The average of 4 distinct prime numbers $a$, $b$, $c$, $d$ is $35$ where $a < b < c < d$. Given that $b$ and $c$ are equidistant from $34$ and; $a$ and $b$ are equidistant from $30$ and; $c$ and $d$ are equidistant from $40$; $a$ and $d$ are equidistant from $36$ . The difference between $a$ and $d$ is
a) $30$ b) $14$ c) $21$ d) can't be determined.
My trial: from given conditions:
$$a+b+c+d=140\tag1$$ $$b+c=2\cdot34=68 \tag2 $$ $$a+b=2\cdot30=60 \tag3 $$ $$c+d=2\cdot40=80 \tag4 $$ $$a+d=2\cdot36=72 \tag5 $$
Solving above equations I couldn't get the value of $d-a$. Can somebody please help me solve this problem? My book says answer says answer is b) $14$ but I didn't get it.