The HCF and LCM of two numbers are 12 and 72 respectively. If the sum of these numbers is 60, then one of the numbers will be

Options

A. 12
B. 24
C. 60
D. 72

Answer

To solve the problem, we need to find two numbers whose HCF is 12, LCM is 72, and their sum is 60. Let's denote the two numbers as \( x \) and \( y \).

### Step 1: Set up the equations
We know from the problem:
- HCF of \( x \) and \( y \) = 12
- LCM of \( x \) and \( y \) = 72
- Sum of \( x \) and \( y \) = 60

From the sum, we can express \( y \) in terms of \( x \):
\[ y = 60 - x \]

### Step 2: Use the relationship between HCF and LCM
We know that:
\[ \text{HCF} \times \text{LCM} = x \times y \]
Substituting the known values:
\[ 12 \times 72 = x \times (60 - x) \]

### Step 3: Simplify the equation
Calculating the left side:
\[ 12 \times 72 = 864 \]
So we have:
\[ 864 = x \times (60 - x) \]

### Step 4: Expand and rearrange the equation
Expanding the right side:
\[ 864 = 60x - x^2 \]
Rearranging gives us a quadratic equation:
\[ x^2 - 60x + 864 = 0 \]

### Step 5: Solve the quadratic equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -60, c = 864 \):
- Calculate the discriminant:
\[ b^2 - 4ac = (-60)^2 - 4 \times 1 \times 864 = 3600 - 3456 = 144 \]

- Now apply the quadratic formula:
\[ x = \frac{60 \pm \sqrt{144}}{2} \]
\[ x = \frac{60 \pm 12}{2} \]

Calculating the two possible values for \( x \):
1. \( x = \frac{72}{2} = 36 \)
2. \( x = \frac{48}{2} = 24 \)

### Step 6: Find the corresponding values for \( y \)
Using \( y = 60 - x \):
1. If \( x = 36 \), then \( y = 60 - 36 = 24 \)
2. If \( x = 24 \), then \( y = 60 - 24 = 36 \)

### Step 7: Verify HCF and LCM
Now we check the HCF and LCM of the pairs \( (36, 24) \):
- HCF(36, 24) = 12
- LCM(36, 24) = 72

Both conditions are satisfied.

### Conclusion
Thus, one of the numbers is **24**.