Real Numbers · Class 10 Mathematics — 176 MCQs with Answers

Q1. Which of these numbers is prime?

Beginner

A 91 B 87 C 51 D 97

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97

$91 = 7 \times 13$, $87 = 3 \times 29$ and $51 = 3 \times 17$. No prime up to 9 divides 97, so 97 is prime.

Q2. Which statement about the number 1 is true?

Beginner

A It is the smallest prime B It is the smallest composite C It is both prime and composite D It is neither prime nor composite

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It is neither prime nor composite

A prime has exactly two factors and a composite more than two. 1 has only one factor, so it is neither.

Q3. How many even prime numbers are there?

Beginner

A None B Exactly one C Exactly two D Infinitely many

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Exactly one

2 is prime. Every larger even number has 1, 2 and itself as factors, so it is composite.

Q4. The prime factorisation of 360 is

Beginner

A $2^2 \times 3^2 \times 10$ B $2^3 \times 3^2 \times 5$ C $2^3 \times 45$ D $2^3 \times 3 \times 15$

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$2^3 \times 3^2 \times 5$

$360 = 36 \times 10 = 2^2 \times 3^2 \times 2 \times 5 = 2^3 \times 3^2 \times 5$. The other options contain the composites 10, 45 or 15.

Q5. The prime factorisation of 140 is

Beginner

A $2 \times 5 \times 14$ B $2^2 \times 35$ C $2 \times 7 \times 10$ D $2^2 \times 5 \times 7$

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$2^2 \times 5 \times 7$

$140 = 2 \times 70 = 2 \times 2 \times 35 = 2^2 \times 5 \times 7$.

Q6. Written as a product of primes, 156 is

Beginner

A $2^2 \times 3 \times 13$ B $2 \times 3 \times 26$ C $2^2 \times 39$ D $2^3 \times 3 \times 13$

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$2^2 \times 3 \times 13$

$156 = 2 \times 78 = 2 \times 2 \times 39 = 2^2 \times 3 \times 13$.

Q7. Which of these is the prime factorisation of a number, written correctly?

Beginner

A $2 \times 9 \times 5$ B $2^2 \times 3 \times 7$ C $4 \times 3 \times 11$ D $3 \times 5 \times 21$

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$2^2 \times 3 \times 7$

Every factor must be prime. 9, 4 and 21 are composite, so only $2^2 \times 3 \times 7$ qualifies.

Q8. A number's prime factorisation is $2^3 \times 3 \times 5$. The number is

Beginner

A 60 B 120 C 240 D 90

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120

$8 \times 3 \times 5 = 120$.

Q9. Which of these is composite?

Beginner

A 221 B 223 C 211 D 227

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221

$221 = 13 \times 17$. The others have no prime factor up to their square roots, so they are prime.

Q10. According to the Fundamental Theorem of Arithmetic, the prime factorisation of a composite number is

Beginner

A unique, including the order of the factors B possible in exactly two ways C unique, apart from the order of the factors D possible only for even numbers

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unique, apart from the order of the factors

The theorem says the factorisation exists and is unique; $2 \times 3$ and $3 \times 2$ count as the same.

Q11. A number ends in the digit 0. Its prime factorisation must contain

Beginner

A 2 but not 5 B both 2 and 5 C 5 but not 2 D at least one 3

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both 2 and 5

Ending in 0 means divisible by $10 = 2 \times 5$, so both primes must appear.

Q12. To test whether 173 is prime, the largest prime you need to try as a divisor is

Beginner

A 13 B 11 C 17 D 7

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13

$13^2 = 169 < 173 < 17^2 = 289$, so the primes to try stop at 13. (None divides 173, so it is prime.)

Q13. In the division ladder for 4620, the numbers are divided in turn by 2, 2, 3, 5 and 7. What is left at the end?

Beginner

A 1 B 7 C 13 D 11

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11

$4620 \div 2 \div 2 \div 3 \div 5 \div 7 = 11$, which is prime, so it is the last factor.

Q14. $72 = 8 \times 9$ and $72 = 6 \times 12$. Both factor trees end at

Beginner

A $2^2 \times 3^3$ B $2^3 \times 3^2$ C $2^3 \times 3^3$ D $2^2 \times 3^2 \times 2$

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$2^3 \times 3^2$

$8 \times 9 = 2^3 \times 3^2$ and $6 \times 12 = (2 \times 3)(2^2 \times 3) = 2^3 \times 3^2$ — the same, as uniqueness promises.

Q15. Assertion (A): 1 is not a prime number. Reason (R): 2 is the only even prime number.

Beginner

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

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Both A and R are true, but R is not the correct explanation of A

Both are true. A fact about 2 does not explain why 1 fails the definition of a prime.

Q16. Which of these numbers has exactly two prime factors (counted without repetition)?

Beginner

A $2^3 \times 7^2$ B $2 \times 3 \times 5$ C $3^4$ D $2 \times 3 \times 5 \times 7$

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$2^3 \times 7^2$

$2^3 \times 7^2$ uses only the primes 2 and 7. The others use three, one and four different primes.

Q17. Express 3150 as a product of its prime factors in power form.

Beginner

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Q18. Check whether $6^n$ can end with the digit 0 for any natural number n.

Beginner

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Q19. Explain why $7 \times 11 \times 13 + 13$ is a composite number.

Beginner

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Q20. Explain why $3 \times 5 \times 7 + 7$ is a composite number.

Beginner

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Q21. The HCF of $2^3 \times 3^2 \times 5$ and $2^2 \times 3^3 \times 7$ is

Beginner

A $2^3 \times 3^3$ B $2^2 \times 3^2 \times 5 \times 7$ C $2^2 \times 3^2$ D $2^3 \times 3^3 \times 5 \times 7$

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$2^2 \times 3^2$

Common primes 2 and 3, smallest powers: $2^2 \times 3^2$. 5 and 7 are not common.

Q22. The LCM of $2^3 \times 3^2 \times 5$ and $2^2 \times 3^3 \times 7$ is

Beginner

A $2^3 \times 3^3 \times 5 \times 7$ B $2^2 \times 3^2$ C $2^3 \times 3^3$ D $2^5 \times 3^5 \times 5 \times 7$

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$2^3 \times 3^3 \times 5 \times 7$

Every prime at its greatest power: $2^3, 3^3, 5, 7$.

Q23. HCF(26, 91) is

Beginner

A 13 B 7 C 1 D 182

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13

$26 = 2 \times 13$, $91 = 7 \times 13$. Only 13 is common.

Q24. LCM(26, 91) is

Beginner

A 182 B 2366 C 91 D 13

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182

$2 \times 7 \times 13 = 182$. Check: $13 \times 182 = 2366 = 26 \times 91$.

Q25. Taking any two positive integers a and b, the product HCF(a, b) × LCM(a, b) comes to

Beginner

A $a + b$ B HCF(a, b) C $a - b$ D $a \times b$

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$a \times b$

Each prime's smaller power goes to the HCF and larger to the LCM, together giving exactly $a \times b$.

