Q: In Euclid's division lemma, for positive integers a and b, there exist unique integers q and r such that a = bq + r. The condition satisfied by r is:

0 \le r < b. Euclid's division lemma states that for positive integers a and b there exist unique integers q and r with a = bq + r and 0 \le r < b. The remainder can be zero (when b divides a exactly) but must always be strictly less than the divisor b.

Q: Using Euclid's division algorithm, the HCF of 135 and 225 is:

45. Apply the algorithm step by step: 225 = 135 \times 1 + 90, then 135 = 90 \times 1 + 45, then 90 = 45 \times 2 + 0. Since the remainder becomes 0 when the divisor is 45, the HCF of 135 and 225 is 45.

Q: By Euclid's division lemma with b = 2, every positive even integer is of the form (where q is some integer):

2q. Taking b = 2 in a = bq + r with 0 \le r < 2, every positive integer is of the form 2q or 2q + 1. The form 2q is divisible by 2, so every positive even integer is of the form 2q, while 2q + 1 gives the odd integers.

Q: The prime factorisation of 140 is:

2^2 \times 5 \times 7. Divide by primes: 140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7. Hence 140 = 2^2 \times 5 \times 7. Multiplying back, 4 \times 5 \times 7 = 140, which confirms the factorisation.

Q: If the HCF of two numbers is 12 and their product is 1800, then their LCM is:

150. For any two positive integers, HCF \times LCM = product of the two numbers. Therefore LCM = \frac{1800}{12} = 150.

Q: The LCM of 12 and 18 is:

36. Write 12 = 2^2 \times 3 and 18 = 2 \times 3^2. The LCM is the product of the highest powers of all primes involved: 2^2 \times 3^2 = 36. Note that 6 is the HCF, a common mix-up.

Q: Which of the following is an irrational number?

\sqrt{7}. \sqrt{4} = 2 and \frac{22}{7} are rational, and 0.\overline{3} = \frac{1}{3} is rational because a non-terminating recurring decimal always represents a rational number. Since 7 is not a perfect square, \sqrt{7} cannot be written as \frac{p}{q} and is irrational.

Q: The sum of a rational number and an irrational number is:

always irrational. Let r be rational and s irrational. If r + s were rational, then s = (r + s) - r would be a difference of two rational numbers and hence rational — a contradiction. So r + s is always irrational; for example, 2 + \sqrt{3} is irrational.

Q: The decimal expansion of \frac{13}{3125} is:

terminating. Here 3125 = 5^5, which is of the form 2^m \times 5^n (with m = 0, n = 5). Hence \frac{13}{3125} has a terminating decimal expansion; in fact \frac{13}{3125} = 0.00416.

Q: Which of the following rational numbers has a non-terminating recurring decimal expansion?

\frac{7}{12}. 8 = 2^3, 125 = 5^3 and 25 = 5^2 are all of the form 2^m \times 5^n, so those three expansions terminate. But 12 = 2^2 \times 3 contains the prime 3, and \frac{7}{12} is already in simplest form, so its decimal expansion is non-terminating recurring.

Question 1

In Euclid's division lemma, for positive integers a and b, there exist unique integers q and r such that a = bq + r. The condition satisfied by r is:

Accepted Answer

0 \le r < b. Euclid's division lemma states that for positive integers a and b there exist unique integers q and r with a = bq + r and 0 \le r < b. The remainder can be zero (when b divides a exactly) but must always be strictly less than the divisor b.

Question 2

Using Euclid's division algorithm, the HCF of 135 and 225 is:

Accepted Answer

45. Apply the algorithm step by step: 225 = 135 	imes 1 + 90, then 135 = 90 	imes 1 + 45, then 90 = 45 	imes 2 + 0. Since the remainder becomes 0 when the divisor is 45, the HCF of 135 and 225 is 45.

Question 3

By Euclid's division lemma with b = 2, every positive even integer is of the form (where q is some integer):

Accepted Answer

2q. Taking b = 2 in a = bq + r with 0 \le r < 2, every positive integer is of the form 2q or 2q + 1. The form 2q is divisible by 2, so every positive even integer is of the form 2q, while 2q + 1 gives the odd integers.

Question 4

The prime factorisation of 140 is:

Accepted Answer

2^2 	imes 5 	imes 7. Divide by primes: 140 = 2 	imes 70 = 2 	imes 2 	imes 35 = 2 	imes 2 	imes 5 	imes 7. Hence 140 = 2^2 	imes 5 	imes 7. Multiplying back, 4 	imes 5 	imes 7 = 140, which confirms the factorisation.

Question 5

If the HCF of two numbers is 12 and their product is 1800, then their LCM is:

Accepted Answer

150. For any two positive integers, HCF 	imes LCM = product of the two numbers. Therefore LCM = \frac{1800}{12} = 150.

Question 6

The LCM of 12 and 18 is:

Accepted Answer

36. Write 12 = 2^2 	imes 3 and 18 = 2 	imes 3^2. The LCM is the product of the highest powers of all primes involved: 2^2 	imes 3^2 = 36. Note that 6 is the HCF, a common mix-up.

Question 7

Which of the following is an irrational number?

Accepted Answer

\sqrt{7}. \sqrt{4} = 2 and \frac{22}{7} are rational, and 0.\overline{3} = \frac{1}{3} is rational because a non-terminating recurring decimal always represents a rational number. Since 7 is not a perfect square, \sqrt{7} cannot be written as \frac{p}{q} and is irrational.

Question 8

The sum of a rational number and an irrational number is:

Accepted Answer

always irrational. Let r be rational and s irrational. If r + s were rational, then s = (r + s) - r would be a difference of two rational numbers and hence rational — a contradiction. So r + s is always irrational; for example, 2 + \sqrt{3} is irrational.

