: Option (C) 24
Explanation:
Prime factorise both numbers:
48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
Take the lowest powers of the common prime factors:
HCF = 2³ × 3 = 8 × 3 = 24
Prime factorise both numbers:
48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
Take the lowest powers of the common prime factors:
HCF = 2³ × 3 = 8 × 3 = 24
Q2. The prime factorisation of 180 is
Correct Answer: Option (A) 2² × 3² × 5
Explanation:
Divide by the smallest prime repeatedly:
180 ÷ 2 = 90
90 ÷ 2 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
Therefore,
180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5
Divide by the smallest prime repeatedly:
180 ÷ 2 = 90
90 ÷ 2 = 45
45 ÷ 3 = 15
15 ÷ 3 = 5
Therefore,
180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5
Q3. Which of the following numbers is irrational?
Correct Answer: Option (B) √11
Explanation:
A number is irrational if it cannot be written as p/q.
0.625 = Rational
7/9 = Rational
3.25 = Rational
11 is not a perfect square, so √11 is irrational.
A number is irrational if it cannot be written as p/q.
0.625 = Rational
7/9 = Rational
3.25 = Rational
11 is not a perfect square, so √11 is irrational.
Q4. The decimal expansion of 9/125 is
Correct Answer: Option (A) Terminating
Explanation:
Prime factorise the denominator:
125 = 5³
Since the denominator contains only 2 and/or 5, the decimal expansion is terminating.
Prime factorise the denominator:
125 = 5³
Since the denominator contains only 2 and/or 5, the decimal expansion is terminating.
Q5. The LCM of 15 and 20 is
Correct Answer: Option (C) 60
Explanation:
Prime factorisation:
15 = 3 × 5
20 = 2² × 5
Take the highest powers:
LCM = 2² × 3 × 5 = 60
Prime factorisation:
15 = 3 × 5
20 = 2² × 5
Take the highest powers:
LCM = 2² × 3 × 5 = 60
Q6. Using Euclid's Division Algorithm, if a = 53, b = 7 then the remainder is
Correct Answer: Option (C) 4
Explanation:
According to Euclid's Division Algorithm:
Dividend = Divisor × Quotient + Remainder
Divide 53 by 7:
53 = 7 × 7 + 4
Remainder = 4
According to Euclid's Division Algorithm:
Dividend = Divisor × Quotient + Remainder
Divide 53 by 7:
53 = 7 × 7 + 4
Remainder = 4
Q7. The HCF of two co-prime numbers is
Correct Answer: Option (B) 1
Explanation:
Co-prime numbers have no common factor except 1.
Example: 8 and 15
Factors of 8: 1, 2, 4, 8
Factors of 15: 1, 3, 5, 15
Common factor = 1
Hence, HCF = 1
Co-prime numbers have no common factor except 1.
Example: 8 and 15
Factors of 8: 1, 2, 4, 8
Factors of 15: 1, 3, 5, 15
Common factor = 1
Hence, HCF = 1
Q8. If a rational number has denominator 2³ × 5² (in lowest form), its decimal expansion is
Correct Answer: Option (A) Terminating
Explanation:
A rational number has a terminating decimal expansion only when the denominator (in lowest form) has prime factors 2 and/or 5 only.
Since 2³ × 5² contains only 2 and 5, the decimal expansion is terminating.
A rational number has a terminating decimal expansion only when the denominator (in lowest form) has prime factors 2 and/or 5 only.
Since 2³ × 5² contains only 2 and 5, the decimal expansion is terminating.
Q9. Assertion–Reason
Assertion (A): Every prime number has exactly two distinct factors.
Reason (R): A prime number cannot be expressed as the product of two smaller natural numbers.
Assertion (A): Every prime number has exactly two distinct factors.
Reason (R): A prime number cannot be expressed as the product of two smaller natural numbers.
Correct Answer: Option (B)
Explanation:
The assertion is true because a prime number has exactly two distinct factors: 1 and itself.
The reason is also true because a prime number cannot be expressed as a product of two smaller natural numbers greater than 1.
However, the reason does not directly explain why a prime number has exactly two factors.
Hence, Option (B).
The assertion is true because a prime number has exactly two distinct factors: 1 and itself.
The reason is also true because a prime number cannot be expressed as a product of two smaller natural numbers greater than 1.
However, the reason does not directly explain why a prime number has exactly two factors.
Hence, Option (B).
