CBSE Class 4 Maths HCF and LCM Question Bank - Free Printable

Let's solve each problem step by step:

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1. Write the first 3 common multiples of 4 and 6 and also find the LCM.

#### Step 1: Find the multiples of 4 and 6.
- Multiples of 4: \(4, 8, 12, 16, 20, 24, 28, 32, \ldots\)
- Multiples of 6: \(6, 12, 18, 24, 30, 36, \ldots\)

#### Step 2: Identify the common multiples.
The common multiples of 4 and 6 are: \(12, 24, 36, \ldots\)

#### Step 3: Write the first 3 common multiples.
The first 3 common multiples are: \(12, 24, 36\).

#### Step 4: Find the Least Common Multiple (LCM).
The LCM is the smallest common multiple, which is \(12\).

Answer:
- First 3 common multiples: \(12, 24, 36\)
- LCM: \(12\)

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2. Write two prime numbers whose sum is also a prime number.

#### Step 1: Recall the definition of a prime number.
A prime number is a number greater than 1 that has no divisors other than 1 and itself.

#### Step 2: Test pairs of prime numbers.
- \(2 + 3 = 5\) (both 2 and 3 are prime, and 5 is prime)
- \(3 + 5 = 8\) (8 is not prime)
- \(5 + 7 = 12\) (12 is not prime)

The pair \(2\) and \(3\) works because their sum \(5\) is also a prime number.

Answer:
- Two prime numbers: \(2\) and \(3\)

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3. Multiply the HCF of 45 and 75 by the HCF of 54 and 81.

#### Step 1: Find the HCF of 45 and 75.
- Prime factorization:
- \(45 = 3^2 \times 5\)
- \(75 = 3 \times 5^2\)
- The HCF is the product of the lowest powers of common prime factors: \(3 \times 5 = 15\).

#### Step 2: Find the HCF of 54 and 81.
- Prime factorization:
- \(54 = 2 \times 3^3\)
- \(81 = 3^4\)
- The HCF is the product of the lowest powers of common prime factors: \(3^3 = 27\).

#### Step 3: Multiply the HCFs.
\[ \text{HCF}(45, 75) \times \text{HCF}(54, 81) = 15 \times 27 = 405 \]

Answer:
- Product: \(405\)

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4. Find the HCF by listing factors.

#### Part (a): HCF of 20 and 28
- Factors of 20: \(1, 2, 4, 5, 10, 20\)
- Factors of 28: \(1, 2, 4, 7, 14, 28\)
- Common factors: \(1, 2, 4\)
- HCF: \(4\)

#### Part (b): HCF of 18 and 24
- Factors of 18: \(1, 2, 3, 6, 9, 18\)
- Factors of 24: \(1, 2, 3, 4, 6, 8, 12, 24\)
- Common factors: \(1, 2, 3, 6\)
- HCF: \(6\)

Answer:
- (a) HCF: \(4\)
- (b) HCF: \(6\)

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5. Find the HCF of the following:

#### Part (a): HCF of 16, 24, 40
- Prime factorization:
- \(16 = 2^4\)
- \(24 = 2^3 \times 3\)
- \(40 = 2^3 \times 5\)
- The HCF is the product of the lowest powers of common prime factors: \(2^3 = 8\).

#### Part (b): HCF of 24, 36, 60
- Prime factorization:
- \(24 = 2^3 \times 3\)
- \(36 = 2^2 \times 3^2\)
- \(60 = 2^2 \times 3 \times 5\)
- The HCF is the product of the lowest powers of common prime factors: \(2^2 \times 3 = 12\).

Answer:
- (a) HCF: \(8\)
- (b) HCF: \(12\)

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6. Find the LCM of 15 and 10 and the LCM of 6 and 20.

#### Part (a): LCM of 15 and 10
- Prime factorization:
- \(15 = 3 \times 5\)
- \(10 = 2 \times 5\)
- The LCM is the product of the highest powers of all prime factors: \(2 \times 3 \times 5 = 30\).

#### Part (b): LCM of 6 and 20
- Prime factorization:
- \(6 = 2 \times 3\)
- \(20 = 2^2 \times 5\)
- The LCM is the product of the highest powers of all prime factors: \(2^2 \times 3 \times 5 = 60\).

Answer:
- LCM of 15 and 10: \(30\)
- LCM of 6 and 20: \(60\)

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7. What would you get on dividing the LCM of 18 and 12 by their HCF?

