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akhil911
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Updated on: 25 Mar 2016, 04:55
Originally posted by akhil911 on 18 May 2014, 05:04.
Last edited by Nevernevergiveup on 25 Mar 2016, 04:55, edited 3 times in total.
Last edited by Nevernevergiveup on 25 Mar 2016, 04:55, edited 3 times in total.
Edited the question and original answer was incorrect.
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A set contains the following 5 numbers: \(a^2, a^5, a, \frac{a}{2}, \frac{a}{5}\). Is the range of the set equal to \(a^2 - a^5\)?
(1) a is negative
(2) \(a^5 < a\)
(1) a is negative
(2) \(a^5 < a\)
25 Mar 2016, 05:25
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A set contains the following 5 numbers: \(a^2, a^5, a, \frac{a}{2}, \frac{a}{5}\). Is the range of the set equal to \(a^2 - a^5\)?
(1) a is negative
(2) \(a^5 < a\)
Concept as per OE is
I tried this way which looks simple enough to me.
Statement 1: a is negative number i.e., a can be -1, -2, ...........so on.
Let a=-1
The 5 no's will be
\(a^2=1,\)
\(a^5=-1,\)
\(a=-1,\)
\(\frac{a}{2}=\frac{-1}{2},\)
\(\frac{a}{5}=\frac{-1}{5}\)
Arranged order will be
\(-1, -1, -1/2, -1/5, 1\) i.e.,
The order is
Since there is some uncertainty, we try the same procedure with a=-2
Arranged order will be
\(-32, -2, -1, -2/5, 4\) i.e.,
The order is
\(a^5, a, \frac{a}{2}, \frac{a}{5}, a^2\) giving us the range \(a^2 - a^5\) and answer YES to the question.
Now we can say Statement 1 is not sufficient.
Statement 2: \(a^5 < a\) this gives us the required hint that a cannot be -1 as above.
But this can be positive as well and be within (0, 1).
Thus Statement 2 is insufficient.
Combining both the statements we get the answer YES for all the values of a<-1.
(1) a is negative
(2) \(a^5 < a\)
Concept as per OE is
Quote:
I tried this way which looks simple enough to me.
Statement 1: a is negative number i.e., a can be -1, -2, ...........so on.
Let a=-1
The 5 no's will be
\(a^2=1,\)
\(a^5=-1,\)
\(a=-1,\)
\(\frac{a}{2}=\frac{-1}{2},\)
\(\frac{a}{5}=\frac{-1}{5}\)
Arranged order will be
\(-1, -1, -1/2, -1/5, 1\) i.e.,
The order is
- either \(a^5, a, \frac{a}{2}, \frac{a}{5}, a^2\) giving us the range \(a^2 - a^5\) and answer YES to the question.
or \(a, a^5, \frac{a}{2}, \frac{a}{5}, a^2\) giving us the range \(a^2 - a\) and answer NO to the question.
Since there is some uncertainty, we try the same procedure with a=-2
Arranged order will be
\(-32, -2, -1, -2/5, 4\) i.e.,
The order is
\(a^5, a, \frac{a}{2}, \frac{a}{5}, a^2\) giving us the range \(a^2 - a^5\) and answer YES to the question.
Now we can say Statement 1 is not sufficient.
Statement 2: \(a^5 < a\) this gives us the required hint that a cannot be -1 as above.
But this can be positive as well and be within (0, 1).
Thus Statement 2 is insufficient.
Combining both the statements we get the answer YES for all the values of a<-1.
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06 Dec 2015, 09:57
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basically this problem boils down to arranging the set in a single way. to do this we need to consider basic number properties. the stem says that the set can consist of any numbers: integers, fractions, positives, negatives - all those inputs would influence the order of the set. so we need to come to one possible interval of values to have an exclusive set
STATEMENT 1: says a is negative, but this cannot be sufficient to arrange our set in one exclusive way because the numbers can still be either integers or fractions (less than -1 or greater) or both and THUS the set could take any order. you can try several numbers to test this or draw a number line.
a<-1 or -1<a<0 produce different sets
STATEMENT 2: says that a is either less than -1 or between 0 and 1, i.e. any number less than -1 or a positive fraction between 0 and 1. again this can produce different sets hence not sufficient.
a^5-a<0 -> a(a-1)(a+1)(a^2+1)<0 -> hence for the inequality to be true -> a<-1 or 0<a<1 give different sets
1+2 = sufficient because now it explicitly says that a is a negative number less than -1 that is a<-1 and hence the set can be organized in a single fashion. you can try different numbers to test this. integers fractions.
STATEMENT 1: says a is negative, but this cannot be sufficient to arrange our set in one exclusive way because the numbers can still be either integers or fractions (less than -1 or greater) or both and THUS the set could take any order. you can try several numbers to test this or draw a number line.
a<-1 or -1<a<0 produce different sets
STATEMENT 2: says that a is either less than -1 or between 0 and 1, i.e. any number less than -1 or a positive fraction between 0 and 1. again this can produce different sets hence not sufficient.
a^5-a<0 -> a(a-1)(a+1)(a^2+1)<0 -> hence for the inequality to be true -> a<-1 or 0<a<1 give different sets
1+2 = sufficient because now it explicitly says that a is a negative number less than -1 that is a<-1 and hence the set can be organized in a single fashion. you can try different numbers to test this. integers fractions.
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