7th grade math help !!!?

Question

thisismyquestion: Marty bought a phonecard for 25$. within the card it costs 10camin to call within his areacode & 20camin to call outside his areacode.He plans to make 1hr of calls within the area code & 2hr outside the area code.Marty has already made 1hr of calls within the areacode. 1. determine how many min of calls outside the areacode he will be able to make with his card. 2. What is the least he could spend for a phonecard that would allow him to make the number of calls he wants to make both inside and outside the area code?
3. Explain how you determined each of the answer. Make sure use mathematical reasons and examples from the problem in your explanation.

Answer 1

1. If it costs 10 c/min to call within the area code, and he made a 1 hr call in the area code, it cost (60 min)*(10 c/min) = 600 c = $6. That means he has $25 - $6 = $19 left on the card. Long distance calls cost 20 c/min. With $19, he can call long distance for (1900 c) / (20 c/min) = 95 min. That's 1 hr, 35 min.

2. The total cost of the calls would be (60 min)*(10 c/min) + (120 min)*(20 c/min) = 600 c + 2400 c = 3000 c = $30.

Answer 2

card = $25.00
in area code = $.10/min
out area code = $0.20/min
# mins in area code = 60
# mins out area code = 120
# mins in area code made = 60

# 1
He made 60 mins of calls at 0.10/min = $6.00 so he has $19.00 left. At 0.20/min he can call 5 mins for $1.00. For $19 he can call 5 mins * 19 = 95 minutes.

# 2
He wants to call 60 mins in the area code and 120 mins outside. At 0.10/min 60 mins = $6.00. At 0.20/min 120 mins = $24.00. To make those calls he would need a $30.00 card

# 3
see above

Answer 3

1. 1 hr. and 35 minutes or 95 minutes
2. $ 30 card.
Try to understand it yourself.

Answer 4

Just let that person run up their minutes.