CBSE Class 10 Mathematics Real Numbers MCQs Set 16

Answer: (a) 170 litres

Explanation:
1. The container must measure both quantities exactly, so its capacity is the HCF of 850 and 680.
2. 850 = 2 × 5² × 17 and 680 = 2³ × 5 × 17, so HCF = 2 × 5 × 17 = 170.

Teacher's Note:
1. "Maximum capacity" that measures exactly means HCF.
2. Use prime factorisation and take the lowest powers of common primes.

Answer: (c) 3

Explanation:
1. 144 = 2⁴ × 3² and 180 = 2² × 3² × 5, so HCF = 2² × 3² = 36.
2. 13m − 3 = 36 gives 13m = 39, so m = 3.

Teacher's Note:
1. First find the HCF, then solve the simple equation.
2. Take the lowest power of every common prime factor.

Answer: (d) 4

Explanation:
1. For two numbers, HCF × LCM = product of the numbers.
2. 2 × 36 = a × 18, so a = 72 ÷ 18 = 4.

Teacher's Note:
1. Use the product rule HCF × LCM = a × b.
2. Check: HCF(4, 18) = 2 and LCM(4, 18) = 36.

Answer: (d) 10:12 am

Explanation:
1. They flash together again after LCM(80, 90, 110) seconds. 80 = 2⁴ × 5, 90 = 2 × 3² × 5, 110 = 2 × 5 × 11, so LCM = 2⁴ × 3² × 5 × 11 = 7920 seconds.
2. 7920 s = 132 min = 2 h 12 min, so the time is 8:00 am + 2 h 12 min = 10:12 am.

Teacher's Note:
1. Events that repeat together again after the LCM of their intervals.
2. Convert seconds into minutes and hours carefully.

Answer: (c) 45

Explanation:
1. Seats per row must divide both 135 and 225, and we want the maximum, so take HCF(135, 225).
2. 135 = 3³ × 5 and 225 = 3² × 5², so HCF = 3² × 5 = 45.

Teacher's Note:
1. Maximum equal group size means HCF.
2. Rows: 135 ÷ 45 = 3 and 225 ÷ 45 = 5.

Answer: (d) 675

Explanation:
1. HCF × LCM = a × b, so LCM = 30375 ÷ 45.
2. 30375 ÷ 45 = 675.

Teacher's Note:
1. Remember HCF × LCM = product of the two numbers.
2. Do the division carefully: 45 × 675 = 30375.

Answer: (d) 720 mins

Explanation:
1. Required time = LCM(48, 60, 72). 48 = 2⁴ × 3, 60 = 2² × 3 × 5, 72 = 2³ × 3².
2. LCM = 2⁴ × 3² × 5 = 720 minutes.

Teacher's Note:
1. Take the highest power of every prime that appears.
2. Bells ringing together again always means LCM.

Answer: (a) 4 cm

Explanation:
1. The greatest equal length that divides both ribbons is HCF(40, 84).
2. 40 = 2³ × 5 and 84 = 2² × 3 × 7, so HCF = 2² = 4 cm.

Teacher's Note:
1. Cutting into equal pieces of the longest possible length is an HCF problem.
2. Pieces: 10 red and 21 blue.

Answer: (a) 33

Explanation:
1. The smallest two-digit prime is 11 and the smallest odd prime is 3.
2. 11 and 3 are co-prime, so LCM = 11 × 3 = 33.

Teacher's Note:
1. The LCM of two co-prime numbers is their product.
2. Note that 2 is the smallest prime but it is even.

Answer: (c) co-prime

Explanation:
1. a and b have no common prime factor.
2. Powers of a and b contain the same primes as a and b, so a² and b³ also have no common prime factor: they are co-prime.

Teacher's Note:
1. Powers do not introduce new prime factors.
2. Co-prime means HCF = 1.

Answer: (c) A is true, but R is false.

Explanation:
1. A is true: 2 has exactly two factors, 1 and 2.
2. R is false: for example (√8)² = 8 is composite and (√3 + 1)² = 4 + 2√3 is irrational, so the square of an irrational number is not always a prime. Hence A is true and R is false.

Teacher's Note:
1. Test a statement with one counter-example.
2. (√8)² = 8 is composite, so R fails.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 2 is not a perfect square, so by the stated rule √2 is irrational.
2. Both A and R are true and R explains A.

Teacher's Note:
1. The reason is the general theorem and the assertion is a special case.
2. This is the result used to prove irrationality of √2, √3, √5.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. LCM = 18144 ÷ 6 = 3024, so A is true.
2. R is the standard property used to get A, so R is true and explains A.

