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What amount (in milliliters) of a 1% sulfuric acid solution must be ad

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What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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29 Mar 2018, 00:25
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What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12
(B) 15
(C) 18
(D) 20
(E) 45

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Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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29 Mar 2018, 01:27
1
1
Bunuel wrote:
What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12
(B) 15
(C) 18
(D) 20
(E) 45

60 ml 6% sulphuric acid has $$\frac{6}{100}*60 = 3.6$$ ml sulphuric acid and rest is water.

x ml of 1% sulphuric acid will have $$\frac{1}{100}*x = x/100$$ ml of sulphuric acid.

so total sulphuric acid will be 3.6 + x/100 in total volume of 60 + x.

This new mixture has to be 5% sulphuric acid

Hence 5% of (60+x) = 3.6 + x/100

$$\frac{5}{100}*(60 + x) = 3.6 + \frac{x}{100}$$
$$3 + \frac{5x}{100} = 3.6 + \frac{x}{100}$$
$$\frac{4x}{100} = 0.6$$
$$x = \frac{60}{4}$$
$$x = 15$$ ml

Hence Option (B) is our answer.

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Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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29 Mar 2018, 03:33
1
Bunuel wrote:
What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12
(B) 15
(C) 18
(D) 20
(E) 45

TIP: For all such questions of Mixtures one must make an equation of actual quantity of one substance.

For instance here the equation can be formed on the quantity of sulfuric acid

Sulfuric acid of 60 ml solution + Sulfuric acid of 1% solution = Sulfuric acid of Final solution

i.e. (6/100)*60 + (1/100)*x = (5/100)*(60+x)

i.e. x = 15

Alternatively: One may use allegation method however I suggest that for starters and not so strong in maths should follow the method of making equation
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Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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29 Mar 2018, 04:13
The best approach shall be using principle of alligation.

Principle: If two ingredients A and B of price x and y respectively are mixed and the price of resultant mixture is M (mean price)then the ratio (R) in which ingredients are mixed is given by, the rule of allegation

R=(M−y)/(x−M)

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Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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29 Mar 2018, 04:17
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Please see the sketch for alligation approach

gmatbusters wrote:
The best approach shall be using principle of alligation.

Principle: If two ingredients A and B of price x and y respectively are mixed and the price of resultant mixture is M (mean price)then the ratio (R) in which ingredients are mixed is given by, the rule of allegation

R=(M−y)/(x−M)

Posted from my mobile device

Posted from my mobile device

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What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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29 Mar 2018, 10:01
Bunuel wrote:
What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12
(B) 15
(C) 18
(D) 20
(E) 45

let x=amount of 1% acid to be added
.01*x+.06*60=.05(60+x)
x=15 milliliters
B
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Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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30 Mar 2018, 10:47
Bunuel wrote:
What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12
(B) 15
(C) 18
(D) 20
(E) 45

We can let the amount of 1% sulfuric acid to be added = n and create the equation:

(0.01n + 0.06 x 60)/(60 + n) = 0.05

0.01n + 3.6 = 0.05(60 + n)

0.01n + 3.6 = 3 + 0.05n

0.6 = 0.04n

15 = n

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What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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31 Mar 2018, 11:38
Bunuel wrote:
What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12
(B) 15
(C) 18
(D) 20
(E) 45

Let x = amount of 1% sulfuric acid to be added
A = solution with 1% sulfuric acid
B = solution with 6% sulfuric acid

(% A)(Vol A) + (% B)(Vol B) = (% of A+B)(Vol of A+B)

$$.01(x) + .06(60) = .05(x + 60)$$
Multiply all by 100 to make math quicker
$$1x + 6(60) = 5(x + 60)$$
$$1x + 360 = 5x + 300$$
$$4x = 60$$
$$x=\frac{60}{4}=15$$

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Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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22 Jul 2018, 20:13
Bunuel wrote:
What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

(A) 12
(B) 15
(C) 18
(D) 20
(E) 45

6% solution: 6 --> 5 = 1
1% solution: 1 --> 5 = 4

Original amount = 60mL

Ratio 6%:1% = 4:1 = 4/1

Original amount / X = Ratio of 6%/1%

60/x = 4/1

x = 15mL
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Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad  [#permalink]

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16 Aug 2018, 11:30
0.06(60)+0.01x=0.05(60+x)
=> 0.04x=0.6
=>x=15
Re: What amount (in milliliters) of a 1% sulfuric acid solution must be ad   [#permalink] 16 Aug 2018, 11:30
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