CBSE Class 10 Mathematics Real Numbers MCQs Set 15

Answer: (a) 13

Explanation:
1. The number must divide 70 − 5 = 65 and 125 − 8 = 117 exactly.
2. H.C.F. (65, 117) = 13.
3. Check: 70 = 5 × 13 + 5 and 125 = 9 × 13 + 8, so the answer is 13.

Teacher's Note:
1. Subtract each remainder from its number first.
2. Then take the H.C.F. of the results.

Answer: (b) 16

Explanation:
1. The number must divide 245 − 5 = 240 and 1029 − 5 = 1024.
2. H.C.F. (240, 1024) = 16, since 240 = 2⁴ × 3 × 5 and 1024 = 2¹⁰.
3. The divisor 16 is greater than the remainder 5, so it is valid. Answer: 16.

Teacher's Note:
1. The divisor must always be larger than the remainder.
2. Find the H.C.F. of the reduced numbers.

Answer: (c) 4, 3 and 5

Explanation:
1. For the largest equal grouping, the number of groups n = H.C.F. (20, 15) = 5.
2. Boys in each group x = 20 ÷ 5 = 4 and girls in each group y = 15 ÷ 5 = 3.
3. So x = 4, y = 3 and n = 5.

Teacher's Note:
1. The number of groups is the H.C.F. of the two counts.
2. Divide each count by the H.C.F. to get the members in each group.

Answer: (c) 1

Explanation:
1. Let q = p + 1. Any common divisor of p and p + 1 also divides their difference, which is 1.
2. So the H.C.F. is 1.

Teacher's Note:
1. Consecutive numbers are always co-prime.
2. Test with 8 and 9 to see it quickly.

Answer: (a) 13

Explanation:
1. Use H.C.F. × L.C.M. = product of the two numbers.
2. H.C.F. = \( \frac{91 \times 26}{182} \) = 13.
3. So the H.C.F. is 13.

Teacher's Note:
1. Remember H.C.F. × L.C.M. = a × b for two numbers.
2. Cancel before multiplying: 182 = 91 × 2.

Answer: (c) \( \frac{22}{7} \) is a rational number.

Explanation:
1. Every point on the number line is a real number, which may be rational or irrational, so (a) is false.
2. Irrational numbers such as \( \sqrt{2} \) can be shown on the number line, so (b) is false.
3. \( \frac{22}{7} \) is of the form p/q with q ≠ 0, so it is rational. Statement (c) is true.

Teacher's Note:
1. A rational number is any number written as p/q with q ≠ 0.
2. Check each statement separately.

Answer: (b) 4

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

Teacher's Note:
1. Divide by the smallest primes first.
2. Add only the exponents, not the primes.

Answer: (c) 67

Explanation:
1. The sum 100 is even, so one prime must be the even prime 2 (otherwise three odd primes would give an odd sum).
2. Let the other two be p and p + 36. Then 2 + p + p + 36 = 100, so p = 31.
3. The primes are 2, 31 and 67, and the largest is 67.

Teacher's Note:
1. Odd + odd + odd is odd, so 2 must be one of the primes.
2. Then form a simple equation.

Answer: (d) \( \frac{133}{990} \)

Explanation:
1. Let x = 0.1343434… Then 10x = 1.343434… and 1000x = 134.3434…
2. Subtract: 990x = 134 − 1 = 133, so x = \( \frac{133}{990} \).
3. The answer is \( \frac{133}{990} \).

Teacher's Note:
1. Multiply so that the repeating block lines up, then subtract.
2. Here the block 34 repeats, so use 1000x − 10x.

Answer: (d) 3600

Explanation:
1. L.C.M. (16, 20, 24) = 240 = 2⁴ × 3 × 5.
2. For a perfect square every prime must have an even power, so multiply by 3 × 5 to get 2⁴ × 3² × 5² = 3600.
3. 3600 = 60² is the least such number: 3600.

