One gallon of soft drink is made of 40% orange juice and 60% water, ho

HarpreetSinghBajwa , if you are looking for an alligation diagram, you will have to look elsewhere. I do not not use the method. But alligation is a form of weighted average. I use straight weighted average, and I believe I am in the minority. I have posted links to other methods below.

In the diagram, there is an original container of Soft Drink A. Soft drink A is 40% orange juice (and 60% water, but we need to track on the concentration / percentage of orange juice - disregard the water). Volume of A is ONE gallon.

Then "juice" is added to A. That juice is 100% orange juice. The question asks us to find how many gallons of the 100% juice, which I've called B, we must add in order to end up with a NEW MIXTURE that is 60% orange juice.

The formula is in black text below the containers. "Concentration" [of orange juice] means: orange juice is 40% of A, 100% of B, and 60% of the final mixture.

We are increasing the concentration of orange juice in the final mixture. But we don't know how many gallons of pure orange juice we need to go from a concentration of 40% orange juice to 60% orange juice. So we use the formula to find out how many gallons. There are many iterations of that formula.

I'll write the equation with and without decimals. I usually use decimals for percentages (concentrations). 100% in decimal form = 1

With decimals: (.40)*(1) + (1)*(x) = (.60)*(1 + x)

Without decimals -- which works just fine:

\((40) (1) + (100) (x) = (60) (1 + x)\)
\((C_A)(V_A) + (C_B)*(V_B) = (C_{final})(V_{final})\)

I'll use this one to finish.

\(40 + 100x = 60 + 60x\)

\(40x = 20\)

\(x =\)\(\frac{20}{40}\)

\(x =\)\(\frac{1}{2}\) gallon

Answer A

Does that help?

A couple of links that might help:

Veritas Prep Karishma is a maestra on mixture problems. She uses what she calls the "scale method":

https://gmatclub.com/forum/weighted-average-and-mixture-problems-on-the-gmat-206999.html

Another:

https://gmatclub.com/forum/mixture-problems-made-easy-49897.html

let x=gallons of orange juice to be added
.4+x=.6(1+x)
x=0.5 gallons

Let x be the quantity to be added

(0.4+x) / 1+x = 60/100
=> 4+10x = 6+6x
=> x = 2/4 = 0.5
.
Answer is A

Using a quick diagram, we can deduce that the quantity of water remains constant but the orange juice % goes from .4 to .4+x because we are adding x amount of orange juice.

The total volume also increases from 1 gallon to 1+x.
% of orange juice =\(\frac{amount of orange juice}{total volume} = \frac{.4+x}{1+x}\)

This value should be = \(\frac{60}{100}\)
40+100x=60+60x
x=0.5

Answer is A

One gallon of soft drink has 0.4 gallons of OJ and 0.6 gallons of water. We can let n = the additional gallons of OJ added to make the soft drink 60% (or 3/5) OJ and create the following equation:

(0.4 + n)/(1 + n) = 3/5

5(0.4 + n) = 3(1 + n)

2 + 5n = 3 + 3n

2n = 1

n = 1/2

Answer: A

given 1 gallon consists of 40% orange juice and 60% water
therefore, orange juice = 0.4 gallon
water = 0.6 gallon

consider new juice volume as x gallon
orange juice content increases to 60% while that of water decreases to 40%(no extra water added)
---> 0.6 = (40/100)x
solving for x = (100 * 0.6)/40
= 1.5

additional juice to be added = 1.5-0.5 = 0.5 gallon

option A

1 gallon contains 0.4 gallon orange juice.

The numbers you have taken: Total --> 5 gallons; Orange juice --> 2 gallons. This step is fine.
'x' is the amount of orange juice added.
You get x as 2.5 gallons. i.e. 2.5 gallons of orange juice is added
If your assumption of 5 gallons corresponds to 1 then 2.5 gallons corresponds to 0.5

Hi nagu - I think genxer123 did a great job illustrating the problem. Nevertheless, here is my attempt to help you understand. Pardon my mess.



