$\{x\}$ is the fractional part of $x$ 
Let $\{x\}\;$ denote the fractional (non-integer) part of $x,\;$ so as examples: $\{3\}= 0,\;$ $\{0.5\}=0.5\;$ and $\{\pi\} = \pi-3.\quad$ IF $\{n\} + \{2n\} = 1.2,\;$ find all possible values of $300\{n\}.$

OKay, so I separated this into cases where $\{n\}< \frac12\quad$ and $\{n\}>\frac12.\quad$ If $\{n\}<\frac12,\quad$ then $\{2n\}\;$ will still be less than 1, so I can just do $\frac{1.2}{3} = 0.4 = \{n\}.\quad$ So $300\{n\} = 120.\quad$ However, I don't know how to approach the second case where $\{n\}>\frac12.$
 A: Write $n=a+b$, where $a$ is integer and $0\le b=\{n\}<1$. You correctly observed that if $b<1/2$, then $2n=2a+2b$ with integer $2a$ and $0\le 2b<1$; therefore $\{2n\}=2b$. Hence
$$
\{n\}+\{2n\}=3b
$$
and $3b=6/5$ implies $300\{n\}=300b=120$.
If $b=1/2$, we have $2n=2a+2b=2a+1$ and so $\{2n\}=0$. So $\{n\}=6/5$, a contradiction.
If $1/2<b<1$, then $2n=2a+2b$ and we know that $1<2b<2$, so $0<2b-1<1$; therefore $2n=(2a+1)+(2b-1)$ and $\{2n\}=2b-1$. Therefore
$$
b+2b-1=\frac{6}{5}
$$
A: 
We can write,$$\{n\}+\{2n\}=1.2$$$$n-[n]+2n-[2n]=1.2$$$$3n-[n]-[2n]=1.2$$$$[n]+[2n]=3n-1.2$$$$[n]+[2n]=3(n-0.4)$$$$[n]+[2n]=3([n]+\{n\}-0.4)$$$$[n]+[2n]=3[n]+3(\{n\}-0.4)$$$$[2n]-2[n]=3(\{n\}-0.4)$$$$2[n]+[2\{n\}]-2[n]=3(\{n\}-0.4)$$$$[2\{n\}]=3(\{n\}-0.4)$$$$3(\{n\}-0.4)=[2\{n\}]$$

$$\bbox[5px,border:2px solid red]{\{n\}=\frac{[2\{n\}]}{3}+0.4}$$

For $2\gt 2\{n\}\gt 1$

$$\{n\}=\frac{1}{3}+0.4$$
$$\{n\}=\frac{11}{15}$$
$$\bbox[yellow,5px]{300\{n\}=220}$$

For $2\{n\}\lt 1$,

$$2\{n\}\lt 1$$
$$\{n\}=\frac{0}{3}+0.4$$
$$\{n\}=0.4$$
Hence,
$$\bbox[yellow,5px]{300\{n\}=120}$$
A: When $n\in (0.5,1) \implies 2n\in (1,2) $ hence $\{ 2n\}$ belongs to $ (0,1) $.  Now since $\{n\}>0.5$ First try to reduce the interval of $n$ . Try with $0.6,0.65,0.7,0.75,0.8$ from this we see $n\in (0.7,0.75)$  hence $2n\in (1.4.1.5) $ .Coming back to the equation and using $x=[x]+\{x\}$ we have $n-[n]+2n-[2n]=1.2$ . As $\{x\}$  is an increasing function between two integers so we have only one solution. Therefore $n-0+2n-1=1.2 \implies n=0.7333.. $ hence $300\{n\}\approx 220$.
A: If $0\le\{n\}<\frac12$, then $\{2n\}=2\{n\}$ so that $300\{n\}=300\cdot\frac{1.2}3=120.$
If $\frac12\le\{n\}<1$, then $\{2n\}=2\{n\}-1$ so that $300\{n\}=300\cdot\frac{2.2}3=220.$
