LCM of two numbers is 28 times their HCF. The sum of the HCF and the LCM is 1740. If one of there numbers is 240, then what is the other number ?
420
Solving the LCM and HCF Problem
This problem involves finding an unknown number when we are given information about its Least Common Multiple (LCM) and Highest Common Factor (HCF) with another known number. We will use the fundamental relationship between two numbers, their LCM, and their HCF to solve this.
Understanding the Given Information
We are given the following conditions:
- The LCM of the two numbers is 28 times their HCF.
- The sum of the HCF and the LCM is 1740.
- One of the two numbers is 240.
Let's use variables to represent these values:
- Let HCF = \(h\)
- Let LCM = \(l\)
- Let the two numbers be \(a\) and \(b\). We are given \(a = 240\). We need to find \(b\).
Setting Up Equations from the Conditions
From the first condition, "LCM of two numbers is 28 times their HCF", we can write:
\(l = 28h\) \(\quad (Equation\; 1)\)
From the second condition, "The sum of the HCF and the LCM is 1740", we can write:
\(h + l = 1740\) \(\quad (Equation\; 2)\)
Finding the Values of HCF and LCM
We have a system of two linear equations with two variables, \(h\) and \(l\). We can substitute Equation 1 into Equation 2 to find the values.
Substitute \(l = 28h\) into \(h + l = 1740\):
\(h + 28h = 1740\)
Combine the terms involving \(h\):
\(29h = 1740\)
Now, solve for \(h\) by dividing 1740 by 29:
\(h = \frac{1740}{29}\)
To calculate this division, we can observe that \(1740 = 174 \times 10\). Let's divide 174 by 29:
\(174 \div 29 = 6\)
So, \(h = 6 \times 10 = 60\).
The HCF of the two numbers is 60.
Now that we have \(h = 60\), we can find \(l\) using Equation 1:
\(l = 28h\)
\(l = 28 \times 60\)
\(l = 1680\)
The LCM of the two numbers is 1680.
Let's quickly check if the sum \(h + l = 1740\): \(60 + 1680 = 1740\). This is correct.
Using the Relationship Between Numbers, HCF, and LCM
A fundamental property relating two positive integers, their HCF, and their LCM is:
Product of the two numbers = Product of their HCF and LCM
In symbols, if the two numbers are \(a\) and \(b\), their HCF is \(h\), and their LCM is \(l\), then:
\(a \times b = h \times l\)
We know \(a = 240\), \(h = 60\), and \(l = 1680\). We need to find \(b\). Substitute these values into the formula:
\(240 \times b = 60 \times 1680\)
To find \(b\), divide the product of HCF and LCM by the known number \(a\):
\(b = \frac{60 \times 1680}{240}\)
We can simplify the calculation. Notice that \(240 = 4 \times 60\).
\(b = \frac{60 \times 1680}{4 \times 60}\)
Cancel out the common factor of 60:
\(b = \frac{1680}{4}\)
Now, perform the division:
\(b = 420\)
So, the other number is 420.
Summary of the Solution Steps
- Identify the given information and represent it using variables for HCF, LCM, and the two numbers.
- Set up equations based on the relationships given between HCF and LCM.
- Solve the equations to find the specific values of HCF and LCM.
- Use the property that the product of two numbers equals the product of their HCF and LCM.
- Substitute the known values (one number, HCF, and LCM) into the property and solve for the unknown number.
Verification
Let's verify if the numbers 240 and 420 fit the original conditions with HCF=60 and LCM=1680.
- HCF(240, 420): \(240 = 60 \times 4\), \(420 = 60 \times 7\). The common factor is 60. HCF is 60. (Correct)
- LCM(240, 420): \(LCM = \frac{product\;of\;numbers}{HCF} = \frac{240 \times 420}{60} = \frac{240}{60} \times 420 = 4 \times 420 = 1680\). LCM is 1680. (Correct)
- Is LCM = 28 times HCF? \(1680 = 28 \times 60\)? \(28 \times 6 = 168\), so \(28 \times 60 = 1680\). Yes, it is. (Correct)
- Is HCF + LCM = 1740? \(60 + 1680 = 1740\). Yes, it is. (Correct)
All conditions are satisfied by the numbers 240 and 420.
Revision Table: Key Concepts
| Concept | Definition | Property Used Here |
|---|---|---|
| HCF (Highest Common Factor) | The largest positive integer that divides two or more numbers without leaving a remainder. Also known as GCD (Greatest Common Divisor). | Used to find the values of HCF and LCM based on given relationships. |
| LCM (Least Common Multiple) | The smallest positive integer that is a multiple of two or more numbers. | Used to find the values of HCF and LCM based on given relationships. |
| Relationship between HCF, LCM, and two numbers | For any two positive integers \(a\) and \(b\), \(a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)\). | Crucial formula used to find the second number when one number, HCF, and LCM are known. |
Additional Information: Finding HCF and LCM
While we found HCF and LCM using algebraic equations in this problem, they can also be found using methods like prime factorization or the division method.
Prime Factorization Method
To find the HCF and LCM of two numbers (e.g., 240 and 420):
- Find the prime factorization of each number:
- \(240 = 24 \times 10 = (2^3 \times 3) \times (2 \times 5) = 2^4 \times 3^1 \times 5^1\)
- \(420 = 42 \times 10 = (2 \times 3 \times 7) \times (2 \times 5) = 2^2 \times 3^1 \times 5^1 \times 7^1\)
- HCF: Identify common prime factors and take the lowest power of each.
- Common factors are 2, 3, 5.
- Lowest power of 2 is \(2^2\).
- Lowest power of 3 is \(3^1\).
- Lowest power of 5 is \(5^1\).
- HCF = \(2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60\).
- LCM: Identify all prime factors from both numbers and take the highest power of each.
- Prime factors are 2, 3, 5, 7.
- Highest power of 2 is \(2^4\).
- Highest power of 3 is \(3^1\).
- Highest power of 5 is \(5^1\).
- Highest power of 7 is \(7^1\).
- LCM = \(2^4 \times 3^1 \times 5^1 \times 7^1 = 16 \times 3 \times 5 \times 7 = 48 \times 35\).
- \(48 \times 35 = 48 \times (30 + 5) = 48 \times 30 + 48 \times 5 = 1440 + 240 = 1680\).
- LCM = 1680.
This confirms our calculated HCF and LCM values are correct for the numbers 240 and 420.
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