Question

A fruit seller sold big, medium and small sized apples for Rs.15, Rs.10 and Rs.5 respectively. The total number of apples sold were in the ratio 3:2:5. Find the average cost of an apple.

Answer 1

Hint: Here we have to find the average cost of the apple to find the average cost. We should use the ratio and find the amount of apples sold and with the general formula of finding average we will find the required average cost.

Complete step by step solution:
It is given that a fruit seller sold big, medium and small sized apples for \[Rs.{\text{ }}15,{\text{ }}Rs.{\text{ }}10{\text{ and }}Rs.{\text{ }}5\].
And it is also given that the total number of apples sold were in the ratio \[3:2:5\].
Now we are going to find the ratio of amount sold using the above ratio, that is nothing but we are going to multiply the respective amount to the respective apple sold,
Let $x$ be the common multiple,
Hence,
The total number of apples sold $ = 3x + 2x + 5x$
We have to find the total cost of apple sold, we are going to multiply the number of apples to its selling cost,
\[ = (15 \times 3x) + (10 \times 2x) + (5 \times 5x)\]
\[ = 45x + 20x + 25x\]
Hence the total cost of apple sold is \[45x + 20x + 25x\]
Our next work is to find the average cost using the above values,
For that we have to find the total amount of apple sold with the help of the ratio \[3:2:5\]
That is, already we have \[3x + 2x + 5x = 10x\]
The total cost of apple sold is \[45x + 20x + 25x = 90x\]
${\text{average = }}\dfrac{{{\text{sum of all the elements}}}}{{{\text{total number of elements}}}}$
Next we are going to find the average cost of the apple,
$\Rightarrow$Average cost \[ = \dfrac{{45x + 20x + 25x}}{{3x + 2x + 5x}}\]
$\Rightarrow$\[ \dfrac{{90x}}{{10x}} = 9\]

$\therefore$ The average cost of apples is Rs.9

Note: We should not add or multiply ratios directly. So we use $x$ as the common multiplier. Without using the common multiply we may get the correct answer. But the use of common multiply makest the solution more clear to understand by students.