A long distance phone company has a monthly fee of $7.95 and charges a rate of $0.05 per minute. Another long distance company has a monthly fee of $9.95 and charges a rate of $0.03 per minute. At how many minutes would the two companies have equal charges?
a) 10 minutes
b) 50 minutes
c) 32 minutes
d) 100 minutes
2007-08-14 02:48:16 · 6 answers · asked by song_fiend 2 in Science & Mathematics ➔ Mathematics
Let C = total charges and m = charge per minute
C = 7.95 +0.05m (for first company)
C = 9.95 + 0.03m (for second company)
Solve this system to find at how many minutes would the two companies have equal charges.
7.95 +0.05m = 9.95 + 0.03m
Subtract 0.03m and 7.95 from each side
0.02m = 2
Divide both sides by 0.02
m = 100
Therefore the two companies would have equal charges at 100 minutes.
2007-08-14 03:11:54 · answer #1 · answered by sigmazee196 2 · 0⤊ 0⤋
Cost equations for the two company would be
C1=7.95+0.05x
C2=9.95+0.03x
where x is the number of minutes
for equal charges C1=C2
7.95+0.05x=9.95+0/03x
0.05x-0.03x=9.95-7.95
0.02x=2
x=2/0.02
=100
so after 100 minutes the two companaies will have the same charge
2007-08-14 10:01:29 · answer #2 · answered by shubham_nath 3 · 0⤊ 0⤋
Difference in plan base rates is 9.95 - 7.95 = 2.00
Difference in plan minute rates is 0.05 - 0.03 = 0.02
So your equation is 0.02x = 2.00
Multiply by 100 to get 2x = 200
Divide by 2 to get x = 100
So answer d) is correct.
2007-08-14 10:02:46 · answer #3 · answered by Don E Knows 6 · 0⤊ 0⤋
let m be the number of minuts:
Company 1 charges 7.95 + 0.05m
Company 2 charges: 9.95 + 0.03m
When do both have equal charges?
7.95 + 0.05m = 9.95 + 0.03m // - 7.95
0.05m = 2 + 0.03m // - 0.03m
0.02m = 2 //*50
m=100
Choose answer d.
2007-08-14 09:58:29 · answer #4 · answered by Amit Y 5 · 0⤊ 0⤋
d) 100 minutes
2007-08-14 10:01:32 · answer #5 · answered by cl3v3r boy 3 · 0⤊ 0⤋
7.95 + .05t = 9.95 + .03t Solve for t.
2007-08-14 09:57:02 · answer #6 · answered by Anonymous · 0⤊ 0⤋