Five times A's money added to B's money is more than Rs. 51.00.Three times A's money minus B's money is Rs. 21.00. If a represents A's money in Rs. a and b represents B's money in Rs. Then,

Options

A. `agt9,bgt6`
B. `agt9,blt6`
C. `agt9,b=6`
D. `agt9`, but we can put no bounds on b

Answer

To solve the problem, we need to set up the equations based on the information given in the question. Let's break it down step by step.

### Step 1: Define the Variables
Let:
- \( a \) = A's money in Rs.
- \( b \) = B's money in Rs.

### Step 2: Formulate the Inequality from the First Statement
The first statement says, "Five times A's money added to B's money is more than Rs. 51." This can be expressed as:
\[ 5a + b > 51 \]

### Step 3: Formulate the Equation from the Second Statement
The second statement says, "Three times A's money minus B's money is Rs. 21." This can be expressed as:
\[ 3a - b = 21 \]

### Step 4: Solve the Second Equation for \( b \)
From the equation \( 3a - b = 21 \), we can isolate \( b \):
\[ b = 3a - 21 \]

### Step 5: Substitute \( b \) in the First Inequality
Now, substitute \( b \) in the first inequality \( 5a + b > 51 \):
\[ 5a + (3a - 21) > 51 \]

### Step 6: Simplify the Inequality
Combine like terms:
\[ 5a + 3a - 21 > 51 \]
\[ 8a - 21 > 51 \]

### Step 7: Add 21 to Both Sides
To isolate \( 8a \), add 21 to both sides:
\[ 8a > 51 + 21 \]
\[ 8a > 72 \]

### Step 8: Divide by 8
Now, divide both sides by 8 to solve for \( a \):
\[ a > 9 \]

### Step 9: Substitute \( a \) Back to Find \( b \)
Now that we have \( a > 9 \), we can substitute this back into the equation for \( b \):
\[ b = 3a - 21 \]
Since \( a > 9 \):
\[ b = 3(9) - 21 = 27 - 21 = 6 \]
Thus, we can conclude:
\[ b > 6 \]

### Final Result
We have found that:
- \( a > 9 \)
- \( b > 6 \)