General Instructions
- The question paper has 21 questions divided into six sections, carrying a total of 40 marks.
- Section A: Questions 1–8, one mark each (8 marks). Section B: Questions 9–10, one mark each (2 marks).
- Section C: Questions 11–14, two marks each (8 marks). Section D: Questions 15–18, three marks each (12 marks).
- Section E: Questions 19–20, four marks each (8 marks). Section F: Question 21, two marks (2 marks).
- In Question 20, attempt either part (a) or part (b). All other questions are compulsory.
- Write the full solution, not just the final answer, wherever marks are awarded for working.
Section A – Multiple Choice Questions
8 questions × 1 mark = 8 marks
Each question below has four options (A, B, C, D). Choose the one correct answer.
Which of the following is an irrational number?
The prime factorisation of 252 is
The HCF of 18 and 30 is
The LCM of 12 and 18 is
Which of the following is a composite number?
If HCF(56, 98) = 14, then LCM(56, 98) is equal to
The number of distinct prime factors of 210 is
Which of the following is an irrational number that lies between 2 and 3?
Section B – Assertion and Reason
2 questions × 1 mark = 2 marks
In each of the following questions, an Assertion (A) and a Reason (R) are given. Choose the correct option from (a) to (d).
- (a) Both A and R are true, and R is the correct explanation of A.
- (b) Both A and R are true, but R is not the correct explanation of A.
- (c) A is true, but R is false.
- (d) A is false, but R is true.
Assertion (A): The HCF of 48 and 60 is 12.
Reason (R): The HCF of two given numbers is the largest number that divides both of them exactly.
Assertion (A): Every irrational number can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
Reason (R): A number that can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\), is called a rational number.
Section C – Short Answer Questions
4 questions × 2 marks = 8 marks
Express 156 as a product of its prime factors.
Find the HCF of 72, 108 and 144.
Classify each of the following numbers as rational or irrational, giving a reason in one line:
(i) \(\sqrt{225}\) (ii) \(3\sqrt{5}\) (iii) \(\frac{\sqrt{8}}{\sqrt{2}}\) (iv) \(4\sqrt{81}\)
The LCM of two numbers is 360 and their HCF is 15. If one of the numbers is 45, find the other number.
Section D – Application-Based Questions
4 questions × 3 marks = 12 marks
These questions test application of concepts. Show your working clearly.
Three alarm bells in a school ring together at 12 noon. They then ring at regular intervals of 24 minutes, 36 minutes and 48 minutes respectively.
(a) After how many minutes will all three bells ring together again? [2]
(b) How many times will the bell that rings every 24 minutes have rung by that time, excluding the ring at 12 noon? [1]
Zoya has 84 red beads, 120 blue beads and 156 green beads. She wants to make identical garlands so that each garland has the same number of beads of each colour and no bead is left over.
(a) Find the maximum number of identical garlands she can make. [2]
(b) How many beads of each colour will be used in one garland? [1]
Two numbers \(A\) and \(B\) have the prime factorisations \(A = 2^3 \times 3^2 \times 5\) and \(B = 2^2 \times 3^3 \times 7\).
(a) Find HCF(\(A\), \(B\)) and LCM(\(A\), \(B\)). [2]
(b) Verify that HCF(\(A\), \(B\)) × LCM(\(A\), \(B\)) = \(A \times B\). [1]
State whether each of the following statements is true or false, and justify each answer in one sentence.
(i) The product of two irrational numbers can be a rational number.
(ii) The square of an irrational number is always irrational.
(iii) The sum of a rational number and an irrational number is always irrational.
Section E – Long Answer Questions
2 questions × 4 marks = 8 marks
A rectangular slab measures 48 cm by 54 cm.
(a) Find the side of the largest square tile that can be used to tile this slab completely without cutting any tile, and the number of such tiles needed. [2]
(b) Find the side of the smallest square that can be tiled completely using whole slabs of size 48 cm by 54 cm without gaps, and the number of slabs required. [2]
Attempt either part (a) or part (b).
(a) Prove that \(\sqrt{5}\) is an irrational number. [4]
(b) Prove that \(5 + 2\sqrt{3}\) is an irrational number. [4]
Section F – Case-Based Question
1 question × 2 marks = 2 marks
96 students and 72 teachers are going on a museum trip. The organiser wants to split everyone into the largest possible number of equal groups, such that every group contains the same number of students and the same number of teachers, and no one is left out of a group.
(a) Find the maximum number of equal groups the organiser can form. [1]
(b) How many people (students and teachers together) will be there in each group? [1]
Check your work: Section A (8) + Section B (2) + Section C (8) + Section D (12) + Section E (8) + Section F (2) = 40 marks. Attempt the paper fully before revealing the answers below.
