Front Brake Pads and Rotor Replacement Cost?

[jamesmulligan]

Hi Everyone,

Couple of weeks ago I purchased a 2000 Ford Explorer 210,000km. Has a service done and they tell me the front pads are 85% worn and the discs are worn to minimum thickness.

The mechanics (Kmart Auto) tell me it will cost around AUS $700 to fix (with a bearing replacement also) and will need to leave it for a couple days due to the procedure.

I've had a little look around the forum and can't find anything specific relating to cost and time so thought to open this thread.

Does anyone think this price and time is realistic? seems like a fair dent in the wallet..I've seen a few threads saying it should take 1.5 hours max?

Cheers

 

[Blee1099]

Front pads $30 for metallic about $60 for ceramic based (heavy duty application)
Front rotors $29 each
Hub assembly/wheel bearing $103 with a Timken bearing

Labor is probably a 1.5 hours-2 hours. Honestly, its not hard to do if you feel like trying it yourself. The prices of the parts may be higher since your in Australia. Sites like Rockauto.com do offer shipping to Australia and other countries so it may be worth buying the stuff and doing it yourself.

 

[pfloydmaster]

Couple of weeks ago I purchased a 2000 Ford Explorer 210,000km. Has a service done and they tell me the front pads are 85% worn and the discs are worn to minimum thickness.

Brake pad and Rotor kit from 1AAuto.com (Part # Part Number: 1ABFS00004). $99.05 (USD) for the kit. Not sure about overseas shipping so send them a message.

 

[jamesmulligan]

Thought about having a go myself. Prices in Australia ended up being around AUS$110-120 for front pads and over $120 each for the front rotors.

Found a guy who will do both for $420 incl labour so....based on the US prices you gave, sounds far better to order from over there.

Cheers

 

[SoNic67]

That 11$ brake pads are junk. Also the cheap rotors will most likely warp in no time.
Brakes are one of the thinks you want to go with higer quality components or at least at OE quality level.