CBSE Class 10 Maths Advanced Real Numbers MCQs Set 02

Answer: (b) 900

Explanation:
1. HCF × LCM = p × q.
2. 12 × LCM = 10800, so LCM = 10800 ÷ 12 = 900.

Teacher's Note:
1. The product of the two numbers equals HCF × LCM.
2. Check with p = 36, q = 300: HCF = 12, LCM = 900.

Answer: (c) 6

Explanation:
1. Divisibility by 18 needs divisibility by 2 and by 9.
2. For divisibility by 2 the last digit d must be even, which rules out 3 and 9.
3. For divisibility by 9 the digit sum 2d + 15 must be a multiple of 9. d = 6 gives 27; d = 4 gives 23, which is not a multiple of 9.
4. So d = 6 (the number is 64566 = 18 × 3587).

Teacher's Note:
1. Test 2 and 9 separately.
2. The digit sum is d + 4 + 5 + 6 + d = 2d + 15.

Answer: (d) 114345

Explanation:
1. A number is divisible by 99 when it is divisible by both 9 and 11.
2. 114345: digit sum 1 + 1 + 4 + 3 + 4 + 5 = 18 (multiple of 9); alternate sum (1 + 4 + 4) − (1 + 3 + 5) = 9 − 9 = 0 (multiple of 11). Indeed 114345 = 99 × 1155.
3. The other three fail the test (their remainders on division by 99 are 88, 63 and 90).
4. The answer is 114345.

Teacher's Note:
1. Use the digit-sum test for 9 and the alternating-sum test for 11.
2. Both conditions must hold.

Answer: (d) 9

Explanation:
1. A number leaves the same remainder on division by 9 as the sum of its digits does.
2. So number − (sum of digits) leaves remainder 0 when divided by 9.
3. It is divisible by 9.

Teacher's Note:
1. Check with 23: 23 − 5 = 18, a multiple of 9.
2. This is the idea behind the divisibility test for 9.

Answer: (c) 6

Explanation:
1. x divides 2x, so x divides 2x + 12 exactly when x divides 12.
2. The positive divisors of 12 are 1, 2, 3, 4, 6, 12.
3. There are 6 possible values.

Teacher's Note:
1. Subtract the multiple 2x to see what x must divide.
2. Count the divisors of 12.

Answer: (b) 3455

Explanation:
1. In each case the divisor exceeds the remainder by 10 (35 − 25 = 45 − 35 = 55 − 45 = 10).
2. So the number plus 10 is a common multiple of 35, 45 and 55. L.C.M.(35, 45, 55) = 5 × 7 × 9 × 11 = 3465.
3. The least number is 3465 − 10 = 3455.

Teacher's Note:
1. Spot the constant gap between divisor and remainder.
2. Check: 3455 = 35 × 98 + 25.

Answer: (a) 1

Explanation:
1. Write the number as 14q + 8 = 7(2q + 1) + 1.
2. On division by 7 the remainder is 1.

Teacher's Note:
1. Since 7 divides 14, only the remainder 8 matters: 8 = 7 + 1.
2. Try 22 = 14 + 8: 22 ÷ 7 leaves 1.

Answer: (d) 83

Explanation:
1. Let the number be 10a + b with a + b = 11. Reversing gives 10b + a.
2. (10a + b) − (10b + a) = 9(a − b) = 45, so a − b = 5.
3. Solving a + b = 11 and a − b = 5 gives a = 8, b = 3. The number is 83.

Teacher's Note:
1. Reversed number is smaller, so the tens digit is bigger.
2. Check: 83 − 38 = 45.

Answer: (b) 25

Explanation:
1. Let there be x adults, so 100 − x children.
2. Apples used: 3x + (100 − x) ÷ 3 = 100. Multiply by 3: 9x + 100 − x = 300, so 8x = 200 and x = 25.

Teacher's Note:
1. Each child gets one-third of an apple.
2. Check: 25 × 3 = 75 apples, and 75 children share 25 apples: total 100.

Answer: (c) \( \frac{7}{48} \)

Explanation:
1. Let 2^x = 4^y = 8^z = 2^t. Since 4 = 2² and 8 = 2³, this means x = t, 2y = t and 3z = t.
2. Then 2x = 2t, 4y = 2t and 6z = 2t, so each of the three terms equals 1/(2t).
3. The sum is 3/(2t) = 24/7, so t = 7/16.
4. Then z = t ÷ 3 = \( \frac{7}{48} \).

