Last visit was: 22 Dec 2024, 16:51 It is currently 22 Dec 2024, 16:51

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3266 [14]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Own Kudos [?]: 377 [1]
Given Kudos: 64
GRE 1: Q160 V152
Send PM
Intern
Intern
Joined: 07 Aug 2021
Posts: 13
Own Kudos [?]: 13 [1]
Given Kudos: 134
Send PM
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3266 [2]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Re: In the country of XYZ, the I-score of a politician is computed [#permalink]
2
Jayz007 wrote:
motion2020 wrote:
revised..

let's check change in \(ab^2c^3\)
\((1+3/5)*(1+2/5)^2*(1-1/5)^3=8/5 * 49/5^2 * 64/5^3 = (8*49*64)/5^6 = 1.6\)
Since I is inversely related with \(d^2\), every increase in I will result in decrease of \(d^2\), unless changes of \(a, b, c\) are outweighing the change in I. Change in I is 150% higher for Flower. Change in I is 2.5 vs changes of \(a, b, c\) equal to 1.6. Change requested is lower or higher, hence sq.root(2.5/1.6)=sq.root(25/10 * 10/16)=sq.root(5/4)=1.25 or 25%

Answer must be A[/It also says that x% is higher or lower what do we conclude from that do we take higher or lower ]


It also says that x% is higher or lower what do we conclude from that do we take higher or lower than the corresponding value of Mayor Flower
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Own Kudos [?]: 377 [0]
Given Kudos: 64
GRE 1: Q160 V152
Send PM
Re: In the country of XYZ, the I-score of a politician is computed [#permalink]
ok, in my previous post there was sq.root left over, though the value was not under sq.root - revised

IMO, higher/lower means ignore the direction of change (increase or decrease) and focus on the percentage value

Jayz007 wrote:
motion2020 wrote:
revised..

let's check change in \(ab^2c^3\)
\((1+3/5)*(1+2/5)^2*(1-1/5)^3=8/5 * 49/5^2 * 64/5^3 = (8*49*64)/5^6 = 1.6\)
Since I is inversely related with \(d^2\), every increase in I will result in decrease of \(d^2\), unless changes of \(a, b, c\) are outweighing the change in I. Change in I is 150% higher for Flower. Change in I is 2.5 vs changes of \(a, b, c\) equal to 1.6. Change requested is lower or higher, hence sq.root(2.5/1.6)=sq.root(25/10 * 10/16)=sq.root(5/4)=1.25 or 25%

Answer must be A[/It also says that x% is higher or lower what do we conclude from that do we take higher or lower ]
avatar
Intern
Intern
Joined: 13 Nov 2021
Posts: 1
Own Kudos [?]: 1 [0]
Given Kudos: 2
Send PM
Re: In the country of XYZ, the I-score of a politician is computed [#permalink]
motion2020 wrote:
revised..

let's check change in \(ab^2c^3\)
\((1+3/5)*(1+2/5)^2*(1-1/5)^3=8/5 * 49/5^2 * 64/5^3 = (8*49*64)/5^6 = 1.6\)
Since I is inversely related with \(d^2\), every increase in I will result in decrease of \(d^2\), unless changes of \(a, b, c\) are outweighing the change in I. Change in I is 150% higher for Flower. Change in I is 2.5 vs changes of \(a, b, c\) equal to 1.6. Change requested is lower or higher, hence sq.root(2.5/1.6)=sq.root(25/10 * 10/16)=5/4=1.25 or 25%

Answer must be A
Manager
Manager
Joined: 23 Sep 2023
Posts: 65
Own Kudos [?]: 15 [0]
Given Kudos: 59
Send PM
In the country of XYZ, the I-score of a politician is computed [#permalink]
motion2020 wrote:
revised..

let's check change in \(ab^2c^3\)
\((1+3/5)*(1+2/5)^2*(1-1/5)^3=8/5 * 49/5^2 * 64/5^3 = (8*49*64)/5^6 = 1.6\)
Since I is inversely related with \(d^2\), every increase in I will result in decrease of \(d^2\), unless changes of \(a, b, c\) are outweighing the change in I. Change in I is 150% higher for Flower. Change in I is 2.5 vs changes of \(a, b, c\) equal to 1.6. Change requested is lower or higher, hence sq.root(2.5/1.6)=sq.root(25/10 * 10/16)=5/4=1.25 or 25%

Answer must be A


motion2020 can you please explain why are we dividing the 2 numbers? Also, how are you getting 64/5^3? Isn't it supposed to be 512/5^3 ??
Verbal Expert
Joined: 18 Apr 2015
Posts: 30475
Own Kudos [?]: 36821 [0]
Given Kudos: 26100
Send PM
Re: In the country of XYZ, the I-score of a politician is computed [#permalink]
Expert Reply
he simply put into formula the decrease pf 20%

\((1-1/5)^3\)

or you can simply write instead of fractions which is more complicated, the direct % decrease

0.80

As for the fraction

4/5^3=64/5. I believe he disregarded to raise to the power of three the 5 because all the fractions are with a denominator of 5. we care about the numerator
Prep Club for GRE Bot
Re: In the country of XYZ, the I-score of a politician is computed [#permalink]
Moderators:
GRE Instructor
88 posts
GRE Forum Moderator
37 posts
Moderator
1115 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne