Solving the Cost Problem: Pens and Pencils

This problem asks us to find the original cost of 2 pens and 2 pencils given information about their combined costs under different conditions. We can solve this using algebra by setting up and solving a system of linear equations.

Defining Variables

Let's use variables to represent the unknown costs:

• Let $p$ be the original cost of one pen in Rupees (₹).
• Let $c$ be the original cost of one pencil in Rupees (₹).

Setting Up the Equations

We are given two scenarios, which we can translate into two equations:

1. Scenario 1: The cost of 12 pens and 7 pencils is ₹285.

   This translates to the equation:

   $12p + 7c = 285 \quad (1)$

2. Scenario 2: The cost of a pen decreases by ₹6 (new cost is $p-6$) and the cost of a pencil increases by ₹5 (new cost is $c+5$). With these new costs, 11 pens and 3 pencils cost ₹118.

   This translates to the equation:

   $11(p-6) + 3(c+5) = 118$

   Let's simplify this second equation:

   $11p - 66 + 3c + 15 = 118$

   Combine the constant terms:

   $11p + 3c - 51 = 118$

   Add 51 to both sides:

   $11p + 3c = 118 + 51$

   $11p + 3c = 169 \quad (2)$

Solving the System of Equations

Now we have a system of two linear equations with two variables:

1. $12p + 7c = 285$
2. $11p + 3c = 169$

We can solve this system using either substitution or elimination. Let's use the elimination method. We want to make the coefficients of either $p$ or $c$ the same (or opposites) in both equations.

Let's eliminate $c$. We can multiply Equation (1) by 3 and Equation (2) by 7:

Multiply Equation (1) by 3:

$(12p + 7c) \times 3 = 285 \times 3$

$36p + 21c = 855 \quad (3)$

Multiply Equation (2) by 7:

$(11p + 3c) \times 7 = 169 \times 7$

$77p + 21c = 1183 \quad (4)$

Now, subtract Equation (3) from Equation (4) to eliminate $c$:

$(77p + 21c) - (36p + 21c) = 1183 - 855$

$77p - 36p + 21c - 21c = 328$

$41p = 328$

Solve for $p$:

$p = \frac{328}{41}$

$p = 8$

So, the original cost of one pen is ₹8.

Finding the Cost of a Pencil

Now substitute the value of $p$ (which is 8) back into either Equation (1) or Equation (2) to find the value of $c$. Let's use Equation (2):

$11p + 3c = 169$

Substitute $p = 8$:

$11(8) + 3c = 169$

$88 + 3c = 169$

Subtract 88 from both sides:

$3c = 169 - 88$

$3c = 81$

Solve for $c$:

$c = \frac{81}{3}$

$c = 27$

So, the original cost of one pencil is ₹27.

Calculating the Final Cost

The question asks for the original cost of 2 pens and 2 pencils.

Cost = $2p + 2c$

Substitute the values $p=8$ and $c=27$:

Cost = $2(8) + 2(27)$

Cost = $16 + 54$

Cost = $70$

Therefore, the original cost of 2 pens and 2 pencils is ₹70.