In the Ruchika's wallet there are only ₹ 16, consisting of 10 paise, 20 paise and ₹ 1 coins. The ratio of no. of coins of 10 paise and 20 paise is 6 : 1. The minimum no. of ₹ 1 coin is :

Options

A. 5
B. 12
C. 4
D. 8

Answer

To solve the problem step by step, we need to analyze the information given and use it to find the minimum number of ₹1 coins in Ruchika's wallet.

### Step 1: Understand the total amount and coin types
Ruchika has a total of ₹16, which is equivalent to 1600 paise (since 1 rupee = 100 paise). The coins in her wallet consist of:
- 10 paise coins
- 20 paise coins
- ₹1 coins

### Step 2: Set up the ratio of coins
The problem states that the ratio of the number of 10 paise coins to 20 paise coins is 6:1.
Let:
- The number of 10 paise coins = 6x
- The number of 20 paise coins = x

### Step 3: Calculate the total value of 10 paise and 20 paise coins
The value of the 10 paise coins:
- Value = Number of coins × Value of each coin = 6x × 10 paise = 60x paise

The value of the 20 paise coins:
- Value = Number of coins × Value of each coin = x × 20 paise = 20x paise

### Step 4: Set up the equation for total value
The total value of the coins (in paise) can be expressed as:
\[ \text{Total value} = \text{Value of 10 paise coins} + \text{Value of 20 paise coins} + \text{Value of ₹1 coins} \]
Converting ₹1 coins to paise:
- Value of ₹1 coins = Number of ₹1 coins × 100 paise

Let the number of ₹1 coins be y. Thus, we have:
\[ 60x + 20x + 100y = 1600 \]
Combining like terms gives:
\[ 80x + 100y = 1600 \]

### Step 5: Simplify the equation
We can simplify the equation by dividing everything by 20:
\[ 4x + 5y = 80 \]

### Step 6: Express y in terms of x
Rearranging the equation gives:
\[ 5y = 80 - 4x \]
\[ y = \frac{80 - 4x}{5} \]

### Step 7: Find integer values for y
For y to be a whole number, \(80 - 4x\) must be divisible by 5. We can check values of x that satisfy this condition.

### Step 8: Check values of x
Let’s check values of x:
- If \(x = 0\), \(y = 16\) (maximum ₹1 coins)
- If \(x = 5\), \(y = 12\)
- If \(x = 10\), \(y = 8\)
- If \(x = 15\), \(y = 4\)
- If \(x = 20\), \(y = 0\) (minimum ₹1 coins)

### Step 9: Minimum number of ₹1 coins
To minimize the number of ₹1 coins, we want to maximize the value of x while ensuring that \(y\) remains non-negative. The maximum value of x that keeps y non-negative is \(x = 20\), which gives us \(y = 0\). However, we need at least one ₹1 coin.

Thus, the minimum value of y that satisfies the conditions is when \(x = 15\), which gives:
\[ y = 4 \]

### Final Answer
The minimum number of ₹1 coins is **4**.

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