Q26. Two numbers have HCF 1. The numbers are called

Beginner

A co-prime B prime C composite D twin primes

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co-prime

HCF 1 means no common prime factor — they are co-prime. 8 and 15 are co-prime though neither is prime.

Q27. The LCM of 8 and 15 is

Beginner

A 120 B 23 C 60 D 1

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120

8 and 15 are co-prime, so LCM = $8 \times 15 = 120$.

Q28. HCF(15, 60) is

Beginner

A 60 B 5 C 15 D 3

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15

15 divides 60, so the HCF is 15 itself (and the LCM is 60).

Q29. Bells ring every 6 and 8 minutes and ring together now. They ring together again after

Beginner

A 24 minutes B 48 minutes C 14 minutes D 2 minutes

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24 minutes

Coinciding again is the LCM: $\text{LCM}(6, 8) = 24$.

Q30. A problem asks for the largest size of equal pieces that two ropes of 48 m and 72 m can be cut into. It needs

Beginner

A the LCM B the HCF C the product D the difference

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the HCF

Largest piece that divides both lengths exactly — the HCF, which is 24 m.

Q31. Assertion (A): HCF(8, 9) = 1. Reason (R): Two consecutive numbers always share the common factor 2.

Beginner

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

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A is true but R is false

Consecutive numbers are co-prime, sharing no factor beyond 1, so the reason is false while the assertion is correct.

Q32. Which statement is always true for two positive integers?

Beginner

A The LCM divides the HCF B The HCF divides the LCM C HCF + LCM = sum of the numbers D HCF = LCM

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The HCF divides the LCM

Every prime power in the HCF is also present in the LCM, at least as large — so the HCF divides the LCM.

Q33. The HCF of three numbers is 6 and they are 12, 18 and 30. Their LCM is

Beginner

A 360 B 180 C 6480 D 90

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180

$2^2 \times 3^2 \times 5 = 180$.

Q34. Find the HCF and LCM of 26 and 91, and verify that HCF × LCM = product of the numbers.

Beginner

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Q35. Find the HCF and LCM of 12, 15 and 21 by prime factorisation.

Beginner

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Q36. Given HCF(306, 657) = 9, find LCM(306, 657).

Beginner

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Q37. Which of these numbers is irrational?

Beginner

A $\sqrt{49}$ B $\sqrt{\frac{9}{16}}$ C $0.25$ D $\sqrt{7}$

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$\sqrt{7}$

$\sqrt{49} = 7$, $\sqrt{\frac{9}{16}} = \frac{3}{4}$ and $0.25 = \frac{1}{4}$ are rational. 7 is not a perfect square, so $\sqrt{7}$ is irrational.

Q38. A number is rational if it can be written as $\frac{p}{q}$ where

Beginner

A p and q are integers and $q \neq 0$ B p and q are any real numbers C p and q are primes D p is an integer and q is irrational

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p and q are integers and $q \neq 0$

That is the definition of a rational number.

Q39. Which of these is rational?

Beginner

A $\sqrt{2} + 1$ B $\sqrt{2} \times \sqrt{2}$ C $3\sqrt{2}$ D $\frac{\sqrt{2}}{3}$

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$\sqrt{2} \times \sqrt{2}$

$\sqrt{2} \times \sqrt{2} = 2$. The other three are a non-zero rational combined with $\sqrt{2}$, so they are irrational.

Q40. The theorem used in proving $\sqrt{3}$ irrational says: if 3 divides $a^2$, then

Beginner

A 9 divides a B 3 divides a C a is prime D 3 divides $a + 1$

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3 divides a

For a prime p, $p \mid a^2 \Rightarrow p \mid a$. With p = 3: 3 divides a.

Q41. A proof by contradiction begins by

Beginner

A assuming what is to be proved B checking examples C drawing a figure D assuming the opposite of what is to be proved

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assuming the opposite of what is to be proved

We assume the statement is false and show this leads to something impossible.

Q42. In the proof that $\sqrt{2}$ is irrational, $\sqrt{2}$ is written as $\frac{a}{b}$ where a and b are

Beginner

A both even B co-prime integers, $b \neq 0$ C both prime D irrational numbers

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co-prime integers, $b \neq 0$

Any common factor is cancelled first, so a and b are co-prime. The contradiction at the end depends on it.

Q43. Squaring $\sqrt{5} = \frac{a}{b}$ and rearranging gives

Beginner

A $5a^2 = b^2$ B $25b^2 = a^2$ C $5b = a^2$ D $5b^2 = a^2$

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$5b^2 = a^2$

$\sqrt{5}\,b = a$, so $5b^2 = a^2$.

Q44. Which of these numbers is NOT irrational?

Beginner

A $\sqrt{0.9}$ B $\sqrt{0.09}$ C $\sqrt{3}$ D $2 + \sqrt{5}$

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$\sqrt{0.09}$

$\sqrt{0.09} = 0.3 = \frac{3}{10}$, rational. $0.9 = \frac{9}{10}$ is not a perfect-square ratio, so $\sqrt{0.9}$ is irrational.

Q45. Assertion (A): $\sqrt{16}$ is a rational number. Reason (R): The square root of every natural number is irrational.

Beginner

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

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A is true but R is false

A is true: $\sqrt{16} = 4$. R is false — the square root of a perfect square is rational.

Q46. Which of these statements is true?

Beginner

A An irrational times an irrational is irrational B An irrational plus an irrational is irrational C A non-zero rational times an irrational is irrational D Zero times an irrational is irrational

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A non-zero rational times an irrational is irrational

$\sqrt{2} \times \sqrt{2} = 2$ and $\sqrt{2} + (-\sqrt{2}) = 0$ break the middle two; $0 \times \sqrt{2} = 0$ breaks the last.

Q47. $\sqrt{2}$ multiplied by which number gives a rational result?

Beginner

A $\sqrt{3}$ B $\sqrt{6}$ C $\sqrt{8}$ D $2$

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$\sqrt{8}$

$\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$. The others give $\sqrt{6}$, $2\sqrt{3}$ and $2\sqrt{2}$.

Q48. Which of these is irrational?

Beginner

A $(\sqrt{3})^2$ B $(1 + \sqrt{2})^2$ C $\frac{\sqrt{18}}{\sqrt{2}}$ D $\sqrt{25} - 1$

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$(1 + \sqrt{2})^2$

$(1 + \sqrt{2})^2 = 3 + 2\sqrt{2}$, irrational. The others are 3, 3 and 4.

Q49. Which of these is an irrational number?

Beginner

A $\sqrt{121}$ B $\frac{22}{7}$ C $\sqrt{11} \times \sqrt{11}$ D $3 + \sqrt{11}$

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$3 + \sqrt{11}$

$\sqrt{121} = 11$, $\frac{22}{7}$ is a fraction and $\sqrt{11} \times \sqrt{11} = 11$. A rational plus an irrational, $3 + \sqrt{11}$, is irrational.

