Question

Anyone like super sweet word problems? Just need to know how to set this up...
A contractor has a 24pound mixture that one-fourth cement and three-fourths sand. How much of a mixture that is half cement and half sand needs to be added to produce a mixture that is one-third cement?

Answer 1

We have 24 pounds of mix. One-fourth of that is cement. So (1/4)*24 = 6 pounds of this mix is pure cement


So we have the fraction

6/24


We want to add some unknown amount of pure cement to the numerator 6 (which we'll call 'x') and add some amount of mixed cement and sand (half and half). So we're adding 2x to the denominator (note: the amounts of cement and sand we're adding is x, so we're adding a total of x+x=2x pounds of mix)


Therefore, after doing this, we get

6 + x
--------
24 + 2x

This is the ratio of cement to the mix. We want this ratio to be 1/3, so set it equal to 1/3 to get


6 + x 1
-------- = -----
24 + 2x 3


So your goal is to solve that equation for x to find the amount of cement you need to add. Then double it to find the total amount of mix to add.

Answer 2

Let me know what you get for your answer

Answer 3

would the common denominator be 3(24+2x)?

Answer 4

yes, you can also just cross multiply

Answer 5

to get

3(6+x) = 1(24+2x)

Answer 6

x=6

Answer 7

good, so you need to add 6 pounds of cement or 2*6 = 12 pounds of total mix

Answer 8

So your final answer is 12 pounds

Answer 9

it seems im supposed to apply this to the quadratic equation somehow...

Answer 10

what do you mean

Answer 11

nevermind, this stuff is like japanese to me

Answer 12

why do you think we need to use the quadratic equation?

Answer 13

did it say you had to?

Answer 14

no but there is a solution manual in the back of the book for all odd problems, the one we were working on is an even but the answers to the questions before and after it had 2 different answers

Answer 15

can you give me a sample problem of where it had two solutions? I'm curious now

Answer 16

one of the odd problems you see

Answer 17

haha im so embarassed, i was looking at the wrong solutions!

Answer 18

oh lol that's alright, you had me going for a sec though

Answer 19

hopefully the explanation I wrote above is valid (I think it is anyway) and it makes sense to you

Answer 20

thanks man, any chance you could help with one more?

Answer 21

sure, go ahead and ask

Answer 22

3x^-2+2x^-1-8=0

Answer 23

\[3x ^{-2}+2x ^{-1}-8=0\]

Answer 24

this looks better, haha

Answer 25

3x^(-2)+2x^(-1)-8=0


3/(x^2)+2/(x^1)-8=0


3/(x^2)+2/(x)-8=0


Now multiply EVERY term by the LCD to clear out the fractions


x^2*( 3/(x^2) )+x^2*( 2/(x) )-8x^2=0*x^2


3 + 2x - 8x^2 = 0


-8x^2 + 2x + 3 = 0


Now use the quadratic formula to solve for x:


x = (-b+-sqrt(b^2-4ac))/(2a)


x = (-(2)+-sqrt((2)^2-4(-8)(3)))/(2(-8))


x = (-2+-sqrt(4-(-96)))/(-16)


x = (-2+-sqrt(100))/(-16)


x = (-2+sqrt(100))/(-16) or x = (-2-sqrt(100))/(-16)


x = (-2+10)/(-16) or x = (-2-10)/(-16)


x = 8/(-16) or x = -12/(-16)


x = -1/2 or x = 3/4


So the solutions are x = -1/2 or x = 3/4

Answer 26

ok, that one i understand a lot better

Answer 27

i appreciate it

Answer 28

yw