A problem on chinese remainder theorem (CSIR NET DEC 2015) 
Which of the following intervals contains an integer satisfying following three congruences
  $$x=2\pmod5\\
x=3\pmod7\\
x=4\pmod{11}$$
  $a) [401,600] \\ b)[601, 800] \\ c)[801,1000] \\ d)[1001,1200]$

(CSIR NET 2015 Dec)
I tried this question and I got answer but it is not in the option.
I applied Chinese remainder theorem.
$$x=2\pmod5\\
x=3\pmod7\\
x=4\pmod {11}$$
$$N_1=7\times11=77\\
N_2=5\times11=55\\
N_3=7\times5=35$$
$77x=1\pmod5\implies b_1=3\\
55x=1\pmod7\implies b_2=6\\
35x=1\pmod {11} \implies b_3=6$
then,
$x=2\times77\times3+3\times55\times6+6\times35\times4=2292$
This answer is not in the option.
If my work is wrong please correct it.
 A: You know that the answer you get applying the CRT is not a unique integer, right?
It is only unique modulo $5\cdot 7\cdot 11$.
In particular, $752$ and $1137$ are solutions.
I'll leave you with the tasks of deciding if this is complete, and how you will use it to answer your question.
A: Here $~5,~7,~11~$ are pairwise prime to each other.
Let $~m=5\times 7\times 11=385.~$
Let $~M_1=\frac m5=77,~~M_2=\frac m7=55,~~M_3=\frac m{11}=35.~$
Then $~~\gcd(M_1,5)=1,~~\gcd(M_2,7)=1,~~\gcd(M_3,11)=1~.$
Now since $~\gcd(M_1,5)=1,~$ the linear congruence $77x\equiv 1~\pmod5~$ has a unique solution and the solution is $~x\equiv 3~\pmod5~.$
Again since $~\gcd(M_2,7)=1,~$ the linear congruence $55x\equiv 1~\pmod7~$ has a unique solution and the solution is $~x\equiv 6~\pmod7~.$
Also since $~\gcd(M_3,11)=1,~$ the linear congruence $35x\equiv 1~\pmod{11}~$ has a unique solution and the solution is $~x\equiv 6~\pmod{11}~.$
Therefore $~x_0=2\cdot(77\cdot3)+3\cdot(55\cdot6)+4\cdot(35\cdot6)=2292~$ is a solution and the solution is unique modulo $~385~.$
Hence the solution of the given system is $~x\equiv 2292~\pmod{385}\equiv 367~\pmod{385}~.$
i.e., the solutions are $$367+385k~,~~~~\text{where}~k=0,~1,~2,\cdots$$
$$=367,~752,~1137,~1422,~\cdots~$$
Answer of the given question:
The intervals contains an integer satisfying the given three congruences are
$b)[601, 800]~~~~\text{and}~~~~ d)[1001,1200]$