Q50. Given that $\sqrt{2}$ is irrational, prove that $\frac{1}{\sqrt{2}}$ is irrational.

Beginner

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Q51. Given that $\sqrt{5}$ is irrational, prove that $7\sqrt{5}$ is irrational.

Beginner

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Q52. Given that $\sqrt{2}$ is irrational, prove that $6 + \sqrt{2}$ is irrational.

Beginner

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Q53. In a factor tree, $x = 2 \times a$, $a = 2 \times b$, $b = 3 \times c$ and $c = 3 \times 7$. The value of x is

Intermediate

A 126 B 504 C 252 D 189

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252

Bottom up: $c = 21$, $b = 63$, $a = 126$, $x = 252$.

Q54. A factor tree reads $x = 5 \times y$, $y = 3 \times z$, $z = 3 \times 11$. The value of y is

Intermediate

A 33 B 495 C 45 D 99

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99

$z = 33$, so $y = 3 \times 33 = 99$ (and $x = 495$).

Q55. The prime factorisation of 3825 is

Intermediate

A $3 \times 5^2 \times 51$ B $3^2 \times 5^2 \times 17$ C $3^2 \times 5 \times 85$ D $3^3 \times 5^2 \times 17$

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$3^2 \times 5^2 \times 17$

$3825 \div 5 = 765$, $\div 5 = 153$, $\div 3 = 51$, $\div 3 = 17$. So $3^2 \times 5^2 \times 17$.

Q56. Which product of primes equals 5005?

Intermediate

A $5 \times 7 \times 11 \times 13$ B $5 \times 7 \times 143$ C $5 \times 11 \times 91$ D $5^2 \times 7 \times 13$

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$5 \times 7 \times 11 \times 13$

$5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$.

Q57. 7429 is the product of three primes. Which are they?

Intermediate

A 13, 19 and 29 B 17, 19 and 23 C 7, 11 and 97 D 11, 17 and 41

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17, 19 and 23

$17 \times 19 = 323$ and $323 \times 23 = 7429$.

Q58. For how many natural numbers n does $4^n$ end with the digit 0?

Intermediate

A Exactly one B Only even n C None D All n

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None

$4^n = 2^{2n}$ has no factor 5. By uniqueness of factorisation, a 5 can never appear, so $4^n$ never ends in 0.

Q59. Which of these can end with the digit 0 for some natural number n?

Intermediate

A $6^n$ B $20^n$ C $15^n$ D $12^n$

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$20^n$

$20^n = 2^{2n} \times 5^n$ has both 2 and 5. $6^n$ and $12^n$ lack 5; $15^n$ lacks 2.

Q60. $15^n$ can never end in 0 because its prime factorisation has no

Intermediate

A 5 B 3 C odd prime D 2

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2

$15^n = 3^n \times 5^n$. It has the 5 but never the 2, so it is never a multiple of 10.

Q61. Which expression shows most directly that $7 \times 11 \times 13 + 13$ is composite?

Intermediate

A $13 \times 78$ B $1014$ C $7 \times 11 \times 14$ D $13 \times 77$

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$13 \times 78$

Take 13 out of both terms: $13(7 \times 11 + 1) = 13 \times 78$ — a product of two numbers bigger than 1.

Q62. $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ equals

Intermediate

A $5 \times 1008$ B $7 \times 721$ C $5 \times 1015$ D $5 \times 1009$

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$5 \times 1009$

$5040 + 5 = 5 \times (1008 + 1) = 5 \times 1009$, so it is composite.

Q63. $N = 2^3 \times 3^2 \times 5 \times 7$. N is NOT divisible by

Intermediate

A 12 B 14 C 25 D 45

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25

$25 = 5^2$ but N has only one 5. $12 = 2^2 \times 3$, $14 = 2 \times 7$ and $45 = 3^2 \times 5$ all fit inside N.

Q64. If $2^a \times 3^b = 72$, then $a + b$ equals

Intermediate

A 6 B 4 C 5 D 7

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5

$72 = 2^3 \times 3^2$, so $a = 3$, $b = 2$, and by uniqueness these are the only values: $a + b = 5$.

Q65. If $2^x \times 5^y = 2000$, the value of $x - y$ is

Intermediate

A 1 B 0 C 2 D −1

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1

$2000 = 2^4 \times 5^3$, so $x = 4$, $y = 3$ and $x - y = 1$.

Q66. The smallest natural number by which 1386 must be multiplied to make it a perfect square is

Intermediate

A 154 B 22 C 77 D 14

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154

$1386 = 2 \times 3^2 \times 7 \times 11$. The primes with odd powers are 2, 7 and 11, so multiply by $2 \times 7 \times 11 = 154$.

Q67. How many distinct prime factors does 2310 have?

Intermediate

A 4 B 5 C 6 D 3

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5

$2310 = 2 \times 3 \times 5 \times 7 \times 11$ — five distinct primes.

Q68. The largest prime factor of 8232 is

Intermediate

A 3 B 2 C 7 D 49

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7

$8232 = 2^3 \times 3 \times 7^3$, so the primes are 2, 3 and 7; the largest is 7. (49 is not prime.)

Q69. Assertion (A): $6^n$ can end with the digit 0 for some natural number n. Reason (R): A number ending in 0 must have both 2 and 5 in its prime factorisation.

Intermediate

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

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A is false but R is true

R is true. A is false: $6^n = 2^n \times 3^n$ never contains 5, so it never ends in 0.

Q70. Assertion (A): $2^3 \times 3^2 \times 5$ and $5 \times 3^2 \times 2^3$ are two different prime factorisations of 360. Reason (R): The prime factorisation of a number is unique apart from the order of its factors.

Intermediate

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

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A is false but R is true

R is the theorem, so it is true. A is false: changing the order does not give a different factorisation.

Q71. A student writes $180 = 2^2 \times 45$ as the prime factorisation of 180. What is wrong?

Intermediate

A $2^2$ should be $2^3$ B 180 has no prime factor 2 C 45 is not prime D Nothing — it is correct

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45 is not prime

In a prime factorisation every factor is prime. $45 = 3^2 \times 5$, so the answer is $2^2 \times 3^2 \times 5$.

Q72. Is 391 prime? Show your working.

Intermediate

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Q73. In a factor tree, $x = 2 \times y$, $y = 3 \times z$, $z = 5 \times w$ and $w = 5 \times 7$. Find w, z, y and x.

Intermediate

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Q74. Without dividing, decide whether $N = 2^4 \times 3^2 \times 7$ is divisible by (i) 36 (ii) 21 (iii) 27. Give reasons.

Intermediate

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Q75. Find the smallest number by which 1386 must be multiplied so that the product is a perfect square.

Intermediate

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Q76. (a) State the Fundamental Theorem of Arithmetic. (b) Find the prime factorisation of 4620 by the division ladder. (c) Hence decide, giving reasons, whether 4620 is divisible by 12, by 25 and by 77.

