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Committees X, Y and Z have at least three members each. No two of thes
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11 Aug 2018, 04:51
Question Stats:
41% (03:00) correct 59% (02:37) wrong based on 78 sessions
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Committees X, Y and Z have at least three members each. No two of these committees, have any common members. The average (arithmetic mean) ages of the members of X, Y and Z are 30 years, 35 years and 40 years respectively. The average age of the members of X, Y and Z together is 35 years. Does X have more members than Y? (1) The average age of the members of Y and Z together is at least 38 years (2) The average age of the members of X and Y together is at most 33 years. Source: Time4education
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Committees X, Y and Z have at least three members each. No two of thes
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11 Aug 2018, 05:26
PKN wrote: Committees X, Y and Z have at least three members each. No two of these committees, have any common members. The average (arithmetic mean) ages of the members of X, Y and Z are 30 years, 35 years and 40 years respectively. The average age of the members of X, Y and Z together is 35 years. Does X have more members than Y? (1) The average age of the members of Y and Z together is at least 38 years (2) The average age of the members of X and Y together is at most 33 years.
Source: Time4education let the number of members in Committees X, Y & Z be \(x, y\) & \(z\) respectively. need to find whether \(x>y\) given \(30x+35y+40z=35(x+y+z) =>30x+35y+40z=35x+35y+35z\) \(=>x=z\) Statement 1: implies \(35y+40z ≥ 38(y+z)=>2z≥3y\) and as \(x=z\), this implies \(=>x≥\frac{3}{2}y\). Hence \(x>y\). SufficientStatement 2: implies \(30x+35y≤33(x+y)\) \(=>2y≤3x => x≥\frac{2}{3}y\). So \(x>y\) or \(x<y\). InsufficientOption A



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Re: Committees X, Y and Z have at least three members each. No two of thes
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11 Nov 2018, 00:00
niks18 wrote: PKN wrote: Committees X, Y and Z have at least three members each. No two of these committees, have any common members. The average (arithmetic mean) ages of the members of X, Y and Z are 30 years, 35 years and 40 years respectively. The average age of the members of X, Y and Z together is 35 years. Does X have more members than Y? (1) The average age of the members of Y and Z together is at least 38 years (2) The average age of the members of X and Y together is at most 33 years.
Source: Time4education let the number of members in Committees X, Y & Z be \(x, y\) & \(z\) respectively. need to find whether \(x>y\) given \(30x+35y+40z=35(x+y+z) =>30x+35y+40z=35x+35y+35z\) \(=>x=z\) Statement 1: implies \(35y+40x ≥ 38(y+z)=>2z≥3y\) \(=>x≥\frac{3}{2}y\). Hence \(x>y\). SufficientStatement 2: implies \(30x+35y≤33(x+y)\) \(=>2y≤3x => x≥\frac{2}{3}y\). So \(x>y\) or \(x<y\). InsufficientOption AHi niks18Loved your approach ! I analyzed it instead and took me three minutes. Can you please explain to me how in statement 1 you were sure that x>y while in statement 2 x<y or x>y I have missed a lot of questions because of this. This is what i havae always applied (is probably wrong) , in an equality x=5y . x>y since y needs to be multiplied by a larger number to balance out the x. For inequalities this does not work at all. This was quite evident with this one. Please guide !!! TIA
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Re: Committees X, Y and Z have at least three members each. No two of thes
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11 Nov 2018, 03:21
A is sufficient to answer, B  atmost can go either ways



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Committees X, Y and Z have at least three members each. No two of thes
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11 Nov 2018, 05:35
ShankSouljaBoi wrote: niks18 wrote: PKN wrote: Committees X, Y and Z have at least three members each. No two of these committees, have any common members. The average (arithmetic mean) ages of the members of X, Y and Z are 30 years, 35 years and 40 years respectively. The average age of the members of X, Y and Z together is 35 years. Does X have more members than Y? (1) The average age of the members of Y and Z together is at least 38 years (2) The average age of the members of X and Y together is at most 33 years.
Source: Time4education let the number of members in Committees X, Y & Z be \(x, y\) & \(z\) respectively. need to find whether \(x>y\) given \(30x+35y+40z=35(x+y+z) =>30x+35y+40z=35x+35y+35z\) \(=>x=z\) Statement 1: implies \(35y+40x ≥ 38(y+z)=>2z≥3y\) \(=>x≥\frac{3}{2}y\). Hence \(x>y\). SufficientStatement 2: implies \(30x+35y≤33(x+y)\) \(=>2y≤3x => x≥\frac{2}{3}y\). So \(x>y\) or \(x<y\). InsufficientOption AHi niks18Loved your approach ! I analyzed it instead and took me three minutes. Can you please explain to me how in statement 1 you were sure that x>y while in statement 2 x<y or x>y I have missed a lot of questions because of this. This is what i havae always applied (is probably wrong) , in an equality x=5y . x>y since y needs to be multiplied by a larger number to balance out the x. For inequalities this does not work at all. This was quite evident with this one. Please guide !!! TIA Hi ShankSouljaBoiif you follow the steps then from statement 1 it is evident that \(=>x≥\frac{3}{2}y => x≥1.5y\). Here x, y & z are all positive. so whatever the value of y will be x will be greater than or equal to 1.5 times y. so clearly x>y similar approach was applied in statement 2. let me know if this helps or you have any other confusion.



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Committees X, Y and Z have at least three members each. No two of thes
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25 Dec 2018, 10:03
niks18 wrote: [
let the number of members in Committees X, Y & Z be \(x, y\) & \(z\) respectively. need to find whether \(x>y\)
given \(30x+35y+40z=35(x+y+z) =>30x+35y+40z=35x+35y+35z\)
\(=>x=z\)
Statement 1: implies \(35y+40x ≥ 38(y+z)=>2z≥3y\)
\(=>x≥\frac{3}{2}y\). Hence \(x>y\). Sufficient
Statement 2: implies \(30x+35y≤33(x+y)\)
\(=>2y≤3x => x≥\frac{2}{3}y\). So \(x>y\) or \(x<y\). Insufficient
Option A Hi niks18, Great way. Small typo highlighted in statement 1. It should be 'z'.



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Re: Committees X, Y and Z have at least three members each. No two of thes
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25 Dec 2018, 11:41
Mo2men wrote: niks18 wrote: [
let the number of members in Committees X, Y & Z be \(x, y\) & \(z\) respectively. need to find whether \(x>y\)
given \(30x+35y+40z=35(x+y+z) =>30x+35y+40z=35x+35y+35z\)
\(=>x=z\)
Statement 1: implies \(35y+40x ≥ 38(y+z)=>2z≥3y\)
\(=>x≥\frac{3}{2}y\). Hence \(x>y\). Sufficient
Statement 2: implies \(30x+35y≤33(x+y)\)
\(=>2y≤3x => x≥\frac{2}{3}y\). So \(x>y\) or \(x<y\). Insufficient
Option A Hi niks18, Great way. Small typo highlighted in statement 1. It should be 'z'. Hi Mo2men, Thanks for highlighting. Edited it




Re: Committees X, Y and Z have at least three members each. No two of thes
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25 Dec 2018, 11:41






