Answer
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Hint: In the above given question, we are given the price of 2 bananas and we are asked to calculate the price of 1 banana. This can be simply done by dividing the price of bananas with the quantity(number) of the bananas.
Complete step-by-step answer:
Let us assume the cost of 1 banana to be $x$.
We know that, there are 12 bananas in a dozen,
so, 2 dozen bananas$ = 2 \times 12$
=24 bananas.
Now, the cost of 24 bananas$ = 24 \times x$
$ = Rs.24x$
Here given that the cost of 2 dozen, that is, 24 bananas is$Rs.30\dfrac{3}{7}$.
Therefore, we can write,
$\Rightarrow$ $24x = 30\dfrac{3}{7}$
$\Rightarrow$ $24x = \dfrac{{213}}{7}$
$\Rightarrow$ $x = \dfrac{{213}}{{7 \times 24}}$
$\Rightarrow$ $x = \dfrac{{213}}{{168}}$
$ \Rightarrow x = \dfrac{{71}}{{56}}$
$\therefore x = 1\dfrac{{15}}{{56}}$
Hence, the cost of 1 banana is Rs.$1\dfrac{{15}}{{56}}$.
Note: When we face such types of problems, first of all convert the mixed fraction terms and then divide the number of bananas given with the price of the banana and convert the price so obtained into the mixed fraction again, as the given price was also in the mixed fraction only.
Complete step-by-step answer:
Let us assume the cost of 1 banana to be $x$.
We know that, there are 12 bananas in a dozen,
so, 2 dozen bananas$ = 2 \times 12$
=24 bananas.
Now, the cost of 24 bananas$ = 24 \times x$
$ = Rs.24x$
Here given that the cost of 2 dozen, that is, 24 bananas is$Rs.30\dfrac{3}{7}$.
Therefore, we can write,
$\Rightarrow$ $24x = 30\dfrac{3}{7}$
$\Rightarrow$ $24x = \dfrac{{213}}{7}$
$\Rightarrow$ $x = \dfrac{{213}}{{7 \times 24}}$
$\Rightarrow$ $x = \dfrac{{213}}{{168}}$
$ \Rightarrow x = \dfrac{{71}}{{56}}$
$\therefore x = 1\dfrac{{15}}{{56}}$
Hence, the cost of 1 banana is Rs.$1\dfrac{{15}}{{56}}$.
Note: When we face such types of problems, first of all convert the mixed fraction terms and then divide the number of bananas given with the price of the banana and convert the price so obtained into the mixed fraction again, as the given price was also in the mixed fraction only.
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