Intermediate

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1 mark — (a) theorem stated with both existence and uniqueness. 1½ marks — (b) correct ladder and power form. ½ mark — (c) 12, with reason. 1 mark — (c) 25, with reason. 1 mark — (c) 77, with reason.

Q77. Given HCF(306, 657) = 9, LCM(306, 657) is

Intermediate

A 201042 B 22338 C 2482 D 11169

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22338

$\frac{306 \times 657}{9} = 34 \times 657 = 22338$.

Q78. The HCF of two numbers is 16 and their product is 3072. Their LCM is

Intermediate

A 3056 B 48 C 192 D 384

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192

LCM = product ÷ HCF = $3072 \div 16 = 192$.

Q79. The LCM of two numbers is 180 and their HCF is 6. If one number is 30, the other is

Intermediate

A 30 B 18 C 36 D 60

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36

Other $= \frac{6 \times 180}{30} = 36$. Check: HCF(30, 36) = 6 and LCM = 180 ✓

Q80. Which pair CANNOT be the HCF and LCM of two numbers?

Intermediate

A HCF 12, LCM 72 B HCF 5, LCM 150 C HCF 1, LCM 77 D HCF 18, LCM 380

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HCF 18, LCM 380

The HCF must divide the LCM. $380 \div 18$ is not whole, so that pair is impossible.

Q81. Sonia takes 18 minutes and Ravi 12 minutes to go once round a track. Starting together at the same point, they next meet at the start after

Intermediate

A 6 minutes B 36 minutes C 216 minutes D 30 minutes

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36 minutes

They meet at the start at the first common multiple: LCM(18, 12) = 36.

Q82. A floor 420 cm by 130 cm is covered with the largest possible square tiles, uncut. The side of a tile is

Intermediate

A 5 cm B 20 cm C 2 cm D 10 cm

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10 cm

HCF(420, 130): $420 = 2^2 \times 3 \times 5 \times 7$, $130 = 2 \times 5 \times 13$, so HCF = 10.

Q83. An army band of 616 members marches behind a contingent of 32, in the same number of columns. The maximum number of columns is

Intermediate

A 8 B 4 C 16 D 32

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8

The columns must divide both 616 and 32, as many as possible: HCF(616, 32) = 8.

Q84. The largest number that divides 245 and 1029 leaving a remainder of 5 in each case is

Intermediate

A 5 B 8 C 16 D 32

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16

It divides $240$ and $1024$ exactly: HCF$(2^4 \times 3 \times 5, 2^{10}) = 2^4 = 16$.

Q85. The smallest number that is divisible by each of 12, 15 and 20 is

Intermediate

A 120 B 60 C 180 D 3600

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60

$\text{LCM} = 2^2 \times 3 \times 5 = 60$.

Q86. If $a = 2^2 \times 3^x$, $b = 2^2 \times 3 \times 5$ and HCF(a, b) = 12, the smallest possible natural number x is

Intermediate

A 0 B 1 C 2 D 3

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1

HCF $= 2^2 \times 3^{\min(x, 1)} = 12 = 2^2 \times 3$ needs $\min(x, 1) = 1$, so $x \ge 1$. The smallest is 1.

Q87. The LCM of $2^x \times 3^2$ and $2^3 \times 3^4 \times 5$ is $2^5 \times 3^4 \times 5$. The value of x is

Intermediate

A 5 B 3 C 2 D 8

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5

The LCM takes the greater power of 2. It is $2^5$, and b has only $2^3$, so $x = 5$.

Q88. Assertion (A): For the numbers 6, 72 and 120, the product of the HCF and the LCM equals the product of the three numbers. Reason (R): For any two positive integers, the product of their HCF and LCM equals their product.

Intermediate

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

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A is false but R is true

R is true for two numbers. A is false: 6 × 360 = 2160, not 51840. The rule does not extend to three numbers.

Q89. Assertion (A): The LCM of 17 and 23 is 391. Reason (R): The LCM of two co-prime numbers is their sum.

Intermediate

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

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A is true but R is false

It is their product, which is why the answer is 391 and not 40. The assertion stands, the reason is false.

Q90. HCF(a, b) = 1. Then LCM(a, b) must be

Intermediate

A 1 B $a \times b$ C $a + b$ D the larger of a and b

Show answer & explanation

$a \times b$

From the product rule, LCM = $\frac{a \times b}{1} = a \times b$.

Q91. Two numbers are in the ratio 3 : 4 and their HCF is 5. Their LCM is

Intermediate

A 35 B 12 C 60 D 20

Show answer & explanation

60

The numbers are 15 and 20 (3 and 4 are co-prime). LCM(15, 20) = 60.

Q92. A trader has 144 litres of one oil and 180 litres of another. He fills them into tins of equal size, as large as possible, without mixing. The number of tins is

Intermediate

A 36 B 4 C 5 D 9

Show answer & explanation

9

Tin size = HCF(144, 180) = 36 litres, so $\frac{144}{36} + \frac{180}{36} = 4 + 5 = 9$ tins.

Q93. Three traffic lights change every 48, 72 and 108 seconds. They change together at 7:00:00. They next change together at

Intermediate

A 7:07:12 B 7:03:36 C 7:14:24 D 7:04:48

Show answer & explanation

7:07:12

LCM(48, 72, 108) $= 2^4 \times 3^3 = 432$ s = 7 min 12 s.

Q94. If HCF(a, b) = a, then LCM(a, b) is

Intermediate

A a B $a \times b$ C 1 D b

Show answer & explanation

b

HCF(a, b) = a means a divides b, so the smallest multiple of both is b.

Q95. Find the HCF and LCM of 510 and 92, and verify the product rule.

Intermediate

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Q96. Find the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively.

Intermediate

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Q97. Two tankers hold 850 litres and 680 litres of petrol. Find the capacity of the largest container that can measure the petrol of each tanker an exact number of times.

Intermediate

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Q98. Three bells toll at intervals of 9, 12 and 15 minutes. If they start together, after how long will they next toll together?

Intermediate

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Q99. A shopkeeper has 120 litres of petrol, 180 litres of diesel and 240 litres of kerosene. He wants to sell them in cans of equal capacity, filling each can completely with one liquid. (a) What is the greatest capacity of such a can? (b) How many cans are needed in all? (c) If he instead wants to buy the smallest drum that can be filled exactly by cans of 120, 180 or 240 litres, what capacity should it be?

Intermediate

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1 mark — factorisations. 1 mark — (a) HCF 60 L, with the reason it is an HCF. 1 mark — (b) 9 cans. 2 marks — (c) recognising LCM and finding 720 L.

Q100. Three runners A, B and C start together from the same point on a circular track and run in the same direction. A completes a lap in 36 s, B in 48 s and C in 60 s. (a) After how many seconds will all three first be together at the start? (b) How many laps will each have completed then? (c) After how many seconds will A and B alone first meet at the start, and what does the product rule give as HCF(36, 48)?

