Algebra word problem for 5th grade

Question

one lady finger and six cookies cost $3.05. At the same prices, two lady fingers and four cookies cost $2.50. How much does one cookie cost?

Answer 1

Let F be the cost of one lady finger

Let C be the cost of on cookie.

 

so,

 

F + 6C = 3.05          eq. 1

2F + 4C  = 2.50       eq. 2

 

use eq. 1 to solve for F:

 

F = 3.05 - 6C          eq. 3

 

subsitute eq. 3 to eq. 2

 

2(3.05 - 6C) + 4C = 2.50

 

6.10 - 12C + 4C = 2.50

 

6.10 - 8C = 2.50

 

6.10 - 2.50 = 8C

 

3.60 = 8C

 

so C = $.45

 

so, each cookie costs 45¢

 

 

 

Answer 2

 

Hey Darlene!

"cut in half" idea: one LF and two C costs $1.25 ... 4 less cookies saves $1.80 

"halving tool" again => 4C is 1.80 ... 2C is 0.90 ... one cookie is 45 cents ... Regards :)

Answer 3

First, set up an algebraic equation:

lady finger = x

cookie = y

 

x+6y=3.05

2x+4y=2.50

 

Then, solve for one variable...

x=3.05-6y

 

Then, plug in your new equation into the other equation...

 

2(3.05-6y)+4y=2.50

 

6.1-12y+4y=2.5

 

3.6=8y

y= 0.45

 

Since Y is the price of a cookie, the price of a cookie is 45 cents/$0.45.