The sum of two numbers is 140. If their LCM is 240 and HCF is 20, then find the smaller number.

Options

A. 60
B. 140
C. 20
D. 80

Answer

To solve the problem step by step, we will use the information given about the sum, LCM, and HCF of the two numbers.

### Step 1: Understand the relationship between the numbers
Let the two numbers be \( a \) and \( b \).
Given:
- Sum of the numbers: \( a + b = 140 \)
- LCM of the numbers: \( LCM(a, b) = 240 \)
- HCF of the numbers: \( HCF(a, b) = 20 \)

### Step 2: Express the numbers in terms of HCF
Since the HCF of the two numbers is 20, we can express the numbers as:
- \( a = 20x \)
- \( b = 20y \)
where \( x \) and \( y \) are co-prime integers (i.e., \( HCF(x, y) = 1 \)).

### Step 3: Substitute into the sum equation
Substituting \( a \) and \( b \) into the sum equation:
\[ 20x + 20y = 140 \]
Dividing the entire equation by 20 gives:
\[ x + y = 7 \quad \text{(Equation 1)} \]

### Step 4: Use the LCM formula
The LCM of the two numbers can be expressed as:
\[ LCM(a, b) = \frac{a \cdot b}{HCF(a, b)} \]
Substituting the values:
\[ 240 = \frac{(20x)(20y)}{20} \]
This simplifies to:
\[ 240 = 20xy \]
Dividing both sides by 20 gives:
\[ xy = 12 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( x + y = 7 \)
2. \( xy = 12 \)

We can solve these equations simultaneously. From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 7 - x \]
Substituting this into Equation 2:
\[ x(7 - x) = 12 \]
Expanding and rearranging gives:
\[ 7x - x^2 = 12 \implies x^2 - 7x + 12 = 0 \]

### Step 6: Factor the quadratic equation
Factoring the quadratic:
\[ (x - 3)(x - 4) = 0 \]
Thus, the solutions are:
\[ x = 3 \quad \text{or} \quad x = 4 \]

### Step 7: Find corresponding values of \( y \)
Using \( x + y = 7 \):
- If \( x = 3 \), then \( y = 7 - 3 = 4 \).
- If \( x = 4 \), then \( y = 7 - 4 = 3 \).

### Step 8: Calculate the original numbers
Now substituting back to find \( a \) and \( b \):
- If \( x = 3 \) and \( y = 4 \):
- \( a = 20 \times 3 = 60 \)
- \( b = 20 \times 4 = 80 \)

- If \( x = 4 \) and \( y = 3 \):
- \( a = 20 \times 4 = 80 \)
- \( b = 20 \times 3 = 60 \)

### Conclusion: Identify the smaller number
In both cases, the smaller number is:
\[ \text{Smaller number} = 60 \]