Q10. Assertion–Reason
Assertion (A): Every terminating decimal number is rational.
Reason (R): Every terminating decimal can be written in the form p/q where q ≠ 0.
Assertion (A): Every terminating decimal number is rational.
Reason (R): Every terminating decimal can be written in the form p/q where q ≠ 0.
Correct Answer: Option (A)
Explanation:
Every terminating decimal can be expressed as a fraction.
Example: 0.25 = 25/100 = 1/4
Since every terminating decimal can be written as p/q, it is rational.
Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.
Every terminating decimal can be expressed as a fraction.
Example: 0.25 = 25/100 = 1/4
Since every terminating decimal can be written as p/q, it is rational.
Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.
SECTION B — Short Answer Questions (5 × 2 = 10 Marks)
Q11. Find the HCF of 252 and 180 using the prime factorisation method.
Solution:
Step 1: Prime factorisation of 252: 252 = 2 × 126 = 2 × 2 × 63 = 2² × 3² × 7
Step 2: Prime factorisation of 180: 180 = 2 × 90 = 2 × 2 × 45 = 2² × 3² × 5
Step 3: Identify common prime factors with the lowest powers:
Common factors: 2², 3²
Therefore, HCF = 2² × 3² = 4 × 9 = 36
Final Answer: 36
Step 1: Prime factorisation of 252: 252 = 2 × 126 = 2 × 2 × 63 = 2² × 3² × 7
Step 2: Prime factorisation of 180: 180 = 2 × 90 = 2 × 2 × 45 = 2² × 3² × 5
Step 3: Identify common prime factors with the lowest powers:
Common factors: 2², 3²
Therefore, HCF = 2² × 3² = 4 × 9 = 36
Final Answer: 36
| Step | Marks |
|---|---|
| Correct prime factorisation of 252 | 0.5 |
| Correct prime factorisation of 180 | 0.5 |
| Identifies common prime factors and lowest powers | 0.5 |
| Correct HCF = 36 | 0.5 |
Q12. Apply Euclid's Division Algorithm to find the HCF of 96 and 36.
Solution:
According to Euclid's Division Algorithm:
Dividend = Divisor × Quotient + Remainder
Divide 96 by 36:
96 = 36 × 2 + 24
Now divide 36 by 24:
36 = 24 × 1 + 12
Now divide 24 by 12:
24 = 12 × 2 + 0
Since the remainder becomes zero, HCF = 12
Final Answer: 12
According to Euclid's Division Algorithm:
Dividend = Divisor × Quotient + Remainder
Divide 96 by 36:
96 = 36 × 2 + 24
Now divide 36 by 24:
36 = 24 × 1 + 12
Now divide 24 by 12:
24 = 12 × 2 + 0
Since the remainder becomes zero, HCF = 12
Final Answer: 12
| Step | Marks |
|---|---|
| First division correctly performed | 0.5 |
| Remaining Euclid steps correct | 0.5 |
| Correct HCF obtained | 0.5 |
| Final answer written clearly | 0.5 |
Q13. Express 540 as a product of prime factors.
Solution:
Prime factorise step by step:
540 = 2 × 270 = 2 × 2 × 135 = 2² × 3 × 45 = 2² × 3² × 15 = 2² × 3³ × 5
Final Answer: 540 = 2² × 3³ × 5
Prime factorise step by step:
540 = 2 × 270 = 2 × 2 × 135 = 2² × 3 × 45 = 2² × 3² × 15 = 2² × 3³ × 5
Final Answer: 540 = 2² × 3³ × 5
| Step | Marks |
|---|---|
| Correct division into prime factors | 1 |
| Correct final prime factorisation | 1 |
Q14. Without performing division, determine whether 17/200 has a terminating decimal expansion. Give reason.
Solution:
Prime factorise the denominator:
200 = 2³ × 5²
Since the denominator contains only the prime factors 2 and 5, the decimal expansion will terminate.
Final Answer: Yes, the decimal expansion is terminating because the denominator has only the prime factors 2 and 5.
Prime factorise the denominator:
200 = 2³ × 5²
Since the denominator contains only the prime factors 2 and 5, the decimal expansion will terminate.
Final Answer: Yes, the decimal expansion is terminating because the denominator has only the prime factors 2 and 5.
| Step | Marks |
|---|---|
| Prime factorisation of denominator | 1 |
| Correct conclusion with reason | 1 |
Q15. Find the LCM of 12, 18 and 24.