Intermediate

Show answer & explanation

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1 mark — factorisations. 1 mark — (a) 720 s. 1 mark — (b) laps. 1 mark — (c) 144 s. 1 mark — (c) HCF 12 by the rule.

Q101. In the proof that $\sqrt{2}$ is irrational, after showing 2 divides a and writing $a = 2c$, the next equation is

Intermediate

A $b^2 = 4c^2$ B $b^2 = 2c^2$ C $a^2 = 2c^2$ D $b = 2c$

Show answer & explanation

$b^2 = 2c^2$

$2b^2 = a^2 = 4c^2$, so $b^2 = 2c^2$.

Q102. Assuming $5 - \sqrt{3} = \frac{a}{b}$, isolating $\sqrt{3}$ gives

Intermediate

A $\frac{a - 5b}{b}$ B $\frac{5 - a}{b}$ C $\frac{5b - a}{b}$ D $\frac{a + 5b}{b}$

Show answer & explanation

$\frac{5b - a}{b}$

$\sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}$.

Q103. Assuming $3 + 2\sqrt{5} = \frac{a}{b}$, isolating $\sqrt{5}$ gives

Intermediate

A $\frac{a - 3}{2}$ B $\frac{a - 3b}{b}$ C $\frac{3b - a}{2b}$ D $\frac{a - 3b}{2b}$

Show answer & explanation

$\frac{a - 3b}{2b}$

$2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$, so $\sqrt{5} = \frac{a - 3b}{2b}$.

Q104. The contradiction reached in the proof that $\sqrt{3}$ is irrational is that

Intermediate

A a and b are both divisible by 3, though co-prime B b = 0 C a is negative D 3 is not prime

Show answer & explanation

a and b are both divisible by 3, though co-prime

The proof shows 3 divides a and then b, contradicting that a and b have no common factor.

Q105. Which of these products is rational?

Intermediate

A $(3 + \sqrt{5})(3 + \sqrt{5})$ B $(\sqrt{3} + \sqrt{5})\sqrt{3}$ C $(3 + \sqrt{5})(3 - \sqrt{5})$ D $(2 + \sqrt{5}) \times 2$

Show answer & explanation

$(3 + \sqrt{5})(3 - \sqrt{5})$

$(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4$. The others are $14 + 6\sqrt{5}$, $3 + \sqrt{15}$ and $4 + 2\sqrt{5}$.

Q106. $\frac{2\sqrt{3}}{\sqrt{12}}$ is

Intermediate

A irrational B rational, equal to 2 C rational, equal to $\frac{1}{2}$ D rational, equal to 1

Show answer & explanation

rational, equal to 1

$\sqrt{12} = 2\sqrt{3}$, so the fraction is 1.

Q107. Which pair of irrational numbers has a rational sum?

Intermediate

A $\sqrt{2}$ and $\sqrt{3}$ B $\sqrt{5}$ and $2\sqrt{5}$ C $1 + \sqrt{2}$ and $\sqrt{2}$ D $2 + \sqrt{3}$ and $2 - \sqrt{3}$

Show answer & explanation

$2 + \sqrt{3}$ and $2 - \sqrt{3}$

$(2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$. The others give $\sqrt{2} + \sqrt{3}$, $3\sqrt{5}$ and $1 + 2\sqrt{2}$.

Q108. For which natural number n is $\sqrt{n}$ rational?

Intermediate

A n = 50 B n = 12 C n = 144 D n = 99

Show answer & explanation

n = 144

$\sqrt{144} = 12$. 50, 12 and 99 are not perfect squares.

Q109. Assertion (A): $5 + \sqrt{3}$ is irrational. Reason (R): $\sqrt{3}$ lies between 1 and 2.

Intermediate

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

Show answer & explanation

Both A and R are true, but R is not the correct explanation of A

Both are true, but knowing where the root lies does not show the sum is irrational.

Q110. Assertion (A): $(2 + \sqrt{3}) + (2 - \sqrt{3})$ is rational. Reason (R): The sum of two irrational numbers is always rational.

Intermediate

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

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A is true but R is false

A is true: the sum is 4. R is false: $\sqrt{2} + \sqrt{3}$ is irrational. The sum of two irrationals can go either way.

Q111. Why is $\frac{a - 3b}{2b}$ rational when a and b are integers, $b \neq 0$?

Intermediate

A Because a and b are co-prime B It is a quotient of integers with non-zero denominator C Because 2 is prime D Because $a > b$

Show answer & explanation

It is a quotient of integers with non-zero denominator

$a - 3b$ and $2b$ are integers, and $2b \neq 0$, so the fraction fits the definition of a rational number.

Q112. If $\frac{\sqrt{2}}{5} = \frac{a}{b}$ with integers a, b, then $\sqrt{2}$ equals

Intermediate

A $\frac{a}{5b}$ B $\frac{b}{5a}$ C $\frac{5a}{b}$ D $\frac{5b}{a}$

Show answer & explanation

$\frac{5a}{b}$

Multiply both sides by 5: $\sqrt{2} = \frac{5a}{b}$, which would be rational — the contradiction.

Q113. Which theorem is the proof of 'p | a² ⇒ p | a' based on?

Intermediate

A The Fundamental Theorem of Arithmetic B Pythagoras' theorem C The Basic Proportionality Theorem D The product rule for HCF and LCM

Show answer & explanation

The Fundamental Theorem of Arithmetic

The primes of $a^2$ are exactly those of a, by uniqueness of prime factorisation.

Q114. $4$ divides $6^2 = 36$ but does not divide 6. This shows that in 'p | a² ⇒ p | a'

Intermediate

A p must be prime B a must be prime C the theorem is false D p must be even

Show answer & explanation

p must be prime

4 is not prime, and the implication fails for it. The theorem holds only for prime p.

Q115. Which of these is equal to a rational number?

Intermediate

A $(\sqrt{5} - 2)^2$ B $\sqrt{5} + 2$ C $(\sqrt{5} - 2)(\sqrt{5} + 2)$ D $\frac{\sqrt{5}}{2}$

Show answer & explanation

$(\sqrt{5} - 2)(\sqrt{5} + 2)$

$(\sqrt{5} - 2)(\sqrt{5} + 2) = 5 - 4 = 1$. $(\sqrt{5} - 2)^2 = 9 - 4\sqrt{5}$ is irrational.

Q116. In the proof that $\sqrt{5}$ is irrational, the conclusion '5 divides a' comes from $5b^2 = a^2$ because

Intermediate

A 5 divides b B 5 is prime and divides $a^2$ C a is odd D a and b are co-prime

Show answer & explanation

5 is prime and divides $a^2$

The theorem 'p prime, p | a² ⇒ p | a' is applied with p = 5.

Q117. $x = 2 + \sqrt{3}$. Which of these is rational?

Intermediate

A $x^2$ B $2x$ C $x + (2 - \sqrt{3})$ D $x - 2$

Show answer & explanation

$x + (2 - \sqrt{3})$

$x + (2 - \sqrt{3}) = 4$. $x^2 = 7 + 4\sqrt{3}$, $2x = 4 + 2\sqrt{3}$ and $x - 2 = \sqrt{3}$ are irrational.

