PDA

View Full Version : help with word problems. systems and inequalities

fiftrombone
07-14-2006, 03:47 AM
hey, im tutoring a lady and im a little rusty on some math. i neeed help with these word problems. if you could just find the equations i need i would be sooo happy. and i need them asap.

1. Ziggys famous yogurt blends regular yogurt that is 3% fat with no fat yougurt to obtain lowfat 1% yogurt. how many pounds of regular yogurt and how many punds of no fat yogurt must be mixed to obtain 60 pounds of 1% yogurt?

2. professor davis counts his midterm as 2/3 of the grade and the final as 1/3 of the grade. jason scored only 64 on the midterm, what range of scores on the final exam would put jason's average between 70 and 79 inconclusive?

3. a furniture maker has a shop that employs 12 workers for for 40 hours a week. they make tables and chairs. if it takes 16 hours to make a table and 8 hours to make a chiar graph the region that shows the possibilities for the number of chiars and tables that could be made in one week.

tkhunny
07-14-2006, 07:26 AM
1. Ziggys famous yogurt blends regular yogurt that is 3% fat with no fat yougurt to obtain lowfat 1% yogurt. how many pounds of regular yogurt and how many punds of no fat yogurt must be mixed to obtain 60 pounds of 1% yogurt?1) Name Stuff
2) Equate Stuff
3) Solve Stuff

1) Write clear, concise, and complete definitions

"how many pounds of regular yogurt"

R = pounds of regular (3%) yogurt

"and how many punds of no fat yogurt"

F = pounds of no fat (0%) yogurt

2) The problem statement should suggest important relationships.

"60 pounds"

R + F = 60 pounds

"1% yogurt"

Equating Milkfat
R*0.03 + F*0.00 = (60 lbs)*0.01

3) Simplify, Substitute, Whatever it takes to solve it.

R*0.03 + F*0.00 = (60 lbs)*0.01
R*0.03 + 0 lbs = (60 lbs)*0.01
R*0.03 = (60 lbs)*0.01 -- Isn't that funny. 'F' disappeared.
R = (60 lbs)*(0.01/0.03) = 20 lbs

Please notice that equating Milkfat was the easier way to tackle this one. It was not the only way. We could have equated NONMilkfat.

R*0.97 + F*1.00 = (60 lbs)*0.99

This leads to the same solution.

Substitution

R*0.97 + (60 lbs - R)*1.00 = (60 lbs)*0.99
R*0.97 + 60 lbs - R = (60 lbs)*0.99
R*0.97 - R = (60 lbs)*0.99 - 60 lbs
-0.03R = -0.01(60 lbs) -- And that looks awfully familiar.

I mention this, not just to confuse, but to suggest there are ALWAYS two ways to do these sorts of problems. Pick the easier. Whe the addition is 100% or 0%, there is always a very easy way.

fiftrombone
07-14-2006, 12:31 PM
thank you very much.

tkhunny
07-14-2006, 01:33 PM
Ah, lovely, the no-feedback-at-all response. You can help me feel a lot better if you will now work the other two and show us what you get.