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A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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Question Stats: 75% (01:49) correct 25% (01:57) wrong based on 186 sessions

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A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26
Retired Moderator V
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Re: A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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vikasp99 wrote:
A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26

In 10 liters of that solution, there is 1 liter nitric acid and 9 liters water.

In the new solution that contains 4% acid or 1 liter acid. This means the total volume of the new solution is 1 / 4% = 25 liter.
The volume of water need to be added is 25 - 10 = 15 liters.

The answer is A.
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A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26

let w=liters of water to be added
.10*10=.04(10+w)
w=15 liters
A

Originally posted by gracie on 24 Feb 2017, 18:23.
Last edited by gracie on 09 Oct 2017, 20:16, edited 2 times in total.
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Re: A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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Here, 10% of 10 liters is 1 liter, which will become 4% :
If 1L ==>4%
Then 24 ==> 96%

And since we already have 9 liters of water we still need to add 15 liter to reach 24 : 24-9 =15
Hence answer : A
Intern  B
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A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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We can solve this by ratios as well.

$$\frac{Original content of Acid}{Original Content of water+Water to be added}=\frac{Acid Content Required}{Water Content Required}$$

$$\frac{.1*10}{.9*10+x}= \frac{4%}{96%}$$

$$\frac{1}{9+x}=\frac{1}{24}$$

24=9+x
Hence, x=15.
Therefore, 15 liters of water must be added.
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Re: A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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vikasp99 wrote:
A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26

We see that 0.1 x 10 = 1 liter is nitric acid. We can let w = amount of water that must be added to dilute the solution to 4%:

4/100 = 1/(10 + w)

4(w + 10) = 100

w + 10 = 25

w = 15

Alternate Solution:

We have 10 liters of 10% nitric acid. To it we add x liters of water (which has 0% nitric acid), and the result is (10 + x) liters of 4% nitric acid. We can express this in an equation as:

10(0.10) + x(0) = (10 + x)(0.04)

1 + 0 = 0.4 + 0.04x

0.6 = 0.04x

60 = 4x

15 = x

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Re: A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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JeffTargetTestPrep wrote:
vikasp99 wrote:
A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26

We see that 0.1 x 10 = 1 liter is nitric acid. We can let w = amount of water that must be added to dilute the solution to 4%:
"
4/100 = 1/(10 + w) Can you explain what is "100" & "1" represent here. I understood "4" % of nitric acid required & "10 + w" is total volume.

4(w + 10) = 100

w + 10 = 25

w = 15

Alternate Solution:

We have 10 liters of 10% nitric acid. To it we add x liters of water (which has 0% nitric acid), and the result is (10 + x) liters of 4% nitric acid. We can express this in an equation as:

10(0.10) + x(0) = (10 + x)(0.04) Please tell me why you didn't use 9 liters of water in the equation

1 + 0 = 0.4 + 0.04x

0.6 = 0.04x

60 = 4x

15 = x

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GMAT 1: 630 Q47 V30 WE: Operations (Non-Profit and Government)
Re: A chemist has 10 liters of a solution that is 10 percent nitric acid b  [#permalink]

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vikasp99 wrote:
A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

○A○ 15
○B○ 18
○C○ 20
○D○ 25
○E○ 26

Total= 10 lt
Acid= 1
Water= 9
Water to add= x

1/(10+x) = 4/100

so x is 15.
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I never gave up what I wanted- Re: A chemist has 10 liters of a solution that is 10 percent nitric acid b   [#permalink] 05 Jun 2019, 06:54
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