Q118. The least natural number k for which $\sqrt{2k}$ is rational is

Intermediate

A 1 B 4 C 2 D 8

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2

$\sqrt{2k}$ is rational when $2k$ is a perfect square. $k = 1$ gives $\sqrt{2}$; $k = 2$ gives $\sqrt{4} = 2$.

Q119. Prove that $\sqrt{5}$ is irrational.

Intermediate

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Q120. Prove that $\sqrt{3}$ is irrational.

Intermediate

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Q121. Given that $\sqrt{5}$ is irrational, prove that $3 + 2\sqrt{5}$ is irrational.

Intermediate

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Q122. Show by an example that (i) the sum of two irrational numbers can be rational, and (ii) the product of two irrational numbers can be rational.

Intermediate

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Q123. Classify as rational or irrational, with working: (i) $(2 - \sqrt{3})(2 + \sqrt{3})$ (ii) $(2 - \sqrt{3})^2$ (iii) $\frac{\sqrt{50}}{\sqrt{2}}$.

Intermediate

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Q124. A tile has the shape of a square of side 1 m, and a triangular tile is cut along its diagonal. (a) Show that the diagonal is $\sqrt{2}$ m and prove that this length is irrational. (b) The perimeter of the triangular piece is $2 + \sqrt{2}$ m. Prove that it is irrational. (c) Two such triangles are joined along their long sides to make the square again. Is the square's perimeter rational? Why does this not contradict (b)?

Intermediate

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½ mark — (a) √2 by Pythagoras. 1½ marks — (a) proof. 1½ marks — (b). 1½ marks — (c) 4 m with explanation.

Q125. Which one of these numbers is prime?

Advanced

A $3 \times 5 \times 7 + 7$ B $2 \times 11 \times 13 + 13$ C $2 \times 3 \times 5 \times 7 + 1$ D $5 \times 7 \times 9 + 9$

Show answer & explanation

$2 \times 3 \times 5 \times 7 + 1$

$2 \times 3 \times 5 \times 7 + 1 = 211$, which is prime. The others have a common factor: $7(15+1)$, $13(22+1)$ and $9(35+1)$.

Q126. $n$ is a natural number. For which base b does $b^n$ end in 0 for EVERY n?

Advanced

A 12 B 25 C 18 D 30

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30

$30 = 2 \times 3 \times 5$ contains both 2 and 5, so $30^n$ always ends in 0. 12 and 18 lack 5; 25 lacks 2.

Q127. If $p$ and $q$ are different primes, how many factors does $p^2 q$ have?

Advanced

A 4 B 3 C 6 D 5

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6

The factors are $1, p, p^2, q, pq, p^2 q$ — six. For example $12 = 2^2 \times 3$ has factors 1, 2, 3, 4, 6, 12.

Q128. If $a = 2^3 \times 3^2$ and $b = 2^2 \times 3^k$, and $a \times b = 2^5 \times 3^6$, the value of k is

Advanced

A 3 B 6 C 4 D 2

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4

Multiply the powers of 3: $3^2 \times 3^k = 3^6$, so $2 + k = 6$ and $k = 4$.

Q129. A number N has prime factorisation $2^2 \times 3 \times 5^2$. Which of these is also a factor of N?

Advanced

A 8 B 60 C 9 D 125

Show answer & explanation

60

$60 = 2^2 \times 3 \times 5$ fits inside N. 8 needs $2^3$, 9 needs $3^2$ and 125 needs $5^3$ — N has too few of each.

Q130. Which of these must be true for every natural number n?

Advanced

A $n^2 + 1$ is prime B $2^n + 1$ is prime C $n(n+1)$ is even D $n^2 + n + 1$ is composite

Show answer & explanation

$n(n+1)$ is even

One of n and n + 1 is even, so the product always has a factor 2. $n = 3$ gives $n^2 + 1 = 10$ (not prime), $n = 3$ gives $2^n + 1 = 9$ (not prime), and $n = 1$ gives $n^2 + n + 1 = 3$ (prime).

Q131. Assertion (A): $5 \times 4 \times 3 \times 2 \times 1 + 5$ is a composite number. Reason (R): Every number that is a product of primes plus a prime is composite.

Advanced

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

Show answer & explanation

A is true but R is false

A is true: $120 + 5 = 5 \times 25$. R is false: $2 \times 3 + 5 = 11$, which is prime. What makes A true is the common factor 5, not a general rule.

Q132. Assertion (A): $4^n$ cannot end with the digit 0 for any natural number n. Reason (R): The only prime in the factorisation of $4^n$ is 2, and by uniqueness of prime factorisation no other prime can appear.

Advanced

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

Show answer & explanation

Both A and R are true, and R is the correct explanation of A

Both are true, and R is exactly the argument that proves A.

Q133. The product of two consecutive natural numbers is 2550. By factorising, the two numbers are

Advanced

A 49 and 50 B 51 and 52 C 45 and 46 D 50 and 51

Show answer & explanation

50 and 51

$2550 = 2 \times 3 \times 5^2 \times 17 = 50 \times 51$ — group the primes into 50 = $2 \times 5^2$ and $51 = 3 \times 17$.

Q134. How many zeros does $2^5 \times 5^3 \times 7$ end with?

Advanced

A 3 B 5 C 8 D 1

Show answer & explanation

3

Each trailing zero needs one pair $2 \times 5$. There are only three 5s, so three pairs: $2^5 \times 5^3 \times 7 = 28000$.

Q135. Which is the smallest number that has 2, 3 and 7 as its only prime factors, each appearing at least once?

Advanced

A 21 B 84 C 126 D 42

Show answer & explanation

42

It must contain each prime at least once and nothing else: $2 \times 3 \times 7 = 42$.

Q136. A number is divisible by both $2^3$ and $3^2$ and by no other prime. Which of these could it be?

Advanced

A 216 B 180 C 120 D 144 × 5

Show answer & explanation

216

$216 = 2^3 \times 3^3$ — only 2 and 3. 180 and 120 contain 5, and so does $144 \times 5$.

Q137. If $2^a \times 3^b \times 5^c = 1800$, find a, b and c. Why can there be no other answer?

Advanced

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Q138. Show that $12^n$ can never end with the digit 0 or 5, for any natural number n.

Advanced

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Q139. (a) Show that $4^n$ cannot end with the digit 0 for any natural number n. (b) Find a natural number k less than 10 such that $k^n$ ends with 0 for every natural number n, and explain why. (c) Can $k^n$ end with 0 for some n but not others? Explain.

Advanced

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2 marks — (a) factorisation $2^{2n}$, need for 5, uniqueness named, conclusion. 1½ marks — (b) reasoning that no k below 10 works; smallest k = 10. 1½ marks — (c) primes of $k^n$ equal primes of k, so all or none.

