Iniya bought 50 kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought ₹1800 worth of apples and ₹600 worth bananas, then how many kg of each fruit did she buy?

Hint: Here first of all we will assume the quantity of apples to ‘a’ kg at the rate of Rs. x per kg and the quantity of bananas to be ‘b’ kg at the rate of Rs. y per kg. Then we will form linear equations in two variables and then then solve them for each of the unknown values to get the desired answer.

Complete step-by-step answer:
Let the quantity of apples be a kg at the rate of Rs. x per kg
This implies the total cost of apples is Rs. \[ax\]
Let the quantity of bananas to be b kg at the rate of Rs. y per kg
This implies the total cost of bananas is Rs. \[by\].
Now it is given that, total quantity of fruits is 50 kg
Therefore,
\[a + b = 50\]……………………… (1)
Now it is given that, Iniya paid twice as much per kg for the apple as she did for the banana.
Therefore,
\[x = 2y\]………………………… (2)
Also, it is given that Iniya bought ₹1800 worth of apples
Therefore, \[ax = 1800\]……………….. (3)
Now it is given that Iniya bought ₹600 worth of bananas
Therefore, \[by = 600\]………………… (4)
Now Putting value from equation 2 in equation 3 we get:-
\[a\left( {2y} \right) = 1800\]
Putting the value of a from equation 1 in above equation we get:-
\[\left( {50 - b} \right)\left( {2y} \right) = 1800\]
Solving it further we get:-
\[100y - 2by = 1800\]
Now putting the value from equation 4 we get:-
\[100y - 2\left( {600} \right) = 1800\]
Simplifying it further we get:-
\[100y - 1200 = 1800\]
\[ \Rightarrow 100y = 3000\]
Solving for the value of y we get:-
\[y = \dfrac{{3000}}{{100}}\]
\[ \Rightarrow y = 30\]
Now Putting this value in equation 2 we get:-
\[x = 2\left( {30} \right)\]
Solving for x we get:-
\[x = 60\]
Now Putting the value of x in equation 3 we get:-
\[a\left( {60} \right) = 1800\]
Solving for the value of a we get:-
\[a = \dfrac{{1800}}{{60}}\]
\[a = 30\]kg
Now putting the value of y in equation 4 we get:-
\[b\left( {30} \right) = 600\]
Solving for the value of b we get:-
\[b = \dfrac{{600}}{{30}}\]
\[b = 20\]kg

Therefore, the quantity of apples is 30kg
The quantity of bananas is 20 kg.

Note: Students should note that the linear equation in one variable has only one variable with highest power 1 while the linear equation in two variables has two variables each with highest power as 1.