Let the original cost of one pen be denoted by $p$ and the original cost of one pencil be denoted by $c$. We can set up a system of linear equations based on the information given.
From the first statement, the cost of 2 pens and 15 pencils is ₹286:
$ 2p + 15c = 286 \quad (1) $
The second scenario involves changes in costs. The cost of a pen decreases by ₹4 (new cost $p-4$) and the cost of a pencil increases by ₹2 (new cost $c+2$). The cost of 17 pens and 3 pencils under these new conditions is ₹128:
$ 17(p - 4) + 3(c + 2) = 128 $
Simplify this equation:
$ 17p - 68 + 3c + 6 = 128 $
$ 17p + 3c - 62 = 128 $
$ 17p + 3c = 128 + 62 $
$ 17p + 3c = 190 \quad (2) $
We have the system:
To eliminate $c$, multiply equation (2) by 5:
$ 5 \times (17p + 3c) = 5 \times 190 $
$ 85p + 15c = 950 \quad (3) $
Subtract equation (1) from equation (3):
$ (85p + 15c) - (2p + 15c) = 950 - 286 $
$ 83p = 664 $
$ p = \frac{664}{83} $
$ p = 8 $
Substitute the value of $p=8$ into equation (1):
$ 2(8) + 15c = 286 $
$ 16 + 15c = 286 $
$ 15c = 286 - 16 $
$ 15c = 270 $
$ c = \frac{270}{15} $
$ c = 18 $
So, the original cost of a pen is ₹8 and the original cost of a pencil is ₹18.
We need to find the original cost of 11 pens and 8 pencils:
Cost = $ 11p + 8c $
Cost = $ 11(8) + 8(18) $
Cost = $ 88 + 144 $
Cost = $ 232 $
The original cost of 11 pens and 8 pencils is ₹232.
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What is the solution of the following equations ?
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