An n-sided polygon has interior angles of which 4 of them are of sizes 125,220,40 and 295 degrees . The remaining interior angles are of sizes 20 degrees each. Find the value of n,

My working:
let the number of sides for the polygon be n and the remaining number of sides be x.
(n-2)x 180 = 125+220+40+295+20x
180n-360=680+20x
180n=680+20x+360
180n=1040+20x

Good idea, but you don't need x. There are n sides and n angles. Since you already have 4 angles, there are n-4 remaining. So,

(n-2)180 = 125+220+40+295+20(n-4)
n=6

Why is the answer n=6 ?

It's alright. Thank you for your help.

To find the value of n, we need to solve the equation 180n = 1040 + 20x.

To do this, we need to find the value of x. We know that the sum of the remaining interior angles is 20 degrees each, and there are n - 4 remaining angles. Therefore, the sum of the remaining angles is (n - 4) * 20.

Substituting this into the equation, we have:
180n = 1040 + 20(n - 4)

Simplifying the equation, we get:
180n = 1040 + 20n - 80
160n = 960
n = 6

Therefore, the polygon has 6 sides.