How many ounces of pure acid must be added to 20 ounces of a solution

Question

How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?

(A) 5/2

(B) 5

(C) 6

(D) 15/2

(E) 10

raj2812 · Accepted Answer

ANS: (B) 5% of 20 ounces=1 so now the question says what quantity of acid must be added to this 1 ounce of acid so that it becomes 24% of the new solution. let the new quantity of acid to be added be x total new quantity of acid=1+x new quantity of solution = 20+x Therefore 1+x= 24/100* (20+x) = solve and the answer is 5 ounces. hope this is of some help!

Bunuel · Answer

18. Mixture Problems 
     Theory   
 Weighted Average and Mixture Problems on the GMAT 
 Tips and Tricks: Mixtures 
     Questions   
 DS Questions 
 PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

generis · Answer

To find the volume of one solution in a resultant mixture, weighted average works well. I use this formula: 
  (Concen_{A})(Vol_{A}) + (Concen_{B})(Vol_{B} = (Concen_{A+B})(Vol_{A+B})

Per the formula, in setting up the equation:
A = the solution that is 5% acid
B = the solution that is pure acid (100% = 1)
Let   x  = # of ounces of pure acid (B) to be added
Desired concentration of resultant mix: 24%=.24
A + B (volume) = 20 + x

.05(20) + 1(x) = .24 (20 + x) 
 1 + x = .24(20) +.24(x) 
 1 + x = 4.8 + .24x 
 .76x = 3.8 
 x =\frac{3.8}{.76}=\frac{380}{76}=5

We need 5 ounces of pure acid.

Answer B

Shobhit7 · Answer

The Equation:
100x + 20*5 = 24(x+20)

Simplifying, we get
76x = (24-5)*20 = 19*20

Therefore, x=5.

Ans B

Posted from my mobile device

JeffTargetTestPrep · Answer

We can let n = the number of ounces of pure (100%) acid added to the solution.

(0.05)(20) + n = 0.24(20 + n)

1 + n = 4.8 + 0.24n

0.76n = 3.8

n = 5

Answer: B

ShekharRavi · Answer

Keep it Simple!

=> Non Acid before = Non Acid After 
=> 95% of 20 = 76% of (20+x)
=> 19 = 76/100*(20+x)
=> 25 = 20+x
=> x = 5

gracie · Answer

let a=ounces of acid to be added
.05*20+a=.24*(20+a)
a=5 ounces
B

Archit3110 · Answer

let soln = x
x+.05*20 = .24*(20+x)
1+x=4.8+.24x
x=5
IMO B

satya2029 · Answer

5% of 20 ounces= 1 ounces
Final quantity=1+x
1+x=0.24(20+x)
x=5 ounces
B:)

MHIKER · Answer

How can the allegation approach be applied here?

adityasuresh · Answer

How many ounces of pure acid must be added to 20 ounces of a solution that is 5% acid to strengthen it to a solution that is 24% acid?

The Question asks how many ounces of *Pure Acid* must be added to a solution that is 5% Acid

Approaching the problem w.r.t concentration of acid.

We can use the weighted average approach. W1/W2 = A2 - Avg/Avg -A1

A1 implies Concentration of acid to be added in a pure acid solution = 100%(pure acid has only acid in its solution) w1 = ?
A2 is given in the stem as 5%, W2 = 20

Avg = 24 %

Substituting values : 5-24/24-100 = w1/20 simplifying W1 = 5

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How many ounces of pure acid must be added to 20 ounces of a solution

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