Answer

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Hint: The given problem is related to percentages. Recall the formulae related to percentage change in the value of an entity and absolute change in the value of the entity. Try to form equations using the information given in the problem statement and solve the equations to find the price of the tractor.

Complete step-by-step answer:

We will consider the original price of the tractor as $Rs.x$ and the price of the tractor after reduction be $Rs.y$. In the question, it is given that the price of the tractor is reduced by 20%.

We know, the formula for percentage change is given as % change $ =\dfrac{(V_f)-(V_i)}{V_i}\times 100$ %.

But the problem is related to reduction of value of an entity. So, the percentage change will be negative. So, we have $-20=\dfrac{y-x}{x}\times 100$.

\[\Rightarrow 20=\dfrac{x-y}{x}\times 100.....(i)\]

It is also given that after reduction, the price of the tractor is$Rs.40000$. So, $y=40000$. On substituting $y=40000$ in equation \[(i)\] , we get $20=\dfrac{x-40000}{x}\times 100$.

$\Rightarrow \dfrac{20}{100}=\dfrac{x-40000}{x}$

$\Rightarrow \dfrac{1}{5}=\dfrac{x-40000}{x}$

Now, we will cross multiply the given equation to form a linear equation in $x$.

On cross multiplication, we get $x=5\left( x-40000 \right)$.

$\Rightarrow x=5x-200000$

$\Rightarrow 5x-x=200000$

$\Rightarrow 4x=200000$

$\Rightarrow x=50000$

So, the original price of the tractor is $Rs.50000$ .

Now, to find the reduction in price, we will subtract the final price after reduction from the original price of the tractor.

So, reduction $=Rs.\left( 50000-40000 \right)=Rs.10000$.

Hence, the original price of the tractor is $Rs.50000$ and the reduction in price is $Rs.10000$ .

Note: While calculating the percentage change, students generally forget the fact that the given percentage is of reduction and hence, it will be negative. Students consider it as positive and end up getting a wrong answer.

Complete step-by-step answer:

We will consider the original price of the tractor as $Rs.x$ and the price of the tractor after reduction be $Rs.y$. In the question, it is given that the price of the tractor is reduced by 20%.

We know, the formula for percentage change is given as % change $ =\dfrac{(V_f)-(V_i)}{V_i}\times 100$ %.

But the problem is related to reduction of value of an entity. So, the percentage change will be negative. So, we have $-20=\dfrac{y-x}{x}\times 100$.

\[\Rightarrow 20=\dfrac{x-y}{x}\times 100.....(i)\]

It is also given that after reduction, the price of the tractor is$Rs.40000$. So, $y=40000$. On substituting $y=40000$ in equation \[(i)\] , we get $20=\dfrac{x-40000}{x}\times 100$.

$\Rightarrow \dfrac{20}{100}=\dfrac{x-40000}{x}$

$\Rightarrow \dfrac{1}{5}=\dfrac{x-40000}{x}$

Now, we will cross multiply the given equation to form a linear equation in $x$.

On cross multiplication, we get $x=5\left( x-40000 \right)$.

$\Rightarrow x=5x-200000$

$\Rightarrow 5x-x=200000$

$\Rightarrow 4x=200000$

$\Rightarrow x=50000$

So, the original price of the tractor is $Rs.50000$ .

Now, to find the reduction in price, we will subtract the final price after reduction from the original price of the tractor.

So, reduction $=Rs.\left( 50000-40000 \right)=Rs.10000$.

Hence, the original price of the tractor is $Rs.50000$ and the reduction in price is $Rs.10000$ .

Note: While calculating the percentage change, students generally forget the fact that the given percentage is of reduction and hence, it will be negative. Students consider it as positive and end up getting a wrong answer.

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