UGCNET Questions

To find the LCM, we factorize the numbers into primes. The number 12 is 2² × 3 and 18 is 2 × 3². The LCM takes the highest power of each prime appearing in either factorization, giving 2² × 3². This product equals 4 × 9 = 36. Hence, 36 is the smallest number that is a multiple of both 12 and 18.

Option A:
Option A, 36, includes both enough factors of 2 and enough factors of 3 to be divisible by the two given numbers. It is also the smallest such number and so satisfies the definition of the LCM.

Option B:
Option B, 24, is divisible by 12 but not by 18 because 24 ÷ 18 is not an integer. It lacks a second factor of 3.

Option C:
Option C, 30, is divisible by neither 12 nor 18 in an exact sense, as it does not contain the required combination of prime powers.

Option D:
Option D, 42, is divisible by 6 and 7, but not by 12 or 18 without leaving a remainder. It is larger than necessary and not a common multiple in this context.

Let the three consecutive even numbers be n − 2, n and n + 2, where n is the middle one. Their sum is (n − 2) + n + (n + 2) = 3n. According to the given condition, 3n = 66, so n = 66 ÷ 3 = 22. Hence, the middle even number is 22.

Option A:
Option A, 20, would give the three numbers 18, 20 and 22, whose sum is 60, not 66. Therefore, it does not satisfy the condition set in the problem.

Option B:
Option B fits perfectly: with 22 as the middle number, the three numbers become 20, 22 and 24, and their sum is 20 + 22 + 24 = 66, matching the given total.

Option C:
Option C, 24, would generate the sequence 22, 24 and 26, whose sum is 72, exceeding the required sum.

Option D:
Option D, 26, would produce 24, 26 and 28, summing to 78, which is also inconsistent with the problem statement.

The highest common factor is the largest number that divides both given numbers exactly. The prime factorization of 18 is 2 × 3 × 3 and of 30 is 2 × 3 × 5. The common prime factors are 2 and 3, and multiplying them gives 2 × 3 = 6. Since no larger number divides both 18 and 30 without remainder, 6 is the highest common factor.

Option A:
Option A correctly uses the idea of common prime factors and combines 2 and 3 to produce 6. It respects both the definition of a factor and the requirement that the factor be as large as possible while still dividing both numbers exactly. This makes 6 the appropriate highest common factor for 18 and 30.

Option B:
Option B, 3, is a common factor but not the greatest one because 6 also divides both numbers and is larger than 3. Choosing 3 ignores the possibility of combining more common prime factors.

Option C:
Option C, 5, divides 30 but does not divide 18, so it fails the basic requirement that the highest common factor must be a factor of both numbers. It is therefore not a common factor at all in this pair.

Option D:
Option D, 9, divides 18 but not 30 because 30 ÷ 9 does not yield an integer. Hence, it also cannot be regarded as a common factor of the two numbers.

To find the smallest number to be added, we first determine the remainder when 625 is divided by 7. Since 7 × 89 = 623, the remainder is 625 − 623 = 2. A number is divisible by 7 when the remainder becomes zero, so we must add a number that brings the remainder 2 up to 7. Therefore, we add 5 (because 2 + 5 = 7), making 625 + 5 = 630, which is divisible by 7.

Option A:
Option A correctly uses the idea that the required adjustment is 7 minus the remainder. With a remainder of 2, adding 5 eliminates the remainder and yields a number (630) that lies exactly on the 7-times table.

Option B:
Option B, 2, would produce 627, and 627 ÷ 7 still leaves a remainder because 7 × 89 = 623 and 7 × 90 = 630. Thus, 627 is not divisible by 7.

Option C:
Option C, 4, gives 629, which also falls between multiples of 7 and leaves a nonzero remainder, so it does not satisfy the condition.

Option D:
Option D, 6, would lead to 631, which again does not align with a multiple of 7, because the nearest multiples are 630 and 637.

To find the LCM, we factor each number into primes. The factorization of 6 is 2 × 3, of 8 is 2³ and of 9 is 3². The LCM must contain the highest powers of each prime present, so it is 2³ × 3² = 8 × 9 = 72. This is the smallest number divisible by 6, 8 and 9 without remainder.

Option A:
Option A, 36, is divisible by 6 and 9 but not by 8, because 36 ÷ 8 is not an integer. Thus, it fails the requirement of being a common multiple of all three numbers.

Option B:
Option B correctly uses the highest powers of primes appearing in the factorization of each number. By combining 2³ and 3², we ensure divisibility by 6, 8 and 9, and 72 is the least such number.

Option C:
Option C, 48, is divisible by 6 and 8 but not by 9, as 48 ÷ 9 is not an integer. Therefore, it cannot be the LCM.

Option D:
Option D, 54, is divisible by 6 and 9 but not by 8, since 54 ÷ 8 is not a whole number. It is a common multiple of only two of the three numbers.

For two positive integers a and b, the product a × b equals HCF(a, b) × LCM(a, b). Here, one number is 24, the HCF is 6 and the LCM is 72. Let the other number be n. Then 24 × n = 6 × 72. The right-hand side is 432, so n = 432 ÷ 24 = 18. Hence, the other number must be 18 to maintain both the given HCF and LCM.

Option A:
Option A, 12, would give 24 × 12 = 288, and since 6 × 72 = 432, the product condition would fail. It also changes the LCM and HCF relationship.

Option B:
Option B satisfies 24 × 18 = 432, matching 6 × 72 exactly. It ensures the pair of numbers has the stated HCF and LCM by the standard product relation.

Option C:
Option C, 36, gives a product of 864, which would require a different LCM or HCF than those given.

Option D:
Option D, 48, yields an even larger product and is inconsistent with the specified values of 6 and 72.