Question

At a certain fruit stand, the price of each apple is \[40\] cents and the price of each orange is $60$ cents. Mary selects a total of $10$ apples and oranges from the fruit stand and the average price (arithmetic mean) of the $10$ pieces of fruit is $56$ cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $52$ cents?
$\left( a \right){\text{ 1}}$
$\left( b \right){\text{ 2}}$
$\left( c \right){\text{ 3}}$
$\left( d \right){\text{ 4}}$
$\left( e \right){\text{ 5}}$

Answer 1

Hint:
So this type of question is solved by making an equation, for this we will assume that the total number of apples are $x$ and that of oranges are $10 - x$, and we will use the condition to find the number of apples and oranges. Also, we will assume that the oranges that were put back were $y$ and use the condition to solve for $y$.

Complete step by step solution:
 In the question, it is given that the price of each apple is \[40\] cents and the price of each orange is $60$ cents. And also it is given that Mary selects a total of $10$ apples and oranges from the fruit stand and the average price of the $10$ pieces of fruit is $56$ cents.
So, let us assume the number of apples selected were $x$
Then the total cost of the apples is $40x$ .
Let us assume the number of oranges selected were $10 - x$
Then the total cost of the apples is $60\left( {10 - x} \right)$ .
According to the question, the average price (arithmetic mean) of the $10$ pieces of fruit is $56$ cents will be equal to
$ \Rightarrow \dfrac{{40x + 60\left( {10 - x} \right)}}{{10}} = 56$
Now on solving the braces of the above equation, we get
$ \Rightarrow \dfrac{{40x + 600 - 60x}}{{10}} = 56$
Now taking the denominator to the right side, we get
$ \Rightarrow 40x + 600 - 60x = 560$
Now taking the like terms together and solving it, we get
$ \Rightarrow 40 = 20x$
And solving for the value of $x$ , we get
$ \Rightarrow x = 2$ , and we will name it equation $1$
So from the equation $1$ , the number of apples were $2$ and the number of oranges was $8$ .
Given that the oranges were Mary put back so that the average price of the pieces of fruit that she keeps is $56$ cents.
Let us assume the oranges that were put back were $y$ . So the equation will be
$ \Rightarrow \dfrac{{40\left( 2 \right) + \left( {8 - y} \right)60}}{{10 - y}} = 52$
Now on solving the braces, we get
$ \Rightarrow \dfrac{{80 + 480 - 60y}}{{10 - y}} = 52$
Taking the denominator to the right side, we get
$ \Rightarrow 560 - 60y = 520 - 52y$
And on solving the above equation, we get
$ \Rightarrow 40 = 8y$
And on dividing it, we get
$ \Rightarrow y = 5$
So, the number of oranges that were put back by Mary was $5$.

Hence, the option $\left( e \right)$ is correct.

Note:
Here, in this question, we should note that when the oranges were put back then at that time it also alters the total number of fruits that were brought. For solving this type of question we always have to try to frame the equation from the information given in the question.