CBSE Class 10 Maths Advanced Real Numbers MCQs Set 01

Answer: (b) \( \frac{21}{2^3 5^6} \)

Explanation:
1. A fraction in lowest terms has a terminating decimal only when its denominator has no prime factor except 2 and 5.
2. (a) 45 = 3² × 5 has the prime 3. (c) 49 = 7² has the prime 7. (d) 2² × 3² has the prime 3.
3. (b) 2³ × 5⁶ has only 2 and 5, and 21 shares no factor with it. So the answer is \( \frac{21}{2^3 5^6} \).

Teacher's Note:
1. Check the denominator after reducing the fraction.
2. Any other prime in the denominator, such as 3 or 7, spoils termination.

Answer: (a) \( 2^m \cdot 5^n \)

Explanation:
1. The decimal of p/q (lowest terms) stops exactly when q can be turned into a power of 10 by multiplying with some number.
2. That happens only when q = 2^m × 5ⁿ, because 10 = 2 × 5.
3. The form is \( 2^m\cdot5^n \).

Teacher's Note:
1. This is the standard test for terminating decimals.
2. Multiplying top and bottom by 2 or 5 as needed gives a power of 10.

Answer: (d) \( \frac{93}{420} \)

Explanation:
1. 80 = 2⁴ × 5 and 320 = 2⁶ × 5 contain only 2 and 5, so (a) and (b) terminate. 20/100 = 1/5 also terminates.
2. 93/420 = 31/140 after dividing by 3, and 140 = 2² × 5 × 7 has the prime 7.
3. So \( \frac{93}{420} \) has a non-terminating repeating expansion.

Teacher's Note:
1. Cancel common factors first: 93/420 = 31/140.
2. The prime 7 in the denominator is the culprit.

Answer: (c) 435

Explanation:
1. Use H.C.F. × L.C.M. = product of the two numbers.
2. 145 × 2175 = 725 × other, so other = (145 × 2175) ÷ 725 = 315375 ÷ 725 = 435.
3. The other number is 435.

Teacher's Note:
1. The product of the two numbers equals the product of their HCF and LCM.
2. Simplify before multiplying: 2175 ÷ 725 = 3, so the answer is 145 × 3.

Answer: (b) 9696

Explanation:
1. LCM = (96 × 404) ÷ HCF = 38784 ÷ 4.
2. 38784 ÷ 4 = 9696.
3. The L.C.M. is 9696.

Teacher's Note:
1. Use the relation HCF × LCM = product of the numbers.
2. Divide 404 by 4 first to keep the numbers small: 96 × 101 = 9696.

Answer: (a) \( \sqrt3 \)

Explanation:
1. √9 = 3, √16 = 4 and √25 = 5 are all whole numbers, so they are rational.
2. 3 is not a perfect square, so √3 cannot be written as a fraction of integers.
3. The number that is not rational is \( \sqrt3 \).

Teacher's Note:
1. A square root of a non-perfect-square natural number is irrational.
2. Simplify each root before deciding.

Answer: (b) 1000

Explanation:
1. HCF × LCM = product of the two numbers.
2. 50 × 20 = 1000.
3. So the value is 1000.

Teacher's Note:
1. No need to find the HCF and LCM separately.
2. Check: HCF = 10, LCM = 100, product = 1000.

Answer: (d) − 19

Explanation:
1. 210 = 55 × 3 + 45, 55 = 45 × 1 + 10, 45 = 10 × 4 + 5, 10 = 5 × 2 + 0, so HCF = 5.
2. Set 210 × 5 + 55p = 5: 1050 + 55p = 5, so 55p = −1045.
3. p = −1045 ÷ 55 = −19.

Teacher's Note:
1. Find the HCF first by Euclid's algorithm.
2. Then treat the given form as an equation in p.

Answer: (b) 17

Explanation:
1. Subtract the remainders: 398 − 7 = 391, 436 − 11 = 425, 542 − 15 = 527. The required number is the H.C.F. of 391, 425 and 527.
2. 391 = 17 × 23, 425 = 5² × 17, 527 = 17 × 31, so the H.C.F. is 17.
3. Also 17 is larger than every remainder, as it must be. The number is 17.

