Find the two number whose L.C.M is 1188 and H.C.F is 9-

Options

A. 27, 396
B. 9, 27
C. 36, 99
D. Data inadequate

Answer

To find the two numbers whose L.C.M is 1188 and H.C.F is 9, we can follow these steps:

### Step 1: Understand the relationship between L.C.M and H.C.F
The relationship between the two numbers (let's call them A and B), their L.C.M, and their H.C.F is given by the formula:
\[ \text{L.C.M} \times \text{H.C.F} = A \times B \]
Given:
- L.C.M = 1188
- H.C.F = 9

### Step 2: Substitute the values into the formula
Substituting the given values into the formula:
\[ 1188 \times 9 = A \times B \]

### Step 3: Calculate the product of A and B
Now, calculate the left side:
\[ 1188 \times 9 = 10692 \]
So, we have:
\[ A \times B = 10692 \]

### Step 4: Express A and B in terms of H.C.F
Since the H.C.F is 9, we can express A and B as:
\[ A = 9x \quad \text{and} \quad B = 9y \]
where x and y are co-prime numbers (i.e., H.C.F of x and y is 1).

### Step 5: Substitute A and B back into the product equation
Substituting A and B into the product equation:
\[ (9x) \times (9y) = 10692 \]
This simplifies to:
\[ 81xy = 10692 \]

### Step 6: Solve for xy
Now, divide both sides by 81:
\[ xy = \frac{10692}{81} \]
Calculating the right side:
\[ xy = 132 \]

### Step 7: Find pairs of co-prime factors of 132
Now, we need to find pairs of co-prime factors of 132. The pairs of factors of 132 are:
- (1, 132)
- (2, 66)
- (3, 44)
- (4, 33)
- (6, 22)
- (11, 12)

From these pairs, we need to identify the co-prime pairs:
- (1, 132) are co-prime.
- (4, 33) are co-prime.
- (11, 12) are co-prime.

### Step 8: Calculate the two numbers
Now, we can calculate the two numbers using the pairs:
1. For (1, 132):
- A = 9 * 1 = 9
- B = 9 * 132 = 1188
2. For (4, 33):
- A = 9 * 4 = 36
- B = 9 * 33 = 297
3. For (11, 12):
- A = 9 * 11 = 99
- B = 9 * 12 = 108

### Conclusion
The pairs of numbers whose L.C.M is 1188 and H.C.F is 9 are:
- (9, 1188)
- (36, 297)
- (99, 108)