One day my assistant Rajesh went to a stationary shop to purchase marker pens.He purchased the pens for ₹52 and he gave a ₹100 note to the shopkeeper The shopkeeper offered Rajesh to take back ₹48in the denominations of only ₹1,₹2 and ₹5 .so Rajesh received total 26 coins from the shopkeeper Minimum number of ₹1 coins which Rajesh has with him if he has atleasst one coin of each denomination offered by the shopkeeper:

Options

A. 4
B. 5
C. 7
D. 10

Answer

To solve the problem step by step, we need to determine the minimum number of ₹1 coins Rajesh can have while ensuring he has at least one coin of each denomination (₹1, ₹2, and ₹5) and that the total number of coins equals 26.

### Step 1: Define the variables
Let:
- \( X \) = number of ₹1 coins
- \( Y \) = number of ₹2 coins
- \( Z \) = number of ₹5 coins

### Step 2: Set up the equations
From the problem, we know:
1. The total number of coins is 26:
\[ X + Y + Z = 26 \]
2. The total value of the coins is ₹48:
\[ 1X + 2Y + 5Z = 48 \]

### Step 3: Solve for one variable
We can express \( Y \) in terms of \( X \) and \( Z \) using the first equation:
\[ Y = 26 - X - Z \]

### Step 4: Substitute \( Y \) in the second equation
Substituting \( Y \) into the second equation gives:
\[ X + 2(26 - X - Z) + 5Z = 48 \]
Expanding this:
\[ X + 52 - 2X - 2Z + 5Z = 48 \]
Simplifying:
\[ - X + 3Z + 52 = 48 \]
\[ - X + 3Z = -4 \]
\[ X = 3Z + 4 \]

### Step 5: Substitute values of \( Z \)
Since Rajesh must have at least one coin of each denomination, \( Z \) must be at least 1. We can substitute values for \( Z \) to find \( X \):

1. **If \( Z = 1 \)**:
\[ X = 3(1) + 4 = 7 \]
Then, substituting \( Z = 1 \) back to find \( Y \):
\[ Y = 26 - 7 - 1 = 18 \]
So, \( (X, Y, Z) = (7, 18, 1) \).

2. **If \( Z = 2 \)**:
\[ X = 3(2) + 4 = 10 \]
Then,
\[ Y = 26 - 10 - 2 = 14 \]
So, \( (X, Y, Z) = (10, 14, 2) \).

3. **If \( Z = 3 \)**:
\[ X = 3(3) + 4 = 13 \]
Then,
\[ Y = 26 - 13 - 3 = 10 \]
So, \( (X, Y, Z) = (13, 10, 3) \).

4. **If \( Z = 4 \)**:
\[ X = 3(4) + 4 = 16 \]
Then,
\[ Y = 26 - 16 - 4 = 6 \]
So, \( (X, Y, Z) = (16, 6, 4) \).

### Step 6: Verify the minimum \( X \)
The minimum number of ₹1 coins occurs at \( Z = 1 \), giving us \( X = 7 \).

### Conclusion
Thus, the minimum number of ₹1 coins Rajesh can have is **7**.