Is x 1/5 ? (1) x 1/23 + 1/24 + 1/25 + 1/26 + 1/27 (2) x

Question

Is  x > \frac{1}{5}  ?

(1)  x > \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27}

(2)  x < \frac{1}{22} + \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26}

chetan2u · Accepted Answer

This basically asks Is x > \frac{1}{5} =Is x > \frac{5}{25} (1) x > \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} Now \frac{1}{23} , \frac{1}{24} is surely greater and \frac{1}{26} , \frac{1}{27} surely lesser than \frac{1}{25} .. If distance of larger terms is GREATER than the distance of smaller terms from 1/25, then YES.. So \frac{1}{25}-\frac{1}{26}=\frac{1}{25*26} and \frac{1}{24}-\frac{1}{25}=\frac{1}{25*24} and \frac{1}{24*25}>\frac{1}{25*26} , when we compare the denominator... Similarly \frac{1}{25}-\frac{1}{27}=\frac{2}{25*27} and \frac{1}{23}-\frac{1}{25}=\frac{1}{25*23} and \frac{1}{23*25}>\frac{1}{25*27} , when we compare the denominator... Since distance of larger terms is GREATER than the distance of smaller terms from 1/25, YES.. (2) x < \frac{1}{22} + \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} here we can say similarly x<5/24, but it can be between 4/25 and 5/25..YES or x could be 0 or <5/25...NO Insuff A

nick1816 · Answer

Statement 1

2a^2>2a^2-2

2a^2>2(a^2-1)

\frac{2a^2}{(a^2-1)}>2 ......for a>1

\frac{2a}{(a^2-1)} > \frac{2}{a}

\frac{2a}{(a-1)(a+1)} > \frac{2}{a}

\frac{1}{a+1} +\frac{1}{a-1}  >  \frac{1}{a}+\frac{1}{a}

Similarly we can prove that
 \frac{1}{a+2} +\frac{1}{a-2}  >  \frac{1}{a}+\frac{1}{a}

Hence
 \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} > \frac{1}{25} + \frac{1}{25} + \frac{1}{25} + \frac{1}{25} + \frac{1}{25}

\frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} >  \frac{5}{25}

\frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} > \frac{1}{5}

Sufficient

Statement 2

x<\frac{5.abc}{25}

Clearly insufficient

Mo2men · Answer

Dear  GMATGuruNY

Ca you please shed light on how to solve such question?

Thanks in advance

GMATGuruNY · Answer

Test an EASY CASE: \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{30+20+15+12+10}{60} = \frac{87}{60} If replace the two greatest values and the two smallest values with the MIDDLE TERM -- \frac{1}{4} -- we get: \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 5*\frac{1}{4} = \frac{5}{4} = \frac{75}{60} Since the top sum is greater than the bottom sum, we can generalize as follows: In each of the statements below, the sum of the 5 fractions will be GREATER THEN 5 TIMES THE MIDDLE TERM. Statement 1: x > a value greater than 5*\frac{1}{25} x > a value greater than \frac{1}{5} Thus, the answer to the question stem is YES. SUFFICIENT Statement 2: x < a value greater than 5*\frac{1}{24} x < a value greater than \frac{5}{24} If x = \frac{5}{24} , the answer to the question stem is YES. If x = 0, the answer to the question stem is NO. INSUFFICIENT A

Is x > 1/5 ? (1) x > 1/23 + 1/24 + 1/25 + 1/26 + 1/27 (2) x < 1/22 +

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