Answer Key
Section A – Multiple Choice Questions
| Question | Answer |
|---|---|
| 1 | C |
| 2 | A |
| 3 | B |
| 4 | C |
| 5 | C |
| 6 | B |
| 7 | C |
| 8 | B |
Section B – Assertion and Reason
| Question | Answer |
|---|---|
| 9 | A |
| 10 | D |
Section C – Short Answer Questions
| Question | Answer |
|---|---|
| 11 | 156 = \(2^2 \times 3 \times 13\) |
| 12 | 36 |
| 13 | (i) rational, (ii) irrational, (iii) rational, (iv) rational |
| 14 | 120 |
Section D – Application-Based Questions
| Question | Answer |
|---|---|
| 15 | (a) 144 minutes (b) 6 times |
| 16 | (a) 12 garlands (b) 7 red, 10 blue, 13 green |
| 17 | (a) HCF = 36, LCM = 7560 (b) verified (\(272160 = 272160\)) |
| 18 | (i) True (ii) False (iii) True |
Section E – Long Answer Questions
| Question | Answer |
|---|---|
| 19 | (a) 6 cm, 72 tiles (b) 432 cm, 72 slabs |
| 20 | Full proof in Detailed Solutions. Attempt only one part. |
Section F – Case-Based Question
| Question | Answer |
|---|---|
| 21 | (a) 24 groups (b) 7 people per group |
Detailed Solutions
Concept Used: A rational number can be written as \(\frac{p}{q}\) with integers \(p, q\) \((q \neq 0)\); a number that cannot is irrational.
3, \(\frac{9}{2}\) and 15 can all be written as ratios of integers. Since 7 is not a perfect square, \(\sqrt{7}\) cannot be written as a ratio of integers.
Concept Used: Writing a number as a product of primes (Fundamental Theorem of Arithmetic).
\(252 = 2 \times 126 = 2 \times 2 \times 63 = 2 \times 2 \times 3 \times 21 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7\).
Concept Used: HCF is the product of the smallest power of each common prime factor.
\(18 = 2 \times 3^2\), \(30 = 2 \times 3 \times 5\). Common primes are 2 and 3. HCF = \(2^1 \times 3^1 = 6\).
Concept Used: LCM is the product of the greatest power of each prime factor appearing in either number.
\(12 = 2^2 \times 3\), \(18 = 2 \times 3^2\). LCM = \(2^2 \times 3^2 = 36\).
Concept Used: A composite number has more than two factors.
23, 29 and 31 are prime (exactly two factors). \(21 = 3 \times 7\) has four factors, so it is composite.
Concept Used: For two positive integers \(a\) and \(b\), HCF(\(a, b\)) × LCM(\(a, b\)) = \(a \times b\).
LCM(56, 98) = \(\frac{56 \times 98}{\text{HCF}} = \frac{56 \times 98}{14} = \frac{5488}{14} = 392\).
Concept Used: Prime factorisation: 210 = \(2 \times 3 \times 5 \times 7\).
The distinct primes present are 2, 3, 5 and 7, i.e. 4 distinct primes.
Concept Used: Identifying irrational numbers and comparing square roots.
\(\sqrt{4} = 2\) is rational, \(\frac{5}{2}\) and \(\frac{9}{4}\) are rational. Since \(4 < 6 < 9\), we have \(2 < \sqrt{6} < 3\), and 6 is not a perfect square, so \(\sqrt{6}\) is an irrational number between 2 and 3.
Concept Used: Meaning and computation of HCF.
\(48 = 2^4 \times 3\) and \(60 = 2^2 \times 3 \times 5\). HCF = \(2^2 \times 3 = 12\), so Assertion is true. The Reason correctly defines HCF and explains why 12 is the answer. Both are true and R is the correct explanation.
Concept Used: Definitions of rational and irrational numbers.
By definition, an irrational number cannot be written as \(\frac{p}{q}\). So the Assertion is false. The Reason correctly defines a rational number, so it is true.
Concept Used: Prime factorisation by repeated division.
\(156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13\).
Concept Used: HCF of three numbers using prime factorisation.
\(72 = 2^3 \times 3^2\), \(108 = 2^2 \times 3^3\), \(144 = 2^4 \times 3^2\).
Common primes: 2 and 3. Smallest power of 2 is \(2^2\) and of 3 is \(3^2\).
HCF = \(2^2 \times 3^2 = 4 \times 9 = 36\).
Concept Used: Perfect squares and surds.
(i) \(\sqrt{225} = 15\), an integer → rational.
(ii) \(3\sqrt{5}\): 5 is not a perfect square, so \(\sqrt{5}\) is irrational and \(3\sqrt{5}\) (non-zero rational × irrational) is irrational.
(iii) \(\frac{\sqrt{8}}{\sqrt{2}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\), an integer → rational.
(iv) \(4\sqrt{81} = 4 \times 9 = 36\), an integer → rational.
Given: LCM = 360, HCF = 15, one number = 45.
Concept Used: HCF × LCM = product of the two numbers.
Other number = \(\frac{\text{HCF} \times \text{LCM}}{45} = \frac{15 \times 360}{45} = \frac{5400}{45} = 120\).
Check: \(45 = 3^2 \times 5\), \(120 = 2^3 \times 3 \times 5\). LCM = \(2^3 \times 3^2 \times 5 = 360\) and HCF = \(3 \times 5 = 15\). ✓
Given: Intervals of 24, 36 and 48 minutes; all ring together at 12 noon.
Concept Used: LCM of the intervals gives the time after which all ring together again.
