Question

If the cost of bananas is increased by Re 1 per dozen, one can get 2 less dozen for Rs. 840. Find the original cost of one dozen bananas.

Answer 1

Hint: Treat it as a word problem where both cost and quantity are unknown. So, consider two variables one for quantity and one for cost. Two variables means at least two equations are to be made by given conditions to solve the question.

Complete step-by-step answer:
Let the original cost of one dozen bananas be ‘y’ and the original number of dozen the person can buy for Rs. 840 be ‘x’.
We know, the quantity (in dozen) we take multiplied by cost per dozen gives the total amount we have to pay for the quantity we take.
For the case before increasing the price;
Cost per dozen of bananas = x.
Dozen of bananas that can be bought for Rs. 840 = y.
So, the equation we get is:
 $xy=840$
$\Rightarrow x=\dfrac{840}{y}......................(i)$
Now, for the case after increasing the price;
Cost per dozen of bananas = x+1.
Dozen of bananas that can be bought for Rs. 840 = y-2.
So, the equation we get is:
 $\left( x+1 \right)\left( y-2 \right)=840....................(ii)$
Now, substituting the value of x from equation (i) in equation (ii):
$\left( x+1 \right)\left( y-2 \right)=840$
$\Rightarrow \left( \dfrac{840}{y}+1 \right)\left( y-2 \right)=840$
Taking LCM we get;
$\left( \dfrac{840+y}{y} \right)\left( y-2 \right)=840$
Taking y to the other side of equal to sign;
$\left( 840+y \right)\left( y-2 \right)=840y$
Further, opening the brackets by multiplying the terms we get:
$840y-1680+{{y}^{2}}-2y=840y$
$\Rightarrow {{y}^{2}}-2y-1680=0$
$\Rightarrow {{y}^{2}}-\left( 42-40 \right)y-1680=0$
$\Rightarrow {{y}^{2}}-42y+40y-1680=0$
$\Rightarrow y\left( y+40 \right)-42(y+40)=0$
$\Rightarrow \left( y+40 \right)(y-42)=0$
Therefore, the values of y are,
$y=-40$ but y is cost and cost cannot be negative so not possible.
$y=42$
So, the original cost of 1 dozen of bananas is y which is equal to Rs. 42.

Note: While substituting try to substitute the variable that is not asked to find, this reduces your effort while you solve the equation. By doing so, the unwanted variable in the equation is already eliminated and the variable for which you are solving is the variable that you need to answer the question.