1. number/solution problem

1.) The sum of the digits of a three digit number is 15. If the tenths digit are interchanged, the number is increased by 360. If the units digit is 1/4 of the sum of the hundreds and tens digits, what is the number?

2.) A 90% acid solution is diluted w/ H2O to make a solution 60% acid. When 3 gal. of H2O is added to dilute it again, the solution becomes 40% acid. How much H2O was added to the solution on the 1st instance? How much 90% acid was available at the start?

These are some of the problems that our teacher gave us to study for our quiz that I cannot solve...help pls.

2. For number 1. No matter how many times I solve it, still I cannot come up with a right solution and the answer.

In number 2. I think I got it. When I try to solve it, I got the answer 5.4 gallons. Is this correct?

3. Originally Posted by ihmth
1.) The sum of the digits of a three digit number is 15. If the tenths digit are interchanged, the number is increased by 360. If the units digit is 1/4 of the sum of the hundreds and tens digits, what is the number?
#1 - Write Clear and complete definitions and look at them often.

H = Hundreds Figit
T = Tens Digit
U = units Digit

#2 - Translate very carefully

"The sum of the digits of a three digit number is 15"

H + T + U = 15

"If the tenths digit are interchanged, the number is increased by 360"

This one is confusing and incomplete. I'll assume it it the hundreds and tens digits that are interchanged. Watch very carefully

100H + 10T + U + 360 = 100T + 10H + U

"If the units digit is 1/4 of the sum of the hundreds and tens digits"

U = (1/4)*(H + T)

That should be enough information. Let's see what you get.

How did you translate the three expressions, particularly the second one. You didn't put the 360 on the wrong side, did you?

4. ah ok, I spotted my mistake...which is U = (T+H)/4....I did something different...hehe

tnx

5. Hello, ihmth!

This is a two-stage mixture problem . . . a bit tricky.
But we can talk our way through it if we don't panic.

2) A 90% acid solution is diluted with water to make a solution 60% acid.
When 3 gal. of water is added to dilute it again, the solution becomes 40% acid.
How much water was added to the solution the first time?
How much 90% acid solution was available at the start?
Since water is being added, consider the amount of water at each stage.

We start with $x$ gallons which is 10% water.
. . It has: . $10\% \times x \:=\:0.1x$ gallons of water.

We add $y$ gallons of water (100% water, of course).
. . This has: . $y$ gallons of water.

This mixture consists of $x+y$ gallons of solution of which $0.1x+y$ gallons is water.
We are told that this mixture is 40% water.
. . We have the equation: . $0.4(x + y) \:=\:0.1x + y \quad\Rightarrow\quad x - 2y \:=\:0\;\;{\color{blue}[1]}$

Three gallons of water are added.
. . The solution contains: . $(0.1x + y) + 3$ gallons of water.
. . The mixture is: . $(x + y) + 3$ gallons of solution.
We are told that this mixture is 60% water.
. . We have the equation: . $0.6(x + y + 3) \:=\:0.1x + y + 3 \quad\Rightarrow\quad 5x-4y \:=\:12\;\;{\color{blue}[2]}$

Solve the system of equations: . $\begin{array}{cccc}x - 2y &=& 0 & {\color{blue}[1]} \\ 5x-4y&=&12 & {\color{blue}[2]} \end{array}$

. . and we get: . $x \:=\:4,\;y \:=\:2$

Therefore: 2 gallons of water were added the first time.
. . . . . . . .There were 4 gallons of solution at the start.