Answer

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Hint: It is given that the merchant has only rupees 1200 for purchase of mango and orange, however the price of mango and orange which he purchase fluctuates every day, so it is given that one day price of mango and orange is 40Rs/Kg so use the concept that kilogram of orange and mango purchased will be Total rupees he has divided by the price per kilogram. Now on another day the relation between the price per kilogram of mango and orange are given so consider the unknown quantity to be a variable and use the same concept to reach to the answer.

Complete step-by-step answer:

Given data

Price of orange on a particular day = 40 Rs/kg.

Price of mango on a particular day = 40 Rs/kg.

Now it is given that a merchant buys orange for 1200 rupees and mango for 1200 rupees.

Now, we have to calculate how much kilogram of orange and mango he bought on that day.

As we see that the price of orange and mango are same on a particular day so he bought same kilogram of orange and mango on that day = $\dfrac{{{\text{Total rupees}}}}{{{\text{Price of 1 kg orange or mango}}}}$

So, the number of kilograms of orange and mango = $\dfrac{{1200}}{{40}} = 30$ kg.

So, the merchant bought 30 kg orange and 30 kg mango on that particular day.

Now on another day the price of mango was 10 rupees less than that of orange. Hence he got 20 kg more mango than orange.

Let the price of orange on another day be x Rs/kg.

So, the price of mango on another day = (x - 10) Rs/kg.

Let the orange he bought = y kg.

So, the mangos he bought = (y + 20) kg.

Now total rupees he has 1200 for mangos and 1200 for oranges.

Now construct the linear equation according to given information

I.e. The price of orange in Rs/kg multiplied by the number of mangoes in kg is equal to total rupees.

$ \Rightarrow xy = 1200$………………… (1)

Similarly,

$ \Rightarrow \left( {x - 10} \right)\left( {y + 20} \right) = 1200$…………………………… (2)

Now simplify equation (2) we have

$ \Rightarrow xy + 20x - 10y - 200 = 1200$

Now from equation (1) we have

$ \Rightarrow 1200 + 20x - 10y - 200 = 1200$

$ \Rightarrow 20x - 10y = 200$

Now divide by 10 in above equation we have,

$ \Rightarrow 2x - y = 20$

Now from equation (1) $y = \dfrac{{1200}}{x}$ so, substitute this value in above equation we have,

$ \Rightarrow 2x - \dfrac{{1200}}{x} = 20$

Now simplify the above equation we have,

$

\Rightarrow 2{x^2} - 20x - 1200 = 0 \\

\Rightarrow {x^2} - 10x - 600 = 0 \\

$

Now factorize the above equation we have,

$ \Rightarrow {x^2} - 30x + 20x - 600 = 0$

$ \Rightarrow x\left( {x - 30} \right) + 20\left( {x - 30} \right) = 0$

$ \Rightarrow \left( {x - 30} \right)\left( {x + 20} \right) = 0$

$ \Rightarrow x = 30,{\text{ & }}x = - 20$ As x cannot be negative.

So, the price of orange on another day is 30 Rs/kg.

So, this is the required answer.

Note: Whenever we face such types of questions the key point is simply to use the given information provided in the question to formulate equations and to use the unitary methods to get the unknown entity. This concept will help you get on the right track to reach the answer.

Complete step-by-step answer:

Given data

Price of orange on a particular day = 40 Rs/kg.

Price of mango on a particular day = 40 Rs/kg.

Now it is given that a merchant buys orange for 1200 rupees and mango for 1200 rupees.

Now, we have to calculate how much kilogram of orange and mango he bought on that day.

As we see that the price of orange and mango are same on a particular day so he bought same kilogram of orange and mango on that day = $\dfrac{{{\text{Total rupees}}}}{{{\text{Price of 1 kg orange or mango}}}}$

So, the number of kilograms of orange and mango = $\dfrac{{1200}}{{40}} = 30$ kg.

So, the merchant bought 30 kg orange and 30 kg mango on that particular day.

Now on another day the price of mango was 10 rupees less than that of orange. Hence he got 20 kg more mango than orange.

Let the price of orange on another day be x Rs/kg.

So, the price of mango on another day = (x - 10) Rs/kg.

Let the orange he bought = y kg.

So, the mangos he bought = (y + 20) kg.

Now total rupees he has 1200 for mangos and 1200 for oranges.

Now construct the linear equation according to given information

I.e. The price of orange in Rs/kg multiplied by the number of mangoes in kg is equal to total rupees.

$ \Rightarrow xy = 1200$………………… (1)

Similarly,

$ \Rightarrow \left( {x - 10} \right)\left( {y + 20} \right) = 1200$…………………………… (2)

Now simplify equation (2) we have

$ \Rightarrow xy + 20x - 10y - 200 = 1200$

Now from equation (1) we have

$ \Rightarrow 1200 + 20x - 10y - 200 = 1200$

$ \Rightarrow 20x - 10y = 200$

Now divide by 10 in above equation we have,

$ \Rightarrow 2x - y = 20$

Now from equation (1) $y = \dfrac{{1200}}{x}$ so, substitute this value in above equation we have,

$ \Rightarrow 2x - \dfrac{{1200}}{x} = 20$

Now simplify the above equation we have,

$

\Rightarrow 2{x^2} - 20x - 1200 = 0 \\

\Rightarrow {x^2} - 10x - 600 = 0 \\

$

Now factorize the above equation we have,

$ \Rightarrow {x^2} - 30x + 20x - 600 = 0$

$ \Rightarrow x\left( {x - 30} \right) + 20\left( {x - 30} \right) = 0$

$ \Rightarrow \left( {x - 30} \right)\left( {x + 20} \right) = 0$

$ \Rightarrow x = 30,{\text{ & }}x = - 20$ As x cannot be negative.

So, the price of orange on another day is 30 Rs/kg.

So, this is the required answer.

Note: Whenever we face such types of questions the key point is simply to use the given information provided in the question to formulate equations and to use the unitary methods to get the unknown entity. This concept will help you get on the right track to reach the answer.

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