JEE Mathematics HCF and LCM MCQs Set 02

Answer: (a) 7200

Explanation:
1. \( 96=2^5\cdot3 \), \( 100=2^2\cdot5^2 \), \( 120=2^3\cdot3\cdot5 \), \( 144=2^4\cdot3^2 \).
2. The L.C.M. takes the highest power of every prime: \( 2^5\cdot3^2\cdot5^2=32\times9\times25=7200 \).

Teacher's Note:
1. Every number divides the L.C.M., so check 7200 / 96 = 75.
2. The highest powers come from different numbers: 2^5 from 96, 3^2 from 144, 5^2 from 100.

Answer: (d) 31

Explanation:
1. \( 1147=2\times527+93 \).
2. \( 527=5\times93+62 \).
3. \( 93=1\times62+31 \).
4. \( 62=2\times31+0 \). The last divisor 31 is the H.C.F.

Teacher's Note:
1. Keep dividing the previous divisor by the remainder until the remainder is 0.
2. Check: 527 = 17 x 31 and 1147 = 37 x 31.

Answer: (b) 2223

Explanation:
1. Each number has digit sum divisible by 9, so 9 is a common factor: 42237 = 9 x 4693, 75582 = 9 x 8398, 8892 = 9 x 988.
2. For 4693, 8398 and 988: 8398 = 8 x 988 + 494 and 988 = 2 x 494, so gcd(988, 8398) = 494; since 4693 is odd the common factor 2 is dropped, giving 247 (494 = 2 x 247), and 247 divides 4693 (4693 = 19 x 247).
3. So the H.C.F. is \( 9\times247=2223 \).

Teacher's Note:
1. Pull out the obvious common factor, then use the division method on the reduced numbers.
2. 247 = 13 x 19 and 2223 = 3^2 x 13 x 19.

Answer: (a) 1260

Explanation:
1. \( 12=2^2\cdot3 \), \( 28=2^2\cdot7 \), \( 63=3^2\cdot7 \), \( 70=2\cdot5\cdot7 \), \( 84=2^2\cdot3\cdot7 \).
2. Highest powers: \( 2^2\cdot3^2\cdot5\cdot7=1260 \).

Teacher's Note:
1. Only 63 contains 3 squared.
2. Check: 1260 / 63 = 20 and 1260 / 70 = 18.

Answer: (c) 2772 seconds

Explanation:
1. They meet at the start after a time that is a common multiple of 252, 308 and 198.
2. The first such time is the L.C.M.: \( 252=2^2\cdot3^2\cdot7 \), \( 308=2^2\cdot7\cdot11 \), \( 198=2\cdot3^2\cdot11 \).
3. L.C.M. \( =2^2\cdot3^2\cdot7\cdot11=2772 \) s (46 min 12 s).

Teacher's Note:
1. The first meeting needs the least common multiple.
2. In 2772 s, A runs 11 rounds, B runs 9 and C runs 14.

Answer: (a) 11

Explanation:
1. They meet after 2772 s.
2. \( \frac{2772}{252}=11 \) rounds.

Teacher's Note:
1. B makes 2772/308 = 9 rounds and C makes 2772/198 = 14 rounds.
2. The number of rounds is the L.C.M. divided by that runner's time per round.

Answer: (b) 45

Explanation:
1. The product of the numbers is \( 9\times270=2430 \).
2. Other number \( =\frac{2430}{54}=45 \).
3. Check: H.C.F.(54, 45) = 9 and L.C.M. = 270.

Teacher's Note:
1. Always verify that the H.C.F. and L.C.M. of the answer pair come out correctly.
2. Both numbers must be multiples of the H.C.F. 9.

Answer: (b) 231

Explanation:
1. A common factor divides the sum and the difference of the numbers.
2. \( 91+140=231=7\times33 \).
3. None of 232, 233, 235 leaves remainder 0 on division by 7.

Teacher's Note:
1. If d divides a and b then d divides ma + nb for any whole numbers m, n.
2. Here 91 = 7 x 13 and 140 = 7 x 20.

Answer: (d) 120

Explanation:
1. The L.C.M. of 8, 12, 15: \( 2^3\cdot3\cdot5=120 \).

Teacher's Note:
1. Check: 120/8 = 15, 120/12 = 10, 120/15 = 8.
2. Any common multiple is a multiple of 120.

Answer: (a) 94

Explanation:
1. L.C.M. of 6, 9 and 15 is 90.
2. Required number \( =90+4=94 \).

Teacher's Note:
1. Add the common remainder to the L.C.M.
2. Check: 94 = 6 x 15 + 4 = 9 x 10 + 4 = 15 x 6 + 4.

Answer: (c) 13

Explanation:
1. The number divides \( 70-5=65 \) and \( 125-8=117 \).
2. \( 65=5\cdot13 \) and \( 117=9\cdot13 \), so the H.C.F. is 13.
3. 13 is greater than the remainders 5 and 8.

