CBSE Class 8 Mathematics Factors and Multiples MCQs Set 30

Answer: (a) 21.6

Explanation:
1. Write all with two decimal places without the decimal point: 54, 180 and 720. Their L.C.M. is 2160 (54 = 2 × 3³, 180 = 2² × 3² × 5, 720 = 2⁴ × 3² × 5, so the L.C.M. = 2⁴ × 3³ × 5 = 2160). Putting back the decimal point with one place gives 21.6.

Teacher's Note:
1. Remove the decimal points after equalising places.
2. Mark the decimal place again in the answer.

Answer: (d) 0.18

Explanation:
1. The numbers without decimal points are 54, 180 and 720 (two decimal places). Their H.C.F. is 18, so the H.C.F. is 0.18.

Teacher's Note:
1. Make the number of decimal places equal.
2. Find the H.C.F. of the whole numbers, then restore the decimal.

Answer: (b) 3/1400

Explanation:
1. H.C.F. of fractions = H.C.F. of numerators / L.C.M. of denominators = H.C.F.(9, 12, 18, 21) / L.C.M.(10, 25, 35, 40) = 3/1400.

Teacher's Note:
1. Use H.C.F. of numerators over L.C.M. of denominators.
2. L.C.M.(10, 25, 35, 40) = 1400.

Answer: (b) 1/84

Explanation:
1. H.C.F. = H.C.F.(3, 5, 6) / L.C.M.(4, 6, 7) = 1/84.

Teacher's Note:
1. H.C.F. of numerators over L.C.M. of denominators.
2. H.C.F.(3, 5, 6) = 1.

Answer: (a) 40

Explanation:
1. Write the numbers as 8/1, 4/5 and 10/7. L.C.M. = L.C.M.(8, 4, 10) / H.C.F.(1, 5, 7) = 40/1 = 40.

Teacher's Note:
1. Convert to improper fractions in lowest terms.
2. L.C.M. of numerators over H.C.F. of denominators.

Answer: (c) 241 and 337 only

Explanation:
1. 522 is even, so it is not prime. 241 and 337 have no divisor among the primes up to 15 and 18 (2, 3, 5, 7, 11, 13, 17), so both are prime.

Teacher's Note:
1. Test divisibility by primes up to the square root.
2. 522 is even.

Answer: (b) 1 m 20 cm

Explanation:
1. In centimetres the lengths are 360, 600, 840 and 1800. Their H.C.F. is 120 cm = 1 m 20 cm.

Teacher's Note:
1. Convert all lengths to the same unit.
2. The greatest common measure is the H.C.F.

Answer: (a) 24

Explanation:
1. 96 = 2⁵ × 3, 528 = 2⁴ × 3 × 11, 792 = 2³ × 3² × 11. The H.C.F. = 2³ × 3 = 24.

Teacher's Note:
1. Use prime factorisation.
2. Take the least power of every common prime.

Answer: (d) 32

Explanation:
1. Subtract the remainder: 256, 928 and 1376 must be exactly divisible. 256 = 2⁸, 928 = 2⁵ × 29, 1376 = 2⁵ × 43, so the H.C.F. is 2⁵ = 32 (greater than the remainder 5).

Teacher's Note:
1. Remove the remainder from each number.
2. Then take the H.C.F.

Answer: (c) 120

Explanation:
1. L.C.M.(3, 4, 5) = 60. The multiples of 60 are 60, 120, 180, ..., and the smallest with three digits is 120.

Teacher's Note:
1. Find the L.C.M.
2. Pick its first 3-digit multiple.

Answer: (a) 540

Explanation:
1. 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5, 27 = 3³. L.C.M. = 2² × 3³ × 5 = 540.

Teacher's Note:
1. Take the highest power of every prime.
2. L.C.M. is the least common multiple.

Answer: (a) 80

Explanation:
1. The product of the numbers = 16 × 240 = 3840. The other number = 3840/48 = 80.

