Answer to Question #253040 in Algebra for rose

Question

Last month, Mrs. Rita bought x kilograms of watermelon for 400 pesos. Today she pay same amount but gets 2 kilograms less due to price increase of 10 pesos per kilogram. find the value of x and the original price of the watermelon

Answer 1

Originally, there are $x$ kilograms of watermelons. Let the price of watermelons per kilogram be $y$. Then, $x\times y=400........(i)$

Currently, the amount of watermelons she buys is $(x-2)$ kilograms. The price she pays per kilogram is $(y+10)$ pesos. The total amount she pays is 400 pesos. Then, $(y+10)\times(x-2)=400.......(ii)$

From equation $(i),$

$x\times y=400\implies x={400\over y}$

Substituting for the value of $x$ given above in equation $(ii)$ we have,

$(y+10)\times({400\over y}-2)=400$

Opening the brackets we have,

$400-2y+{4000\over y}-20=400$

${4000\over y}-2y=20$

Multiplying through by $y$ and dividing through by 2,

$2000-y^2=10y\implies -y^2-10y+2000=0.......(iii)$

Equation $(iii)$ is a quadratic equation which can be solved by using quadratic formula given as,

$y={-b\pm\sqrt{b^2-4ac}\over2a}$

where, $a=-1, \space b=-10,\space c=2000$

Therefore,

$y={10\pm\sqrt{(-10)^2-4(-1\times 2000)}\over 2(-1)}={10\pm\sqrt{8100}\over-2}={10\pm90\over-2}$

Solving for the value of $y$,

$y={(10+90)\over -2}={100\over-2}=-50 \space or \space {(10-90)\over-2}={-80\over -2}=40$

Since the value of $y$ can not be negative, we have that $y=40$

From equation $(i)$, the value of $x$ is,

$x\times y=400\implies x={400\over y}={400\over 40}=10$

Therefore, $x=10$ and the original price of watermelon is $y=40$ pesos per kilogram.