Teacher's Note:
1. Subtract the remainders, then take the H.C.F.
2. The divisor has to be larger than both remainders.

Answer: (a) 5 minutes

Explanation:
1. The L.C.M. of 15, 20, 25 is \( 2^2\cdot3\cdot5^2=300 \) s.
2. 300 s = 5 minutes.

Teacher's Note:
1. Convert the L.C.M. from seconds to minutes at the end.
2. In 300 s the bells toll 20, 15 and 12 times.

Answer: (d) 900

Explanation:
1. L.C.M. \( =180=2^2\cdot3^2\cdot5 \).
2. For a perfect square every prime must have an even power, so multiply by 5: \( 900=2^2\cdot3^2\cdot5^2=30^2 \).

Teacher's Note:
1. Smallest common multiple is not necessarily a square.
2. Multiply the L.C.M. by the smallest factor that makes all indices even.

Answer: (b) \( \frac{2}{27} \)

Explanation:
1. H.C.F. of fractions \( =\frac{\text{H.C.F. of numerators}}{\text{L.C.M. of denominators}} \).
2. Numerators 2, 8, 10 have H.C.F. 2; denominators 3, 9, 27 have L.C.M. 27.
3. H.C.F. \( =\frac{2}{27} \).

Teacher's Note:
1. \( \frac{40}{3} \) is the L.C.M. of the fractions: L.C.M. of numerators over H.C.F. of denominators.
2. Check: 2/3 = 9 x 2/27, 8/9 = 12 x 2/27, 10/27 = 5 x 2/27.

Answer: (b) 36

Explanation:
1. Common primes are 2 and 3.
2. Take the lowest powers: \( 2^2\cdot3^2=36 \).

Teacher's Note:
1. 22680 is the L.C.M.
2. 5 and 7 are not common, so they are excluded.

Answer: (d) 22680

Explanation:
1. Take every prime that appears with its highest power.
2. \( 2^3\cdot3^4\cdot5\cdot7=8\times81\times35=22680 \).

Teacher's Note:
1. Include primes that appear in only one of the numbers.
2. Check: 22680 / 360 = 63 and 22680 / 2268 = 10.

Answer: (a) 37

Explanation:
1. \( 2701=1\times2257+444 \).
2. \( 2257=5\times444+37 \).
3. \( 444=12\times37+0 \).
4. The last divisor is 37.

Teacher's Note:
1. Divide the larger by the smaller, then the divisor by the remainder.
2. Check: 2257 = 37 x 61 and 2701 = 37 x 73.

Answer: (c) 1540

Explanation:
1. \( 35=5\cdot7 \) and \( 44=2^2\cdot11 \) have no common factor, so H.C.F. = 1.
2. L.C.M. \( =35\times44=1540 \).

Teacher's Note:
1. For co-prime numbers, L.C.M. is the product.
2. Check: 1540 = 35 x 44.

Answer: (d) 140

Explanation:
1. Let the numbers be 3k and 4k. Since 3 and 4 are co-prime, the H.C.F. is k and the L.C.M. is 12k.
2. \( 12k=240 \), so k = 20.
3. Numbers: 60 and 80, sum 140.

Teacher's Note:
1. For numbers in ratio a : b with co-prime a and b, the L.C.M. is abk.
2. Check: L.C.M.(60, 80) = 240.

Answer: (b) 2520

Explanation:
1. Highest powers needed: \( 2^3 \) (from 8), \( 3^2 \) (from 9), 5 and 7.
2. L.C.M. \( =8\times9\times5\times7=2520 \).

Teacher's Note:
1. The L.C.M. of 1 to 10 uses the highest prime powers up to 10.
2. 2520 is also divisible by 12, 14 and 15.

Answer: (b) 6 m

Explanation:
1. The piece length must divide each rope length, so it is a common factor.
2. The greatest is the H.C.F. of 12, 18, 30, which is 6.

Teacher's Note:
1. Cutting with no waste means the length divides all three lengths.
2. Greatest length means H.C.F., not L.C.M.

Answer: (d) 77

Explanation:
1. Work in centimetres: 660 cm by 420 cm.
2. Side of the largest tile is the H.C.F. of 660 and 420, which is 60 cm.
3. Number of tiles \( =\frac{660}{60}\times\frac{420}{60}=11\times7=77 \).

Teacher's Note:
1. The side of the tile must divide both dimensions.
2. Count tiles along each edge and multiply.

Answer: (c) 30

Explanation:
1. Product of the numbers \( =6\times1260=7560 \).
2. Other number \( =\frac{7560}{252}=30 \).
3. Check: H.C.F.(252, 30) = 6 and L.C.M. = 1260.

Teacher's Note:
1. Use H.C.F. x L.C.M. = product, then verify.
2. 210 would give H.C.F. 42.