#### Step 1: Find the LCM of 18 and 12.
- Prime factorization:
- \(18 = 2 \times 3^2\)
- \(12 = 2^2 \times 3\)
- The LCM is the product of the highest powers of all prime factors: \(2^2 \times 3^2 = 36\).

#### Step 2: Find the HCF of 18 and 12.
- The HCF is the product of the lowest powers of common prime factors: \(2 \times 3 = 6\).

#### Step 3: Divide the LCM by the HCF.
\[ \frac{\text{LCM}}{\text{HCF}} = \frac{36}{6} = 6 \]

Answer:
- Result: \(6\)

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8. The LCM of 2 numbers is 143 and the HCF is 1. If one of the numbers is 11, what is the other number?

#### Step 1: Use the relationship between LCM, HCF, and the product of the numbers.
For any two numbers \(a\) and \(b\):
\[ \text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b \]

Given:
- \(\text{LCM} = 143\)
- \(\text{HCF} = 1\)
- One number, \(a = 11\)

Let the other number be \(b\). Then:
\[ 143 \times 1 = 11 \times b \]
\[ 143 = 11 \times b \]
\[ b = \frac{143}{11} = 13 \]

Answer:
- The other number: \(13\)

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9. Find the LCM of the following:

#### Part (a): LCM of 32, 112, 8
- Prime factorization:
- \(32 = 2^5\)
- \(112 = 2^4 \times 7\)
- \(8 = 2^3\)
- The LCM is the product of the highest powers of all prime factors: \(2^5 \times 7 = 224\).

#### Part (b): LCM of 15, 225, 45
- Prime factorization:
- \(15 = 3 \times 5\)
- \(225 = 3^2 \times 5^2\)
- \(45 = 3^2 \times 5\)
- The LCM is the product of the highest powers of all prime factors: \(3^2 \times 5^2 = 225\).

Answer:
- (a) LCM: \(224\)
- (b) LCM: \(225\)

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10. Put the following sets of numbers in order of their HCF from lowest to highest.

#### Part (a): 12, 15, and 18
- Prime factorization:
- \(12 = 2^2 \times 3\)
- \(15 = 3 \times 5\)
- \(18 = 2 \times 3^2\)
- The HCF is the product of the lowest powers of common prime factors: \(3\).

#### Part (b): 15, 10, and 6
- Prime factorization:
- \(15 = 3 \times 5\)
- \(10 = 2 \times 5\)
- \(6 = 2 \times 3\)
- The HCF is the product of the lowest powers of common prime factors: \(1\).

#### Part (c): 8, 16, and 24
- Prime factorization:
- \(8 = 2^3\)
- \(16 = 2^4\)
- \(24 = 2^3 \times 3\)
- The HCF is the product of the lowest powers of common prime factors: \(8\).

#### Part (d): 4, 6, and 24
- Prime factorization:
- \(4 = 2^2\)
- \(6 = 2 \times 3\)
- \(24 = 2^3 \times 3\)
- The HCF is the product of the lowest powers of common prime factors: \(2\).

#### Part (e): 27, 18, and 45
- Prime factorization:
- \(27 = 3^3\)
- \(18 = 2 \times 3^2\)
- \(45 = 3^2 \times 5\)
- The HCF is the product of the lowest powers of common prime factors: \(9\).

#### Order the HCFs from lowest to highest:
- HCF(15, 10, 6) = 1
- HCF(4, 6, 24) = 2
- HCF(12, 15, 18) = 3
- HCF(8, 16, 24) = 8
- HCF(27, 18, 45) = 9

Answer:
- Order: (b), (d), (a), (c), (e)

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Final Answers:

1. First 3 common multiples: \(12, 24, 36\); LCM: \(12\)
2. Two prime numbers: \(2\) and \(3\)
3. Product: \(405\)
4. (a) HCF: \(4\); (b) HCF: \(6\)
5. (a) HCF: \(8\); (b) HCF: \(12\)
6. LCM of 15 and 10: \(30\); LCM of 6 and 20: \(60\)
7. Result: \(6\)
8. Other number: \(13\)
9. (a) LCM: \(224\); (b) LCM: \(225\)
10. Order: (b), (d), (a), (c), (e)

\[
\boxed{12, 24, 36; 12; 2, 3; 405; 4, 6; 8, 12; 30, 60; 6; 13; 224, 225; (b), (d), (a), (c), (e)}
\]