CBSE Class 10 Mathematics Real Numbers MCQs Set 21

Answer: (c) The sum of a rational number and an irrational number is irrational

Explanation:
1. If a rational r and an irrational x had a rational sum, then x = (r + x) − r would be rational, a contradiction. So option (c) holds.
2. Option (a) fails for \( \sqrt{2} + (-\sqrt{2}) = 0 \); option (b) fails for \( \sqrt{2}\times\sqrt{2} = 2 \).
3. Option (d) fails for the rational number 0, since \( 0 \times \sqrt{2} = 0 \).

Teacher's Note:
1. Rational plus irrational is always irrational.
2. Sums and products of two irrationals can be rational, and zero times an irrational is zero.

Answer: (c) The product of two irrational numbers is always irrational

Explanation:
1. Rationals are closed under addition and multiplication, so (a) and (b) are true.
2. (c) is true because a rational plus an irrational can never be rational.
3. (d) is false: \( \sqrt{3}\times\sqrt{3} = 3 \), which is rational.

Teacher's Note:
1. Irrational numbers are not closed under addition or multiplication.
2. One counter-example is enough to show a statement is false.

Answer: (b) \( \sqrt{21} \)

Explanation:
1. 22/7 is a fraction of integers, 3.1416 terminates and \( 5.\overline{63} \) repeats, so all three are rational.
2. 21 is not a perfect square, so \( \sqrt{21} \) is irrational.

Teacher's Note:
1. The square root of a non-square positive integer is irrational.
2. The value 22/7 is only an approximation of the irrational number \( \pi \), but 22/7 itself is rational.

Answer: (c) \( 3.\overline{142857} \)

Explanation:
1. \( 3.\overline{142857} \) repeats a block of digits, so it equals \( 3 + \frac{142857}{999999} = \frac{22}{7} \).
2. The other three are non-terminating and non-repeating, so they are irrational.

Teacher's Note:
1. A repeating decimal can always be written as a fraction.
2. Look for a repeating block to decide if a decimal is rational.

Answer: (c) 3

Explanation:
1. \( \pi \) is irrational.
2. 21 is not a perfect square, so \( \sqrt{21} \) is irrational; 3 is not a perfect cube, so \( \sqrt[3]{3} \) is irrational.
3. 22/7 and 3.1416 are rational. So 3 numbers are irrational.

Teacher's Note:
1. Check each number: a fraction of integers or a terminating decimal is rational.
2. Roots of numbers that are not perfect powers are irrational.

Answer: (b) 1.535335333… (the block of 3s grows each time)

Explanation:
1. The first two decimals repeat a block, and 22/7 is a fraction, so they are rational.
2. In 1.535335333… no block repeats, as it never ends and has no pattern of repetition, so it is irrational.

Teacher's Note:
1. Irrational decimals neither terminate nor repeat.
2. A growing pattern of digits is not a repeating block.

Answer: (a) \( (\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5}) \)

Explanation:
1. \( (\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5}) = 3 - 5 = -2 \), which is rational.
2. \( 5+3\sqrt{2} \), \( 3\sqrt{7} \) and \( \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \) are all irrational.

Teacher's Note:
1. Use \( (a+b)(a-b) = a^2 - b^2 \) to remove the roots.
2. Expressions with a surd that does not cancel stay irrational.

Answer: (c) \( \frac{2q-p}{q} \)

Explanation:
1. \( 2 - \sqrt{3} = \frac{p}{q} \Rightarrow \sqrt{3} = 2 - \frac{p}{q} = \frac{2q-p}{q} \).
2. This would make \( \sqrt{3} \) rational, contradicting the fact that it is irrational.

Teacher's Note:
1. Isolate the surd first, then show it becomes a ratio of integers.
2. The contradiction proves the original number is irrational.

Answer: (a) \( \frac{1}{\sqrt{3}} \)

Explanation:
1. \( \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \), and \( \sqrt{3} \) is irrational, so this number is irrational.
2. \( \frac{\sqrt{12}}{\sqrt{3}} = 2 \), \( \sqrt{16} = 4 \), \( \sqrt{3}\times\sqrt{27} = 9 \); all rational.

Teacher's Note:
1. Simplify each surd before deciding.
2. Rational times irrational (non-zero) is irrational.

Answer: (c) \( \frac{11}{2^3\times 3} \)

Explanation:
1. A fraction in lowest terms terminates only if its denominator has no prime factors other than 2 and 5.
2. The denominators \( 2^3\times5^2 \), 125 and \( 320 = 2^6\times 5 \) fit this, but \( 2^3\times 3 \) contains the prime 3.

Teacher's Note:
1. Factorise the denominator of the fraction in lowest terms.
2. Any prime other than 2 or 5 makes the decimal non-terminating.

Answer: (d) non-terminating and repeating

Explanation:
1. 9/35 is in lowest terms and \( 35 = 5\times 7 \).
2. The prime 7 is not 2 or 5, so the decimal does not terminate; being rational, it repeats.

Teacher's Note:
1. Every rational number has a terminating or a repeating decimal.
2. The factor 7 in the denominator forces repetition.

