Star Wars VS Marvel

Let the number of Star Wars comics be $x$ and the number of Marvel comics be $y$ .

The first condition is that $12 x + 5y \leq 100$ .
The second condition is that $x + y \geq 12$ .
We want to find the maximum value of $x$ .

Algebraic approach:

$7x + 5 \times 12 \leq 7x + 5 ( x +y ) = 12x + 5y \leq 100 \Rightarrow 7x \leq 40 \Rightarrow x \leq 5$

$(x,y) = (5, 8 )$ satisfies the given inequalities, so the maximum number that he could buy is 5.

12 books cost at least 12 × $5 = $60. Buying a Star Wars book costs us $12 - $5 = $7 extra each. We can spend $100 - $60 = $40 on this "extra".

$\lfloor 40 ÷ 7 \rfloor = \boxed {5}$

Indeed, 5 Star Wars books and 7 Marvel comics cost:

5 × $12 + 7 × $5 = $95

We are looking for that cost of star wars books which ends with a 0 or 5 so that, the remaining money is a multiple of 5, so as to buy integral number of Marvel books. Keeping in mind the spend limit of 100$, the only possibility meeting our needs is $12\times 5=60$ , so the number of star wars books he could buy is $\boxed{5}$ .

Let the number of Star Wars comic be 'x' and Marvel comic be 'y'.

According to the question, $x+y$ $\ge$ $12$

For $100,we can buy atmost 8 Starwars comic irrespective of that eqn.

So,only for less number of books, he can buy maximum number of Starwars comic.

$12x+5y$ $\le$ $100$ and $x,y$ $\in$ $N$

Now the maximum value for x satisfying both the equations is the answer. $i.e$ 5

We can write the equation 12s+5m=100, s being the amount of Star Wars comics and m the amount of Marvel comics.

Now, we can proceed in many ways, one of them is the following: We isolate s wich leaves s=(100-5m)/12=5(20-m)/12, now we move the 5 to the left side, leaving us with s/5=(20-m)/12, which means s must be a multiple of 5. 0, 5, 10, 15... are all possible options, but we notice that from 10 on the equation exceeds 100 and therefore are not answers. So 0 and 5 are the only options, obviously the greatest is 5.