HCF & LCM- 2 - Free MCQ Practice Test with solutions, Class 5 Mathematics

To determine which number does not divide 3630, follow these steps:

Option A: Check 5: A number is divisible by 5 if it ends in 0 or 5. 3630 ends in 0, so it is divisible by 5.

Option B: Check 11: Use the rule for 11, which states that the difference between the sum of the digits in odd positions and the sum of the digits in even positions should be a multiple of 11. For 3630: (3 + 3) - (6 + 0) = 6 - 6 = 0, which is divisible by 11.

Option D: Check 6: A number is divisible by 6 if it is divisible by both 2 and 3. 3630 is even, so it's divisible by 2. The sum of its digits is 12, which is divisible by 3. Thus, 3630 is divisible by 6.
Option C: Check 9: A number is divisible by 9 if the sum of its digits is a multiple of 9.

The sum of the digits of 3630 is 3 + 6 + 3 + 0 = 12,

which is not divisible by 9.

The number 9 does not divide 3630. So Option C is the correct answer

• Composite numbers are numbers with more than two factors. 
• 4, 6, 8, 9, 10, 12 have more than two factors, hence they are composite numbers.

To find the HCF (Highest Common Factor) of 190, 171, and 152:
Break down each number into its prime factors.
190: 2 × 5 × 19
171: 3 × 3 × 19
152: 2 × 2 × 2 × 19
The common factor across all three numbers is 19.
Therefore, the HCF is 19.

To determine if 5695 is divisible by a number, consider the following:
Divisibility by 2: A number must be even. Since 5695 ends in 5, it is not divisible by 2.
Divisibility by 3: The sum of the digits must be divisible by 3. Calculate 5 + 6 + 9 + 5 = 25, which is not divisible by 3.
Divisibility by 4: The last two digits must form a number divisible by 4. The number 95 is not divisible by 4.
Divisibility by 5: A number must end in 0 or 5. Since 5695 ends in 5, it is divisible by 5.
Therefore, 5695 is divisible by 5.

Odd numbers are whole numbers that cannot be divided exactly into pairs.
3, 7, 9, 11, 13 cannot be divided into pairs, hence they are odd numbers.

To determine which number cannot be the LCM of two numbers with a given HCF of 9, consider the following:

The HCF (Highest Common Factor) is the largest number that divides both numbers. Here, it is 9.
The LCM (Lowest Common Multiple) must be a multiple of the HCF.
Thus, the LCM must also be divisible by 9.

Now, let's check each option:
Option A: 27: Divisible by 9.
Option B: 36: Divisible by 9.
Option C: 54: Divisible by 9.
Option D: 48: Not divisible by 9.
Therefore, the number that cannot be the LCM is 48.

The value of LCM(3, 4, 5) will be the smallest number that is exactly divisible by 3, 4, and 5.
Multiples of 3, 4, and 5:
Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, . . . ., 48, 51, 54, 57, 60, . . . .
Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, . . . ., 52, 56, 60, . . . .
Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, . . . ., 45, 50, 55, 60, . . . .
Therefore, the LCM of 3, 4, and 5 is 60.

To determine the least number of candies that can be bought, we need to figure out how to distribute the candies equally among both groups. The key here is that we need to find the Least Common Multiple (LCM) of 8 and 12.
Step 1: Find the multiples of 8 and 12.
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, ...
Multiples of 12:
12, 24, 36, 48, 60, 72, ...
Step 2: Find the LCM of 8 and 12.
The LCM is the smallest number that appears in both lists of multiples.
The common multiples of 8 and 12 are:
24, 48, 72, ...
The smallest common multiple is 24.
The least number of candies that Neelam must buy is 24, as it is the smallest number divisible by both 8 (the number of children in Group A) and 12 (the number of children in Group B).

To determine which rooms all three will visit, we need to find the common rooms that are multiples of both 4 and 6. The key is to calculate the Least Common Multiple (LCM) of 4 and 6 and then find all multiples of that LCM within the range from 10 to 50.
Step 1: Find the LCM of 4 and 6.
Multiples of 4:
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, ...
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, ...
The LCM of 4 and 6 is 12, as 12 is the smallest number that appears in both lists of multiples.
Step 2: Find the common multiples of 4 and 6 within the range of 10 to 50.
The multiples of 12 between 10 and 50 are:
12, 24, 36, 48

Ravi has marbles that leave a remainder of 2 when divided by 4, 6, and 8. This means that Ravi’s total number of marbles is 2 more than a multiple of 4, 6, and 8. To find the number of marbles, we need to calculate the Least Common Multiple (LCM) of 4, 6, and 8, and then add 2.
Step 1: Find the LCM of 4, 6, and 8.
Multiples of 4:
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, ...
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, ...
The LCM of 4, 6, and 8 is 24 because 24 is the smallest number that appears in all three lists.
Step 2: Add 2 to the LCM.
Since Ravi is left with 2 marbles when grouped by 4, 6, and 8, we add 2 to the LCM:
24+2=26