Teacher's Note:
1. Write every power as a power of 2 and compare exponents.
2. The three denominators 2x, 4y and 6z are all equal.

Answer: (b) 2

Explanation:
1. Since b/a = (a/b)⁻¹, the right side equals (a/b)^−(x−3).
2. Equal bases (a/b) require equal exponents: x − 1 = −(x − 3) = 3 − x.
3. So 2x = 4 and x = 2.

Teacher's Note:
1. Turn the reciprocal into a negative exponent.
2. Check: (a/b)¹ on the left and (b/a)⁻¹ = a/b on the right.

Answer: (a) 1

Explanation:
1. a = c^z = (b^y)^z = b^yz = (a^x)^yz = a^xyz.
2. So a^xyz = a¹, giving xyz = 1 (for a ≠ 1).

Teacher's Note:
1. Chain the three equations back to a.
2. The product of exponents equals 1.

Answer: (b) \( \left(-\frac52\right)(\sqrt3+\sqrt5) \)

Explanation:
1. Multiply the top and bottom by the conjugate √3 + √5.
2. Denominator: (√3 − √5)(√3 + √5) = 3 − 5 = −2.
3. So the fraction becomes \( \frac{5(\sqrt3+\sqrt5)}{-2} \) = \( \left(-\frac52\right)(\sqrt3+\sqrt5) \)

Teacher's Note:
1. The conjugate of a − b is a + b.
2. Watch the sign: 3 − 5 is negative.

Answer: (d) \( 2^{99} \)

Explanation:
1. \( \frac{2^{100}}{2} = 2^{100-1} = 2^{99} \).
2. The value is \( 2^{99} \).

Teacher's Note:
1. Subtract exponents when dividing powers of the same base.
2. Do not halve the exponent.

Answer: (b) 9

Explanation:
1. Trailing zeros come from pairs of factors 2 × 5, so count the 5s and 2s.
2. 5s: 5 → 1, 10 → 1, 25 → 2, 40 → 1, 50 → 2, 55 → 1, 65 → 1, 125 → 3, 80 → 1; total 13.
3. 2s: 10 → 1, 40 → 3, 50 → 1, 80 → 4; total 9.
4. Number of zeros = min(13, 9) = 9.

Teacher's Note:
1. Zeros are limited by the smaller of the counts of 2 and 5.
2. Do not stop at counting 5s alone.

Answer: (b) 140

Explanation:
1. The first three sons get x/2 + x/4 + x/5 = (10x + 5x + 4x)/20 = 19x/20.
2. The rest, x − 19x/20 = x/20, goes to the fourth son: x/20 = 7.
3. So x = 140.

Teacher's Note:
1. Add the three fractions with common denominator 20.
2. Check: 70 + 35 + 28 + 7 = 140.

Answer: (c) 7

Explanation:
1. 28 = 2² × 7, 35 = 5 × 7, 91 = 7 × 13.
2. The only common prime is 7, so H.C.F. = 7.

Teacher's Note:
1. Factorise each number.
2. Every number here is a multiple of 7.

Answer: (d) \( \sqrt[3]{9} \)

Explanation:
1. √11 ≈ 3.32 and √8 ≈ 2.83 are both above 2.5. ∛7 ≈ 1.91 is below 2.
2. ∛9 ≈ 2.08. Check exactly: 2³ = 8 < 9 < 15.625 = 2.5³, so 2 < ∛9 < 2.5.
3. The number is \( \sqrt[3]{9} \).

Teacher's Note:
1. Cube 2 and 2.5 to compare with the number under the cube root.
2. 2³ = 8 and (2.5)³ = 15.625.

Answer: (a) 15 minute

Explanation:
1. 1 hour = 60 minutes.
2. One quarter of 60 minutes is 60 ÷ 4 = 15 minutes.

Teacher's Note:
1. Convert the fraction of an hour into minutes.
2. Half an hour is 30 minutes, so a quarter is half of that.

Answer: (d) 1

Explanation:
1. Two different primes share no common factor except 1.
2. So their H.C.F. is 1 (they are co-prime).

