Please Explain

Thank you for the explanation Sir

@Sujoykumardatta please explain how 30th percentile and 70th percentile terms have been calculated

Set A contains 350 values
30% of 350 = 105
So, a 30th percentile score is greater than 105 values in the set.
So, if we list all 350 values in ascending order, the 106th value is in the 30th percentile

Set B contains 200 values
70% of 200 = 140
So, a 70th percentile score is greater than 140 values in the set.
So, if we list all 200 values in ascending order, the 141st value is in the 70th percentile

Cheers,
Brent

A different approach. Assume the median for set B = k, the median for A will be k + 200 (round 199.5)
30th in A: k + 200 - 350*(.5 - .3)*2 = k + 60
70th in B: k + 200*(.7 - .5)*3 = k + 120
Therefore the difference is about 60. Caution that this is a rough calculation.

Sir, I think you've made a typo. The first case is for Set 'A', but you have made it 'B'.

Fixed - thanks for the heads up!!
Kudos for you!!!!

The median of set A is the 175th value = x + 199.5
The median of set B is the 100th value = x

Since set A contains 350 terms, the 30th percentile is the 350*.3 = 105th term in the list
This is 70 terms below the median (175th value - 105th value = 70 terms away).
We are told that set A counts by twos, so to get this 30th percentile value, which is below the median, we calculate
x + 199.5 - 70*2 = x + 59.5
If you wanted to be really precise, you would subtract one more term, so that we're defining what BELOW the 105th term, but as we'll see below, the two calculations aren't even close, so I won't bother getting this detailed.

Since set B contains 200 items, the 70th percentile is the 200*.7 = 140th term in the list.
This is 40 terms away form the median (140th value - 100th value = 40 terms away)
We are told that set B counts by threes, so to get to this 70th percentile value, which is above the median, we calculate
x + 40*3 = x +120
Again, we could add one less term just to be precise about what is BELOW the 140th term, but since the values aren't even close between A and B, we don't need to be this close in our calculation.

x + 59.5 < x + 120, so quantity B is bigger.

Hi there sujoykrdatta! I was wondering where you got 210 and 420.

The 30th percentile of 350 terms would be 0.3(350) = 105. As it's a multiple of 2, we get 2(105) = 210.

The 70th percentile of 200 terms would be 0.7(200) = 140. As it's a multiple of 3, we get 3(140) = 420.

Hope this helps.

The median of Set A is 199.5 greater than the median of Set B.

means that [(a175+a176)/2] - [(b100+b101)/2] = 199.5
-> a175 + a176 - b100 - b101 = 399

A Set: 30% of 350 = 105
B Set: 70% of 200 = 140

a175 + a176 - b100 - b101 = 399
-> (a105 + 2x70) + (a105 + 2x71) - (b140 - 3x40) - (b140 - 39x3) = 399
-> 2(a105) - 2(b140) + 140 + 142 + 120 + 117 = 399
-> 2(a105) - 2(b140) = -120
-> a105 - b140 = -60

Therefore, b140 > a105
The answer is B

I think, there's a simpler solution than what's been given. Please feel free to correct me:

A: 30th percentile of 350 => the term that's >= 30% of 350 = 106th term. => (2)^106
Similarly, B: 150th term => (3)^150

Clearly, 3^150 >> 2^106 !

However, I haven't used the second the fact (difference of the medians) at all.

Here, the 30th percentile of A should be 105, so the equation should be 2(x+104) = 2x + 210 and 70th percentile of B should be the 140th term which equals 3(y + 139) = 3y + 420 according to your spec. Please clarify if my interpretation is missing something and thank you for your explanation

The 30th percentile score is greater than 105 values in the set. Therefore should 106

Also when we have 2(x+104) x is at least 1 so we have 106 per logic

Is this method a good way?

this is best

https://gre.myprepclub.com/forum/set-a- ... tml#p51028

Lets assume there are only 2 grades in exam record pool. 50th percentile is the grade of the lower one which is first right its not the second one. why do we make 105+1 i dont get it