Posted by Nick on Thursday, October 9, 2008 at 10:51pm.
Your problem would be clearer if you placed periods at the end of each sentence, started each sentence with a capital letter, and added commas where needed. We could live without the commas but not the other two.
you are buying beads and string to make a necklace. The string coast $1.50; A pack of 10 decorative beads costs $0.50; And a pack of 25 plain beads costs $0.75. You can only spend $7.00 and you need 150 beads. You wish the necklace to be as decrative as possiable. How many packs of each type of bead should you buy?
Here is what I would do.
We need the string regardless. Therefore, 7.00 - 1.50 for the string leaves 5.50 for the two kinds of beads.
Let d = decorative beads. They cost 0.50 for a pack of 10 which makes them cost 0.05 (a nickel) each.
Let p = plain beads. They cost 0.75 for a pack of 25 or 0.03 (3 cents) each. Then set up two equations for the two unknowns.
=================
total beads = 150 so
d + p = 150
cost of beads = 5.50 so
0.05d + 0.03p = 5.50
Solve the two equations for d and p. Then divide them up into packets; i.e., total d beads divided by 10 = # packets and total p beads divided by 25 = # packets. Calculate the price of the packets to make sure the total is 5.50. Then, to make sure this is the best that can be done, take 1 pack of plain beads and replace with pack(s) of d beads and see if you can end up with 150 beads at a cost of 5.50.
WHAT IS
X2+289
simplify. Assume that no variable equals 0.
30y4power over -5y2power
if you buy 3 sets of the decorative beads at .50 a piece and 5 sets of the plain beads at .75 a piece plus the string 1.50 that would give you a grand total of 6.75 and with a quarter left over and 5 beads that is more than what you need.
5 packs of decorative beads,
4 packs of plain beads