My textbook gave me a different answer and I don't know what I did wrong.

Question

Q. Your tires are adjusted to 227.5kPa at 10 degrees Celsius in the mechanic's garage. You then take your car home and park it outside. The overnight temperature drops to -5 degree celsius. Determine the new tire pressure?

I did Gay-Lussac Law and I got 215.45kPa. My texbook answer is 2.2 x 10*2kPa

Answer 1

2.2*10^2 = 220 close enough
but anyway
P V = n RT
V, n and R are the same before and after so
agree P2 = P1 (T2/T1)
T1 = 10 C = 293 K
T2 = -5 C = 268 K
so
P2 = (268/293)* 227.5 k pascals
= 223.3 k Pa which rounds to 2.2*10^2