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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  Committees X, Y and Z have at least three members each. No two of thes

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VP  D
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1009
WE: Supply Chain Management (Energy and Utilities)
Committees X, Y and Z have at least three members each. No two of thes  [#permalink]

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5 00:00

Difficulty:   95% (hard)

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
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1210
Location: India
GPA: 3.82
Committees X, Y and Z have at least three members each. No two of thes  [#permalink]

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1
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 $$x>y$$. 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
Manager  G
Joined: 21 Jun 2017
Posts: 234
Concentration: Finance, Economics
WE: Corporate Finance (Commercial Banking)
Re: Committees X, Y and Z have at least three members each. No two of thes  [#permalink]

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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 $$x>y$$. 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

Loved 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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Intern  B
Joined: 29 Apr 2017
Posts: 29
Location: India
Concentration: Finance, Accounting
Re: Committees X, Y and Z have at least three members each. No two of thes  [#permalink]

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A is sufficient to answer, B - atmost can go either ways
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1210
Location: India
GPA: 3.82
Committees X, Y and Z have at least three members each. No two of thes  [#permalink]

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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 $$x>y$$. 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

Loved 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 ShankSouljaBoi

if 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.
SVP  V
Joined: 26 Mar 2013
Posts: 2229
Committees X, Y and Z have at least three members each. No two of thes  [#permalink]

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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'.
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1210
Location: India
GPA: 3.82
Re: Committees X, Y and Z have at least three members each. No two of thes  [#permalink]

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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   [#permalink] 25 Dec 2018, 11:41
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