Question 9

The decimal expansion of \frac{13}{3125} is:

Accepted Answer

terminating. Here 3125 = 5^5, which is of the form 2^m 	imes 5^n (with m = 0, n = 5). Hence \frac{13}{3125} has a terminating decimal expansion; in fact \frac{13}{3125} = 0.00416.

Question 10

Which of the following rational numbers has a non-terminating recurring decimal expansion?

Accepted Answer

\frac{7}{12}. 8 = 2^3, 125 = 5^3 and 25 = 5^2 are all of the form 2^m 	imes 5^n, so those three expansions terminate. But 12 = 2^2 	imes 3 contains the prime 3, and \frac{7}{12} is already in simplest form, so its decimal expansion is non-terminating recurring.

Question 11

The square of any positive integer can never be of the form (where m is an integer):

Accepted Answer

3m + 2. By Euclid's lemma any integer is 3q, 3q + 1 or 3q + 2. Squaring each: (3q)^2 = 3(3q^2), (3q+1)^2 = 3(3q^2 + 2q) + 1, and (3q+2)^2 = 3(3q^2 + 4q + 1) + 1. So a perfect square is always of the form 3m or 3m + 1, never 3m + 2; note that 4m is possible, e.g. 6^2 = 36 = 4 	imes 9.

Question 12

The largest number which divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively is:

Accepted Answer

625. The number must exactly divide 1251 - 1 = 1250, 9377 - 2 = 9375 and 15628 - 3 = 15625. Factorising: 1250 = 2 	imes 5^4, 9375 = 3 	imes 5^5 and 15625 = 5^6. The common part is 5^4 = 625, so the required largest number is 625.

Question 13

If 	ext{HCF}(x, 18) = 2 and 	ext{LCM}(x, 18) = 36, then x equals:

Accepted Answer

4. Since HCF 	imes LCM = x 	imes 18, we get x = \frac{2 	imes 36}{18} = 4. Check: HCF(4, 18) = 2 and LCM(4, 18) = 36, so x = 4 satisfies both conditions; no other option does.

Question 14

The smallest number which when divided by 12, 16 and 24 leaves a remainder of 7 in each case is:

Accepted Answer

55. The required number is 7 more than the smallest common multiple of 12, 16 and 24. Since 12 = 2^2 	imes 3, 16 = 2^4 and 24 = 2^3 	imes 3, the LCM is 2^4 	imes 3 = 48, so the answer is 48 + 7 = 55. (103 = 2 	imes 48 + 7 also leaves remainder 7, but it is not the smallest.)

Question 15

Consider the two statements: (I) The sum of two irrational numbers is always irrational. (II) The product of two irrational numbers is always irrational. Then:

Accepted Answer

neither (I) nor (II) is true. Counterexamples defeat both statements: \sqrt{2} + (-\sqrt{2}) = 0 is rational, so (I) fails, and \sqrt{2} 	imes \sqrt{2} = 2 is rational, so (II) fails. The sum or product of two irrationals is only sometimes irrational, e.g. \sqrt{2} + \sqrt{3} and \sqrt{2} 	imes \sqrt{3} = \sqrt{6} are irrational.

Question 16

The number \frac{7\sqrt{3}}{\sqrt{75}} is:

Accepted Answer

rational, equal to \frac{7}{5}. Simplify the denominator: \sqrt{75} = \sqrt{25 	imes 3} = 5\sqrt{3}. Therefore \frac{7\sqrt{3}}{\sqrt{75}} = \frac{7\sqrt{3}}{5\sqrt{3}} = \frac{7}{5}, a rational number. This shows a quotient of two irrational numbers need not be irrational.

Question 17

To prove that 5 - 2\sqrt{3} is irrational, we assume it is a rational number r. This leads to \sqrt{3} = \frac{5 - r}{2}. The contradiction is that:

Accepted Answer

the right-hand side is rational while \sqrt{3} is known to be irrational. If r is rational, then \frac{5 - r}{2} is also rational, since subtracting rationals and dividing by 2 keeps the result rational. But this expression equals \sqrt{3}, which is irrational — a contradiction. Hence the assumption was wrong and 5 - 2\sqrt{3} is irrational.

Question 18

The smallest rational number by which \frac{1}{3} should be multiplied so that its decimal expansion terminates after one place of decimal is:

Accepted Answer

\frac{3}{10}. \frac{1}{3} 	imes \frac{3}{10} = \frac{1}{10} = 0.1, which terminates after exactly one decimal place. Multiplying by \frac{1}{10} gives \frac{1}{30}, which is still non-terminating, while \frac{3}{100} gives 0.01, terminating after two places, not one.

Question 19

The decimal expansion of \frac{77}{2^3 	imes 5^7} terminates after how many places of decimal?

Accepted Answer

7. Multiply numerator and denominator by 2^4 to equalise the powers: \frac{77 	imes 2^4}{2^7 	imes 5^7} = \frac{1232}{10^7} = 0.0001232. Since 77 = 7 	imes 11 shares no factor with 10, the expansion terminates after \max(3, 7) = 7 decimal places.

Question 20

The number 43.\overline{123456789} (the block of digits repeating forever) can be written in the form \frac{p}{q} in simplest form. Which statement about q is correct?

Accepted Answer

q has at least one prime factor other than 2 and 5. A non-terminating recurring decimal always represents a rational number, so it can be written as \frac{p}{q}. A fraction in simplest form has a terminating expansion exactly when q = 2^m 	imes 5^n. Since this expansion does not terminate, q must contain at least one prime factor other than 2 and 5.

Real Numbers — Class 10 CBSE Mathematics MCQs with Solutions

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Sample questions with solutions