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Comment on Three Flasks
I did find the solution in
Unfortunately, that approach
Unfortunately, that approach is incorrect; it just happened to work with the numbers given in the question.
If we were to change the question so that one flask was 1/6 full, one flask was 1/7 full, and one flask was 1/12, the correct answer would still be 1/6. However, your approach would yield an incorrect answer.
I used this approach: since
That's a valid approach -
That's a valid approach - nice work!
This seems to be an another
1.) Since we know that all three have equal quantities of water initially and we have removed 1/2 of 1st two flasks.
2.) 2 halfs = 1/12 of 3rd flask.
3.) New water level in the 3rd flask = 1/12(newly added) + 1/12 (previous level) = 2* 1/2 = 1/6.
Thanks
There are several ways to
There are several ways to solve this question. Your solution is perfect.
hi sir,
i have a doubt in this question. actually u said with highest denominator is the smallest value. but, in the above question it is incorrect. 1/12 is the largest flask in above question.
I'm not sure what part of the
I'm not sure what part of the video you're referring to. I re-watched the video twice, and I couldn't find any part where I said (or even suggested) that "the highest denominator is the smallest value."
What I did was I realized
That works for me!
That works for me!
I dont understand why you say
They key idea is that we're
They key idea is that we're pouring an EQUAL VOLUME of water into all three containers. The greater the capacity of a container, the smaller the fraction that it is full.
For example, if you pour 1 liter of water into a certain vase, it becomes 1/2 full. If you pour 1 liter of water into a certain swimming pool, it becomes 1/10,000 full.
Since the pool has a much greater capacity than the vase, the fraction 1/10,000 is much less than 1/2
Hi Brent, I understand your
As I thought 1/12 is the smallest contatainer here compare to the other two? Therefore not sure why are we pouring 1/2 of water from the 1/6 and 1/8 into 1/12? Thanks Brent
We are told that each flask
We are told that each flask contains the same volume of water.
So, let's say that each flask contains 1 liter of water.
So, if the CAPACITY of flask A is 6 liters, then flask A will be 1/6 full.
If the CAPACITY of flask B is 8 liters, then flask B will be 1/8 full.
If the CAPACITY of flask C is 12 liters, then flask C will be 1/12 full.
Among the three flasks, which one is the biggest (i.e., which flask has the greatest capacity?)
The flask with a capacity of 12 liters will be the biggest flask.
Oh my god...this is such a
Noted now thanks Brent
hey just a question for a
we cannot just simply add 50% of 1/6 and 50% of 1/8 to 1/12 since we do not know the individual capacities of the three flasks, right ?
No that won't work. Keep in
No that won't work. Keep in mind that the two smaller flasks each contain the SAME volume of water. So, 50% of the volume in one small flask must be EQUAL to 50% of the volume in one small flask.
So, based on your suggestion, it would have to be the case that 50% of 1/6 = 50% of 1/8, however those two values are NOT equal.
Hi Brent! Thanks for the
I'd place it around 700-level
I'd place it around 700-level.
Hey, please help me to
1/6 = 8/48 = Flask one
1/8 full = 6/48 = Flask two
1/12 ful = 4/48 = Flask three (the biggest)
When i poure now one half from 14/48 (8/48 + 6/48) into the biggest one, my solution is the following: 7/48 + 4/48 = 11/48
Thanks !
The problem is that you are
The problem is that you are adding fractions when you need to be adding actual volumes. When you add fractions, you are assuming that each vase has the same capacity, which they don't.
Here's what I mean:
Let's say that a vase can hold 2 liters in total and a pool can hold 2,000,000 liters in total.
Now let's say that both the vase and the pool are half full of water. That is, the vase contains 1 liter of water, and the pool contains 1,000,000 liters of water.
Now, what happens if we pour half of the water from the vase into the pool?
CORRECT SOLUTION: Adding volumes
Half of 2 liters = 1 liter. So, I'm pouring 1 liter of water into the pool, which means the pool now holds 1,000,001 liters of water.
So, 1,000,001/2,000,000 = the fraction of the pool that is now full.
INCORRECT SOLUTION: Adding fractions
The vase is 1/2 full and the pool is 1/2 full.
We can also say that the vase is 2/4 full
So, HALF of the vase's water = 1/4
When we add the vase's 1/4 to the pool's 1/2, we get: 1/2 + 1/4 = 3/4
Can we now say that the pool is 3/4 full? No.
Does that help?
Cheers,
Brent
my way of answering this
I assume that the total capacity of each flask is the denominator
So in the 1/12, I assume that the capacity is 12 liters (for example) and it contains 1 liter of water.
In the 1/8, I assume that the capacity is 8 liters (for example) and it contains 1 liter of water.
In the 1/6, I assume that the capacity is 6 liters (for example) and it contains 1 liter of water.
If we take half of both tanks 1/8 and 1/6, we get 1 liter only
Adding this 1 liter to 1/12, we got 2/12 which is equal to 1/6
What do you think about this approach?
That's a perfectly valid
That's a perfectly valid approach. Nice work!
Best approach for my
Thank You
https://gmatclub.com/forum/in
pls help with this sir
Here's my full solution:
Here's my full solution: https://gmatclub.com/forum/in-a-200-member-association-consisting-of-men...
