Kids on a $45$-seat bus (I'm only a Year $7$ so please explain clearly how you found the solution.)
A bus has $45$ seats. Each seat can fit two children or one child with his backpack.
If $2/3$ of the children have  backpacks, how many children can be seated on the bus?
I started by saying that $45 \times 2= 90$ if no one was wearing a backpack. But since $2/3$ of the kids are, $45/3 \times 2= 30$ children would need their own seat and $90-30= 60$. So I thought $60$ children would fit in the bus. Apparently the answer was $54$, but why?
P.S. I don't know Markdown, so somebody please edit this question
 A: If you know basic equations, here's my explanation:
Let $N$ be the number of children we want to know. 
Then we know, that 2/3 of then (i.e. 2N/3) need their own seat. The rest of them (N/3) needs only half a seat (since two children fit on one seat). So, we have
$$45=\frac{2}{3}N\cdot 1+\frac{1}{3}N\cdot \frac{1}{2} $$
If we summarize, we get
$$45=\frac{5}{6}N$$
And hence $N=\frac{6}{5}\cdot 45=54$.
A: Subtracting 30 from 90 means that you're removing some children from the bus -- but those you are kicking out all have backpacks. Therefore, among those that are left on the bus, those with backpacks are more than two thirds.
In other words, you now have a bus with 30 kids without backpacks and 30 kids with -- and those 60 kids do excatly fill out all the seats, but now only half of them have backpacks.
A better way to proceed would be to note that every group of three children (two with backpacks, one without) will need $2\frac12$ seats, which is $5$ half-seats. There are $90$ half-seats in the bus, so there must be $\frac{90}{5}=18$ groups of $3$ children, for a total of $18\times 3=54$ children.
