There are two containers A and B filled with mentha oil with different

Question

There are two containers A and B filled with mentha oil with different prices and with volumes 140 and 60 liters respectively. Equal quantities are drawn from both A and B in such a manner that the oil drawn from A is poured in into B and oil drawn from B is poured into A. After doing so, the price per liter becomes equal in both containers. What is the (equal) quantity that was drawn?

A. 21 liters
B. 27 liters
C. 35 liters
D. 42 liters
E. 45 liters

A. 21 liters
B. 27 liters
C. 35 liters
D. 42 liters
E. 45 liters

sumitkrocks · Accepted Answer

Let us assume price per liter of the mentha oil in container A and B be a and b respectively 
From the given information we can say that quantity of oil in the container A and B will be 140 and 60 each respectively; However, concentration or mix/composition will change to achieve an average price which will be between a and b with more aligned towards content of A as A is higher in quantity i.e 140 compared to 60

Average price of the mix =  \frac{140*a + 60*b}{ 200}  Equation ---1 which will be equal to the average price of the content in each container

Say x Ltr of oil is drawn from each container and put in the other one

Average price of content of container A =  \frac{(140-x)*a + x*b }{140}  Equation ---2

Average price of content of container A =  \frac{(60-x)*b + x*a}{ 60}  Equation----3

Equation 1 and 2 and 3 will be equal

From 2 and 3

\frac{140*a + x*(b-a) }{ 140} = \frac{60*b + x*(a-b) }{ 60}

Solving further ...  3*14(a-b) = x*(a-b)  
Since we are given a is not equal to b; a-b can not be zero 
x= 3*14 =42

(D) is the answer IMO

andreagonzalez2k · Answer

In order to have the same price ratio in both mixtures must be the same.

\frac{140-x}{x}=\frac{x}{60-x}

x^2=140*60-200x+x^2

x=42

zanto · Answer

I don't know if my approach is right. I probably got a lucky guess.  Bunuel , do you mind giving input / providing the official solution?

Price / Liter = P(total) / Q(total)

A container: P  / (Qtotal - a + b)
a= mixture leaving a
b = mixture added from b

B container: P  / (Qtotal -b + a)
b = mixture leaving b
a = mixture added from a

set both equations equal to each other:
P  / (Qtotal - a + b) = P  / (Qtotal -b + a)

The Ps cancel.

Plug in 140 and 60.

Side note: notice both are integers that are even, all the answers are odd except for 42. That was my thinking..

Thunderbird03 · Answer

P is the quantity removed.
Q is the cost of the first liquid.
R is the cost of the second liquid.

(140-P)Q+PR/140= (60-P)R+PQ/40

On solving the above equation you get => 42Q-PR=42R-PQ

42-P*Q-R=0

Either P has to be 0 or A & B should be equal. But it's already said that prices are different, so P is 42

vipulgoel · Answer

can some one please elaborate this ??

andreagonzalez2k · Answer

You remove the same amount of liters from A container and B container. Let it be 'x'.

After that you have:
In A container:
- '140 - x' of A mentha oil
- 'x' of B mentha oil
In B container:
- 'x' of A mentha oil
- '60 - x' of B mentha oil

Ratio of A mentha oil to B mentha oil must be the same in both containers in order that the price is the same.

\frac{140-x}{x}=\frac{x}{60-x}

x^2=140*60-200x+x^2

x=42

KarishmaB · Answer

Say the oil in 140 lts container is cheaper and the oil in 60 ltr container is more expensive. 
After mutual replacement, overall price becomes the same. This means that the concentration of either ingredient (one kind of oil) is the same in both mixtures.

Say, we remove x litres from each and replace in the other.

Concentration of expensive oil in both containers should be equal. So 
 \frac{x}{140} = \frac{(60 - x)}{60}  
x = 42 liters

Answer (D)

Note that we could have done the same thing with the concentration of cheaper oil and we would have got the same answer. In mixtures, we need to deal with the concentration of any one ingredient e.g. with concentration of sugar in sugar solution.

SoHaaM · Answer

Ok, there are two containers A and B. Suppose there are two types(price) of mentha oil:

- X is in A, which is 140 L.
- Y is in B, which is 60 L.

Since they became equal in price, we can assume that both have the same proportion of X and Y in them. we would option test

------------ A ------------

option(a)

x=140-21=119
y=21

now, x/y = 119/21

------------ B ------------

x=21
y=60-21=39

≠ 21/39

(b)(C)(E) options don't work too.

----------------------------------------------------------------

test option(d)

------------ A ------------

x=140-42=98
y=42

x/y=98/42=7/3

------------ B ------------

x=42
y=60-42=18

= 42/18=7/3

----------------------------------------------------------------

so answer is (D)

Gmatiseasyforall · Answer

Hi Bunuel! Thanks for posting this question! This is one of those mixture problems that confuse me a lot!

The Answer provided by andreogon is right on the dots! The answer to this type of mixture problems is always 
  \frac{ ab}{a+b }   where   a   and   b   are the respective volumes of the two containers. Notice that this is the HALF of the Harmonic Mean formula i.e  \frac{ Harmonic Mean}{2 }  . My question is why is Harmonic mean is applicable to this type of mixture problems? Is there any specific structure of the question that make Harmonic mean appropriate for this type of question? Or is there any specific property of Harmonic Mean which makes it a fit for this type of question?

Will much appreciate your insights in this regard! Thanks in advance!

Bunuel · Answer

I’d say there is a real reason, but it is not a general rule that mixture problems use the harmonic mean.

If x liters are exchanged, then x/a of container A is replaced and x/b of container B is replaced. For the final prices to become equal, these two fractions must together account for the whole original difference, so:

x/a + x/b = 1

Solving gives:

x = ab/(a+b)

This expression is half the harmonic mean of a and b. So the harmonic mean appears because this problem naturally creates a sum of reciprocals, not because harmonic mean is generally used in mixture problems.