Teacher's Note:
1. Use H.C.F. × L.C.M. = product.
2. Divide by the given number.

Answer: (d) 0.21

Explanation:
1. With two decimal places the numbers are 63, 105 and 210, whose H.C.F. is 21. Marking two decimal places gives 0.21.

Teacher's Note:
1. Equalise the decimal places first.
2. Restore two decimal places at the end.

Answer: (c) 195 and 143

Explanation:
1. Let the numbers be 15x and 11x. Their H.C.F. is x, so x = 13 and the numbers are 15 × 13 = 195 and 11 × 13 = 143.

Teacher's Note:
1. The common factor of the ratio is the H.C.F.
2. Multiply the ratio terms by 13.

Answer: (b) 45 cm

Explanation:
1. In cm: 495 = 3² × 5 × 11, 900 = 2² × 3² × 5², 1665 = 3² × 5 × 37. The H.C.F. = 3² × 5 = 45 cm.

Teacher's Note:
1. Change all to centimetres.
2. Greatest common measure = H.C.F.

Answer: (a) 221 and 293

Explanation:
1. L.C.M.(6, 8, 9) = 72. The multiples of 72 between 200 and 300 are 216 and 288. Adding the remainder 5 gives 221 and 293.

Teacher's Note:
1. Find multiples of the L.C.M. first.
2. Add the common remainder.

Answer: (b) 12, 18, 20 and 24

Explanation:
1. Sum of proper factors: 12 gives 1 + 2 + 3 + 4 + 6 = 16 > 12; 18 gives 1 + 2 + 3 + 6 + 9 = 21; 20 gives 1 + 2 + 4 + 5 + 10 = 22; 24 gives 1 + 2 + 3 + 4 + 6 + 8 + 12 = 36. All other numbers below 25 have a proper-factor sum not above the number.

Teacher's Note:
1. Add the proper factors (all factors except the number itself).
2. Compare the sum with the number.

Answer: (a) 6

Explanation:
1. The proper factors of 6 are 1, 2 and 3, and 1 + 2 + 3 = 6. No other number below 25 has this property (28 is the next perfect number).

Teacher's Note:
1. Add the proper factors.
2. The sum must equal the number.

Answer: (d) the H.C.F. of two numbers always divides their L.C.M., and 16 does not divide 145

Explanation:
1. Every common factor divides both numbers, so the H.C.F. is a factor of the L.C.M. Since 145 ÷ 16 is not a whole number, 16 and 145 cannot be an H.C.F. and L.C.M. pair.

Teacher's Note:
1. The H.C.F. is always a factor of the L.C.M.
2. Check whether 16 divides 145.

Answer: (d) 3

Explanation:
1. A number is divisible by 3 if the sum of its digits is divisible by 3. The sum 8 + 3 + 5 + 8 + 6 = 30 is already divisible by 3, so * must be 0, 3, 6 or 9. Among the options only 3 qualifies.

Teacher's Note:
1. Add the known digits.
2. The digit * must keep the sum a multiple of 3.

Answer: (c) 2

Explanation:
1. 2 divides every even number, and 2 itself is the largest factor of 2. So the H.C.F. is 2.

Teacher's Note:
1. Every even number is a multiple of 2.
2. The H.C.F. of 2 and any number cannot exceed 2.

Answer: (a) All three statements are true

Explanation:
1. (i) 24 = 2³ × 3, 36 = 2² × 3², 90 = 2 × 3² × 5, so L.C.M. = 2³ × 3² × 5 = 360. (ii) Other number = (6 × 72)/18 = 24. (iii) L.C.M. = 2560/16 = 160. All are true.

Teacher's Note:
1. Use H.C.F. × L.C.M. = product.
2. Use prime factors for (i).

Answer: (a) 2

Explanation:
1. A prime has exactly two factors. 2 has factors 1 and 2, and every other even number also has 2 as a factor, so 2 is the only even prime.

Teacher's Note:
1. Every even number greater than 2 is divisible by 2.
2. A prime has exactly two factors.