Answer: (c) 0.115

Explanation:
1. \( \frac{23}{2^3\times5^2} = \frac{23}{200} \).
2. Multiply top and bottom by 5 to get \( \frac{115}{1000} = 0.115 \).

Teacher's Note:
1. Make the denominator a power of 10 by supplying the missing factors of 2 or 5.
2. Check by dividing 23 by 200.

Answer: (a) 0.192

Explanation:
1. \( 125 = 5^3 \), so multiply the numerator and the denominator by \( 2^3 = 8 \).
2. \( \frac{24\times 8}{1000} = \frac{192}{1000} = 0.192 \).

Teacher's Note:
1. Three factors of 5 need three factors of 2 to make 1000.
2. 24 × 8 = 192.

Answer: (a) 0.009375

Explanation:
1. \( \frac{15}{1600} = \frac{3}{320} \) and \( 320 = 2^6\times 5 \).
2. \( \frac{3}{320} = \frac{3\times 5^5}{10^6} = \frac{9375}{1000000} = 0.009375 \).

Teacher's Note:
1. First reduce the fraction, then equalise the powers of 2 and 5.
2. Six decimal places are needed because the larger power is \( 2^6 \).

Answer: (a) 6

Explanation:
1. \( 320 = 2^6\times 5^1 \), so the denominator is made a power of 10 by multiplying by \( 5^5 \).
2. \( \frac{17}{320} = \frac{17\times 3125}{10^6} = \frac{53125}{10^6} = 0.053125 \), which has 6 decimal places.

Teacher's Note:
1. The number of decimal places is the larger of the two exponents of 2 and 5.
2. Here the larger exponent is 6.

Answer: (c) \( \frac{8}{9} \)

Explanation:
1. Let x = 0.888… Then 10x = 8.888…
2. Subtract: 9x = 8, so \( x = \frac{8}{9} \).

Teacher's Note:
1. Multiply by a power of 10 equal to the length of the repeating block.
2. Subtracting removes the repeating tail.

Answer: (a) \( \frac{11}{90} \)

Explanation:
1. Let x = 0.1222… Then 10x = 1.222… and 100x = 12.222…
2. Subtract: 90x = 11, so \( x = \frac{11}{90} \).

Teacher's Note:
1. When a non-repeating digit comes first, use both 10x and 100x.
2. The difference of the two multiples removes the repeating part.

Answer: (a) \( \frac{8}{33} \)

Explanation:
1. Let x = 0.2424… The block has 2 digits, so 100x = 24.2424…
2. Subtract: 99x = 24, so \( x = \frac{24}{99} = \frac{8}{33} \).

Teacher's Note:
1. A two-digit repeating block means multiplying by 100.
2. Always cancel common factors at the end.

Answer: (c) \( \frac{101}{45} \)

Explanation:
1. Let x = 2.2444… Then 10x = 22.444… and 100x = 224.444…
2. Subtract: 90x = 202, so \( x = \frac{202}{90} = \frac{101}{45} \).

Teacher's Note:
1. Reduce 202/90 by dividing by 2.
2. Check: 101 ÷ 45 = 2.2444…

Answer: (c) \( \frac{181}{495} \)

Explanation:
1. Let x = 0.36565… Then 10x = 3.6565… and 1000x = 365.6565…
2. Subtract: 990x = 362, so \( x = \frac{362}{990} = \frac{181}{495} \).

Teacher's Note:
1. The block 65 repeats after one non-repeating digit, so use 10x and 1000x.
2. Cancel the common factor 2.

Answer: (b) 1679

Explanation:
1. By Euclid's division lemma, dividend = divisor × quotient + remainder.
2. Number = 61 × 27 + 32 = 1647 + 32 = 1679.

Teacher's Note:
1. Use \( a = bq + r \).
2. Check that the remainder is smaller than the divisor: 32 < 61.

Answer: (d) 43

Explanation:
1. 1365 = d × 31 + 32, so d × 31 = 1333.
2. d = 1333 ÷ 31 = 43. The remainder 32 is smaller than 43, as it should be.

Teacher's Note:
1. Rearrange \( a = bq + r \) to find the divisor.
2. The remainder must be less than the divisor.

Answer: (d) 45

Explanation:
1. 2520 = 405 × 6 + 90
2. 405 = 90 × 4 + 45
3. 90 = 45 × 2 + 0, so the HCF is 45.

Teacher's Note:
1. Divide the larger number by the smaller and repeat with the remainder.
2. The last non-zero remainder is the HCF.

Answer: (d) 36

Explanation:
1. 1188 = 504 × 2 + 180
2. 504 = 180 × 2 + 144
3. 180 = 144 × 1 + 36, then 144 = 36 × 4 + 0. So the HCF is 36.

Teacher's Note:
1. Keep dividing until the remainder is zero.
2. The last divisor is the HCF.

Answer: (a) 18 and 1584

Explanation:
1. \( 144 = 2^4\times 3^2 \) and \( 198 = 2\times 3^2\times 11 \).
2. HCF = \( 2\times 3^2 = 18 \); LCM = \( 2^4\times 3^2\times 11 = 1584 \).

Teacher's Note:
1. HCF uses the lowest powers of common primes; LCM uses the highest powers of all primes.
2. Check: 18 × 1584 = 144 × 198.