50 + Problems on H.C.F and L.C.M Questions and Answers | Quantitative Aptitude📖

Solution: Let the numbers be 5m and 11m. Since 5:11 is already the reduced ratio, ‘m’ has to be the HCF. So, the numbers are 5 x 7 = 35 and 11 x 7 = 77.

Solution: Let us first convert each length to cm. So, the lengths are 450 cm, 990 cm, and 1620 cm. Now, we need to find the length of the largest plank that can be used to measure these lengths as the largest plank will take the least time. For this, we need to take the HCF of 450, 990, and 1620. 450 = 2 x 3 x 3 x 5 x 5 = 2 x 3² x 5² 990 = 2 x 3 x 3 x 5 x 11 = 2 x 3² x 5 x 11 1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5 = 2² x 3⁴ x 5 Therefore, HCF (450, 990, 1620) = 2 x 3 x 3 x 5 = 90 Thus, we need a plank of length 90 cm to measure the given lengths in the least time.

Solution: The required number leaves remainders 1 and 4 on dividing 70 and 50 respectively. This means that the number exactly divides 69 and 46. So, we need to find the HCF of 69 (3 x 23) and 46 (2 x 23). HCF (69, 46) = 23 Thus, 23 is the required number. 

Solution: To find the required number, we need to find the HCF of (136-64), (238-136), and (238-64), i.e., HCF (72, 102, 174). 72 = 2³ x 3² 102 = 2 x 3 x 17 174 = 2 x 3 x 29 Therefore, HCF (72, 102, 174) = 2 x 3 = 6 hence, 6 is the required number. 

Solution: In these types of questions, we need to find the LCM of the divisors and add the common remainder (3) to it. So, LCM (5, 7, 9, 12) = 1260 Therefore, required number = 1260 + 3 = 1263 

Solution: To find: LCM (6, 8).
The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, …
The multiples of 8 are 8, 16, 24, 32, 40, 48, ….
Thus, the smallest common multiple of 6 and 8 is 24.
Therefore, the LCM of 6 and 8 is 24.

Solution: To find the LCM of 4 and 12 using the prime factorisation method, follow the below steps.
Step 1: Find the prime factorization of given numbers:
The prime factorisation of 4 is 2 × 2
The prime factorisation of 12 is 2 × 2 × 3.
Step 2: The LCM of given numbers is found by multiplying the product of all factors. (Note: The common factor is included only once)
Hence, the product of prime factors = 2 × 2 × 3 = 12.
Therefore, the LCM of 4 and 12 is 12.

Solution: The prime factorisation of 54 is 2 × 3 × 3 × 3.
The prime factorisation of 60 is 2 × 2 × 3 × 5.
Thus, the product of prime factors = 2 × 2 × 3 × 3 × 3 × 5 = 540
Hence, the LCM of 54 and 60 is 540.

Solution: Given: LCM(12, 7) × HCF(12, 7) = Product(12, 7) …(1)
Finding LCM (12, 7):
The multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, ..
The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 120, …
Hence, the LCM of 12 and 7 is 84.
Finding HCF (12, 7):
The factors of 7 are 1 and 7.
The factors of 12 are 1, 2, 3, 4, 6 and 12.
Hence, HCF of 12 and 7 is 1.
Finding Product of 12 and 7:
The product of 12 and 7 = 12 × 7 = 84.
Now, substitute the obtained values in (1), we get
84 × 1 = 84
84 = 84
Hence, LHS = RHS
Therefore, LCM(12, 7) × HCF(12, 7) = Product(12, 7) is proved.

Solution: Given fractions: 2/5, 4/7 and 6/11
As we know, the formula to find the LCM of fractions is:
LCM of fractions = LCM of Numerators/HCF of Denominators .. (1)
Thus, the LCM of Numerators = LCM (2, 4, 6) = 12.
HCF of denominators = HCF (5, 7, 11) = 1
Now, substitute the values in (1), we get
LCM of fractions = 12/1 = 12
Hence, the LCM of the fractions 2/5, 4/7 and 6/11 is 12.

1. 3800
2. 4200
3. 4400
4. 3200
5. None of these
Solution: Option 2
25 = 5 × 5, 30 = 5 × 3 × 2, 35 = 5 × 7, 40 = 2 × 2 × 2 × 5
Required LCM = 2 × 2 × 2 × 5 × 5 × 3 × 7 = 4200

1. 193
2. 183
3. 223
4. 213
5. 233
Solution: Option 4
H.C.F. of 639 and 1065 is 213. H.C.F. of 213 and 1491 is 213.

1. 4/189
2. 6/63
3. 2/63
4. 20/21
5. None of these
Solution: Option 3
H.C.F of 4/9, 10/21 and 20/63 = H.C.F of 4,10 and 20 / L.C.M of 9,21 and 63
= H.C.F of 4, 10 and 20 = 2 & L.C.M. of 9, 21 and 63 = 63. Required H.C.F. = 2/63

1. 66, 77
2. 70, 84
3. 94, 108
4. 84, 96
5. 66, 106
Solution: Option 4
The difference of requisite numbers must be 12 and each should be divisible by 12. Checking the options given, only the fourth option satisfies.

1. 1,174
2. 74,100
3. 29, 154
4. 29, 145
5. None of these
Solution: Option 4
Let the numbers be 29a and 29b. Then, 29a + 29b= 174 or 29(a + b) = 174 or, a + b = 174/29 = 6. Values of co-primes (with sum 6) is(1, 5).
So, the possible pairs of numbers is (29 x 1, 29 x 5) i.e. 29 and 145, which will become the answer.

Solution: The largest four-digit number is 9999. Now, LCM (15, 21, 28) = 420 On dividing 9999 by 420, we get 339 as the remainder. Therefore, the required number is 9999-339 = 9660

Solution: They all will whistle again at the same time after an interval that is equal to the LCM of their individual whistle-blowing cycles. So, LCM (42, 60, 78) = 2 x 3 x 7 x 10 x 13 = 5460 Therefore, they will blow the whistle again simultaneously after 5460 sec, i.e., after 1 hour 31 minutes, i.e., at 11:01:00 hours.

Solution: LCM (6, 7, 8) = 168 So, the number is of the form 168m + 3. Now, 168m + 3 should be divisible by 9. We know that a number is divisible by 9 if the sum of its digits is a multiple of 9. For m = 1, the number is 168 + 3 = 171, the sum of whose digits is 9. Therefore, the required number is 171.

Solution: Let the common ratio be ‘m’. So, the numbers are 2m and 3m. Now, we know that the Product of numbers is = Product of LCM and HCF. => 2m x 3m = 294 => m² = 49 => m = 7 Therefore, the numbers are 14 and 21.

Solution: We need to find the size of a square tile such that a number of tiles cover the field exactly, leaving no area unpaved. For this, we find the HCF of the length and breadth of the field. HCF (180, 105) = 15 Therefore, size of each tile = 15m x 15m Also, number of tiles = area of field / area of each tile => Number of tiles = (180 x 105) / (15 x 15) => Number of tiles = 84 Hence, we need 84 tiles, each of size 15m x 15m.