CBSE Class 10 Mathematics Real Numbers MCQs Set 25

Answer: (b) The sum of two irrational numbers is always irrational

Explanation:
1. Take √2 and −√2, both irrational; their sum is 0, which is rational. So the statement about two irrational numbers is false. The other three statements are always true.

Teacher's Note:
1. Test a general statement with a clever counter-example such as a number and its negative.
2. Remember that closure of rationals under addition and multiplication is a basic fact.

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

Explanation:
1. If a rational r and an irrational x had a rational sum, then x would equal that sum minus r, which is rational, a contradiction. So the sum is always irrational. For the others: √2 × √2 = 2 is rational, √2 + (−√2) = 0 is rational, and 0 × √2 = 0 is rational.

Teacher's Note:
1. A single counter-example is enough to reject a statement that says 'always'.
2. The product of a non-zero rational and an irrational number is irrational, but zero is an exception.

Answer: (d) √3 and 5√3

Explanation:
1. The product √3 × 5√3 = 5 × 3 = 15, which is rational. The other products are √6, √35 and 2√15, all irrational.

Teacher's Note:
1. Multiply the numbers under the roots first and see whether the result is a perfect square.
2. Numbers of the form a√n and b√n give a rational product.

Answer: (c) √21

Explanation:
1. 21 is not a perfect square, so √21 is irrational. 22/7 is a fraction, 3.1416 is a terminating decimal and 5.636363... is a repeating decimal, so all three are rational.

Teacher's Note:
1. A number is irrational if its decimal expansion neither ends nor repeats.
2. The square root of a non-square positive integer is irrational.

Answer: (a) ∛3

Explanation:
1. ∛8 = 2, ∛27 = 3 and ∛64 = 4 are integers. 3 is not a perfect cube, so ∛3 is not rational.

Teacher's Note:
1. Check whether the number under the cube root is a perfect cube.
2. Roots of numbers that are not perfect powers are irrational.

Answer: (d) 3.142857142857... (the block 142857 repeats)

Explanation:
1. A decimal that repeats a fixed block, such as 142857, is rational (it equals 22/7). π and √21 are irrational, and the decimal 1.535335333... has a growing block, so it is irrational too.

Teacher's Note:
1. Rational numbers have terminating or repeating decimals.
2. If the repeating block keeps changing length, the number is irrational.

Answer: (c) 5.636363... (the block 63 repeats)

Explanation:
1. The block 63 repeats, so 5.636363... = 5 + 7/11 = 62/11, a rational number. The other three have no repeating pattern or are roots of non-perfect powers, so they are irrational.

Teacher's Note:
1. Convert a repeating decimal into a fraction to prove it is rational.
2. A pattern with growing blocks is not a repeating decimal.

Answer: (b) 3√7

Explanation:
1. √7 is irrational, and the product of a non-zero rational (3) and an irrational number is irrational. √49 = 7, √(16/25) = 4/5 and 0.375 = 3/8 are rational.

Teacher's Note:
1. Simplify roots of perfect squares first.
2. A non-zero rational times an irrational is irrational.

Answer: (d) 1/√3

Explanation:
1. 1/√3 = √3/3, which is an irrational number divided by a non-zero rational, so it is irrational. (√3)² = 3, √3 × √27 = √81 = 9 and 0.5 are rational.

Teacher's Note:
1. Rationalise the denominator to see the nature of a number.
2. Dividing an irrational number by a non-zero rational gives an irrational number.

Answer: (c) √3 + √5

Explanation:
1. If √3 + √5 were rational, squaring would give 8 + 2√15 rational, so √15 would be rational, which is false. So √3 + √5 is irrational. The others equal 9, −1 and 5.

Teacher's Note:
1. Square a sum of two roots to test rationality.
2. Use (a + b)(a − b) = a² − b² to remove roots.

Answer: (b) 0.115

Explanation:
1. 200 = 2³ × 5², which has only the prime factors 2 and 5, so the decimal terminates. Divide: 23/200 = 0.115.

Teacher's Note:
1. A fraction in lowest terms terminates only if the denominator has no prime factor other than 2 or 5.
2. Multiply the numerator and denominator to make the denominator a power of 10 if that is easier.

