A shopkeepr sells a table at 12% loss and a book 19% profit and he get 160 profit. If he sells a table at 12% profit and book at 16% loss then he inverse 40 loss. Then find the price of book.

Options

A. Rs 6000
B. Rs 4000
C. Rs 5000
D. Rs 5500

Answer

Let's solve the problem step by step.

### Step 1: Define Variables
Let:
- \( T \) = Cost Price (CP) of the table
- \( B \) = Cost Price (CP) of the book

### Step 2: Set Up the Equations
1. When the table is sold at a 12% loss:
\[ \text{Selling Price of Table} = T - 0.12T = 0.88T \]

2. When the book is sold at a 19% profit:
\[ \text{Selling Price of Book} = B + 0.19B = 1.19B \]

3. The total profit from selling both items is given as 160:
\[ \text{Total Selling Price} = \text{Selling Price of Table} + \text{Selling Price of Book} \]
\[ 0.88T + 1.19B - (T + B) = 160 \]
Simplifying this gives:
\[ 0.88T + 1.19B - T - B = 160 \]
\[ -0.12T + 0.19B = 160 \quad \text{(Equation 1)} \]

### Step 3: Set Up the Second Scenario
1. When the table is sold at a 12% profit:
\[ \text{Selling Price of Table} = T + 0.12T = 1.12T \]

2. When the book is sold at a 16% loss:
\[ \text{Selling Price of Book} = B - 0.16B = 0.84B \]

3. The total loss from selling both items is given as 40:
\[ 1.12T + 0.84B - (T + B) = -40 \]
Simplifying this gives:
\[ 1.12T + 0.84B - T - B = -40 \]
\[ 0.12T - 0.16B = -40 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( -0.12T + 0.19B = 160 \) (Equation 1)
2. \( 0.12T - 0.16B = -40 \) (Equation 2)

We can add both equations to eliminate \( T \):
\[ (-0.12T + 0.19B) + (0.12T - 0.16B) = 160 - 40 \]
This simplifies to:
\[ 0.03B = 120 \]
Thus,
\[ B = \frac{120}{0.03} = 4000 \]

### Step 5: Conclusion
The cost price of the book is \( \text{Rs. } 4000 \).

---