NET10 Cell Phone Costs?

Several years ago we started using NET10 cell phones. They offer a plan for 10 cents a minute. However, your total minutes are lost if you don't buy more in some period. Two (600 min) and six (1000 min) months plans are available. I use it sporadically and only when I'm away from our small town. I might use 400 minutes in an entire year. However, to keep the service, I always purchasing more than I need. As a consequence, I currently have 1600 min. In one month I will have to buy more. This in my view is a losing proposition.

Is there a plan with another company that's more comfortable for my situation? Comments?

Reply to
W. eWatson
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Don't buy "minute cards"; buy extended activation.

It might be cheaper to change to a Tracfone plan. Tracfone sells by the minute (i.e. a 60 minute card) but also offers 400 minutes valid for a year for $100 (unused minutes carry over with the next year's purchase). That translates to $8/month.

If you have cable TV, you pay a flat fee each month regrdless of actual usage. I just look at the $8/month cost as my "flat fee" for cell phone usage and let the minutes accumulate (currently somewhere above 800 minutes). My cell usage is usually between 20-30 minutes/month - but the extra minutes are a cushion for travelling or an emergency.

You can get one of Tracfone's Motorola flip phones with camera for about $20 at Kmart or Walmart.

John

Reply to
news

Yes, I think you have it right. My wife and I looked if over. We are really paying $180/year, and accumulating minutes far beyond what we might use. It's sort of like who cares. I am heading for a tracphone when I cancel the current phone.

Another thought on this is why not take advantage of the web connection? At the moment, there is no useful description for how to use it. I called NET10, and they offered to lead me through the basic steps. I'd be better off with a booklet that explained it. Then too my wife thinks web connections might be slow. At 25 cents a minute, one can afford to do some experimenting.

Reply to
W. eWatson

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