# What will be the difference?

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:

$$\frac{a+b+c+d}{4}=35$$

$$a+b+c+d=140\tag1$$ $$b+c=2\cdot34=68 \tag2$$ $$a+b=2\cdot30=60 \tag3$$ $$c+d=2\cdot40=80 \tag4$$ $$a+d=2\cdot36=72 \tag5$$

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.

Thank you.

• Well, you know $30<b<34$ so there are aren't a lot of candidates for $b$...
– lulu
Jan 19, 2018 at 23:12
• You aren't using the fact that these are prime numbers. Jan 19, 2018 at 23:14

As lulu points out, $30 < b < 34$ and $b$ is a prime number, so $b = 31$. Now, $a = 29$, $c = 37$ and $d = 43$, so $d - a = 43 - 29 = 14$.