I had Rs. 14.40 in one rupee coins and 20 paise coins when I went out shopping. When I returned, I had as many one rupee coins as I originally had 20 paise coins and as many 20 paise coins as I originally had one rupee coins.Briefly, I cameback with about one-third of what I had started out with. How many one rupee coins did I have initially?

Options

A. 10
B. 12
C. 14
D. 16

Answer

To solve the problem step by step, we will define the variables and set up equations based on the information provided.

### Step 1: Define Variables
Let:
- \( x \) = number of one rupee coins
- \( y \) = number of 20 paise coins

### Step 2: Set Up the First Equation
The total amount of money initially is Rs. 14.40. Since one rupee coins contribute Rs. 1 each and 20 paise coins contribute Rs. 0.20 each, we can express this as:
\[ x + 0.2y = 14.40 \]
To eliminate the decimal, we can multiply the entire equation by 10:
\[ 10x + 2y = 144 \]
This is our **Equation 1**.

### Step 3: Set Up the Second Scenario
After shopping, the problem states that the number of one rupee coins equals the original number of 20 paise coins, and the number of 20 paise coins equals the original number of one rupee coins. Thus:
- Number of one rupee coins after shopping = \( y \)
- Number of 20 paise coins after shopping = \( x \)

### Step 4: Calculate the Total Money Left
It is mentioned that after shopping, the total money left is about one-third of what was initially there. One-third of Rs. 14.40 is:
\[ \frac{14.40}{3} = 4.80 \]
So, the equation for the money left can be expressed as:
\[ y + 0.2x = 4.80 \]
Again, to eliminate the decimal, we multiply the entire equation by 10:
\[ 10y + 2x = 48 \]
This is our **Equation 2**.

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( 10x + 2y = 144 \) (Equation 1)
2. \( 10y + 2x = 48 \) (Equation 2)

We can rearrange Equation 1 to express \( y \) in terms of \( x \):
\[ 2y = 144 - 10x \implies y = 72 - 5x \]

### Step 6: Substitute \( y \) in Equation 2
Now we substitute \( y \) in Equation 2:
\[ 10(72 - 5x) + 2x = 48 \]
Expanding this gives:
\[ 720 - 50x + 2x = 48 \]
Combining like terms:
\[ 720 - 48 = 50x - 2x \implies 672 = 48x \]
Now, divide both sides by 48:
\[ x = \frac{672}{48} = 14 \]

### Step 7: Find \( y \)
Now we can find \( y \) using the value of \( x \):
\[ y = 72 - 5(14) = 72 - 70 = 2 \]

### Conclusion
Thus, the number of one rupee coins I had initially is \( \boxed{14} \).