Since Orange juice does not have any water in it (at least from the question's point of view), the percentage of water in orange juice is assumed to be 0. For ease, I will be using Water Proportion in my steps below. You could also use Orange juice to calculate the proportion. Using weighted average formula:

Proportion of Water in the resultant = \(\frac{(Proportion of Water in Soft Drink) (Volume of Soft Drink)+ (Proportion of Water in Orange Juice)(Volume of Orange Juice)}{ Volume of Soft Drink + Volume of Orange Juice}\)

\(\frac{40}{100}\) = \(\frac{\frac{60}{100}. 1 gallon + \frac{0}{100}. x gallon(s)}{x+1}\)

Notice, in the above equation,

the LHS represents the proportion of water in the final resultant mixture =>\(\frac{Water}{Total}\)= \(\frac{40}{100}\)

and the RHS represents

Proportion of water in the initial soft drink\(\frac{Water}{Total}\) = \(\frac{40}{100}\) * Volume of solution
+
Proportion of water in the orange juice\(\frac{Water}{Total}\) = \(\frac{0}{100}\) * Volume of solution

Hello Friends, can someone please let me know what am i doing wrong?
I did similar to what Senthil7 did but instead of using initially o:t = 0.4:1 i used o:t = 2/5 & finally o:t = 60:100 i used o:t = 3/5
(2+x)/(5+x) = 3/5
5*(2+x) = 3*(5+x)
10 + 5x = 15 + 3x
2x = 5
x = 5/2 = 2.5
Therefore, additional gallons added = 2.5 - 1 = 1.5???
I know that the answer is A but i am not able to figure out what i am doing wrong? Can someone please help!!!

Let x = amount of orange juice to be added to the original one gallon, where concentration of added OJ is 100% or, in decimal form, 1

.40(1 gal) + x = .60(1 + x)

.4 + x = .6 + .6x

.4x = .2 ----> 4x = 2

x = \(\frac{1}{2}\) , Answer A

can someone please solve this problem with diagram?

According to question after adding extra orange juice the water content is 40% of new quantity.
Original water content = 60% of 1 = 0.6
Using option A
New quantity is 1+0.5 = 1.5
Water content in new quantity is 40% of 1.5 = 0.6 which is the water content originally available.
Thus option A

4:6 is present ratio i.e., 2:3
we need to make it 6:4 i.e., 3:2

juice water total
0.4 0.6 1
add 0.2 juice
0.6 0.6 1.2 ratio= 1:1
add 0.2 juice again
0.8 0.6 1.4 ratio= 4:3
now 0.1,
0.9 0.6 1.5 ratio=3:2

so total juice added= 0.2+0.2+0.1= 0.5
Hence, A

genxer123 - Very nicely done! I am starting to like weighted averages method as well. IMO, it is a fail safe method compared to Alligation, which is only applicable when the resultant mixture's information is given (e.g., ratio or volume).

p.s. On a side note, why does M$oft draw a red squiggly under the word Alligation? It seems to be a correct spelling

Blackbox : Levity is always welcome. As is graciousness. Thank you.

I have ascertained the spelling of "alligation" more times than I care to admit.
Who knew that red squigglies could have that much power?

I gave alligation a very determined try. I know it helps many.
Alligation lost me somewhere near the "numbers play tug of war backwards" analogy.

I learned the weighted average method for the GRE.
For me, it is quick and accurate.
Best of all, as you point out, it is applicable consistently.

BTW, the best part of your link's content, my emphasis: "Alligation can also refer to the actions of alligators."

Cheers!

I agree. Well, to me, alligation method is not the savior of the day but it is good to know as a fallback plan and I meant to say "ratio" not ration in my earlier post :D

Weighted Avg. is still new to me and I am watching a lot of videos on youtube. Do you recommend any specific resource(s)? Also, where can I find more mixtures questions besides on GMAT Club, do you know?

Can any body please explain this problem how can it be solved using weighted average method?