Teacher's Note:
1. Find the L.C.M. first, then fix the odd powers.
2. A perfect square has all even exponents.

Answer: (d) 24

Explanation:
1. Of the even numbers n and n + 2, one is a multiple of 4, so n(n + 2) is divisible by 2 × 4 = 8.
2. Among three consecutive numbers one is a multiple of 3.
3. So the product is divisible by 8 × 3 = 24. For n = 2 the product is 24, so no larger number works: 24.

Teacher's Note:
1. Test small cases such as n = 2 to find the upper limit.
2. Combine the factors 8 and 3 for the general proof.

Answer: (a) 515

Explanation:
1. In each case the divisor minus the remainder is 10 (15 − 5 = 25 − 15 = 35 − 25 = 10).
2. So the number is 10 less than a common multiple. L.C.M. (15, 25, 35) = 525.
3. Required number = 525 − 10 = 515.

Teacher's Note:
1. Look for a constant gap between divisor and remainder.
2. Subtract that gap from the L.C.M.

Answer: (d) 2, 3 and 13

Explanation:
1. 3¹³ − 3¹⁰ = 3¹⁰ (3³ − 1) = 3¹⁰ × 26 = 3¹⁰ × 2 × 13.
2. So it is divisible by 2, 3 and 13.
3. It has no factor 5, so it is not divisible by 5 or 10. Answer: 2, 3 and 13.

Teacher's Note:
1. Take out the common power of 3 first.
2. Factorise the number that is left.

Answer: (d) 345

Explanation:
1. Let the number be 3tu. Reversing the digits and adding: 303 + 20t + 101u = 888. Swapping the tens and unit digits adds 9u − 9t = 9, so u = t + 1.
2. Substituting gives 121t + 101 = 585, so t = 4 and u = 5. The number is 345.
3. Check: 345 + 543 = 888 and 354 − 345 = 9, so the number is 345.

Teacher's Note:
1. Test each option against both conditions.
2. Only one option satisfies both.

Answer: (d) \( \frac{23}{8} \)

Explanation:
1. A fraction in lowest terms terminates only if its denominator has no prime factors other than 2 and 5.
2. 210 = 2 × 3 × 5 × 7, 30 = 2 × 3 × 5 and 441 = 3² × 7² all contain other primes (after cancelling, 77/210 = 11/30).
3. 8 = 2³, so \( \frac{23}{8} \) = 2.875 terminates. Answer: (d).

Teacher's Note:
1. Reduce the fraction, then factorise the denominator.
2. Only 2s and 5s are allowed.

Answer: (d) 4

Explanation:
1. Use H.C.F. × L.C.M. = x × 18, so 2 × 36 = 18x and x = 4.
2. Check: L.C.M. (4, 18) = 36 and H.C.F. (4, 18) = 2.
3. So x = 4.

Teacher's Note:
1. Use the product rule for two numbers.
2. Always verify with both conditions.

Answer: (b) 2

Explanation:
1. The L.C.M. takes the highest power of each prime.
2. The highest power of 3 among a, b, c must be 3². Only c has a variable power of 3, so n = 2.
3. So n = 2.

Teacher's Note:
1. L.C.M. uses the greatest power of every prime.
2. Compare the powers of 3 in all three numbers.

Answer: (a) an even number

Explanation:
1. p₁² − p₂² = (p₁ + p₂)(p₁ − p₂).
2. The sum and the difference of two odd numbers are both even, so the product is even.
3. So it is an even number (in fact it is a multiple of 8).

Teacher's Note:
1. Odd ± odd = even.
2. Factorise the difference of squares.

Answer: (a) −1

Explanation:
1. Write x = 5m + 2 and y = 5n + 4. Then x + y = 5(m + n) + 6 = 5(m + n + 1) + 1.
2. So the remainder is z = 1.
3. Then \( \frac{2z - 5}{3} \) = \( \frac{2 - 5}{3} \) = −1.

