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

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Re: The volume of water in a certain tank is x percent greater  [#permalink]

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New post 08 Apr 2017, 20:35
thinktank wrote:
Hey!
I chose volume = 100 , x=5% , r=10%
Help plz? :(
Thanks
Aj



If the initial volume was 100 and you increased by 5%, that's 105. So, now you need to go from 105 to 90. That'percentage (r) is 15/105*100 = (1/7)*100 = 14%.

So, D gives you 14: 100*(15/105).

Not the easiest numbers that you picked though.

Kudos if you agree!! Comment if you have a better method.
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Re: The volume of water in a certain tank is x percent greater  [#permalink]

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New post 18 Apr 2017, 15:16
2
fozzzy wrote:
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. \(100( \frac{x+10}{x+100})\)

E. \(100 ( \frac{10x+1}{10x+10})\)


We can let the volume of water in the tank one week ago = n.

Since the volume is now x percent greater than it was one week ago, the volume today is:

(1 + x/100)n

Next we are given that r percent of the current volume is removed. Thus we now have:

(1 - r/100)(1 + x/100)n

Since the resulting volume will be 90 percent of what it was a week ago, we can create the following equation and isolate r:

(1 - r/100)(1 + x/100)n = 0.9n

(1 - r/100)(1 + x/100) = 0.9

[(100 - r)/100][(100 + x)/100] = 0.9

[10,000 - rx + 100x - 100r]/10,000 = 0.9

10,000 - rx + 100x - 100r = 9,000

1,000 + 100x = rx + 100r

1,000 + 100x = r(x + 100)

100(10 + x) = r(x + 100)

100(10 + x)/(x + 100) = r

Answer: D
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Re: The volume of water in a certain tank is x percent greater  [#permalink]

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Re: The volume of water in a certain tank is x percent greater &nbs [#permalink] 02 May 2018, 08:09

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