The LCM of two numbers is 40 and their HCF is 4. If the difference between 

the two numbers is 12, then the sum of the numbers is : 

 (A) 20 (B) 24 

 (C) 28 (D) 32

Given the problem, let's use the following information:

- LCM (Least Common Multiple) of two numbers = 40
- HCF (Highest Common Factor) of two numbers = 4
- The difference between the two numbers = 12

Let the two numbers be \( a \) and \( b \). Since their HCF is 4, we can express them as:

\[
a = 4x \quad \text{and} \quad b = 4y
\]

where \( x \) and \( y \) are coprime (i.e., their HCF is 1). We are also given that:

\[
|a - b| = 12 \implies |4x - 4y| = 12 \implies |x - y| = 3
\]

Now, the LCM of \( a \) and \( b \) is given by:

\[
\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)} = \frac{4x \times 4y}{4} = 4xy = 40
\]

Simplifying this, we have:

\[
xy = 10
\]

From the absolute difference condition, we have two cases:

1. \( x - y = 3 \)
2. \( y - x = 3 \)

### Case 1: \( x - y = 3 \)

From \( x - y = 3 \) and \( xy = 10 \), we can set up a system of equations:

\[
x = y + 3
\]

Substituting in the product equation:

\[
(y + 3) \times y = 10
\]

\[
y^2 + 3y - 10 = 0
\]

Solving this quadratic equation:

\[
y = \frac{-3 \pm \sqrt{3^2 + 4 \times 10}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}
\]

\[
y = 2 \quad \text{or} \quad y = -5
\]

Since we need positive integers (as \( a \) and \( b \) are positive), \( y = 2 \).

Thus, \( x = y + 3 = 5 \).

So, the numbers are:

\[
a = 4x = 4 \times 5 = 20, \quad b = 4y = 4 \times 2 = 8
\]

### Sum of the Numbers:

\[
a + b = 20 + 8 = 28
\]

Thus, the answer is:

**(C) 28**

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you find the HCF and LCM of two numbers?
2. Can the LCM of two numbers be smaller than either of the numbers?
3. What are other methods to solve problems involving LCM, HCF, and number properties?
4. How does the relationship between HCF and LCM relate to coprime numbers?
5. What are some real-world applications of using LCM and HCF?

**Tip:** The product of the LCM and HCF of two numbers is equal to the product of the numbers themselves.

The LCM of two numbers is 40 and their HCF is 4. If the difference between the two numbers is 12, then the sum of the numbers is : (A) 20 (B) 24 (C) 28 (D) 32

Learn how to solve a problem involving the LCM of 40, HCF of 4, and a difference of 12 between two numbers. The detailed solution explores key concepts in number theory, including the relationship between LCM and HCF. Suitable for students in Grades 8-10.

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