# Algebra II

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Could someone please help me? I'm super confused by these questions and i just keep making a mess all over my paper!

1) A box of coins contains nickels, dimes, and quarters. The total value of the coins is \$4.00. If there were 3 times as many nickels, half as many dimes, and 6 more quarters, the value of the coins would be \$5.50. How many quarters are in the box?

I'm just not sure how you would set up a system with amounts of money and amounts of coins.

2) A truck radiator contains 14 quarts of a mixture that is 75% antifreeze and 25% water. How much should be drained and replaced with pure water for the radiator to contain 14 quarts of a mixture that is 60% antifreeze and 40% water?

This question seems impossible! I might be reading it incorrectly though. I have no clue how to set this up!

Okay last one...
3)The average of three numbers is 18. If the first number is increased by 5, the second number is doubled, and the third number is tripled, the average becomes 34. If the first number is decreased by 5, the second number is tripled, and the third number is doubled, the average becomes 48. Find the largest of the three numbers.

I think I have the systems but there are three and I don't know how to figure those out..

X + Y + Z= 54
(X + 5) + 2Y + 3Z= 102
(X - 5) + 3Y + 2Z= 144

I hope those are right..

• Algebra II (#3 of 3) - ,

Please post questions like this separately, not in groups.
Rewrite your equations, which are correct, as
X + Y + Z = 54
X + 2Y + 3Z= 97
X + 3Y + 2Z= 149
Subtract the first eq. from the second.
Y + 2Z = 43
Subtract the first eq. from the third
2Y + Z = 95
Double the last one
4Y + 2Z = 190
Get rid of the Z terms
3Y = 190 - 43 = 147
Y = 49
49 + 2Z = 43
Z = -3
X + 49 -3 = 54
X = 8
The largest number is Y.

• Algebra II (first question) - ,

let D, N and Q represent the number of dimes, nickels and quarters respectively.

then 10D + 5N + 25Q = 400 and
10(3N) + 5(N/2) + 25(Q+6) = 550

there has to be more information given to continue with this question, e.g perhaps you made a typo and the first part included the total number of coins in the box.

• Algebra II (#2) - ,

Let the amount of the current liquid to be drained and replaced by pure water be x quarts

Looking only at the water content in the rad...

.25(14-x) + x = .4(14)

solving this I got x = 2.8 quarts

• Algebra II - ,

#2 of 3
The original 14 gallons is 1/4 water, so it contains 3.5 gallons of water.
When you are done removing and replacing x gallons of the mix with water, you still have 14 gallons, but it is 40% water, so that makes 5.6 gallons of water in the radiator afterwards. The equation you have to solve is:
Initial water present + water removed -water added = final amount of water
3.5 - 0.25 x + x = 5.6
0.75 x = 2.1
x = 2.8 gallons

• Algebra II - ,

10d +5n divided which expression represents the money about among 5 in cents each persons receives

• Algebra II - ,

• Algebra II - ,

en 100 platos cuantos litro de agua dentran