Teacher's Note:
1. Use LCM = product ÷ HCF.
2. 6 × 3024 = 18144.

Answer: (a) prime, co-prime

Explanation:
1. Two numbers whose HCF is 1 are called relatively prime numbers or co-prime numbers.
2. So the blanks are filled by prime, co-prime.

Teacher's Note:
1. Co-prime numbers need not be prime themselves, for example 8 and 9.
2. Learn this definition by heart.

Answer: (c) a³b²

Explanation:
1. p = a¹b² and q = a³b¹.
2. LCM takes the highest power of each prime: a³ and b², so LCM = a³b².

Teacher's Note:
1. LCM uses the higher power of each prime.
2. HCF would use the lower powers: ab.

Answer: (d) 2520

Explanation:
1. Take the highest powers of primes up to 10: 2³, 3², 5 and 7.
2. LCM = 8 × 9 × 5 × 7 = 2520.

Teacher's Note:
1. Find the LCM of 1, 2, ..., 10 using highest prime powers.
2. 2520 is divisible by 1 to 10: 2520 ÷ 8 = 315, 2520 ÷ 9 = 280.

Answer: (b) 2

Explanation:
1. 65 = 5 × 13 and 117 = 3² × 13, so HCF = 13.
2. 65m − 117 = 13 gives 65m = 130, so m = 2.

Teacher's Note:
1. Find the HCF first and then solve for m.
2. Check: 65 × 2 − 117 = 130 − 117 = 13.

Answer: (a) 45 minutes

Explanation:
1. LCM(30, p) = 90 and HCF(30, p) = 15.
2. HCF × LCM = 30 × p, so 15 × 90 = 30p and p = 1350 ÷ 30 = 45 minutes.

Teacher's Note:
1. Use HCF × LCM = product.
2. Verify: HCF(30, 45) = 15 and LCM(30, 45) = 90.

Answer: (c) 20

Explanation:
1. The smallest two-digit composite number is 10 and the smallest composite number is 4.
2. 10 = 2 × 5 and 4 = 2², so LCM = 2² × 5 = 20.

Teacher's Note:
1. Smallest composite number is 4 (not 1 or 2).
2. Smallest two-digit composite number is 10.

Answer: (b) 4

Explanation:
1. 196 = 2 × 98 = 2 × 2 × 49 = 2² × 7².
2. Sum of exponents = 2 + 2 = 4.

Teacher's Note:
1. Write the factorisation in exponent form first.
2. Add exponents, not the prime numbers.

Answer: (d) √5

Explanation:
1. √20 = 2√5.
2. 2√5 × √5 = 2 × 5 = 10, a rational number, so the smallest such irrational number is √5.

Teacher's Note:
1. Simplify the surd first.
2. Multiply by the irrational part to remove the root.

Answer: (a) 1 : 2

Explanation:
1. Least composite = 4 and least prime = 2.
2. HCF(4, 2) = 2 and LCM(4, 2) = 4, so the ratio is 2 : 4 = 1 : 2.

Teacher's Note:
1. Write both numbers, then find HCF and LCM.
2. Reduce the ratio to its lowest form.

Answer: (c) 5:00 pm

Explanation:
1. LCM(20, 25, 30): 20 = 2² × 5, 25 = 5², 30 = 2 × 3 × 5, so LCM = 2² × 3 × 5² = 300 min.
2. 300 min = 5 hours, so they beep together at 12 noon + 5 h = 5:00 pm.

Teacher's Note:
1. Convert 300 minutes to hours.
2. The first common time is after the LCM.

Answer: (c) 38

Explanation:
1. LCM of two primes is their product, so pq = 221 = 13 × 17.
2. Since p > q, p = 17 and q = 13, so 3p − q = 51 − 13 = 38.

Teacher's Note:
1. Factorise 221 into primes.
2. Check which prime is larger as given: p > q.

Answer: (c) (√3, √27)

Explanation:
1. √16 = 4 and √4 = 2 are rational, so (a) is not a pair of irrational numbers.
2. √3 × √27 = √81 = 9 is rational and both √3 and √27 = 3√3 are irrational, so the answer is (√3, √27). In (b) the product is √10 and in (d) it is 6√2, both irrational.

Teacher's Note:
1. First check that both numbers in the pair are irrational.
2. Then multiply and see whether the result is rational.

Answer: (a) 550

Explanation:
1. HCF × LCM = 2520 × 6600, so LCM = 2520 × 6600 ÷ 120 = 138600.
2. 252 × k = 138600, so k = 550.

Teacher's Note:
1. Use HCF × LCM = product of the numbers.
2. Divide 138600 by 252 carefully.

