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# If the price of sugar decreased by Rs. 5, one can buy 1 kg more sugar in Rs. 150, what is the price of sugar?  Hint: Treat it as a word problem where both Cost and quantity are unknown. So, consider two variables one for quantity and one for cost. Two variables means at least two equations are to be made to solve the question. So, use the data given in the question to form two equations and solve them to get the answer.

Let the original cost of 1 kg of sugar be ‘y’ and the original quantity that can be bought for Rs. 150 be x kg.
We know, quantity (quantity) we take multiplied by cost per kg gives the total amount we have to pay for the quantity we take.
For the case before increasing the price;
Cost per kg of sugar = y.
Quantity that can be bought for Rs. 150 = x.
So, the equation we get is:
$xy=150$
$\Rightarrow x=\dfrac{150}{y}......................(i)$
Now, for the case after decreasing the price;
Cost per kg of sugar = y-5.
Quantity of sugar that can be bought = x+1.
So, the equation we get is:
$\left( x+1 \right)\left( y-5 \right)=150....................(ii)$
Now, substituting the value of x from equation (i) in equation (ii):
$\left( x+1 \right)\left( y-5 \right)=150$
$\Rightarrow \left( \dfrac{150}{y}+1 \right)\left( y-5 \right)=150$
Taking LCM we get;
$\left( \dfrac{150+y}{y} \right)\left( y-5 \right)=150$
Taking y to the other side of equal to sign;
$\left( 150+y \right)\left( y-5 \right)=150y$
Further, opening the brackets by multiplying the terms we get:
$150y-750+{{y}^{2}}-5y=150y$
$\Rightarrow {{y}^{2}}-5y-750=0$
We know that 5y can be written as (30y-25y). On doing so, we get
${{y}^{2}}-\left( 30-25 \right)y-750=0$
$\Rightarrow {{y}^{2}}-30y+25y-750=0$
$\Rightarrow y\left( y-30 \right)+25(y-30)=0$
$\Rightarrow \left( y+25 \right)(y-30)=0$
Therefore, the values of y are,
$y=-25$ but y is cost and cost cannot be negative so not possible.
$y=30$
So, the original cost of 1 kg of sugar is y which is equal to Rs. 30.

Note: While substituting try to substitute the variable that is not asked to find, this reduces your effort while you solve the equation. By doing so, the unwanted variable in the equation is already eliminated and the variable for which you are solving is the variable that you need to answer the question.
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