Solution:
Prime factorisation:
12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3
Take the highest powers of each prime factor:
LCM = 2³ × 3² = 8 × 9 = 72
Final Answer: 72
Prime factorisation:
12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3
Take the highest powers of each prime factor:
LCM = 2³ × 3² = 8 × 9 = 72
Final Answer: 72
| Step | Marks |
|---|---|
| Correct prime factorisation | 1 |
| Correct LCM = 72 | 1 |
SECTION C — Competency Based Questions (4 × 3 = 12 Marks)
Q16. A school purchased:
144 Mathematics books
216 Science books
The books are to be packed into identical cartons so that each carton contains books of only one subject and the maximum possible number of books.
Find:
a) Number of books in each carton
b) Number of cartons required for each subject.
144 Mathematics books
216 Science books
The books are to be packed into identical cartons so that each carton contains books of only one subject and the maximum possible number of books.
Find:
a) Number of books in each carton
b) Number of cartons required for each subject.
Solution:
Step 1: Find the HCF of 144 and 216
Prime factorisation:
144 = 2⁴ × 3²
216 = 2³ × 3³
Take the common prime factors with the lowest powers:
HCF = 2³ × 3² = 8 × 9 = 72
Therefore, each carton can contain 72 books.
Step 2: Find the number of cartons
Mathematics Books: 144 / 72 = 2 cartons
Science Books: 216 / 72 = 3 cartons
Step 1: Find the HCF of 144 and 216
Prime factorisation:
144 = 2⁴ × 3²
216 = 2³ × 3³
Take the common prime factors with the lowest powers:
HCF = 2³ × 3² = 8 × 9 = 72
Therefore, each carton can contain 72 books.
Step 2: Find the number of cartons
Mathematics Books: 144 / 72 = 2 cartons
Science Books: 216 / 72 = 3 cartons
| Step | Marks |
|---|---|
| Finds HCF = 72 | 1 |
| Calculates number of cartons correctly | 1 |
| Writes complete final answer | 1 |
Q17. A factory packs chocolates in boxes. One machine packs 45 chocolates every minute while another packs 60 chocolates every minute. If both machines start together, after how many chocolates packed will both complete a cycle together? (Hint: Find LCM.)
Solution:
Prime factorisation:
45 = 3² × 5
60 = 2² × 3 × 5
Take the highest powers of all prime factors:
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180
Therefore, both machines complete a cycle together after 180 chocolates.
Final Answer: 180
Prime factorisation:
45 = 3² × 5
60 = 2² × 3 × 5
Take the highest powers of all prime factors:
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180
Therefore, both machines complete a cycle together after 180 chocolates.
Final Answer: 180
| Step | Marks |
|---|---|
| Prime factorisation | 1 |
| Correct LCM = 180 | 1 |
| Correct interpretation and conclusion | 1 |
Q18. Find the prime factorisation of 1260. Hence determine:
a) HCF of 1260 and 420
b) LCM of 1260 and 420
a) HCF of 1260 and 420
b) LCM of 1260 and 420
Solution:
Step 1: Prime factorisation of 1260:
1260 = 2 × 630 = 2² × 315 = 2² × 3² × 5 × 7
Prime factorisation of 420:
420 = 2² × 3 × 5 × 7
Step 2: HCF (Take the lowest powers)
HCF = 2² × 3 × 5 × 7 = 420
Step 3: LCM (Take the highest powers)
LCM = 2² × 3² × 5 × 7 = 1260
Final Answer:
Prime factorisation of 1260: 2² × 3² × 5 × 7
HCF: 420
LCM: 1260
Step 1: Prime factorisation of 1260:
1260 = 2 × 630 = 2² × 315 = 2² × 3² × 5 × 7
Prime factorisation of 420:
420 = 2² × 3 × 5 × 7
Step 2: HCF (Take the lowest powers)
HCF = 2² × 3 × 5 × 7 = 420
Step 3: LCM (Take the highest powers)
LCM = 2² × 3² × 5 × 7 = 1260
Final Answer:
Prime factorisation of 1260: 2² × 3² × 5 × 7
HCF: 420
LCM: 1260
| Step | Marks |
|---|---|
| Correct prime factorisation of 1260 | 1 |
| Correct HCF | 1 |
| Correct LCM | 1 |
Q19. Given that √7 is irrational, prove that 4 + √7 is also irrational.
