Understanding the Cost Problem

This problem involves finding the original costs of pens and pencils given two different scenarios relating their quantities and total costs. We need to use algebra to solve for the unknown costs.

Setting Up Algebraic Equations

Let's define the original cost of one pen as $p$ rupees and the original cost of one pencil as $c$ rupees.

From the first statement: "The cost of 8 pens and 10 pencils is ₹132." This translates to the equation:

$8p + 10c = 132 \quad (1)$

From the second statement: "If the cost of a pen decreases by ₹2 (new cost is $p-2$) and the cost of a pencil increases by ₹7 (new cost is $c+7$), then the cost of 17 pens and 6 pencils is ₹136." This translates to the equation:

$17(p-2) + 6(c+7) = 136$

Let's simplify this second equation:

$17p - 34 + 6c + 42 = 136$

$17p + 6c + 8 = 136$

$17p + 6c = 136 - 8$

$17p + 6c = 128 \quad (2)$

Solving the System of Equations

We now have a system of two linear equations with two variables:

1. $8p + 10c = 132$
2. $17p + 6c = 128$

We can simplify Equation (1) by dividing the entire equation by 2:

$4p + 5c = 66 \quad (1a)$

Now, let's use the elimination method. We can multiply Equation (1a) by 6 and Equation (2) by 5 to make the coefficients of $c$ the same:

Multiply Equation (1a) by 6: $6 \times (4p + 5c = 66) \implies 24p + 30c = 396$

Multiply Equation (2) by 5: $5 \times (17p + 6c = 128) \implies 85p + 30c = 640$

Now, subtract the first modified equation from the second modified equation to eliminate $c$:

$(85p + 30c) - (24p + 30c) = 640 - 396$

$61p = 244$

Solve for $p$: $p = \frac{244}{61}$

$p = 4$

So, the original cost of one pen is ₹4.

Now, substitute the value of $p$ back into Equation (1a) to find $c$: $4p + 5c = 66$

$4(4) + 5c = 66$

$16 + 5c = 66$

$5c = 66 - 16$

$5c = 50$

Solve for $c$: $c = \frac{50}{5}$

$c = 10$

So, the original cost of one pencil is ₹10.

Calculating the Original Cost of 12 Pens and 9 Pencils

The question asks for the original cost of 12 pens and 9 pencils. Using the original costs we found ($p=4$ and $c=10$):

Cost = $12p + 9c$

Cost = $12(4) + 9(10)$

Cost = $48 + 90$

Cost = $138$

Final Answer Verification

Let's quickly check our values:

• Original Scenario: 8 pens (8 * ₹4 = ₹32) + 10 pencils (10 * ₹10 = ₹100) = ₹32 + ₹100 = ₹132. (Matches the problem statement)
• New Prices Scenario: Pen cost = ₹4 - ₹2 = ₹2. Pencil cost = ₹10 + ₹7 = ₹17. 17 pens (17 * ₹2 = ₹34) + 6 pencils (6 * ₹17 = ₹102) = ₹34 + ₹102 = ₹136. (Matches the problem statement)

Our calculated original costs for a pen and a pencil are correct. Therefore, the calculated cost for 12 pens and 9 pencils is also correct.