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

Question

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

Gladiator59 · Answer

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.

Best,
Gladi

GMATinsight · Answer

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

Answer: Option B

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

GMATBusters · Answer

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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GMATBusters · Answer

Please see the sketch for alligation approach

gracie · Answer

let x=amount of 1% acid to be added
.01*x+.06*60=.05(60+x)
x=15 milliliters
B

ScottTargetTestPrep · Answer

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

Answer: B

generis · Answer

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

Answer B

jch103020 · Answer

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

KaulMeAnkit · Answer

0.06(60)+0.01x=0.05(60+x)
=> 0.04x=0.6
=>x=15

MHIKER · Answer

Would please differentiate this PS and the following PS and help in solving

Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?

A) 18
B) 20
C) 24
D) 36
E) 42

Rainman91 · Answer

2--------------------12

----------5-----------

7---------------------3

Total is 10.

The question says it produces 60 liters. 
So 10 is 60. We know that 2% is in the proportion of 7. Hence, 7x6 = 42. 
Why x6? Because 10x6 = 60 liters. So multiply each proportion by 6. voila!

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

Prep Toolkit

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