# concentration problem

• Oct 3rd 2008, 07:28 AM
Amelia
concentration problem
Freezing weather is coming!
is full with 18 quarts of fluid that is 15% antifreeze.
Question: How much must I drain out into a disposable container,
Then retighten drain plug, and pour back 100% pure
antifreeze into radiator to make 18 quarts of 40%
antifreeze.
• Oct 3rd 2008, 08:04 AM
galactus
Let x=amount to be drained.

Original mixture less amount drained: (18-x)

Amount of pure antifreeze=.15(18-x)

plus

pure antifreeze is x quarts.

The new mixture is 40%= .4(18)=7.2

So, we have the equation to solve:

.15(18-x)+x=7.2
• Oct 3rd 2008, 08:37 AM
Amelia
i think i get it... so would the answer be -5.3 qts... that doesnt make sense
• Oct 3rd 2008, 08:52 AM
janvdl
Quote:

Originally Posted by Amelia
i think i get it... so would the answer be -5.3 qts... that doesnt make sense

Check again. I get +5.29
• Oct 3rd 2008, 09:00 AM
Amelia
thank you so much
is there ne way i could help on one more problem... it has to do with quarters, nickles, and dimes for some reason i cant get it:

I have a collection of nickels, dimes, and quarters
in a money bag worth \$7.95. If the number of
quarters is one more than the number of nickels
and the number of dimes is five more than the number
of nickels, find the number of each type of coin.
# Nickels ________
# Dimes_________¬
# Quarters________
• Oct 3rd 2008, 09:05 AM
janvdl
Quote:

Originally Posted by Amelia
is there ne way i could help on one more problem... it has to do with quarters, nickles, and dimes for some reason i cant get it:

I have a collection of nickels, dimes, and quarters
in a money bag worth \$7.95. If the number of
quarters is one more than the number of nickels
and the number of dimes is five more than the number
of nickels, find the number of each type of coin.
# Nickels ________
# Dimes_________¬
# Quarters________

Okay we have 3 different types of coins.

So:

x + y + z = 7,95

Nickels = x
Quarters = y = x + 1
Dimes = z = x + 5

So you'll just need to substitute the values of these coins. And solve.

a(x) + b(x + 1) + c(x + 5) = 7,95

Where a;b;c are the monetary values of nickels, dimes, and quarters (not respectively!). I don't know the American system.
• Oct 3rd 2008, 09:28 AM
Amelia
would that make x 8.4 ? im confused :( a little: in america dime=.10 nickels=.5 and quarters=.25

i'm confused
• Oct 3rd 2008, 09:36 AM
janvdl
Quote:

Originally Posted by Amelia
would that make x 8.4 ? im confused :( a little: in america dime=.10 nickels=.5 and quarters=.25

i'm confused

Can you double check the question?
• Oct 3rd 2008, 09:44 AM
Amelia
yea :
I have a collection of nickels, dimes, and quarters
in a money bag worth \$7.95. If the number of
quarters is one more than the number of nickels
and the number of dimes is five more than the number
of nickels, find the number of each type of coin.
# Nickels ________
# Dimes_________¬
# Quarters________

nickels are worth .5
dimes are worth .10
quarters are worth .25
a dollare is 1.00
• Oct 3rd 2008, 09:47 AM
janvdl
Quote:

Originally Posted by Amelia
yea :
I have a collection of nickels, dimes, and quarters
in a money bag worth \$7.95. If the number of
quarters is one more than the number of nickels
and the number of dimes is five more than the number
of nickels, find the number of each type of coin.
# Nickels ________
# Dimes_________¬
# Quarters________

nickels are worth .5
dimes are worth .10
quarters are worth .25
a dollare is 1.00

Either the question has a fault, or I made a very stupid mistake in my calculations. The latter being very likely.

You have the idea though, that should carry you through.
• Oct 3rd 2008, 10:01 AM
earboth
Quote:

Originally Posted by Amelia
is there ne way i could help on one more problem... it has to do with quarters, nickles, and dimes for some reason i cant get it:

I have a collection of nickels, dimes, and quarters
in a money bag worth \$7.95. If the number of
quarters is one more than the number of nickels
and the number of dimes is five more than the number
of nickels, find the number of each type of coin.
# Nickels ________
# Dimes_________¬
# Quarters________

Probably this comes too late but I'll post it nevertheless:

Let x denote the number of nickel coins, then the value of these coins is 0.05x
Let x+1 denote the number of quarter coins, then the value of these coins is 0.25(x+1)
Let x+5 denote the number of dime coins, then the value of these coins is 0.1(x+5)

Sum up the values

$0.05x+0.25(x+1)+0.1(x+5)=7.95$

Expand the brackets, collect like terms and solve for x.

I've got x =18.
• Oct 3rd 2008, 10:01 AM
Amelia
oh so then what do i do with the x value?
• Oct 3rd 2008, 10:03 AM
janvdl
Quote:

Originally Posted by Amelia
oh so then what do i do with the x value?

Wait I get it. A nickel is 5 cents, and not 50 cents like you said. That's why Earboth's solution works out so nicely.
• Oct 3rd 2008, 10:13 AM
Amelia
oh yea.. lol stupid me.. :(

but i took x=18
then that means their are 18 nickels
then 18+1 which is 19 quarters
then 18+5 which is 23..
omg your right!!!!!! THANKK YOU SO MUCH!!! YOU GUYS ARE AWESOME!!!!!