Answer: (a) 0.192

Explanation:
1. 125 = 5³, which has only the prime factors 2 and 5, so the decimal terminates. Divide: 24/125 = 0.192.

Teacher's Note:
1. A fraction in lowest terms terminates only if the denominator has no prime factor other than 2 or 5.
2. Multiply the numerator and denominator to make the denominator a power of 10 if that is easier.

Answer: (c) 0.21375

Explanation:
1. 800 = 2⁵ × 5², which has only the prime factors 2 and 5, so the decimal terminates. Divide: 171/800 = 0.21375.

Teacher's Note:
1. A fraction in lowest terms terminates only if the denominator has no prime factor other than 2 or 5.
2. Multiply the numerator and denominator to make the denominator a power of 10 if that is easier.

Answer: (b) 0.009375

Explanation:
1. 1600 = 2⁶ × 5², which has only the prime factors 2 and 5, so the decimal terminates. Divide: 15/1600 = 0.009375.

Teacher's Note:
1. A fraction in lowest terms terminates only if the denominator has no prime factor other than 2 or 5.
2. Multiply the numerator and denominator to make the denominator a power of 10 if that is easier.

Answer: (a) 0.00608

Explanation:
1. 3125 = 5⁵, which has only the prime factors 2 and 5, so the decimal terminates. Divide: 19/3125 = 0.00608.

Teacher's Note:
1. A fraction in lowest terms terminates only if the denominator has no prime factor other than 2 or 5.
2. Multiply the numerator and denominator to make the denominator a power of 10 if that is easier.

Answer: (d) 8/9

Explanation:
1. Let x = 0.888... Then 10x = 8.888..., so 9x = 8 and x = 8/9. The simplest form is 8/9.

Teacher's Note:
1. Multiply by a power of 10 so that the repeating part lines up, then subtract.
2. Reduce the fraction to its lowest terms at the end.

Answer: (b) 8/33

Explanation:
1. Let x = 0.2424... Then 100x = 24.2424..., so 99x = 24 and x = 24/99 = 8/33. The simplest form is 8/33.

Teacher's Note:
1. Multiply by a power of 10 so that the repeating part lines up, then subtract.
2. Reduce the fraction to its lowest terms at the end.

Answer: (a) 11/90

Explanation:
1. Let x = 0.1222... Then 10x = 1.222... and 100x = 12.222..., so 90x = 11 and x = 11/90. The simplest form is 11/90.

Teacher's Note:
1. Multiply by a power of 10 so that the repeating part lines up, then subtract.
2. Reduce the fraction to its lowest terms at the end.

Answer: (d) 101/45

Explanation:
1. Let x = 2.2444... Then 100x = 224.444... and 10x = 22.444..., so 90x = 202 and x = 202/90 = 101/45. The simplest form is 101/45.

Teacher's Note:
1. Multiply by a power of 10 so that the repeating part lines up, then subtract.
2. Reduce the fraction to its lowest terms at the end.

Answer: (c) 11/24

Explanation:
1. 24 = 2³ × 3 contains the prime factor 3, so 11/24 does not terminate; it repeats. The other fractions have denominators with only 2 and 5 as prime factors, so they terminate.

Teacher's Note:
1. Reduce the fraction first, then factorise the denominator.
2. Only the prime factors 2 and 5 allow a decimal to terminate.

Answer: (a) 77/210

Explanation:
1. 77/210 = 11/30 and 30 = 2 × 3 × 5 has the factor 3, so the decimal is non-terminating and repeating. The other fractions have denominators with only 2 and 5 as prime factors, so they terminate.

Teacher's Note:
1. Reduce the fraction first, then factorise the denominator.
2. Only the prime factors 2 and 5 allow a decimal to terminate.

Answer: (d) 129/(2² × 5⁷ × 7⁵)

Explanation:
1. The denominator contains 7⁵, and 129 = 3 × 43 has no factor 7, so the fraction is in lowest terms with a factor other than 2 and 5 in the denominator. It is non-terminating and repeating. The other fractions have denominators with only 2 and 5 as prime factors, so they terminate.

Teacher's Note:
1. Reduce the fraction first, then factorise the denominator.
2. Only the prime factors 2 and 5 allow a decimal to terminate.