The problem is as follows:
Jenny goes to the groceries supermarket and she notices the following; In the fruit section, for one kilogram of oranges, six small or four large ones are obtained, in addition the price of one kilogram of oranges varies from $4$ usd to $6$ usd. Find the minimum cost for Jenny to buy a dozen large oranges and two dozen small oranges.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\textrm{27 USD}\\ 2.&\textrm{28 USD}\\ 3.&\textrm{30 USD}\\ 4.&\textrm{29 USD}\\ \end{array}$
How exactly should this problem be solved?. My initial thought goes as this. In order to minimize the cost, I would want to get the least price and maximize the number of oranges. But in this case the number is fixed because it mentions that a dozen of the larger ones is required and two dozen of the small ones are needed.
Then I made the following convertion factors: I'm using $l$ for large oranges and $s$ for small oranges.
$1\,kg=4l=6s$
Since it is requested that it is needed two dozen of the small ones then:
$2\times 12\times\frac{6s}{1\,kg}=144\frac{s}{kg}$
For the bigger ones it is needed just one dozen:
$12\times \frac{4l}{kg}=48\frac{l}{kg}$
Then this means in order to ensure the least prize is to use the small prize.
Thus:
$144\times 4 + 48 \times 6 = 576+288=864\,\textrm{usd}$
But this answer does not appear in the alternatives. What exactly went wrong here and more importantly why? it would help me a lot a wordy explanation or the logic which should be used in this scenario because I'm confused.
Which part did not went right in my strategy?. Can someone help me here?.
As you have stated, we have $ 1kg = 4l = 6s$, where $s$ denotes the small oranges, $l$ denotes the large ones. Now to buy two dozen ($2 \cdot 12$) small oranges, we need to buy $24s$, that is $4$kg. Similarly, to obtain $1$ dozen of large oranges $l$, we need to buy $12l$, i.e. $ 3$kg. If we assume we get the best price, that is $4$ usd per kg, we will pay $7 \cdot 4 = 28$ usd minimum.