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Re: In the country of XYZ, the I-score of a politician is computed [#permalink]
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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
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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 ]
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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
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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 ??
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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
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Re: In the country of XYZ, the I-score of a politician is computed [#permalink]
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