A tarder bought number of article for …
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Sia ? 4 years, 10 months ago
Let number of articles be x, and cost of each article be y.
Therefore cost of x articles = xy = 900 {tex}\Rightarrow{/tex} y = {tex}\frac{900}{x}{/tex}
According to given information we have,
(x - 5)({tex}\frac{900}{x}{/tex} + 2) = 900 + 80
{tex}\Rightarrow{/tex}(x-5) ({tex}\frac{900}{x}{/tex} + 2x) = 980
{tex}\Rightarrow{/tex}{tex}\frac { 900 x + 2 x ^ { 2 } - 4500 - 10 x } { x }{/tex} = 980
{tex}\Rightarrow{/tex} 2x2 + 890x - 4500 = 980x
{tex}\Rightarrow{/tex}2x2 - 90x - 4500 = 0
{tex}\Rightarrow{/tex}x2 - 45x - 2250 = 0
By factorization we have,
x2 - 15x + 30x - 2250 = 0
{tex}\Rightarrow{/tex} x(x - 75) + 30(x - 75) = 0
{tex}\Rightarrow{/tex}(x + 30)(x - 75) = 0
Here, x = - 30 (rejected), therefore x = 75
{tex}\therefore{/tex} Number of articles = 75
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