Equal amounts of water were poured into two empty jars of different ca

1/2 seems straight forward. If x amount of water fills up the jar upto 1/4th of its capacity, 2x will fill up 1/2.

My take on this (although Mavens like Bunuel can always expound in 2 lines)

Let v be the volume of water added to both :

v = x/4

v = y/3

x = 4v (total capacity of larger jar)

y = 3v (total capacity of smaller jar)

So v is added again to larger jar, hence total water = v+v
2v
--- = 1/2
4v

Capacity larger jar = x
Let 20 litters water poured into each jar
so x*1/4 = 20
x = 80
Total water of larger jar after pouring the water of smaller jar = 20+20 = 40
40/80 = 1/2
Ans. C

Hi Experts
Why below answer is wrong.....

Let the capacity of both jug X(small) & Y(big)
After pouring water... capacity of water both jug x/4 and y/3... as given x/4=y/3 so x=4y/3

now given smaller jug water poured into bigger made the equ- x/4+y/3=>3x+4y/12..putting x=4y/3...we got 2y/3 which mean 2/3 rd of y

Rgds
Prasannajeet

x/4=y/3, thus x is larger jar and y is smaller jar.

First determine which is the big one?

\(\frac{1}{4} th\) is the big container as equal quantities of water poured made it "less full"

Same quantity of water added twice, so multiply by 2

\(2 * \frac{1}{4} = \frac{1}{2}\)

Answer = C

I actually drew the jars to solve it, without calculating anything...

Just to give you a good laugh, I will describe how:

So, I first created one jar, and marked its 1/4th and 1/3rd. This showed me that if these were 2 jars, the one filled up to its 1/4th must be the bigger one.

Then I just drew 2 same jars, next to eac other, and marked 1/4th in one and 1/3rd in the other. I pulled the line that marked the 1/4th of one jar on the second jar, and saw that there was only a little it lower than the 1/3rd. The I realised that if I moved the rest to jar one, it would be filled half way... So, 1/2.

Mind that I am a psychologist... Sometimes, it is easier for me not to do the math!

Let’s say that one jar has a capacity of 16 liters and the other, 12 liters. So, when equal parts water are poured into each jar, one jar has 4 liters and the other has 4 liters. When the contents of the small jar (4 liters from the 12-liter jar) are poured into the larger jar (4 litters in the 16-liter jar), the larger jar is now 8/16 = 1/2 full.

Alternate Solution:

Consider only the larger jar. When the first amount of water is poured into it, the jar becomes ¼ full. Now, a second amount of water, which is actually equal to the first amount, is poured into the larger jar. This second amount of water is also equivalent to ¼ of the jar’s capacity. Thus, the jar is now ¼ + ¼ = ½ full.

Answer: C

One way to solve this is by assigning values that are divisible by both 3 and 4, let one jar be 36, the other one be 48.

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Equal amounts of water were poured into two empty jars of different capacities, which made one jar 1/4 full and other jar 1/3 full. If the water in the jar with lesser capacity is then poured into the jar with greater capacity, what fraction of the larger jar will be filled with water?

To ease the problem, let us first get rid of fractions and take the volume of water poured = LCM {3,4} = 12

For first jar, filled = (1/4)*V1 = 12 => V1 = 48
For second jar, filled = (1/3)*V2 = 12 => V2 = 36

If the water in the jar with lesser capacity is then poured into the jar with greater capacity, water in 1st jar = 12+12=24
fraction filled = 24/48=1/2

Hence C