Teacher's Note:
1. Add the two forms and rewrite the remainder as less than 5.
2. Then substitute z.

Answer: (a) 4

Explanation:
1. 13 − 3 = 21 − 11 = 10, so the number is 10 less than a multiple of L.C.M. (13, 21) = 273.
2. Multiples of 273 are 273, 546, 819; the number between 500 and 600 is 546 − 10 = 536.
3. 536 = 19 × 28 + 4, so the remainder is 4.

Teacher's Note:
1. Find the common gap between divisor and remainder.
2. Use the L.C.M. and pick the value in the range.

Answer: (c) \( \frac{235}{999} \)

Explanation:
1. Let x = 0.235235… Then 1000x = 235.235235…
2. Subtracting, 999x = 235, so x = \( \frac{235}{999} \).

Teacher's Note:
1. Three digits repeat, so multiply by 1000.
2. The denominator is 999 for a three-digit block.

Answer: (c) 2m

Explanation:
1. An even integer is divisible by 2, so it can be written as 2m for some integer m.
2. The form 2m + 1 gives odd integers. So the answer is 2m.

Teacher's Note:
1. Even means a multiple of 2.
2. Odd numbers leave remainder 1.

Answer: (d) 2q + 1

Explanation:
1. An odd integer leaves remainder 1 when divided by 2 (Euclid's division lemma).
2. So it has the form 2q + 1.

Teacher's Note:
1. Use a = 2q + r with r = 0 or 1.
2. r = 1 gives odd numbers.

Answer: (b) two decimal places

Explanation:
1. The denominator 2² × 5 has only the factors 2 and 5, so the expansion terminates.
2. \( \frac{33}{20} \) = 1.65, which has two decimal places.
3. So it terminates after two decimal places.

Teacher's Note:
1. The number of decimal places is the larger of the powers of 2 and 5.
2. Here that is 2.

Answer: (b) 117

Explanation:
1. The H.C.F. of co-prime numbers is 1.
2. Product = H.C.F. × L.C.M., so L.C.M. = 117 ÷ 1 = 117.

Teacher's Note:
1. H.C.F. of co-prime numbers is 1.
2. Use H.C.F. × L.C.M. = product.

Answer: (d) None of these.

Explanation:
1. (a) is false: \( \sqrt{2} + (3 - \sqrt{2}) = 3 \) is rational, but \( \sqrt{2} + \sqrt{3} \) is irrational, so the sum is not always rational.
2. (b) is false: both \( \sqrt{2} \) and \( 2\sqrt{2} \) rationalise \( \sqrt{2} \), so the factor is not unique.
3. (c) is false: \( \pi \) is irrational but not a surd.
4. So the answer is none of these.

Teacher's Note:
1. Look for one counter-example for each statement.
2. Use pi for (c).

Answer: (a) \( \frac{7^3}{5^4} \) is a non-terminating repeating decimal.

Explanation:
1. (a) The denominator 5⁴ has only the prime factor 5, so the decimal terminates. This statement is false.
2. (b) a + b = 2 + \( \sqrt{2} \), which is irrational, so (b) is true.
3. (c) is true because 19 is prime, and (d) is true as 25 × 625 = 15625.
4. The statement that is not correct is (a).

Teacher's Note:
1. Check the denominator for primes other than 2 and 5.
2. Test each statement one by one.

Answer: (c) Every real number is rational.

Explanation:
1. Every integer n = n/1 is rational, so (a) is true.
2. Rational + irrational is always irrational, so (b) is true. Every number-line point is a real number, so (d) is true.
3. Real numbers include irrational numbers such as \( \sqrt{2} \), so (c) is not correct.

Teacher's Note:
1. Real numbers are rational numbers together with irrational numbers.
2. One irrational number is enough to disprove (c).

Answer: (a) There are infinitely many even primes.

