maths question =]?

Question

ace car rentals advertises that a rental car costs $25 per day plus the charge of $0.10 per km . for the same car, Better car rental advertises a price of $40 per day plus $0.05 per km. for what number of kilometers is the cost of renting a car the same for both companies?

add how you did it

Answer 1

25 + .10x = 40 + .05x
-25 -.05x -25 - .05x

.05x/.05=15/.05
x= 300

my bad..i didnt check it..lol

Answer 2

Now calculate like this

Let x be the distance covered.
for x kilo meters ACE car rentals cost $(25+0.10x)
for the same distance BETTER car rentals cost $(40+0.05x)

Now for the same cost equate both
25 + 0.10x = 40 + 0.05x

This gives the value of x as 300

Therefore 300 Kilometers is the cost of renting a car the same for both companies.

Answer 3

The question can be modeled by the equation

25+0.1k=40+0.05k

(where k is the number of kilometers)

rearranging:
0.1k=15+0.05k

0.05k=15

k=15/0.05=300

therefore the cost of renting a car is the same at 300km.

Answer 4

Use simuataneous equations.

Let A be cost with Ace.
Let B be cost with Better.
Let d be distance travelled.

Then A=25+0.1d, B=40+.05d

Find when A=B.

Answer 5

$25 + .10X = $40 + .05X
or .05X = $15
or X = $15/.05
or X = 300 Kms

Answer 6

let X be the number of kilometers.
=> 25 + 0.10X = 40 + 0.05X
0.10X - 0.05X = 40 - 25

so, 0.05X = 15
and X= 15/0.05 = 300
its 300 km

Answer 7

Assuming that you only rent the car for one day then 300km.

The differential day rate is $15. The differential travel rate is $0.05.

15/0.05=300

Answer 8

25+.1x=40+.05x
25-25+.1x=40-25+.05x
.1x=15+.05x
.1-.05=15+.05x-.05x
.05x=15

.05x/.05=15/.05=300

so plug back in...

ace..

25+.1*300=
.1*300=30
25+30=55

better...

40+.05*300=
.05*300=15
15+40=55

55=55

check ya later ♥

Answer 9

15.7894736842