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There is 40g of 8% salty water in a glass. To make it 20%, how much mo

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Intern
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There is 40g of 8% salty water in a glass. To make it 20%, how much mo  [#permalink]

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New post 14 Feb 2019, 13:00
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Question Stats:

82% (01:58) correct 18% (01:40) wrong based on 33 sessions

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There is 40g of 8% salty water in a glass. To make it 20%, how much more salt should be added?

A- 12g
B- 10g
C- 9g
D- 8g
E- 6g
Intern
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Joined: 25 Sep 2017
Posts: 3
Re: There is 40g of 8% salty water in a glass. To make it 20%, how much mo  [#permalink]

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New post 14 Feb 2019, 13:30
1
This question is trying to bait you into executing a bunch of arithmetic. Testtakers get tricked on questions such as this because they totally know how to solve this question using math.

BUT, think before you math. This question may be answered much faster using estimation and backsolving. How much is currently salt? a 10 percent mixture would have 4 grams and a 5 percent mixture would have 2 grams. So, about 3 grams exist - 3/40.

Now, add the answer choices to both the top and the bottom and see which works out to 20 percent. Adding 9 to the top and the bottom makes 12/49. That's greater than 20 percent. Adding 6 to the top and bottom makes 9/46. Thats very close to 20 percent and probably right. Check 8. It works out to 11/48. That's further from 20 percent. 6 is the answer.

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Re: There is 40g of 8% salty water in a glass. To make it 20%, how much mo  [#permalink]

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New post 14 Feb 2019, 13:54
1
I don't care for the wording of the problem, but it's just a weighted average question: we're mixing an 8% salt solution with pure salt, so a 100% salt solution, and we want to arrive at a 20% mixture. Using the number line method (sometimes called 'alligation' -- I won't explain the method fully here, but if anyone wants to learn more about it, just google that term) :

---8%-----20%---------------------100%----

The ratio of the distances to the middle must be the same as the ratio of the two components of the mixture, so we are mixing them in a 12 to 80 ratio, or in a 3 to 20 ratio, where the pure salt is the smaller of the two components. Since we have 40g of the 8% solution, we must have 6g of pure salt.
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Re: There is 40g of 8% salty water in a glass. To make it 20%, how much mo  [#permalink]

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New post 14 Feb 2019, 20:27
WABL wrote:
Hello,

There is 40g of 8% salty water in a glass. To make it 20%, how much more salt should be added?

A- 12g
B- 10g
C- 9g
D- 8g
E- 6g

please help!


let s=salt to be added
.08*40+s=.2(40+s)
s=6g
E
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Re: There is 40g of 8% salty water in a glass. To make it 20%, how much mo  [#permalink]

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New post 14 Feb 2019, 20:30
8% of 40g is 3.2g of salt. In other words, [3.2]/[40] = 0.08

We want 20% of concentration and X extra salt. So:

[3.2+X]/[40+X] = 0.2

Solving for X:
3.2+X = 8 +0.2X
0.8X = 4.8
X = 6

Answer is E
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Re: There is 40g of 8% salty water in a glass. To make it 20%, how much mo  [#permalink]

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New post 18 Feb 2019, 18:12
WABL wrote:
There is 40g of 8% salty water in a glass. To make it 20%, how much more salt should be added?

A- 12g
B- 10g
C- 9g
D- 8g
E- 6g


There are 40 x 0.08 = 3.2 grams of salt and 36.8 grams of water. To make it 20%, or 1/5, salt, we have:

(3.2 + n)/(40 + n) = 1/5

5(3.2 + n) = 40 + n

16 + 5n = 40 + n

4n = 24

n = 6

Alternate Solution:

To 40 grams of 8% salt, we will add n grams of 100% salt, which will yield (40 + n) grams of 20% salt. We can express this in an equation as:

40 x 0.08 + n x 1.0 = (40 + n) x 0.2

3.2 + n = 8 + 0.2n

0.8n = 4.8

n = 6

Answer: E
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There is 40g of 8% salty water in a glass. To make it 20%, how much mo  [#permalink]

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New post 22 Feb 2019, 11:20
WABL wrote:
There is 40g of 8% salty water in a glass. To make it 20%, how much more salt should be added?

A- 12g
B- 10g
C- 9g
D- 8g
E- 6g

Using weighted average, track on salt.

Let \(x\) = B below = amt of salt needed to achieve desired result
\(x\) is 100% salt = decimal form 1
-- 100% = \(\frac{100}{100} = 1\)
Below, A = the 40 g of solution at 8% salt concentration
(A + B) = resultant solution

Original + unknown = desired resultant solution

(A %)(A volume) + (B %)(B vol) = (% A+B)(vol A+B)

\((.08)(40) + (1)(x) = (.20)(40+x)\)

\(3.2 + 1.0x = 8 + 0.20x\)

\(0.80x = 4.8\)

\(x = \frac{4.8}{0.8} = \frac{48}{8} = 6\)

Answer E
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There is 40g of 8% salty water in a glass. To make it 20%, how much mo   [#permalink] 22 Feb 2019, 11:20
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