Q140. (a) Explain why $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ and $11 \times 13 \times 17 + 17$ are composite. (b) Is $2 \times 3 \times 5 + 1$ composite? Justify. (c) What does (b) tell you about the method used in (a)?

Advanced

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1 mark — first number in (a) factorised. 1 mark — second number in (a) factorised. 1½ marks — (b) 31 shown to be prime by testing primes up to √31. 1½ marks — (c) the common factor is what makes the method work.

Q141. (a) Find the prime factorisation of 2550. (b) Hence find two consecutive natural numbers whose product is 2550. (c) Is 2550 a perfect square? If not, find the smallest number it must be multiplied by to become one.

Advanced

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1½ marks — (a) factorisation. 1½ marks — (b) grouping to 50 and 51. 1 mark — (c) reason it is not a square. 1 mark — (c) multiplier 102.

Q142. How many pairs of natural numbers have HCF 81 and sum 1215?

Advanced

A 1 B 3 C 4 D 2

Show answer & explanation

4

Write the numbers $81m$ and $81n$ with m, n co-prime: $m + n = 15$. Co-prime pairs with sum 15 (m < n): (1,14), (2,13), (4,11), (7,8) — that is four pairs.

Q143. The LCM of three numbers $2^3 \times 5$, $2^2 \times 3^2$ and $2 \times 3 \times 5^2$ is

Advanced

A 360 B 3600 C 900 D 1800

Show answer & explanation

1800

Greatest powers: $2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$.

Q144. The HCF and LCM of two numbers are 9 and 360. If one number is 45, the other is

Advanced

A 72 B 81 C 40 D 36

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72

$\frac{9 \times 360}{45} = 72$. Check: HCF(45, 72) = 9, LCM = 360 ✓

Q145. The smallest number which, when divided by 12, 16 and 24, leaves remainder 7 each time is

Advanced

A 48 B 55 C 41 D 103

Show answer & explanation

55

It is 7 more than a common multiple: LCM(12, 16, 24) + 7 = 48 + 7 = 55.

Q146. If p and q are distinct primes, HCF$(p^3 q^2, p^2 q^4)$ × LCM of the same two numbers equals

Advanced

A $p^2 q^2$ B $p^5 q^6$ C $p^3 q^4$ D $p^6 q^8$

Show answer & explanation

$p^5 q^6$

Product rule: HCF × LCM = $p^3 q^2 \times p^2 q^4 = p^5 q^6$.

Q147. Which is true for every pair of natural numbers a and b with a < b?

Advanced

A LCM(a, b) ≤ b B HCF(a, b) ≥ a C HCF(a, b) = LCM(a, b) D HCF(a, b) ≤ a ≤ b ≤ LCM(a, b)

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HCF(a, b) ≤ a ≤ b ≤ LCM(a, b)

The HCF divides a, so it cannot exceed a; the LCM is a multiple of b, so it cannot be below b.

Q148. The HCF of two consecutive even numbers is

Advanced

A 1 B 2 C 4 D their difference squared

Show answer & explanation

2

Consecutive even numbers are 2k and 2k + 2 = 2(k + 1). k and k + 1 are co-prime, so the HCF is 2.

Q149. Assertion (A): The HCF of two numbers is 12 and their LCM is 72, so their product is 864. Reason (R): HCF × LCM = product holds for any number of positive integers.

Advanced

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

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A is true but R is false

A is true: $12 \times 72 = 864$ for two numbers. R is false — the rule holds for two numbers only.

Q150. Assertion (A): There are two numbers with HCF 16 and LCM 380. Reason (R): The HCF of two numbers always divides their LCM.

Advanced

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

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A is false but R is true

R is true. A is false: 16 does not divide 380, so no such pair exists.

Q151. A rectangular garden 18 m by 24 m is to have trees at equal distances along its boundary, one at each corner, with the distance as large as possible. How many trees are planted?

Advanced

A 7 B 12 C 84 D 14

Show answer & explanation

14

Gap = HCF(18, 24) = 6 m. Perimeter $= 84$ m, so $84 \div 6 = 14$ trees on a closed boundary.

Q152. The greatest 4-digit number exactly divisible by 12, 15 and 18 is

Advanced

A 9999 B 9720 C 9900 D 9180

Show answer & explanation

9900

LCM(12, 15, 18) = 180. $9999 \div 180 = 55$ remainder 99, so the answer is $180 \times 55 = 9900$.

Q153. The HCF of two numbers is 23 and their LCM is 1449. If one number is 161, find the other.

Advanced

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Q154. Explain why there cannot be two natural numbers with HCF 14 and LCM 204.

Advanced

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Q155. Find the smallest number which leaves remainder 3 when divided by 8, 12 and 18.

Advanced

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Q156. (a) Find the HCF and LCM of 72, 120 and 180 by prime factorisation. (b) Show that HCF × LCM is not equal to the product of the three numbers, and explain why the two-number rule fails. (c) Find the HCF and LCM of 72 and 120 alone, and verify the rule for this pair.

Advanced

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1 mark — three factorisations. 1 mark — HCF 12 and LCM 360. 1½ marks — (b) numbers compared and the middle-power reason. 1½ marks — (c) HCF 24, LCM 360 and verification.

Q157. (a) The HCF of two numbers is 81 and their sum is 1215. Show that there are four possible pairs, and list them. (b) The LCM of $2^3 \times 3 \times 5^y$ and $2^x \times 3^2 \times 5$ is $2^4 \times 3^2 \times 5^3$. Find x and y.

Advanced

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1 mark — setting up 81m, 81n with m, n co-prime. 1 mark — m + n = 15. 1 mark — four pairs listed, with the excluded pairs explained. 2 marks — (b) x = 4 and y = 3 with reasoning.

Q158. $(\sqrt{2} + \sqrt{3})^2$ is

Advanced

A irrational, equal to $5 + 2\sqrt{6}$ B rational, equal to 5 C rational, equal to 25 D irrational, equal to $5 + \sqrt{6}$

Show answer & explanation

irrational, equal to $5 + 2\sqrt{6}$

$(\sqrt{2})^2 + 2\sqrt{2}\sqrt{3} + (\sqrt{3})^2 = 5 + 2\sqrt{6}$.

Q159. x is irrational and $x^2$ is rational. Which could x be?

Advanced

A $1 + \sqrt{3}$ B $\sqrt{2} + \sqrt{3}$ C $\frac{1}{2}$ D $\sqrt{3}$

Show answer & explanation

$\sqrt{3}$

$(\sqrt{3})^2 = 3$. $(1 + \sqrt{3})^2 = 4 + 2\sqrt{3}$ and $(\sqrt{2} + \sqrt{3})^2 = 5 + 2\sqrt{6}$ are irrational; $\frac{1}{2}$ is not irrational.

Q160. If $a$ is a non-zero rational and $\sqrt{b}$ is irrational, which of these MUST be irrational?

