Question

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?

Answer 1

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.