4. A university bookstore recently sold a wire bound graph-paper notebook for $2.50 and a college-ruled notebook for $2.30. At the start of the spring semester, a combination of 50 of these notebooks was sold for a total of $118.60. How many of each type were sold?
2007-10-17 15:15:17 · 4 answers · asked by Hope C 1 in Science & Mathematics ➔ Mathematics
[14]
Let x no of graph=paper notebook and 50-x no of ruled notebook were sold
According to the problem,
2.50x+2.30(50-x)=118.60
2.50x+115-2.30x=118.60
0.20x=118.60-115=3.60
x=3.60/0.20=18
Therefore 18 nos of first type and (50-18) or 32 nos of second type notebooks were sold
2007-10-17 15:24:11 · answer #1 · answered by alpha 7 · 0⤊ 0⤋
g = number of notebooks of graph paper
c = number of college notebooks
$2.50g + $2.30c = $118.60
g + c = 50 total notebooks
g = 50 - c (just turn the equation around)
Now substitute for g in the first equation
2.50(50-c) + 2.3c = 118.6
125 - 2.5c + 2.3c = 118.6
125 - 118.6 = .2c
6.4 = .2c
32 = c
therefore 50-32 = 18 = g
2007-10-17 22:26:50 · answer #2 · answered by Steve A 7 · 0⤊ 0⤋
let x= graph paper notebook
y= college ruled notebook
x + y= 50
x= 50- y
2.5x + 2.3y= 118.60
2.5(50-y) + 2.3y= 118.60
125 -2.5y+2.3y= 118.60
-0.2y= -7
y= 35
50-35= x
x= 15
2007-10-17 22:22:42 · answer #3 · answered by bob b 2 · 0⤊ 1⤋
2.50x + 2.30y = 118.60
x + y = 50
There's your formulas, think you can figure it out from there?
2007-10-17 22:19:21 · answer #4 · answered by Anonymous · 0⤊ 0⤋