Quantitative Aptitude — HCF and LCM

To find the least number exactly divisible by 12, 15, and 20, we need to find their Least Common Multiple (LCM). Prime factorization of the numbers: 12 = 2^2 * 3 15 = 3 * 5 20 = 2^2 * 5 LCM is the product of the highest powers of all prime factors involved: LCM(12, 15, 20) = 2^2 * 3^1 * 5^1 = 4 * 3 * 5 = 60. Therefore, the least number is 60.

To find the greatest number that divides 48 and 72 exactly, we need to find their Highest Common Factor (HCF). Prime factorization of the numbers: 48 = 2^4 * 3 72 = 2^3 * 3^2 HCF is the product of the lowest powers of all common prime factors: HCF(48, 72) = 2^3 * 3^1 = 8 * 3 = 24. Therefore, the greatest number is 24.

Formula: Product of two numbers = HCF * LCM. Let the two numbers be A and B. We are given A = 275, HCF = 11, LCM = 7700. So, A * B = HCF * LCM 275 * B = 11 * 7700 B = (11 * 7700) / 275 B = (11 * 7700) / (11 * 25) (Since 275 = 11 * 25) B = 7700 / 25 B = 308. Therefore, the other number is 308.

The smallest number which when divided by 8, 9, 12, and 15 leaves a remainder of 1 in each case is (LCM of 8, 9, 12, 15) + 1. Prime factorization of the numbers: 8 = 2^3 9 = 3^2 12 = 2^2 * 3 15 = 3 * 5 LCM is the product of the highest powers of all prime factors involved: LCM(8, 9, 12, 15) = 2^3 * 3^2 * 5^1 = 8 * 9 * 5 = 72 * 5 = 360. Required number = LCM + 1 = 360 + 1 = 361. Therefore, the smallest number is 361.

Formula for HCF of fractions: HCF(a/b, c/d, e/f) = (HCF of numerators) / (LCM of denominators). Numerators are 2, 4, 6. HCF(2, 4, 6): 2 = 2 4 = 2^2 6 = 2 * 3 HCF(2, 4, 6) = 2. Denominators are 3, 5, 7. LCM(3, 5, 7): Since 3, 5, and 7 are prime numbers, their LCM is their product. LCM(3, 5, 7) = 3 * 5 * 7 = 105. HCF of fractions = 2/105. Therefore, the HCF of 2/3, 4/5, and 6/7 is 2/105.

Formula for LCM of fractions: LCM(a/b, c/d, e/f) = (LCM of numerators) / (HCF of denominators). Numerators are 1, 3, 5. LCM(1, 3, 5): Since 1, 3, and 5 are coprime (or 1 is a factor of all), their LCM is their product. LCM(1, 3, 5) = 1 * 3 * 5 = 15. Denominators are 2, 4, 6. HCF(2, 4, 6): 2 = 2 4 = 2^2 6 = 2 * 3 HCF(2, 4, 6) = 2. LCM of fractions = 15/2. Therefore, the LCM of 1/2, 3/4, and 5/6 is 15/2.

If a number N divides a, b, and c leaving the same remainder, then N must be a common factor of the differences (b-a), (c-b), and (c-a). We need to find the HCF of the absolute differences between the numbers: Difference 1: |305 - 130| = 175 Difference 2: |455 - 305| = 150 Difference 3: |455 - 130| = 325 Now, find the HCF of 175, 150, and 325. Prime factorization of the differences: 175 = 5^2 * 7 150 = 2 * 3 * 5^2 325 = 5^2 * 13 HCF is the product of the lowest powers of all common prime factors: HCF(175, 150, 325) = 5^2 = 25. Therefore, the greatest number is 25.

To find when the bells will ring together again, we need to find the Least Common Multiple (LCM) of their ringing intervals. Intervals are 6, 8, 12, and 18 seconds. Prime factorization of the intervals: 6 = 2 * 3 8 = 2^3 12 = 2^2 * 3 18 = 2 * 3^2 LCM is the product of the highest powers of all prime factors involved: LCM(6, 8, 12, 18) = 2^3 * 3^2 = 8 * 9 = 72. So, the bells will ring together again after 72 seconds. Convert 72 seconds to minutes and seconds: 72 seconds = 1 minute and 12 seconds. If they started ringing together at 10:00 AM, they will ring together again at 10:00 AM + 1 minute 12 seconds = 10:01:12 AM. Therefore, they will ring together again at 10:01:12 AM.

Let the two numbers be A and B. Given HCF(A, B) = 23. We can express the numbers as A = 23a and B = 23b, where 'a' and 'b' are coprime integers (i.e., HCF(a, b) = 1). Given their sum is 207: A + B = 207 23a + 23b = 207 23(a + b) = 207 a + b = 207 / 23 a + b = 9. Now, we need to find pairs of coprime integers (a, b) whose sum is 9. We assume a < b to avoid duplicate pairs. 1. If a = 1, then b = 9 - 1 = 8. HCF(1, 8) = 1. This is a valid pair (23*1, 23*8) = (23, 184). 2. If a = 2, then b = 9 - 2 = 7. HCF(2, 7) = 1. This is a valid pair (23*2, 23*7) = (46, 161). 3. If a = 3, then b = 9 - 3 = 6. HCF(3, 6) = 3. This is NOT a coprime pair. Invalid. 4. If a = 4, then b = 9 - 4 = 5. HCF(4, 5) = 1. This is a valid pair (23*4, 23*5) = (92, 115). If a = 5, then b = 4, which is the same pair as (4,5). So, there are 3 such pairs of numbers possible. Therefore, 3 pairs of numbers are possible.

Observe the difference between each divisor and its corresponding remainder: 15 - 10 = 5 20 - 15 = 5 25 - 20 = 5 30 - 25 = 5 Since the difference is constant (5) in all cases, the required number will be (LCM of divisors) - (constant difference). First, find the LCM of 15, 20, 25, and 30. Prime factorization of the numbers: 15 = 3 * 5 20 = 2^2 * 5 25 = 5^2 30 = 2 * 3 * 5 LCM is the product of the highest powers of all prime factors involved: LCM(15, 20, 25, 30) = 2^2 * 3^1 * 5^2 = 4 * 3 * 25 = 12 * 25 = 300. Now, subtract the constant difference from the LCM: Required number = LCM - 5 = 300 - 5 = 295. Therefore, the smallest number is 295.