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 Rs. $1800$ worth apples and Rs. $600$ worth bananas, then how many kgs of each fruit did she buy?

Answer 1

Hint:
First, assume the total kg of apples bought be x and the total kg of bananas bought be y. Let the cost of per kg apple be a and the cost of per kg banana be b. Then the total cost of apples is ax and the total cost of bananas is yb. Then make equations according to the statement given in the question. Then put $a = 2b$ in ax. Then put x=$\left( {50 - y} \right)$ and solve for b. Then find the value of a and x and y.

Complete step by step solution:
Given the total number of fruits bought by Iniya=$50$kg
Let the total kg of apples bought be x and the total kg of bananas bought be y. Let the cost of per kg apple be a and the cost of per kg banana be b.
Then the total cost of apples is ax and the total cost of bananas is yb.
Then we can write-
$ \Rightarrow x + y = 50$ --- (i)
Now it is given that the price of apples is twice the price of banana then we can write-
$ \Rightarrow a = 2b$ --- (ii)
Also given that the total cost of apples is Rs. $1800$and the total cost of bananas is Rs. $600$. Then we can write-
$ \Rightarrow $ ax= Rs. $1800$ and yb= Rs. $600$--- (iii)
Then on substituting the value of a from eq. (i) in eq. (iii), we get-
$ \Rightarrow 2bx = 1800$
On putting the value of x from eq. (i) in the above equation we get-
$ \Rightarrow 2b\left( {50 - y} \right) = 1800$
Then on simplifying we get-
$ \Rightarrow 100b - yb = 600$
On substituting value of yb, we get-
$ \Rightarrow 100b - 2 \times 600 = 1800$
On simplifying, we get-
$ \Rightarrow 100b - 1200 = 1800$
On transferring the constant on right side we get-
$ \Rightarrow 100b = 1800 + 1200 = 3000$
So on solving, further, we get-
$ \Rightarrow b = \dfrac{{3000}}{{100}} = 30$
Then from eq. (ii), we get,
$ \Rightarrow a = 2 \times 30 = 60$
Now on putting values of a and b in eq. (iii), we get-
$ \Rightarrow 60x = 1800$ and $20y = 600$
On solving we get-
$ \Rightarrow x = \dfrac{{1800}}{{60}} = 30$ and $y = \dfrac{{600}}{{30}} = 20$

Hence apples bought by Iniya are $30{\text{ kg}}$ and bananas are $20{\text{ kg}}$.

Note:
The student may go wrong if he or she tries to substitute the value of y from eq. (i) in yb because then we get-
$ \Rightarrow \left( {50 - x} \right)b = 600$
On multiplication we get-
$ \Rightarrow 50b - xb = 600$
Now here we do not know the value of xb so again we will have to substitute another value which will make the calculation complex. Hence we substituted the value of a from eq. (ii) in eq. (iii) and then we substituted the value of x from eq. (i) to obtain yb.