One can of Bajtra liqueur is added to a jug of alcohol-free

Question

One can of Bajtra liqueur is added to a jug of alcohol-free lemonade, and the resulting mixture is 10% alcohol by volume. After a second can of Bajtra is added, the resulting mixture is 16% alcohol. Successive cans of Bajtra are added until the resulting mixture is at least 30% alcohol. How many cans of Bajtra in total are added to the jug?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

Show Answer

Official Answer

E

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

Riyaagoell · Answer

Solution and how to approach?

atharvadixit · Answer

Man idk if my approach is right or if it was just a fluke -

so i did 1/10 = 10% (given)
2/12.5 = 16% (given)

then extrapolate

3/15 = 20%
4/17.5 = 22.8%
.
.
.
.
9/30 = 30%

kevincan  - is the logic correct?

Riyaagoell · Answer

But where did you get 12.5 from?

kevincan · Answer

That’s absolutely fantastic as long as you understand the reasoning behind it !

atharvadixit · Answer

I back calculated. So... 2/x = 0.16... x= 12.5

HarshavardhanR · Answer

This is a great question! One way to solve this -> https://gmatclub.com/forum/download/file.php?id=88519 Hope this helps. --- Harsha GMAT-Club-Forum-ifqdgpk6.png

GG27 · Answer

Hi  kevincan . Could you please send a solution to this question?

SomyaMBA2025 · Answer

Please share a proper solution

kevincan · Answer

The above solutions are brilliant, and the one below is more textbook!

**1. Define the Variables:**
Let:
*  L  = Volume of the alcohol-free lemonade in the jug.
*  V  = Total volume of one can of Bajra liqueur.
*  A  = Volume of pure alcohol in one can of Bajra liqueur.

**2. Translate Scenarios into Equations:**
* **First Can Added (10% alcohol):**
 \frac{A}{L + V} = 0.10 \implies A = 0.1L + 0.1V  — (Eq. 1)

* **Second Can Added (16% alcohol):**
 \frac{2A}{L + 2V} = 0.16 \implies 2A = 0.16L + 0.32V  — (Eq. 2)

**3. Solve for the Relationships:**
To eliminate  A , multiply Equation 1 by 2:
 2A = 0.2L + 0.2V

Set the two expressions for  2A  equal to each other:
 0.2L + 0.2V = 0.16L + 0.32V 
 0.04L = 0.12V \implies L = 3V

Substitute  L = 3V  back into Equation 1 to find  A  in terms of  V :
 A = 0.1(3V) + 0.1V = 0.4V  
*(This means each can of Bajra is 40% alcohol by volume).*

**4. Calculate the Total Number of Cans ( n ):**
We want the final concentration to be at least 30%:
 \frac{nA}{L + nV} \geq 0.30

Substitute  L = 3V  and  A = 0.4V :
 \frac{n(0.4V)}{3V + nV} \geq 0.30

Cancel out the common volume term  V :
 \frac{0.4n}{3 + n} \geq 0.30

Multiply both sides by  (3 + n)  and solve:
 0.4n \geq 0.30(3 + n) 
 0.4n \geq 0.9 + 0.3n 
 0.1n \geq 0.9 
 n \geq 9

One can of Bajtra liqueur is added to a jug of alcohol-free

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One can of Bajtra liqueur is added to a jug of alcohol-free

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