algrebra word problem help?
you are choosing between two long distant telephone plans. Plan A has a monthly fee of 20$ with a charge of $0.05 per minute for all long distant calls. Plan B has a monthly fee of $5 with charge of $0.10 per minute for all long distant calls
A. for how many minutes of long distance calls will the costs for the two plans be the same ? what will be the cost for each plan?
B. if you make approximately 10 long distances calls per month, each averaging 20 minutes, which plan should you select explain the answer.
how do i set up this problems i am having trouble solving this type of word problems the answers is plan A= (300min $35) and plan B is it will vary
Comments
A)
20 + .05x = 5 + .10x
Solve for x
A = 20 + .05x
B = 5 + .10x
See which one is less.
You let x = the number of minutes and determine an equation for each scenario (Plan A &
Then, since you want to know for how many minutes the cost of the plans will be equal, you set the two equations up so that they are equal and solve. Like this:
Plan A = $20 + .05 (x)
Plan B = $5 + .10(x)
20 + .05x = 5 + .10x (Solve for x)
-5 -5
_____________
15 + .05x = .10x
- .05x - .05x
____________
15/.05 = .05x/.05
300 = x
When 300 minutes of long distance calls are made the cost of the two plans will be equal. The bill will be $35 for 300 minutes in either plan.
CHECK: $20 + .05(300) = $5 + .10(300)
20 + 15 + 5 + 30
35 = 35
If 10 calls lasting 20 minutes each are made in one month the total minutes used that month are 20 x 10 = 200
Plan A: $20 + .05(200) =
20 + 10 = $30
Plan B: $5 + .10(200) =
5 + 20 = $25
If only 200 minutes are used per month, Plan B is cheaper by $5.
P.S. When setting up your equations think about what is known, or what is "fixed", in this case the monthly fees are fixed but the cost per minute varies - that's why you use the number of minutes as your unknown (x).
You're taking one dollar amount and adding another dollar amount that will fluctuate depending on the variable. You'll see a lot of problems like this one. GOOD LUCK!!
plan A=20+.05m
plan B=5+.10m
for the plans to be equal:
plan A=plan B means 20+.05m=5+.10m
which is 15=.05m
so m=15/.05=1500/5=300
so at 300 min the plans are the same.
if you make 10 LD calls per month, at 20 minutes apiece, thats 200 minutes.
plan A at 200 LD minutes is 20+.05(200)=20+10=30 dollars
plan B at 200 LD minutes is 5+.10(200)=5+20=25 dollars
so plan B would benefit you.
i'm uncertain that your first analogy is right yet 8+8=2(8) are the two equivalent to sixteen yet would not clarify the great project. you should endure in innovations that one key to fixing the project is what a while will the the two be in 3 yrs. With numbers it isn't any longer undemanding. Barry will constantly be 8 yrs. older than his sister yet I he won't constantly be two times as previous. you go with an knowledgeable wager. i presumed that Barry could be 13 yrs. previous now & this might make his sister 5 yrs. previous & this works on account that 13-5=8 & that's what share yrs. older Barry is then his sister. So now in the event that they'e 5 & 13 yrs. previous, in 3 yrs., Barry would be sixteen & sister would be 8. !6 is two times 8 so because it particularly is terrific suited after 3 yrs. you should endure in innovations that their difference in age is 8 yrs. & a splash to this project is that 2x8=sixteen. This 2x8=sixteen is the considerable to answering the two issues. merely endure in innovations that their difference in age is 8 yrs. & on account that he would be two times her age in 3 yrs., then merely 2x8 & it particularly is the way you positioned up the respond. You have been close yet this 2x8 anwers the two issues. they're 8 yrs. aside in age so which you merely double 8 & it particularly is sixteen so she would be 8 yrs. previous that's a million/2 his age besides as 8 yrs. aside in age on an identical time. Now on account that this 2x8 is in 3 yrs. & this makes Barry sixteen & his sister 8, merely subtract 3 from the two sixteen & 8 this offers you with 13 &5 which provides you with their a while now. to examine your artwork, 13 +3=sixteen & 5+3=8. the two issues can merely be anwered via doubling the version in age that's 8. 2x8=sixteen so then this sixteen must be two times as much as something else. the project then must be set up wisely as 2x=sixteen now 2 cases her age of 8 provides you along with his age of sixteen. x then is the version in age & you merely double that. So very solid for you on account which you have been getting in the terrific suited course. solid success!!