Teacher's Note:
1. The divisor must exceed each remainder (here 17 > 15).
2. Subtract remainders first, then take the HCF.

Answer: (c) All real numbers can be represented on a number line.

Explanation:
1. π is irrational, so (a) is false.
2. Rational and irrational numbers are different kinds of numbers, so (b) is false.
3. √7 ÷ 8 is an irrational number divided by a non-zero rational, so it is irrational and (d) is false.
4. Every real number corresponds to a point on the number line, so the true statement is (c).

Teacher's Note:
1. The real numbers fill the number line completely.
2. An irrational number divided by a non-zero rational stays irrational.

Answer: (d) \( 6+\sqrt9 \)

Explanation:
1. 5 − √3, √2 + √5 and 4 + √2 are all irrational.
2. √9 = 3, so 6 + √9 = 9, a rational number.
3. The number which is not irrational is \( 6+\sqrt9 \).

Teacher's Note:
1. Simplify any root of a perfect square first.
2. Rational + irrational is irrational.

Answer: (a) 2

Explanation:
1. The least prime factor of a is 3, so a is odd. Similarly b is odd.
2. The sum of two odd numbers is even.
3. So 2 divides a + b, and the least prime factor is 2.

Teacher's Note:
1. A least prime factor of 3 or more tells you the number is odd.
2. Odd + odd = even.

Answer: (c) 0 ≤ r < b

Explanation:
1. The remainder can be 0 (when b divides a) but must be smaller than the divisor.
2. Hence the condition is 0 ≤ r < b.

Teacher's Note:
1. Remember: remainder is at least 0 and strictly less than b.
2. r = b would mean the quotient could be increased by 1.

Answer: (c) any odd positive integer

Explanation:
1. Let n be odd: n = 2k + 1. Then n² − 1 = 4k² + 4k = 4k(k + 1).
2. k(k + 1) is a product of consecutive integers, hence even, so 4k(k + 1) is divisible by 8.
3. For even n, n² − 1 is odd, so it is never divisible by 8. The answer is any odd positive integer.

Teacher's Note:
1. Test with n = 3 (8) and n = 5 (24); both are multiples of 8.
2. Even n gives an odd value of n² − 1.

Answer: (b) 2

Explanation:
1. Write n = 9q + 7.
2. 3n − 1 = 27q + 21 − 1 = 27q + 20 = 9(3q + 2) + 2.
3. The remainder is 2.

Teacher's Note:
1. Substitute n = 9q + 7 and divide again.
2. Try n = 7: 3n − 1 = 20 = 9 × 2 + 2.

Answer: (b) 2

Explanation:
1. 153 = 85 × 1 + 68, 85 = 68 × 1 + 17, 68 = 17 × 4 + 0, so HCF = 17.
2. Set 85n − 153 = 17, so 85n = 170 and n = 2.

Teacher's Note:
1. Find the HCF with Euclid's algorithm.
2. Then solve the linear equation for n.

Answer: (c) 5

Explanation:
1. 97 = 79 × 1 + 18, 79 = 18 × 4 + 7, 18 = 7 × 2 + 4, 7 = 4 × 1 + 3, 4 = 3 × 1 + 1, so HCF = 1.
2. Back-substitute: 1 = 4 − 3 = 2 × 4 − 7 = 2 × 18 − 5 × 7 = 22 × 18 − 5 × 79 = 22 × 97 − 27 × 79.
3. So n = 22 and m = 27, giving m − n = 5.

Teacher's Note:
1. Other pairs (n, m) also work, but m − n takes values 5, 23, ... and only 5 is among the options.
2. Work backwards through the Euclid steps to get the form.

Answer: (c) 5

Explanation:
1. 144 = 90 × 1 + 54, 90 = 54 × 1 + 36, 54 = 36 × 1 + 18, 36 = 18 × 2 + 0, so HCF = 18.
2. 18 = 54 − 36 = 54 − (90 − 54) = 2 × 54 − 90 = 2(144 − 90) − 90 = 2 × 144 − 3 × 90.
3. So x = 2, y = −3 and x − y = 5.

Teacher's Note:
1. Express the HCF by going back through the division steps.
2. Other solutions give x − y = 18, −8, ...; only 5 is in the options.