Teacher's Note:
1. Any two different primes are co-prime.
2. Their LCM, by contrast, is ab.

Answer: (b) 1 \( \frac{3}{5} \)

Explanation:
1. The opposite (additive inverse) of −5/8 is 5/8.
2. The inverse (reciprocal) of 5/8 is 8/5 = 1 3/5.
3. The answer is 1 \( \frac{3}{5} \).

Teacher's Note:
1. First take the opposite, then the reciprocal.
2. Taking the reciprocal first would give the wrong sign.

Answer: (a) 7

Explanation:
1. ⁴√16 = 2 because 2⁴ = 16.
2. ⁴√625 = 5 because 5⁴ = 625.
3. x = 2 + 5 = 7.

Teacher's Note:
1. Check each root by raising to the 4th power.
2. 16 = 2⁴ and 625 = 5⁴.

Answer: (b) \( \frac{2}{15} \)

Explanation:
1. For fractions, H.C.F. = H.C.F. of numerators ÷ L.C.M. of denominators.
2. H.C.F.(6, 4, 2) = 2 and L.C.M.(5, 15, 5) = 15.
3. So the H.C.F. is \( \frac{2}{15} \).

Teacher's Note:
1. Check: 6/5 = 18 × (1/15)... each fraction is a whole-number multiple of 2/15: 9, 2 and 3 times.
2. Numerators go to HCF, denominators go to LCM.

Answer: (d) \( \frac{1}{\sqrt2} \)

Explanation:
1. Let u = √2 − √5, so the denominators are u + √3 and u − √3. Their sum of reciprocals is (u − √3 + u + √3) ÷ (u² − 3) = 2u ÷ (u² − 3).
2. u² = 2 + 5 − 2√10 = 7 − 2√10, so u² − 3 = 4 − 2√10 = 2(2 − √10).
3. The value is 2(√2 − √5) ÷ [2(2 − √10)] = (√2 − √5) ÷ (2 − √10). Since 2 − √10 = √2(√2 − √5), this equals 1/√2.
4. The answer is \( \frac{1}{\sqrt2} \).

Teacher's Note:
1. Group √2 − √5 together to use (a + b)(a − b).
2. A numerical check: both sides come to 0.7071.

Answer: (b) 492

Explanation:
1. Count digits: 1 to 9 use 9 digits; 10 to 99 are 90 numbers with 2 digits each, i.e. 180 digits; 100 to 200 are 101 numbers with 3 digits each, i.e. 303 digits.
2. Total = 9 + 180 + 303 = 492.

Teacher's Note:
1. Split the numbers by number of digits.
2. Do not forget that 100 to 200 contains 101 numbers.

Answer: (c) \( \sqrt[3]{12} \)

Explanation:
1. Bring all three to the same order 6: √3 = ⁶√(3³) = ⁶√27, ∛12 = ⁶√(12²) = ⁶√144 and ⁶√100.
2. Compare 27, 144 and 100: the largest is 144.
3. So the greatest is \( \sqrt[3]{12} \).

Teacher's Note:
1. The LCM of the orders 6, 3, 2 is 6.
2. A bigger radicand with the same order means a bigger number.

Answer: (b) 4

Explanation:
1. They change together again after L.C.M.(48, 72, 108) seconds. 48 = 2⁴ × 3, 72 = 2³ × 3², 108 = 2² × 3³, so L.C.M. = 2⁴ × 3³ = 432 seconds.
2. 7:00 to 7:30 is 30 minutes = 1800 seconds. 1800 ÷ 432 = 4.17, so the multiples 432, 864, 1296 and 1728 fall within the half hour.
3. They change together 4 times after 7 a.m.

Teacher's Note:
1. Convert 30 minutes to seconds first.
2. Do not count the starting moment 7 a.m. itself.

Answer: (a) \( \sqrt2 \)

Explanation:
1. y = 1/x = 1/(2 + √3) = 2 − √3. Note that x = 2 + √3 = (√3 + 1)²/2, so √x = (√3 + 1)/√2, and similarly √y = (√3 − 1)/√2.
2. Then √2 + √x = (2 + √3 + 1)/√2 = (3 + √3)/√2 and √2 − √y = (2 − √3 + 1)/√2 = (3 − √3)/√2.
3. The sum is √2[(2 + √3)/(3 + √3) + (2 − √3)/(3 − √3)] = √2 × [(2 + √3)(3 − √3) + (2 − √3)(3 + √3)]/6 = √2 × 6/6 = \( \sqrt{2} \).