\(24 = 2^3 \times 3\), \(36 = 2^2 \times 3^2\), \(48 = 2^4 \times 3\).
LCM = \(2^4 \times 3^2 = 16 \times 9 = 144\) minutes.
(a) They ring together again after 144 minutes (2 hours 24 minutes).
(b) Number of times the 24-minute bell rings in 144 minutes = \(\frac{144}{24} = 6\) times (excluding 12 noon).
Given: 84 red, 120 blue, 156 green beads.
Concept Used: The maximum number of identical garlands = HCF of the three quantities.
\(84 = 2^2 \times 3 \times 7\), \(120 = 2^3 \times 3 \times 5\), \(156 = 2^2 \times 3 \times 13\).
HCF = \(2^2 \times 3 = 12\).
(a) Maximum number of garlands = 12.
(b) Red beads per garland = \({84}/{12} = 7\), blue = \({120}/{12} = 10\), green = \({156}/{12} = 13\).
Given: \(A = 2^3 \times 3^2 \times 5\), \(B = 2^2 \times 3^3 \times 7\).
Concept Used: HCF takes the smallest power of each common prime; LCM takes the greatest power of every prime present.
(a) HCF = \(2^2 \times 3^2 = 36\).
LCM = \(2^3 \times 3^3 \times 5 \times 7 = 8 \times 27 \times 35 = 7560\).
(b) \(A \times B = (8 \times 9 \times 5) \times (4 \times 27 \times 7) = 360 \times 756 = 272160\).
HCF × LCM = \(36 \times 7560 = 272160\).
Therefore HCF × LCM = \(A \times B\). ✓
Concept Used: Closure properties of rational and irrational numbers.
(i) True. Example: \(\sqrt{2} \times \sqrt{2} = 2\), a rational number.
(ii) False. Example: \(\left(\sqrt{2}\right)^2 = 2\), which is rational.
(iii) True. If \(r + s\) were rational (where \(r\) is rational and \(s\) irrational), then \(s = (r + s) - r\) would be rational (difference of rationals), which contradicts \(s\) being irrational.
Given: Rectangular slab 48 cm by 54 cm.
Concept Used: (a) HCF decides the largest square tile; (b) LCM decides the smallest square covered by whole slabs.
(a) \(48 = 2^4 \times 3\), \(54 = 2 \times 3^3\). HCF = \(2 \times 3 = 6\).
Largest square tile side = 6 cm.
Number of tiles = area of slab ÷ area of tile = \(\frac{48 \times 54}{6 \times 6} = \frac{2592}{36} = 72\).
(Alternatively, \(8 \times 9 = 72\) tiles.)
(b) LCM(48, 54) = \(2^4 \times 3^3 = 16 \times 27 = 432\).
Smallest square side = 432 cm.
Number of slabs = \(\frac{432 \times 432}{48 \times 54} = \frac{186624}{2592} = 72\) slabs.
(Alternatively, \(9 \times 8 = 72\) slabs.)
Given: To prove \(\sqrt{5}\) is irrational.
Concept Used: Proof by contradiction.
Solution:
Suppose, to the contrary, that \(\sqrt{5}\) is rational. Then \(\sqrt{5} = \frac{p}{q}\), where \(p\) and \(q\) are coprime integers and \(q \neq 0\).
Squaring both sides: \(5 = \frac{p^2}{q^2}\) ⇒ \(p^2 = 5q^2\).
So 5 divides \(p^2\). Since 5 is prime, 5 must divide \(p\). Let \(p = 5k\).
Then \(25k^2 = 5q^2\) ⇒ \(q^2 = 5k^2\), so 5 divides \(q^2\), and hence 5 divides \(q\).
Thus 5 divides both \(p\) and \(q\), contradicting that \(p\) and \(q\) are coprime.
Therefore our assumption was wrong: \(\sqrt{5}\) is irrational.
Given: To prove \(5 + 2\sqrt{3}\) is irrational.
Concept Used: Rational + irrational = irrational; \(\sqrt{3}\) is irrational (proved by the same method as \(\sqrt{5}\)).
Solution:
Suppose, to the contrary, that \(5 + 2\sqrt{3}\) is rational, say equal to \(r\).
Then \(2\sqrt{3} = r - 5\). Since \(r\) and 5 are rational, \(r - 5\) is rational, so \(2\sqrt{3}\) is rational.
But the product of the non-zero rational number 2 and \(\sqrt{3}\) would then be rational, forcing \(\sqrt{3}\) to be rational; this contradicts the fact that \(\sqrt{3}\) is irrational.
Hence our assumption was wrong: \(5 + 2\sqrt{3}\) is irrational.
Given: 96 students and 72 teachers; largest possible equal groups, same number of students and teachers per group, nobody left out.
Concept Used: HCF of 96 and 72 gives the maximum number of equal groups.
\(96 = 2^5 \times 3\), \(72 = 2^3 \times 3^2\). HCF = \(2^3 \times 3 = 24\).
(a) Maximum number of groups = 24.
(b) Students per group = \(\frac{96}{24} = 4\), teachers per group = \(\frac{72}{24} = 3\).
Total people per group = \(4 + 3 = 7\).