Best way to solve this math problem?

Factual Questions
1

I was comparing the rates on a couple of pay-as-you-go cell phone plans and wanted to figure out how many minutes would have to be used before one plan would be the better value.

Basically, “Plan A” is .25 for the first 5 minutes, and .10 each minute thereafter. “Plan B” is just .15 a minute. I figured out that Plan A starts to cost less after the fifteenth minute, but I had to compare to columns to get the answer.

I’m guessing that the difference between the prices at minute 5, which is .50 has something to do with it. At minute 5, Plan A is advancing .10/min and B is at .15/min, at the .05 difference it would take 10 more minutes to account for the .50 difference, hence minute 15 is when they become equal.

What type of problem is this, and how would a math bad-ass solve it?

2

First note that Plan B is cheaper for the first 5 minutes.

Then A = .255 + .10t

B = .155 + .15t

Set A = B and solve for t. The answer is that B is cheaper if calls are less than 10 minutes. A is cheaper for calls more than ten minutes. And they are equal for calls of 10 minutes.

3

After 10 minutes, Plan A costs1.75, and B costs 1.50. B is cheaper untill 15 minutes when they are equal, after that A is cheaper.

4

Maybe it should be solved for t plus 5?

5

The first five minutes is part of the total time and has to expressed as such The correct way would be:

Plan A --> 5*.25 + (t-5)*.1
Plan B --> 0.15 t

To reduce this problem we can say that we want to find the time(t) at which both plans are equal (as we know Plan A will increase at a slower rate after this).

To solve for t:

5*.25 +(t-5)*.1 = 0.15t
1.25 + .1t - .5 = 0.15t
1.25 - .5 = 0.15t - 0.1t
0.75 = 0.05t
0.75/0.05 = t
t = 15

So both plans are equal at 15 minutes and Plan A will obviously be cheaper for each additional minute ($0.10 per minute instead of $0.15/minute).

6

I sure can’t argue with your arithmetic. My equations must be faulty. :smack:

7

Bah! My equations were OK but the time I computed was the time after the first 5 minutes and so my answer is also 15 minutes. :smack: :smack:

8

Also - :o :o :o

9

It’s one of those word problems that you solve with elementary algebra.

10

A “word problem” in math?!?! What’ll they think of next? :dubious:

11

I would call the problem an exercise in “breakeven analysis”. You are trying
to find the point the two plans are of equal cost, below that point of usage
one plan is cheaper and above that point the other.

12

I’m tempted to give this very problem to one of my algebra classes, and say, “See? Algebra can be used to solve problems that somebody would actually care about!”

Shall I credit you, dnooman?

13

Problems like this are standard fare in most current elementary algebra books with an applications bent.

14

Yes; this very problem could easily appear in a math textbook. But if that was where you encountered it, it would be a hypothetical question, not a real situation that a real person wanted to know about.

15

Here’s a slightly different approach for t > 5 minutes:

The cost of a call on plan A equals Mt, where M is the mean price and t is the length of the call in minutes.

M = % of time spent at 5 minutes * .25, % of time spent > 5 minutes * .1
= (5/t).25 + ((t-5)/t).1

The cost of a call on plan Plan B is simply .15t.

We are looking for the situation when Plan A costs less:

Plan A < Plan B

((5/t).25 + ((t-5)/t).1)t < .15t

1.25/t + .1 - .5/t < .15

(1.25 - .5)/t < .15 - .1

0.75 < .05t

15 < t

As you already figured out!

16

Another slightly more intuitive approach would be to graph the two equations and see where they cross. This works well for slightly more complicated formulas like linear vs quadratic costs where it’s possible to have 2 or more crossing points.

17

Sure. For added reality, Plan A is Virgin mobile’s everyday rate, and Plan B is Boost mobile’s “discounted” weekend rate. The kids might be more interested since they may also be considering either plan.