Teacher's Note:
1. Write x and y as perfect squares of nested surds.
2. A numerical check gives 1.41421 on both sides.

Answer: (b) 492

Explanation:
1. 1 to 9: 9 digits. 10 to 99: 90 numbers × 2 = 180 digits. 100 to 200: 101 numbers × 3 = 303 digits.
2. Total = 9 + 180 + 303 = 492.

Teacher's Note:
1. Same method as before: group by digit-length.
2. Check that 9 + 90 + 101 = 200 numbers.

Answer: (a) 1

Explanation:
1. m = 5a + 2 and n = 5b + 4, so m + n = 5(a + b) + 6 = 5(a + b + 1) + 1.
2. The remainder is 1.

Teacher's Note:
1. Add the remainders: 2 + 4 = 6, then reduce mod 5.
2. 6 is bigger than the divisor, so it gives one more 5 and remainder 1.

Answer: (c) \( \sqrt2+\sqrt7 \)

Explanation:
1. Look for numbers whose sum is 9 and whose product is 14: these are 2 and 7.
2. 9 + 2√14 = 2 + 7 + 2√(2 × 7) = (√2 + √7)².
3. The square root is \( \sqrt2+\sqrt7 \).

Teacher's Note:
1. Use (√a + √b)² = a + b + 2√(ab).
2. Option (b) fails: (√3 + √6)² = 9 + 6√2.

Answer: (b) \( 1-\frac17 \)

Explanation:
1. 1 − 127/343 = 216/343.
2. Since 216 = 6³ and 343 = 7³, the cube root is 6/7.
3. 6/7 = 1 − 1/7, so the answer is \( 1-\frac17 \).

Teacher's Note:
1. Subtract first, then take the cube root of numerator and denominator.
2. Know cubes: 6³ = 216 and 7³ = 343.

Answer: (a) \( 0.\overline{6} \)

Explanation:
1. 2.666... = 2 + 2/3 = 8/3, and 1.999... = 2 (since 0.999... = 1).
2. The difference is 8/3 − 2 = 2/3 = 0.666... = \( 0.\overline{6} \).

Teacher's Note:
1. Convert the repeating decimals to fractions.
2. 0.999... equals 1 exactly.

Answer: (a) \( \frac{20\sqrt5-35}{31} \)

Explanation:
1. Multiply by the conjugate 7 − 4√5: the denominator becomes 49 − 16 × 5 = 49 − 80 = −31.
2. The numerator is 5(7 − 4√5) = 35 − 20√5.
3. So the value is (35 − 20√5)/(−31) = \( \frac{20\sqrt5-35}{31} \).

Teacher's Note:
1. (4√5)² = 16 × 5 = 80, not 20.
2. A negative denominator can be moved into the numerator by changing signs.

Answer: (a) 16, 24, 5, 80

Explanation:
1. Let the common value be k. The numbers are k − 4, k + 4, k ÷ 4 and 4k.
2. Their sum: k − 4 + k + 4 + k/4 + 4k = 6.25k = 125, so k = 20.
3. The integers are 16, 24, 5, 80, i.e. option (a).

Teacher's Note:
1. Express all four numbers in terms of the common value.
2. Check: 16 + 4 = 24 − 4 = 5 × 4 = 80 ÷ 4 = 20.

Answer: (c) 39

Explanation:
1. a² + b² = 2(6 + 5) = 22 and ab = 6 − 5 = 1.
2. 2a² − 5ab + 2b² = 2(a² + b²) − 5ab = 44 − 5 = 39.

Teacher's Note:
1. Use a² + b² = (a + b)² − 2ab or expand directly.
2. ab = 1 makes the second term easy.

Answer: (b) 3

Explanation:
1. √(m⁴n⁴) = (mn)².
2. ⁶√(m²n²) = (mn)^2/6 = (mn)^1/3 and ∛(m²n²) = (mn)^2/3.
3. Multiplying: (mn)^{2 + 1/3 + 2/3} = (mn)³, so k = 3.

Teacher's Note:
1. Write each root as a fractional power of (mn).
2. Add the exponents when multiplying.