Answer: (d) 9

Explanation:
1. 420 = 130 × 3 + 30, 130 = 30 × 4 + 10, 30 = 10 × 3 + 0, so HCF = 10.
2. 10 = 130 − 4 × 30 = 130 − 4(420 − 3 × 130) = 13 × 130 − 4 × 420.
3. So p = −4 and q = 13, giving p + q = 9.

Teacher's Note:
1. Back-substitution is the key skill here.
2. Check: 420 × (−4) + 130 × 13 = −1680 + 1690 = 10.

Answer: (c) every common divisor of ‘a’ and ‘b’ is a common divisor of ‘b’ and ‘r’, and vice-versa.

Explanation:
1. From a = bq + r, any d dividing a and b divides r = a − bq, so it divides b and r.
2. Conversely any d dividing b and r divides bq + r = a, so it divides a and b.
3. Hence the statement is (c); this is why H.C.F.(a, b) = H.C.F.(b, r).

Teacher's Note:
1. This is the logic behind Euclid's algorithm.
2. Common divisors of a and b are exactly the common divisors of b and r.

Answer: (c) 10000

Explanation:
1. HCF × LCM = 125 × 80.
2. 125 × 80 = 10000.
3. The value is 10000.

Teacher's Note:
1. Just multiply the two given numbers.
2. 125 × 8 = 1000, then multiply by 10.

Answer: (a) 12

Explanation:
1. Take each common prime with its smaller exponent: 2 → min(3, 2) = 2, 3 → min(1, 3) = 1. The prime 5 is not in y.
2. HCF = 2² × 3 = 12.

Teacher's Note:
1. Smaller power of each common prime gives the HCF; the larger power gives the LCM.
2. 108 would be the LCM-style choice for the prime 3 and 2.

Answer: (b) 40

Explanation:
1. HCF × LCM = 253 × 440, so 11 × 253R = 253 × 440.
2. Dividing by 253: 11R = 440, so R = 40.

Teacher's Note:
1. Cancel 253 from both sides to save work.
2. LCM = 253 × 40 = 10120.

Answer: (a) 2

Explanation:
1. Every prime factor of a is at least 5, so a is odd; similarly b is odd.
2. Odd + odd is even, so a + b is divisible by 2.
3. The least prime factor is 2.

Teacher's Note:
1. The same idea works whenever neither number is even.
2. The sum of two odd numbers is always even.

Answer: (b) 4

Explanation:
1. 5005 ÷ 5 = 1001 and 1001 = 7 × 11 × 13.
2. So 5005 = 5 × 7 × 11 × 13, which has four prime factors.

Teacher's Note:
1. Divide by 5 first since 5005 ends with 5.
2. 1001 = 7 × 11 × 13 is worth remembering.

Answer: (d) 8

Explanation:
1. The smallest prime is 2 and the smallest composite number is 4.
2. HCF × LCM = 2 × 4 = 8.
3. The product is 8.

Teacher's Note:
1. The smallest composite number is 4, not 1 or 2.
2. Use HCF × LCM = product of the numbers.

Answer: (b) two decimal places

Explanation:
1. \( \frac{31}{2^2\times5} = \frac{31}{20} \).
2. Multiply top and bottom by 5 to get a power of 10: 31 × 5 ÷ 100 = 155 ÷ 100 = 1.55.
3. The decimal ends after two decimal places.

Teacher's Note:
1. The number of decimal places is the larger of the exponents of 2 and 5.
2. Here the larger exponent is 2 (from 2²).

Answer: (c) \( \frac{1}{7} \)

Explanation:
1. 35/14 = 5/2 = 2.5 and 14/35 = 2/5 = 0.4 both terminate. 7/8 = 0.875 also terminates.
2. 1/7 has denominator 7, which is not of the form 2^m5ⁿ, so 1/7 = 0.142857... repeats.
3. The answer is \( \frac{1}{7} \).

Teacher's Note:
1. Reduce each fraction before judging.
2. 35/14 and 14/35 look scary but simplify to 5/2 and 2/5.

Answer: (a) Terminating

Explanation:
1. 400 = 2⁴ × 5², which has only the primes 2 and 5.
2. So 27/400 terminates; in fact 27/400 = 0.0675.
3. The decimal representation is terminating.