Answer: (c) 35

Explanation:
1. HCF takes lower powers: p²q³, so m = 2, n = 3. LCM takes higher powers: p³q⁴, so r = 3, s = 4.
2. (m + n)(r + s) = 5 × 7 = 35.

Teacher's Note:
1. HCF: smaller exponents. LCM: larger exponents.
2. Then just add and multiply.

Answer: (d) A is false, but R is true.

Explanation:
1. q = 3² × 7² × b, where b is a prime other than 3 and 7. If b = 2, then q is even, so A is false.
2. 3² × 7² = 63 × 7 = 441 is odd, so R is true. Hence A is false but R is true.

Teacher's Note:
1. A product is even if even one factor is even.
2. The prime b can be 2, which was not excluded.

Answer: (c) Only B, C, D

Explanation:
1. If a prime r divides y², then r divides y. So r divides 10y, 8y and y³.
2. But y/2 need not be divisible by r (take r = 2, y = 2). So only B, C, D.

Teacher's Note:
1. Use: if a prime divides a², it divides a.
2. Test A with r = 2, y = 2: y/2 = 1.

Answer: (d) All of these

Explanation:
1. 24 + 96 = 120 with HCF 24; 96 + 24 = 120 with HCF 24; 72 + 48 = 120 with HCF(72, 48) = 24.
2. All three pairs satisfy both conditions, so the answer is all of these.

Teacher's Note:
1. Check sum and HCF for every option.
2. (96, 24) is the same pair as (24, 96) in a different order.

Answer: (d) Can be precisely located

Explanation:
1. Every rational number p/q corresponds to an exact point on the number line.
2. So rational numbers can be precisely located.

Teacher's Note:
1. Rational numbers have terminating or repeating decimals and exact positions.
2. Irrational numbers also lie on the line, but cannot be written as p/q.

Answer: (a) If p divides a², then p must divide a as well.

Explanation:
1. By the fundamental theorem of arithmetic, every prime factor of a² is a prime factor of a.
2. So if p divides a², then p must divide a.

Teacher's Note:
1. This statement is the key step in the proof that √2 is irrational.
2. It works only for primes: 4 divides 6² but 4 does not divide 6.

Answer: (c) A is true, but R is false.

Explanation:
1. 0 = 0/1, so A is true.
2. Division by 0 is not defined, so R is false.

Teacher's Note:
1. Never divide by zero.
2. A rational number p/q needs q ≠ 0.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are the Fundamental Theorem of Arithmetic and are true.
2. The uniqueness of the factorisation is the reason we can describe a composite number by its prime factors, so R explains A.

Teacher's Note:
1. Existence and uniqueness are the two parts of the theorem.
2. The order of the factors does not matter.

Answer: (a) 3² × 5² × 17

Explanation:
1. 3825 ÷ 3 = 1275, 1275 ÷ 3 = 425, 425 ÷ 5 = 85, 85 ÷ 5 = 17.
2. So 3825 = 3² × 5² × 17.

Teacher's Note:
1. Divide by the smallest primes one by one.
2. Multiply back to check: 9 × 25 × 17 = 3825.

Answer: (a) 22338

Explanation:
1. LCM = (306 × 657) ÷ HCF = 201042 ÷ 9.
2. 201042 ÷ 9 = 22338.

Teacher's Note:
1. Use LCM = product ÷ HCF.
2. 306 × 657 = 201042.

Answer: (a) 3 and 420

Explanation:
1. 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7.
2. HCF = 3 and LCM = 2² × 3 × 5 × 7 = 420.

Teacher's Note:
1. HCF: common primes with lowest powers. LCM: all primes with highest powers.
2. Here only 3 is common to all three numbers.

Answer: (a) 30

Explanation:
1. Let the numbers be 2x and 3x. Their HCF is x and their LCM is 6x.
2. 6x = 180, so x = 30, and the numbers are 60 and 90 with HCF 30.

Teacher's Note:
1. For numbers in ratio a : b (co-prime), HCF = x and LCM = abx.
2. Check the product: 60 × 90 = 30 × 180.

Answer: (a) 60

Explanation:
1. LCM = xy ÷ HCF = 180 ÷ 3.
2. LCM = 60.

Teacher's Note:
1. Use HCF × LCM = xy.
2. The LCM can never be smaller than either number.

Answer: (b) 28 min

Explanation:
1. The time = LCM(4, 7, 14). 4 = 2², 7 = 7, 14 = 2 × 7.
2. LCM = 2² × 7 = 28 minutes, so they ring together at 6:28 am.

Teacher's Note:
1. Use the highest power of each prime in the LCM.
2. Add the answer to the starting time.