Gmatiseasyforall · Answer

Yes, like you mentioned, I was expecting that we are averaging rates/ratios which involve reciprocals relationships but I wasn't able to figure out the reciprocal terms themselves. There is only one point in your equation that I don't understand:- Why does $\frac{x}{a}$+$\frac{x}{b}$ equals 1 and not any other number? Where does this 1 come from? How is this 1 related to the expression on the left hand side of the equation?

I would much appreciate your insights in this regard! I won't be dragging this question much further! Appreciate your patience! Thanks in advance!

I would much appreciate your insights in this regard! I won't be dragging this question much further! Appreciate your patience! Thanks in advance!

Bunuel · Answer

The 1 represents the entire original difference between the two prices per liter.

When x liters are exchanged, the price per liter in A changes by x/a of that original difference, and the price per liter in B changes by x/b of that original difference. For the two final prices to become equal, these two changes together must eliminate the whole original difference. Therefore:

x/a + x/b = 1

Now suppose, just for illustration, that the oil in A costs $20 per liter and the oil in B costs $30 per liter. The original difference is $10 per liter.

If x liters are exchanged, A gains oil that is $10 more expensive per liter, so the total value of the oil in A increases by 10x dollars. Since A contains 140 liters, its price per liter increases by:

10x/140

Similarly, B loses 10x dollars in total value, so its price per liter decreases by:

10x/60

For the two final prices per liter to become equal, these two changes must together eliminate the original $10 difference:

10x/140 + 10x/60 = 10

Dividing by 10:

x/140 + x/60 = 1

So the 1 simply represents 100% of the original price difference. Solving gives x = 42 liters.

Gmatiseasyforall · Answer

Simply a Killer answer! No words can explain my gratitude!

I was thinking this on different lines that the main property of harmonic mean is that: if a,x,b are in harmonic progression then \frac{(a-x)}{(x-b)} = \frac{a}{b} , which is also how Greeks used to define harmonic mean.
This is also similar to the alligation equation. On solving we get x= \frac{2ab}{(a+b)}  which is the harmonic mean. Then  \frac{x}{2}  will be the required quantity drawn from each container. But, now I am not sure if that line of thinking was correct! Would you like to add any words on that? But that's solely your discretion!

egmat · Answer

Hi Gmatiseasyforall,

Great instinct spotting the pattern, and you're right that the answer here is  140 · 60 /( 140 + 60 ) =  8400 / 200  =  42 , which is exactly half the harmonic mean of the two volumes.

But here's the key idea:  the harmonic mean isn't actually the  reason  the problem works  - it's just the  shape  the answer happens to take once you do the algebra. There's no special "harmonic mean property" being invoked. Let me show you where the formula really comes from.

The real structure

Look back at KarishmaB's step in the thread - the one that reads x/ 140  = ( 60 −x)/ 60 . That single equation is the heart of it. Two facts force it:

- After you draw x and swap,  each container's total volume is unchanged  ( 140  stays  140 ,  60  stays  60 ).
- "Equal price per liter" is the same as  equal concentration  of either oil in both containers.

Set the concentration of A's oil equal in both containers, using general volumes a and b:

(a-x)/a = x/b

Cross-multiply:

- b(a - x) = x·a
- ab - xb = xa
- ab = x(a + b)
- x = ab/(a + b)

That's it. The formula ab/(a+b)  falls straight out of the equal-concentration condition  - no harmonic mean needed to derive it. It only  looks  like half the harmonic mean because ab/(a+b) and  2 ab/(a+b) share the same algebraic form.

When the shortcut is safe to use

The formula x = ab/(a+b) applies whenever the problem has this exact structure:

- Equal amounts drawn from each container and  swapped ,
- Volumes end up unchanged,
- The final condition is  equal price/concentration ,
- The two starting prices are different.

Strip away any of those and the derivation breaks - so lean on the  structure , not on remembering "harmonic mean."

Answer: D

Gmatiseasyforall · Answer

With all due respect, Kindly provide the source of this question!

Asking, because my friend also gave me this question too from the same math quiz, that he gave me the other similar question that I posted yesterday and which was unfortunately archived. Probably both question are from the same math magazine!

I have never seen these type of questions on any GMAT book or forums or any relevant GMAT material ever!

Regards

Gmatiseasyforall · Answer

Hi e-gmat! Thanks for your answer! This question is actually built around the primary property of Harmonic mean with slight modifications.

A similar question can be found on the Hall and Knight Higher algebra book- miscellaneous examples section- Question no. 54. KarishmaB provided the same solution as provided by the official solutions written by the same authors. Maybe she used the same book.

But Hall and knight were not the actual creators of this question. This question is from the Cambridge college examination series around late 1840's. Hall and knight just appended the question in their book.

This question is build around the same question as may be found in the Harmonical progressions chapter of Hall and knight book
Q:9If b is the harmonic mean between a and c, prove that:- \frac{1}{b-a} + \frac{1}{b-c} = \frac{1}{a} + \frac{1}{c}

Even this question wasn't created by Hall and knight: this question already appeared in Cambridge Examination series around 1820s.

Cambridge just created a word problem out of the Fundamental property of Harmonic Mean. These are those type of question that requires analysis of the structure of the question and recognize the underlying pattern and logic involved in between those words.

Nevertheless, your insights were helpful too! Cause under examination conditions and time pressure when the mind doesn't work that deep, your approach might be useful which is btw a rote based learning, no offense! I have seen students, many IITians included, who get top ranks using that rote based approach. But I personally prefer an in depth look at the underlying logic and structure of the question, which btw is very time consuming and slower approach, but trust me, it pays!

Best

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