Teacher's Note:
1. 27 and 400 are co-prime, so the fraction is already in lowest terms.
2. Always factorise the denominator.

Answer: (d) Infinite

Explanation:
1. The average of two rational numbers is rational and lies between them.
2. Repeating the averaging with each new pair never ends.
3. So there are infinitely many rational numbers between any two.

Teacher's Note:
1. There is no 'next' rational number.
2. This density property is special to rational (and real) numbers.

Answer: (b) A rational number

Explanation:
1. The digits 63 repeat endlessly, so the decimal is non-terminating and repeating.
2. Every non-terminating repeating decimal can be written as a fraction p/q, so it is rational.

Teacher's Note:
1. A repeating pattern means the number is rational.
2. It is not an integer because it has a fractional part.

Answer: (d) 2 or 5

Explanation:
1. The expansion terminates exactly when the denominator is 2^m × 5ⁿ.
2. So the only allowed prime factors are 2 or 5.

Teacher's Note:
1. Think of what is needed to turn the denominator into 10, 100, 1000, ...
2. The primes of 10 are 2 and 5.

Answer: (d) prime

Explanation:
1. A − B is a prime, hence at least 2, so A > B. If A and B were both odd, A − B and A + B would both be even primes, i.e. both equal 2, which is impossible. So one of A, B is 2, and since A > B, B = 2.
2. Now A − 2, A, A + 2 must all be prime. Of three numbers spaced by 2, one is divisible by 3, so that number must be 3: A − 2 = 3, giving A = 5 (and A + 2 = 7).
3. The four primes are 5, 2, 3, 7 and their sum is 17, which is prime.

Teacher's Note:
1. Parity (odd/even) quickly forces B = 2.
2. Among A − 2, A, A + 2 exactly one is a multiple of 3.

Answer: (d) 0

Explanation:
1. V = 41! is divisible by every number from 1 to 41.
2. For k = 1, 2, ..., 40 we have A + k = V + (k + 1), and (k + 1) is one of 2, 3, ..., 41. So (k + 1) divides both V and (k + 1), hence divides A + k.
3. Each A + k is therefore divisible by a number between 2 and 41 that is smaller than A + k itself, so it is composite.
4. The number of primes is 0.

Teacher's Note:
1. Think of the famous idea: n! + k is divisible by k for 2 ≤ k ≤ n.
2. This shows that long runs of consecutive composite numbers exist.

Answer: (b) 7

Explanation:
1. a² − b² = (a − b)(a + b) = 13. Since 13 is prime and a + b > a − b, we need a − b = 1 and a + b = 13.
2. Adding: 2a = 14, so a = 7 (and b = 6).

Teacher's Note:
1. Factorise the difference of squares.
2. 13 is prime, so its only factor pair is 1 × 13.

Answer: (a) \( 2^2\times3^5\times7^2 \)

Explanation:
1. 3240 = 2³ × 3⁴ × 5 and 3600 = 2⁴ × 3² × 5². Also 36 = 2² × 3².
2. The L.C.M. has 3⁵ and 7², which neither given number has, so the third number must contain 3⁵ × 7².
3. The H.C.F. has no factor 5, but 3240 and 3600 both have 5, so the third number has no 5. For the power of 2, the H.C.F. needs 2² and the two given numbers already have 2³ and 2⁴, so the third number has exactly 2².
4. The third number is \( 2^2\times3^5\times7^2 \).

Teacher's Note:
1. Compare exponents prime by prime with the HCF and LCM.
2. Check: min exponent of 3 is 2 and max is 5, matching 36 and the LCM.

Answer: (c) 8

Explanation:
1. Since the G.C.D. is 6, write a = 6x and b = 6y where x and y are co-prime.
2. Then 6(x + y) = 90, so x + y = 15 with H.C.F.(x, y) = 1.
3. x and y are co-prime exactly when x is co-prime to 15. For x = 1, ..., 14 the numbers co-prime to 15 are 1, 2, 4, 7, 8, 11, 13, 14, which is 8 values.
4. There are 8 ordered pairs.

Teacher's Note:
1. gcd(x, 15 − x) = gcd(x, 15).
2. Count numbers below 15 that share no factor 3 or 5 with 15.