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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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Please Explain
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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Thank you for the explanation Sir
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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@Sujoykumardatta please explain how 30th percentile and 70th percentile terms have been calculated
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Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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upretidu wrote:
@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
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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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.
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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GreenlightTestPrep wrote:
upretidu wrote:
@Sujoykumardatta please explain how 30th percentile and 70th percentile terms have been calculated


Set B 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


Sir, I think you've made a typo. The first case is for Set 'A', but you have made it 'B'.
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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SivhHarish wrote:
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!!
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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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.
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
Hi there sujoykrdatta! I was wondering where you got 210 and 420.
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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ashelyt wrote:
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.
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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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
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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Carcass wrote:
Set A consists of 350 consecutive multiples of 2.

Set B consists of 200 consecutive multiples of 3.

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


Quantity A
Quantity B
The \(30^{th}\) percentile of Set A
The \(70^{th}\) percentile of Set B



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.



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.
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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Re: Set A consists of 350 consecutive multiples of 2. Set B cons [#permalink]
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