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# How many ounces of pure water must be added to 50 ounces of

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VP
Joined: 09 Jul 2007
Posts: 1100
Location: London
How many ounces of pure water must be added to 50 ounces of [#permalink]

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02 Nov 2007, 16:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?
Director
Joined: 11 Jun 2007
Posts: 915
Re: solution:is there any simplest method? [#permalink]

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02 Nov 2007, 17:37
Ravshonbek wrote:
How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?

.15* 50 / 50+x = .10

7.5 = .10 (50+x)
7.5 = 5 + .10x
2.5 = .10x
x = 25
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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02 Nov 2007, 23:17
Hi,
50*15%=10%*X then X=75 and that means that 25 ounces hould be added
Director
Joined: 09 Aug 2006
Posts: 755
Re: solution:is there any simplest method? [#permalink]

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02 Nov 2007, 23:31
Ravshonbek wrote:
How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?

25 oz.

Currently there is 15/100*50 = 7.5 salt and 42.5 water.
Let the amt. of water that should be added = x

42.5 + x = 90/100(50+x)
x= 25
VP
Joined: 08 Jun 2005
Posts: 1145

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03 Nov 2007, 00:12
since the new ratio is 2:1.

50:x --> ratio in oz

2/1 = 50/x

x = 25 ---> amount of water to use

see attachment.

:)
Attachments

Mixture 3.JPG [ 6.53 KiB | Viewed 2858 times ]

CEO
Joined: 29 Mar 2007
Posts: 2560
Re: solution:is there any simplest method? [#permalink]

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03 Nov 2007, 08:59
Ravshonbek wrote:
How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?

I believe you can just do:

7.5 (15% of 50) --->

7.5/(50+x)=.1 ---> x= 25
Re: solution:is there any simplest method?   [#permalink] 03 Nov 2007, 08:59
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