I absolutely LOVE this
Thank you Brent. Thank you very much!
Hey, Could you please guide
Here's my full solution:
Here's my full solution: https://gmatclub.com/forum/the-only-articles-of-clothing-in-a-certain-cl...
Cheers,
Brent
Hi Brent,
The way I tried to solve it in the first place is as follows:
1/6 1/8 1/12
1/2 * 1/6 = 1/12
1/2 * 1/8 = 1/16
Then I added these fractions to the larger flask (1/12)
1/12 + 1/16 + 1/12 = 4/48 + 3/48 + 4/48 = 11/48
As you can see, I did not find a solution.
Now your solution makes perfect sense, I was just curious where I went wrong and if the way I tried is possible as well?
Thanks!
Kind Regards,
Glenn
Hi Glenn,
Hi Glenn,
That's a good idea.
However, this approach will require us the assign some variables to the capacity of each flask.
For example, we might says that:
x = capacity of smallest flask
y = capacity of medium-sized flask
z = capacity of largest flask
At first, each flask contains EQUAL VOLUMES.
So, we can write: (1/6 of x) = (1/8 of y) = (1/12 of z)
So, one-half of the liquid in the smallest flask = (1/2) of (1/6 of x)
= (1/2) of (1/12 of z)
= (1/24) of z
And, one-half of the liquid in the medium flask = (1/2) of (1/8 of y)
= (1/2) of (1/12 of z)
= (1/24) of z
------------------------
(1/24) of z + (1/24) of z = (1/12) of z
So, we are adding (1/12) of z to (1/12) of z
We get (1/6) of z
Does that help?
Cheers,
Brent
Hi Brent,
Yes, all clear!
Thanks!
Kind regards,
Glenn
Yeah I did the exact same
Agreed - that's the key
Agreed - that's the key
A certain library assesses
$0.60
$0.70
$0.80
$0.90
$1.00
How is the answer 0.7?
Here's my full solution:
Here's my full solution: https://gmatclub.com/forum/a-certain-library-assesses-fines-for-overdue-...
Cheers,
Brent
Hi Brent,
I understand that since equal volumes of water was poured into three flasks, the fraction of the flasks filled with water didn't matter. So it was just half + half = one.
But what if the flasks were filled with unequal amount of water ? Would this be a valid question and if so how would be solve this?
Good question. If the flasks
Good question. If the flasks were filled with UNEQUAL amounts of water, then the question would be unsolvable.
I found another way of
A= 24/144 (1/6)
B= 24/192 (1/8)
C= 24/288 (1/12)
I took half of the two smaller flasks (24) in the largest flask and got 48/288 which is 1/6
That's a perfectly valid
That's a perfectly valid solution. Nice work!!!
This was a fun thought
Since one equal portion of water raises the level to 1/12ths full, if all of the water was poured into the flask it would be (1+1+1)/12ths or 3/12ths full. But since we're only adding one full portion and two half portions, its (1+.5+.5)/12ths or 2/12ths full, which simplifies to 1/6th.
Perfectly reasoned, Hyord.
Perfectly reasoned, Hyord.
Am I wrong with the below
Since the volume remains same so,
1/6+1/8+1/12 = 1/12+1/16(since halved) +(?)
Now I shall solve for ?
although I am not getting the reqd. ans. let me know if this approach seems to be right or wrong?
The problem is that 1/6, 1/8
The problem is that 1/6, 1/8 and 1/12 aren't individual fractions; they're fractions OF some capacity.
If we let A, B and C be the CAPACITIES of the 3 flasks, then we can write:
(1/6 of A) + (1/8 of B) + (1/12 of C) = (1/12 of A) + (1/16 of B) + (x of C).
Unfortunately, I don't believe this equation can be solved for x.
I get a different answer.
I understood from the problem that 1/2 of the water from each of the two smallest flasks were supposed to be 0.5/6 or 4/24 from the 1/6 and 1.5/24 from 1/8. Is it not correct to pull out half of the water from each flask by calculating half of the total water in each flask? Why can we assume that the problem would like us to use half of the water the largest flask here?
That's a good idea, but there
That's a good idea, but there are issues.
We can't say that half the water in the first flask = 4/24 and half the water in the second flask is 1.5/24.
Remember, each flask contains the same amound of water.
So, half the volume in the first flask must equal half the water in the second flask.
Using your values it must be true that 4/24 = 1.5/24,(which is clearly false.)
Using your approach will require us the assign some variables to the capacity of each flask.
For example, we might says that:
x = capacity of smallest flask
y = capacity of medium-sized flask
z = capacity of largest flask
At first, each flask contains EQUAL VOLUMES.
So, we can write: (1/6 of x) = (1/8 of y) = (1/12 of z)
So, one-half of the liquid in the smallest flask = (1/2) of (1/6 of x)
= (1/2) of (1/12 of z)
= (1/24) of z
And, one-half of the liquid in the medium flask = (1/2) of (1/8 of y)
= (1/2) of (1/12 of z)
= (1/24) of z
------------------------
(1/24) of z + (1/24) of z = (1/12) of z
So, we are adding (1/12) of z to (1/12) of z
We get (1/6) of z
Does that help?
Cheers,
Brent