Explanation:
1. 2 is the only even prime, so (a) is false.
2. (b) is a standard theorem used to prove irrationality, so it is true.
3. (c) is true, for example 12 = 2² × 3 and 7 = 2⁰ × 7. (d) is true as well.
4. The incorrect statement is (a).

Teacher's Note:
1. Every even number above 2 has the factor 2 and is composite.
2. Recall the theorem on p dividing a².

Answer: (c) 288

Explanation:
1. The number of books must be a multiple of both 32 and 36, and we want the least one.
2. L.C.M. (32, 36) = 2⁵ × 3² = 288.
3. So the minimum number of books is 288.

Teacher's Note:
1. Equal distribution to both sections means a common multiple.
2. The minimum is the L.C.M.

Answer: (b) 4

Explanation:
1. 32 = 2⁵ and 36 = 2² × 3².
2. The common factor with the lowest power is 2², so H.C.F. (32, 36) = 4.

Teacher's Note:
1. H.C.F. uses the smallest power of common primes.
2. Check: 4 divides both 32 and 36.

Answer: (a) 2² × 3²

Explanation:
1. 36 = 2 × 2 × 3 × 3.
2. So 36 = 2² × 3².

Teacher's Note:
1. Divide by 2 until the number turns odd, then by 3.
2. Check: 4 × 9 = 36.

Answer: (b) composite number

Explanation:
1. 7 × 11 × 13 × 15 + 15 = 15 × (7 × 11 × 13 + 1) = 15 × 1002.
2. It has factors other than 1 and itself, so it is composite.

Teacher's Note:
1. Take 15 out as a common factor.
2. A number with a proper factor is composite.

Answer: (b) a²b²

Explanation:
1. p = a × b × b and q = a × a × b.
2. L.C.M. takes the highest power of each prime: a² and b².
3. So L.C.M. (p, q) = a²b².

Teacher's Note:
1. Take the greatest power of each prime for L.C.M.
2. Take the smallest power for H.C.F.

Answer: (b) 12

Explanation:
1. The room size must divide 60, 84 and 108, and we want the largest such number.
2. H.C.F. (60, 84, 108) = 12.
3. So each room holds at most 12 participants.

Teacher's Note:
1. Same subject in each room means a common divisor.
2. The maximum is the H.C.F.

Answer: (d) 21

Explanation:
1. With 12 participants per room: Hindi 60 ÷ 12 = 5, English 84 ÷ 12 = 7, Mathematics 108 ÷ 12 = 9.
2. Rooms needed = 5 + 7 + 9 = 21.

Teacher's Note:
1. Use the H.C.F. found in the previous step.
2. Add the rooms for each subject.

Answer: (a) 3780

Explanation:
1. 60 = 2² × 3 × 5, 84 = 2² × 3 × 7 and 108 = 2² × 3³.
2. L.C.M. = 2² × 3³ × 5 × 7 = 4 × 27 × 35 = 3780.

Teacher's Note:
1. Write each number in prime factors.
2. Use the highest powers.

Answer: (d) 45360

Explanation:
1. H.C.F. = 12 and L.C.M. = 3780.
2. Product = 12 × 3780 = 45360.

Teacher's Note:
1. Find the H.C.F. and the L.C.M. separately.
2. Multiply carefully.

Answer: (d) 2² × 3³

Explanation:
1. 108 = 2 × 2 × 3 × 3 × 3.
2. So 108 = 2² × 3³.

Teacher's Note:
1. Keep dividing by 2 and then by 3.
2. Check: 4 × 27 = 108.

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

Explanation:
1. 3125 = 5⁵ = 2⁰ × 5⁵, which is of the form 2ⁿ × 5ᵐ, so the Reason is true.
2. Therefore 13/3125 terminates, so the Assertion is true.
3. The Reason explains the Assertion: option (a).

Teacher's Note:
1. Check the denominator for primes other than 2 and 5.
2. The Reason is exactly the rule that proves the Assertion.