A man spends one-third of his monthly income on food, one-fourth on children's education, and one-sixth on clothing. He keeps the remaining `Rs 7,000` for miscellaneous expenses of the month. Find his monthly income.

Options

A. Rs `35,000`
B. Rs `7,000`
C. Rs `28,000`
D. Rs `21,000`

Answer

To solve the problem step by step, let's denote the man's monthly income as \( x \).

### Step 1: Define the expenditures
- The amount spent on food is \( \frac{1}{3}x \).
- The amount spent on children's education is \( \frac{1}{4}x \).
- The amount spent on clothing is \( \frac{1}{6}x \).

### Step 2: Define the remaining amount
According to the problem, the remaining amount after these expenditures is \( 7000 \) rupees.

### Step 3: Set up the equation
The total amount spent plus the remaining amount equals the total income:
\[ \frac{1}{3}x + \frac{1}{4}x + \frac{1}{6}x + 7000 = x \]

### Step 4: Find a common denominator
To combine the fractions, we need a common denominator. The least common multiple of 3, 4, and 6 is 12. We can rewrite the fractions:
- \( \frac{1}{3}x = \frac{4}{12}x \)
- \( \frac{1}{4}x = \frac{3}{12}x \)
- \( \frac{1}{6}x = \frac{2}{12}x \)

Now, substitute these back into the equation:
\[ \frac{4}{12}x + \frac{3}{12}x + \frac{2}{12}x + 7000 = x \]

### Step 5: Combine the fractions
Combine the fractions on the left side:
\[ \frac{4 + 3 + 2}{12}x + 7000 = x \]
This simplifies to:
\[ \frac{9}{12}x + 7000 = x \]

### Step 6: Simplify the equation
We can simplify \( \frac{9}{12}x \) to \( \frac{3}{4}x \):
\[ \frac{3}{4}x + 7000 = x \]

### Step 7: Isolate \( x \)
Subtract \( \frac{3}{4}x \) from both sides:
\[ 7000 = x - \frac{3}{4}x \]
This simplifies to:
\[ 7000 = \frac{1}{4}x \]

### Step 8: Solve for \( x \)
Multiply both sides by 4 to solve for \( x \):
\[ x = 7000 \times 4 \]
\[ x = 28000 \]

### Conclusion
The man's monthly income is \( \text{Rs } 28,000 \).

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