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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.

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