Answer :
Sure, let's break down each problem step by step:
12. If the product of two numbers is 4032 and their HCF is 12, find their LCM.
Step-by-Step Solution:
1. Given:
- The product of the two numbers is 4032.
- The HCF (Highest Common Factor) of the two numbers is 12.
2. We know the relationship between HCF, LCM (Least Common Multiple), and the product of the two numbers:
[tex]\[ \text{HCF} \times \text{LCM} = \text{Product of the two numbers} \][/tex]
3. Substitute the given values into the equation:
[tex]\[ 12 \times \text{LCM} = 4032 \][/tex]
4. To find the LCM, divide the product by the HCF:
[tex]\[ \text{LCM} = \frac{4032}{12} = 336 \][/tex]
So, the LCM of the two numbers is 336.
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13. The HCF and LCM of two numbers are 9 and 270 respectively. If one of the numbers is 45, find the other number.
Step-by-Step Solution:
1. Given:
- The HCF is 9.
- The LCM is 270.
- One of the numbers is 45.
2. Let the other number be \( x \).
3. We know the relationship between the numbers, their HCF, and LCM:
[tex]\[ \text{HCF} \times \text{LCM} = \text{Product of the two numbers} \][/tex]
4. Substitute the given values into the equation:
[tex]\[ 9 \times 270 = 45 \times x \][/tex]
5. Solve for \( x \):
[tex]\[ 2430 = 45 \times x \][/tex]
[tex]\[ x = \frac{2430}{45} = 54 \][/tex]
So, the other number is 54.
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14. Find the HCF of 180 and 336.
Step-by-Step Solution:
1. To find the HCF (Highest Common Factor) of 180 and 336, we can use the Euclidean algorithm:
2. Apply the Euclidean algorithm:
- Divide 336 by 180, and find the remainder:
[tex]\[ 336 \div 180 = 1 \, \text{remainder} \, 156 \][/tex]
- Now, divide 180 by the remainder 156:
[tex]\[ 180 \div 156 = 1 \, \text{remainder} \, 24 \][/tex]
- Finally, divide 156 by 24:
[tex]\[ 156 \div 24 = 6 \, \text{remainder} \, 0 \][/tex]
3. Since the remainder is now 0, the last non-zero remainder is the HCF.
So, the HCF of 180 and 336 is 24.
12. If the product of two numbers is 4032 and their HCF is 12, find their LCM.
Step-by-Step Solution:
1. Given:
- The product of the two numbers is 4032.
- The HCF (Highest Common Factor) of the two numbers is 12.
2. We know the relationship between HCF, LCM (Least Common Multiple), and the product of the two numbers:
[tex]\[ \text{HCF} \times \text{LCM} = \text{Product of the two numbers} \][/tex]
3. Substitute the given values into the equation:
[tex]\[ 12 \times \text{LCM} = 4032 \][/tex]
4. To find the LCM, divide the product by the HCF:
[tex]\[ \text{LCM} = \frac{4032}{12} = 336 \][/tex]
So, the LCM of the two numbers is 336.
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13. The HCF and LCM of two numbers are 9 and 270 respectively. If one of the numbers is 45, find the other number.
Step-by-Step Solution:
1. Given:
- The HCF is 9.
- The LCM is 270.
- One of the numbers is 45.
2. Let the other number be \( x \).
3. We know the relationship between the numbers, their HCF, and LCM:
[tex]\[ \text{HCF} \times \text{LCM} = \text{Product of the two numbers} \][/tex]
4. Substitute the given values into the equation:
[tex]\[ 9 \times 270 = 45 \times x \][/tex]
5. Solve for \( x \):
[tex]\[ 2430 = 45 \times x \][/tex]
[tex]\[ x = \frac{2430}{45} = 54 \][/tex]
So, the other number is 54.
---
14. Find the HCF of 180 and 336.
Step-by-Step Solution:
1. To find the HCF (Highest Common Factor) of 180 and 336, we can use the Euclidean algorithm:
2. Apply the Euclidean algorithm:
- Divide 336 by 180, and find the remainder:
[tex]\[ 336 \div 180 = 1 \, \text{remainder} \, 156 \][/tex]
- Now, divide 180 by the remainder 156:
[tex]\[ 180 \div 156 = 1 \, \text{remainder} \, 24 \][/tex]
- Finally, divide 156 by 24:
[tex]\[ 156 \div 24 = 6 \, \text{remainder} \, 0 \][/tex]
3. Since the remainder is now 0, the last non-zero remainder is the HCF.
So, the HCF of 180 and 336 is 24.
