Understanding the Problem

This problem involves finding the original prices of pens and notebooks given two different scenarios involving their costs and changes in price. We need to set up and solve a system of linear equations to find the values.

Setting Up the Equations

Let's define the variables:

• Let $p$ represent the original price of one pen.
• Let $n$ represent the original price of one notebook.

From the first statement, "The price of 4 pens and 5 notebooks is ₹130", we can write the first equation:

Equation 1: $4p + 5n = 130$

Next, let's consider the changes in price:

• The price of a pen increases by 25%. The new price of a pen is $p + 0.25p = 1.25p$.
• The price of a notebook decreases by 20%. The new price of a notebook is $n - 0.20n = 0.80n$.

The second statement says, "the price of 2 pens and 4 notebooks becomes ₹82" after these changes. This gives us the second equation:

Equation 2: $2(1.25p) + 4(0.80n) = 82$

Let's simplify Equation 2:

$2.5p + 3.2n = 82$

To make it easier to work with, we can multiply this equation by 10 to eliminate the decimals:

$25p + 32n = 820$

Solving the System of Equations

Now we have a system of two linear equations:

1. $4p + 5n = 130$
2. $25p + 32n = 820$

We can use the method of elimination or substitution. Let's use elimination. We can multiply Equation 1 by 25 and Equation 2 by 4 to make the coefficients of $p$ the same:

• Multiply Equation 1 by 25: $25 \times (4p + 5n) = 25 \times 130 \implies 100p + 125n = 3250$
• Multiply Equation 2 by 4: $4 \times (25p + 32n) = 4 \times 820 \implies 100p + 128n = 3280$

Now, subtract the modified Equation 1 from the modified Equation 2:

$(100p + 128n) - (100p + 125n) = 3280 - 3250$

$3n = 30$

Divide by 3 to find the value of $n$:

$n = \frac{30}{3} = 10$

So, the original price of one notebook ($n$) is ₹10.

Now substitute the value of $n$ back into Equation 1 to find $p$:

$4p + 5(10) = 130$

$4p + 50 = 130$

Subtract 50 from both sides:

$4p = 130 - 50$

$4p = 80$

Divide by 4 to find the value of $p$:

$p = \frac{80}{4} = 20$

So, the original price of one pen ($p$) is ₹20.

Calculating the Final Price

The question asks for the original price of 1 pen and 2 notebooks. This can be represented as $p + 2n$.

Substitute the values we found for $p$ and $n$:

$Original Price = $p + 2n = 20 + 2(10)$

$Original Price = $20 + 20 = 40$

Therefore, the original price of 1 pen and 2 notebooks is ₹40.