A tank contains x gallons of antifreeze that is, by volume, y% of

Question

A tank contains x gallons of antifreeze that is, by volume, y% of propylene glycol and (100-y)% water, where y < 60. Shilah wishes to strengthen the mixture to 60% propylene glycol and 40% water. How many gallons of propylene glycol must Shilah add to make the stronger mixture?

(1) xy = 3200

(2) 0.6x - \(\frac{xy}{100}\) = 16

(1)  xy = 3200

(2)  0.6x - \frac{xy}{100} = 16

egmat · Accepted Answer

Here is a  detailed video solution  for this interesting question.
See how this question can be solved in the test environment using " Owning the Dataset " approach.

https://www.youtube.com/watch?v=VPHaKpLHswY

Let us know if you have any questions.

gmatophobia · Answer

Before Strengthening

Volume of propylene glycol in the mixture = y% of x =  \frac{xy}{100} 
 
 Volume of water in the mixture =  x - \frac{xy}{100}

After Strengthening

The volume of water represents 40% of the volume of the liquid.

Assume Shilah needs to add  V  volume of propylene glycol to strengthen the mixture to 60% propylene glycol and 40% water

\frac{x - \frac{xy}{100}}{x+V} = \frac{40}{100}

x - \frac{xy}{100} = 0.4x + 0.4V

x - 0.4x - \frac{xy}{100} = 0.4V

0.6x - \frac{xy}{100} = 0.4V

Hence, if we know the value of  0.6x - \frac{xy}{100} , we can find the value of  V .

Statement 1

(1) xy = 3200

From the above analysis, we know that  0.6x - \frac{xy}{100} = 0.4V .

While this statement provides us with the value of  xy , we don't have information on the value of  x . Therefore, by just knowing the value of  xy , we will not be able to obtain the value of  0.6x - \frac{xy}{100}

Hence, this statement alone is not sufficient. Eliminate A and D.

Statement 2

(2) 0.6x -  \frac{xy}{100}  = 16

From the pre-analysis we know that  0.6x - \frac{xy}{100} = 0.4V

16 = 0.4V

\frac{16}{0.4} = V

V = 40

We have a unique value of  V . Hence, Statement 2 alone is sufficient to answer the question.

Option B

chetan2u · Answer

A clear C trap. Don't expect the answer to be given on a platter in most occasions. So, do check for that C trap.
If not C, the next answer can only be B.

The best method would be to equate water quantity as it does not change, only the % changes. Beautifully shown by  gmatophobia  above.

Let z be the quantity of added gallons of glycol.

WATER: 
 (100-y)% of x becomes 40% of x+z or  (100-y)*x=40(x+z).....100x-xy=40x+40z......60x-xy=40z 
OR
Initial  =(100-y)*\frac{x}{100 } 
Final =  \frac{40}{100}*\frac{(x+z)}{100} 
 Initial = Final 
 =>  (100-y)*\frac{x}{100 }= \frac{40}{100}*\frac{(x+z)}{100} 
 x(100-y)=40(x+z).....60x-xy=40z

If you had taken Glycol:
 Initial  =(y)*\frac{x}{100 } 
Final =  \frac{60}{100}*\frac{(x+z)}{100} 
 Initial + z = Final 
 =>  y*\frac{x}{100 }+z= \frac{60}{100}*\frac{(x+z)}{100} 
 xy+100z=60(x+z)........xy+100z=60x+60z..........40z=60x-xy

(1) xy = 3200
 We require x and y separately 
Insufficient

(2) 0.6x -  \frac{xy}{100}  = 16
0.6x*100-xy=1600
60x-xy=1600
we know 60x-xy=40z=1600.......z=40
Suff

B

KarishmaB · Answer

Tricky one. I deep dived in it and actually did the work because (C) seemed too easy. Sometimes we may not be able to get the individual values of the variables but we are given the value of the expression which is asked.

At the heart of it, it is a simple mixtures question. We can use our weighted averages formula. We need to find w2.

\(\frac{x}{w2} = \frac{(1 - .6)}{(.6 - y/100)}\)

\(\frac{x(.6-y/100)}{.4} = w2\)

\(w2 = \frac{(.6x - xy/100)}{.4}\)

(1) xy = 3200

Gives us the value of xy but that is not sufficient. We can also not find individual values of x and y. Not sufficient.

(2) \(0.6x - \frac{xy}{100} = 16\)

To get w2, we needed this value. When we plug it in the numerator, we get w2 as 40.

Sufficient alone.

Answer (B)

Here are posts and videos on weighted avgs and mixtures:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/

Gg153 · Answer

Can you tell me what this part in the weighted average formula is? 
   \frac{(1 - .6)}{(.6 - y/100)}

Bunuel · Answer

Here is a short and direct approach.

Let t be gallons to add.

x * \frac{y}{100} + t = 0.6(x + t)

\frac{xy}{100} + t = 0.6x + 0.6t

0.4t = 0.6x - \frac{xy}{100}

t = \frac{0.6x - \frac{xy}{100}}{0.4}

(1) xy = 3200.

This gives  t = \frac{0.6x - 32}{0.4} , the value of which depends on x. Not sufficient.

(2)  0.6x - \frac{xy}{100} = 16

This gives  t = \frac{16}{0.4}=40 . Sufficient.

Answer B.

KarishmaB · Answer

I have explained it here: https://anaprep.com/arithmetic-weighted-averages/

MartyMurray · Answer

A tank contains x gallons of antifreeze that is, by volume, y% of propylene glycol and (100-y)% water, where y < 60. Shilah wishes to strengthen the mixture to 60% propylene glycol and 40% water. How many gallons of propylene glycol must Shilah add to make the stronger mixture?

(1) xy = 3200

(2) 0.6x - \(\frac{xy}{100}\) = 16

(1)  xy = 3200

Given what this statement says and what the passage says, x and y could be any numbers such that y < 60 and x > 3200/60.

So, an infinite number of different scenarios are possible, each of which involves adding a different amount of propylene glycol to make the stronger mixture.

Insufficient.

(2)  0.6x - \frac{xy}{100} = 16

Notice that  0.6x  is 60% of the current mixture and  \frac{xy}{100}  is the amount of propylene glycol currently in the mixture.

So,  0.6x -  xy/100   is the amount of water in the current mixture that would have be removed and replaced with propylene glycol for the x gallons Shilah already has to become 60% propylene glycol.

So, to make the x gallons a 60% mixture, lets add 16 gallons of propylene glycol.

Of course, we didn't remove any water. So, all the water from the original mixture is still there. So, we now have x gallons of a 60% mixture plus 16 gallons of water.

So, we now need to add enough propylene glycol for that extra 16 gallons of water to also be part of a 60% mixture.

16/0.4 = 40. So, we need to add 0.60 × 40 = 24 more gallons of propylene glycol to have a 24:16 = 60:40 mixture of propylene glycol and water.

The total amount of propylene glycol we need to add is thus 16 + 24 = 40 gallons.

Sufficient.

Correct answer:   B

DonBosco7 · Answer

https://gmatclub.com/forum/download/file.php?id=86704 GMAT-Club-Forum-pjlwsjhd.png

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