The volume of water in a certain tank is x percent greater t

[ghong14]

The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. what is the value of r in terms of x?

A) x+10
b) 10x+1
c) 100(x+10)
d) 100( (x+10)/(x+100))
e) 100 ( (10x+1)/(10x+10))

I was trying to solve this problem using multiple methods. I think I have the algebraic approach down pretty solid. However, I was also trying to figure out a number picking method.

The numbers I used were 100 for the original. 10 for X and then solved for R which came to be 200/11 which is 18.1818181818%. Obviously this is not a pretty number and plugging 100((10+10)/(10+100)) that pretty either. Any suggestions?

[RonPurewal]

Any suggestions?

Just one: Quit putting obstacles in your own path. Seriously, just pick up the proverbial shovel and start digging.

You're looking for 18 point blah blah blah ugly decimal thing. Or, alternatively, 200/11 (there's not really any good solid reason to decimal-ize this in the first place).
So...
(A) This will be an integer, so it's wrong.
(B) This will be an integer, so it's wrong.
(C) This will be an integer, so it's wrong.
(D) is 100 x 20/110 = 2000/110. No need to divide anything, that's definitely the same as 200/11.
(E) is 100 x 101/110, which is going to be awfully close to 100, and therefore nowhere close to the goal number.

There, you're done. That took... what, ten seconds to plug those numbers? Maybe fifteen seconds, if you're really sleepy, tired, and distracted.

Seriously, the test already puts enough obstacles in your way; the last thing you need is more obstacles.
DO NOT EVER forgo a problem-solving method because you're thinking, "oh, that's not pretty" or "oh, that might take too long".
Just DO IT.

That's it. Really, I promise you. Don't make this any more complicated than it needs to be.

In the extremely unlikely event that your method does turn out to take a really long time, then you can always quit; no one is holding a gun to your head.
But, in most cases -- like this one -- you'll find that the "un-pretty" things that you were so afraid of won't take more than a small fraction of a minute.
In fact, in most cases, sitting there and worrying about how long something will take will actually take longer than just doing the thing in the first place!
Yeah.

[rohandisingh]

Not sure I fully understand how to do this problem using algebra. I set up an equation but it doesn't lead me to the right answer:

v = original amount of water

v*(100+x/100)(100-r/100)=(90/100)*v


Can anyone explain to me what I am doing wrong???

thanks!

[jnelson0612]

Not sure I fully understand how to do this problem using algebra. I set up an equation but it doesn't lead me to the right answer:

v = original amount of water

v*(100+x/100)(100-r/100)=(90/100)*v


Can anyone explain to me what I am doing wrong???

thanks!

Let me give you a hint and maybe you can fix your entire equation. The problem says that the volume of water in the tank is x percent more than it was a week ago.

Let's play for a second, forgetting the rest of the information. I'm not going to worry about making this example match the entire problem; I'm just going to demonstrate one big mistake that you are making.

Let's pretend that I had 5 gallons of water in the tank (the volume, V), and then it increased by x percent. I choose x=20.

So I did have 5 gallons, but with the 20% increase I now have 6 gallons.

Let's set up an equation for that that makes sense:
5(1+20/100) = 6
V(1+P/100) = new volume

Can you see your error? See if you can use that to fix your entire equation, and please report back if you need more help.