Solution:
We will use the method of contradiction.
Assume that 4 + √7 is rational.
Subtracting 4 from both sides:
√7 = (4 + √7) - 4
Since 4 is a rational number and the difference of two rational numbers is rational, this would imply that √7 is rational.
But this contradicts the given fact that √7 is irrational.
Therefore, our assumption is false.
Hence, 4 + √7 is irrational.
We will use the method of contradiction.
Assume that 4 + √7 is rational.
Subtracting 4 from both sides:
√7 = (4 + √7) - 4
Since 4 is a rational number and the difference of two rational numbers is rational, this would imply that √7 is rational.
But this contradicts the given fact that √7 is irrational.
Therefore, our assumption is false.
Hence, 4 + √7 is irrational.
| Step | Marks |
|---|---|
| Assumes is rational | 1 |
| Derives contradiction by subtracting 4 | 1 |
| Correct conclusion | 1 |
SECTION D — Long Answer Questions (3 × 6 = 18 Marks)
Q20. Apply Euclid's Division Algorithm to determine the HCF of 868 and 372. Verify your answer using prime factorisation.
Solution:
Part A: Using Euclid's Division Algorithm
According to Euclid's Division Algorithm:
Dividend = Divisor × Quotient + Remainder
Step 1: Divide 868 by 372.
868 = 372 × 2 + 124
(Quotient = 2, Remainder = 124)
Step 2: Divide 372 by 124.
372 = 124 × 3 + 0
Since the remainder is 0, the last divisor is the HCF.
HCF = 124
Part B: Verification by Prime Factorisation
Prime factorise 868:
868 = 2 × 434 = 2 × 2 × 217 = 2² × 7 × 31
Prime factorise 372:
372 = 2 × 186 = 2 × 2 × 93 = 2² × 3 × 31
Take the common prime factors with the smallest powers:
HCF = 2² × 31 = 4 × 31 = 124
The answer obtained by prime factorisation matches the answer obtained using Euclid's Division Algorithm.
Hence, the verification is complete.
Final Answer: HCF = 124
Part A: Using Euclid's Division Algorithm
According to Euclid's Division Algorithm:
Dividend = Divisor × Quotient + Remainder
Step 1: Divide 868 by 372.
868 = 372 × 2 + 124
(Quotient = 2, Remainder = 124)
Step 2: Divide 372 by 124.
372 = 124 × 3 + 0
Since the remainder is 0, the last divisor is the HCF.
HCF = 124
Part B: Verification by Prime Factorisation
Prime factorise 868:
868 = 2 × 434 = 2 × 2 × 217 = 2² × 7 × 31
Prime factorise 372:
372 = 2 × 186 = 2 × 2 × 93 = 2² × 3 × 31
Take the common prime factors with the smallest powers:
HCF = 2² × 31 = 4 × 31 = 124
The answer obtained by prime factorisation matches the answer obtained using Euclid's Division Algorithm.
Hence, the verification is complete.
Final Answer: HCF = 124
| Step | Marks |
|---|---|
| Correct Euclid's Division Algorithm steps | 2 |
| Correct HCF = 124 | 1 |
| Correct prime factorisation of both numbers | 1 |
| Correct verification | 1 |
| Proper mathematical conclusion | 1 |
Notes for Evaluators:
• Arithmetic error with correct method: Deduct 0.5 mark.
• If only prime factorisation is used without Euclid's Algorithm, award a maximum of 3 marks, as the specified method was not followed.
• Arithmetic error with correct method: Deduct 0.5 mark.
• If only prime factorisation is used without Euclid's Algorithm, award a maximum of 3 marks, as the specified method was not followed.
Q21. Three school bells ring at intervals of 24 minutes, 36 minutes, and 48 minutes. If they ring together at 8:00 a.m., at what time will they next ring together? Explain using LCM.
Solution:
Since the bells ring at fixed intervals, we find the Least Common Multiple (LCM) of 24, 36, and 48.
Step 1: Prime Factorisation
24 = 2³ × 3
36 = 2² × 3²
48 = 2⁴ × 3
Step 2: Find the LCM
Take the highest powers of each prime factor:
LCM = 2⁴ × 3² = 16 × 9 = 144
Thus, the bells will ring together again after 144 minutes.
Step 3: Convert Minutes into Hours
144 minutes = 2 hours 24 minutes
Step 4: Add to the Starting Time
Starting time: 8:00 a.m.
