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A researcher combined x ounces of a saline solution that contained 20%

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A researcher combined x ounces of a saline solution that contained 20%  [#permalink]

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New post 01 Jul 2018, 08:37
5
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

60% (01:23) correct 40% (01:48) wrong based on 118 sessions

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A researcher combined x ounces of a saline solution that contained 20 percent saline, by volume, with y ounces of a saline solution that contained 5 percent saline, by volume, to produce z ounces of a saline solution that was 12 percent saline, by volume. What is the value of x?

(1) y = 16
(2) z = 30
Manager
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A researcher combined x ounces of a saline solution that contained 20%  [#permalink]

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New post 01 Jul 2018, 17:07
Bump -- could someone please explain why the answer isn't A? This question comes from Jeff Sackman's Word Problem Challenge, btw.
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Re: A researcher combined x ounces of a saline solution that contained 20%  [#permalink]

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New post 01 Jul 2018, 17:30
(.2x+.05y)/z=.12
Z=x+y
(1) (.2x+.8)/(x+16)=.12
(2) (.2x+.05(30-x))/30=.12
IMO answer D


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Re: A researcher combined x ounces of a saline solution that contained 20%  [#permalink]

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New post 05 Sep 2018, 23:29
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TippingPoint93 wrote:
Bump -- could someone please explain why the answer isn't A? This question comes from Jeff Sackman's Word Problem Challenge, btw.


The first statement is sufficient to get x,

Solving by alligation method,

x y
20 5
12 (z = x+y)
7 8

Hence x and y will be in 7:8 ratio.

Statement 1: y = 16
If y = 16, to maintain the ratio of 7:8, x will have to be 14.
Sufficient.

Statement 2: z=30
z is nothing but x+y,
so x = \(\frac{7}{15}* 30\)
Which is 7*2 = 14.
Sufficient.

Hence answer is D.
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Re: A researcher combined x ounces of a saline solution that contained 20%  [#permalink]

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New post 11 Nov 2018, 08:46
20 x+ 5y=12x+12y

solving
x=7/8y

x+y=z

so stm1 or stmt 2 rithe of them is sufficient
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Re: A researcher combined x ounces of a saline solution that contained 20%   [#permalink] 11 Nov 2018, 08:46
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