Advanced

A $(\sqrt{b})^2$ B $a\sqrt{b}$ C $\sqrt{b} - \sqrt{b}$ D $a \times 0$

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$a\sqrt{b}$

A non-zero rational times an irrational is irrational. $(\sqrt{b})^2 = b$, $\sqrt{b} - \sqrt{b} = 0$ and $a \times 0 = 0$ are rational.

Q161. A student proves $\sqrt{6}$ irrational by copying the $\sqrt{2}$ proof with p = 6. Which step is not justified?

Advanced

A Assuming $\sqrt{6} = \frac{a}{b}$ B Squaring to get $6b^2 = a^2$ C 6 | a² ⇒ 6 | a, since 6 is not prime D Concluding from the contradiction

Show answer & explanation

6 | a² ⇒ 6 | a, since 6 is not prime

The theorem 'p | a² ⇒ p | a' is only for primes. (It happens to hold for 6, but it needs a separate reason — e.g. use 2 | a² ⇒ 2 | a.)

Q162. Given that $\sqrt{3}$ is irrational, which of these is NOT proved irrational by the 'isolate the root' method directly?

Advanced

A $4 - \sqrt{3}$ B $\sqrt{3} + \sqrt{5}$ C $\frac{\sqrt{3}}{7}$ D $2 + 5\sqrt{3}$

Show answer & explanation

$\sqrt{3} + \sqrt{5}$

The method needs one root and rationals: isolating $\sqrt{3}$ in $\sqrt{3} + \sqrt{5}$ leaves another irrational, so no contradiction appears directly.

Q163. Assertion (A): $\frac{1}{\sqrt{5}}$ is irrational. Reason (R): If $\frac{1}{\sqrt{5}} = \frac{a}{b}$ with $a \neq 0$, then $\sqrt{5} = \frac{b}{a}$ would be rational.

Advanced

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

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Both A and R are true, and R is the correct explanation of A

Both true, and R is precisely the proof of A.

Q164. Assertion (A): The product of two irrational numbers can be rational. Reason (R): $\sqrt{3} \times \sqrt{12} = 6$.

Advanced

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

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Both A and R are true, and R is the correct explanation of A

A is true, and R gives a correct example that establishes it: $\sqrt{36} = 6$.

Q165. Which of these numbers lies between 1 and 2 AND is irrational?

Advanced

A $\sqrt{3}$ B $\frac{3}{2}$ C $\sqrt{\frac{9}{4}}$ D $\sqrt{5}$

Show answer & explanation

$\sqrt{3}$

$\sqrt{3} \approx 1.73$ is irrational. $\frac{3}{2}$ and $\sqrt{\frac{9}{4}} = \frac{3}{2}$ are rational; $\sqrt{5} \approx 2.24$ is above 2.

Q166. If $\sqrt{2} = \frac{a}{b}$ in lowest terms, the proof shows that a is even. What does it then show about b?

Advanced

A b is also even B b is odd C b = 1 D b is prime

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b is also even

With $a = 2c$: $b^2 = 2c^2$, so b is even too — which contradicts lowest terms.

Q167. Which is the correct conclusion if assuming '$7 - 2\sqrt{3}$ is rational' leads to '$\sqrt{3}$ is rational'?

Advanced

A $\sqrt{3}$ is rational B $7 - 2\sqrt{3}$ is rational C No conclusion is possible D $7 - 2\sqrt{3}$ is irrational

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$7 - 2\sqrt{3}$ is irrational

$\sqrt{3}$ is known to be irrational, so the assumption must be false: $7 - 2\sqrt{3}$ is irrational.

Q168. $p$ is prime. Which number is irrational for EVERY prime p?

Advanced

A $\sqrt{p^2}$ B $\sqrt{4p^2}$ C $\frac{p}{2}$ D $\sqrt{p}$

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$\sqrt{p}$

A prime is never a perfect square, so $\sqrt{p}$ is irrational. $\sqrt{p^2} = p$, $\sqrt{4p^2} = 2p$ and $\frac{p}{2}$ are rational.

Q169. How many of these are irrational: $\sqrt{2}$, $\sqrt{4}$, $\sqrt{8}$, $\sqrt{16}$, $\sqrt{32}$?

Advanced

A 3 B 2 C 4 D 5

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3

$\sqrt{4} = 2$ and $\sqrt{16} = 4$ are rational; $\sqrt{2}$, $\sqrt{8} = 2\sqrt{2}$ and $\sqrt{32} = 4\sqrt{2}$ are irrational.

Q170. Which statement correctly completes: 'If $a + b\sqrt{2}$ is rational, where a, b are rational, then…'

Advanced

A a = 0 B a = b C a + b = 0 D b = 0

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b = 0

If $b \neq 0$, then $\sqrt{2} = \frac{(a + b\sqrt{2}) - a}{b}$ would be rational. So b must be 0.

Q171. Assertion (A): $\frac{\sqrt{45}}{\sqrt{5}}$ is rational. Reason (R): $\sqrt{45} = 3\sqrt{5}$.

Advanced

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

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Both A and R are true, and R is the correct explanation of A

Both are true, and R explains A: $\frac{3\sqrt{5}}{\sqrt{5}} = 3$, a rational number.

Q172. Given that $\sqrt{3}$ is irrational, prove that $5 - 2\sqrt{3}$ is irrational.

Advanced

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Q173. Prove that $\sqrt{2} + \sqrt{3}$ is irrational. (Hint: assume it equals r, and square $\sqrt{3} = r - \sqrt{2}$.)

Advanced

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Q174. (a) State the theorem about a prime dividing a square that is used in irrationality proofs. (b) Using it, prove that $\sqrt{2}$ is irrational. (c) Hence prove that $4 - 3\sqrt{2}$ is irrational.

Advanced

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1 mark — (a) theorem stated with p prime. ½ mark — (b) assumption with a, b co-prime. 1 mark — (b) 2 | a from $2b^2 = a^2$ using the theorem. 1 mark — (b) 2 | b and the contradiction. 1½ marks — (c) isolation of √2 and conclusion.

Q175. (a) Prove that $\sqrt{3}$ is irrational. (b) Show that $(\sqrt{3} + 1)^2$ is irrational but $(\sqrt{3} + 1)(\sqrt{3} - 1)$ is rational. (c) What does (b) show about the product of two irrational numbers?

Advanced

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2 marks — (a) complete proof. 1 mark — (b) $4 + 2\sqrt{3}$ shown irrational. 1 mark — (b) product = 2. 1 mark — (c) conclusion.

Q176. (a) Prove that $\sqrt{5}$ is irrational. (b) Hence show that $\frac{2}{\sqrt{5}}$ and $3 + \sqrt{5}$ are irrational. (c) Give an irrational number x such that $x + (3 + \sqrt{5})$ is rational.

Advanced

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2 marks — (a). 1 mark — (b) $\frac{2}{\sqrt{5}}$. 1 mark — (b) $3 + \sqrt{5}$. 1 mark — (c) a valid x with the check.

Real Numbers · Class 10 Mathematics: 176 MCQs with answers

Warm-up · 52 questions

Building up · 72 questions

Topper level · 52 questions

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