After 2 hours 24 minutes: 8:00 + 2:24 = 10:24 a.m.
Final Answer: The three bells will ring together again at 10:24 a.m.
Since the bells ring at fixed intervals, we find the Least Common Multiple (LCM) of 24, 36, and 48.
Step 1: Prime Factorisation
24 = 2³ × 3
36 = 2² × 3²
48 = 2⁴ × 3
Step 2: Find the LCM
Take the highest powers of each prime factor:
LCM = 2⁴ × 3² = 16 × 9 = 144
Thus, the bells will ring together again after 144 minutes.
Step 3: Convert Minutes into Hours
144 minutes = 2 hours 24 minutes
Step 4: Add to the Starting Time
Starting time: 8:00 a.m.
After 2 hours 24 minutes: 8:00 + 2:24 = 10:24 a.m.
Final Answer: The three bells will ring together again at 10:24 a.m.
| Step | Marks |
|---|---|
| Correct prime factorisation | 2 |
| Correct LCM = 144 minutes | 1 |
| Correct conversion to 2 hours 24 minutes | 1 |
| Correct time = 10:24 a.m. | 1 |
| Final concluding statement | 1 |
Q22. Case Study (6 Marks)
A stationery shop receives:
360 notebooks
540 pens
720 pencils
The owner wants to arrange them into identical gift packs. Each pack should contain only one type of item and have the maximum possible number of items.
Answer the following:
(a) Find the maximum number of items in each pack. (2)
(b) Number of notebook packs. (1)
(c) Number of pen packs. (1)
(d) Number of pencil packs. (1)
(e) Total number of packs prepared. (1)
A stationery shop receives:
360 notebooks
540 pens
720 pencils
The owner wants to arrange them into identical gift packs. Each pack should contain only one type of item and have the maximum possible number of items.
Answer the following:
(a) Find the maximum number of items in each pack. (2)
(b) Number of notebook packs. (1)
(c) Number of pen packs. (1)
(d) Number of pencil packs. (1)
(e) Total number of packs prepared. (1)
Solution:
Step 1: Find the HCF of 360, 540, and 720.
Prime factorisation:
360 = 2³ × 3² × 5
540 = 2² × 3³ × 5
720 = 2⁴ × 3² × 5
Take the common prime factors with the lowest powers:
HCF = 2² × 3² × 5 = 4 × 9 × 5 = 180
Hence, each gift pack can contain 180 items.
(b) Number of notebook packs = 360 / 180 = 2
(c) Number of pen packs = 540 / 180 = 3
(d) Number of pencil packs = 720 / 180 = 4
(e) Total number of packs = 2 + 3 + 4 = 9
Total packs = 9
Step 1: Find the HCF of 360, 540, and 720.
Prime factorisation:
360 = 2³ × 3² × 5
540 = 2² × 3³ × 5
720 = 2⁴ × 3² × 5
Take the common prime factors with the lowest powers:
HCF = 2² × 3² × 5 = 4 × 9 × 5 = 180
Hence, each gift pack can contain 180 items.
(b) Number of notebook packs = 360 / 180 = 2
(c) Number of pen packs = 540 / 180 = 3
(d) Number of pencil packs = 720 / 180 = 4
(e) Total number of packs = 2 + 3 + 4 = 9
Total packs = 9
| Step | Marks |
|---|---|
| Correct HCF = 180 | 2 |
| Correct number of notebook, pen, and pencil packs | 2 |
| Correct total number of packs | 1 |
| Final conclusion | 1 |
General Evaluation Guidelines
Method Marks: Award method marks if the correct procedure is followed, even if there is a minor arithmetic error.
Step Marks: Do not deduct all marks for a single calculation mistake if the remaining work follows logically.
Presentation: Award full marks when the solution is clear, logically organised, and the final answer is correctly highlighted.
Alternative Methods: Accept any mathematically valid method unless the question explicitly asks for a specific method (e.g., Euclid's Division Algorithm or Prime Factorisation).
Step Marks: Do not deduct all marks for a single calculation mistake if the remaining work follows logically.
Presentation: Award full marks when the solution is clear, logically organised, and the final answer is correctly highlighted.
Alternative Methods: Accept any mathematically valid method unless the question explicitly asks for a specific method (e.g., Euclid's Division Algorithm or Prime Factorisation).