A set contains the following 5 numbers: a^2, a^5, a, a/2, a/5. Is the

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

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.

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.

Hi Akhil

IMO it should be B.

As a^5 is less than a, it means that a is between 0 and 1. Hence range of the set would be "a-a^5", which makes it sufficient to answer. Can you please explain your reasoning.

Thanks in advance.

For Statement 2, What if a < -1 In that case as well Range can be a^2> a^5 ??

Since there are 5 values it's important to make a table below



Combining both statement you see that a is negative and a<-1

Ans is C
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Hi Ankit,
It would be
There will be two cases for
a^5 < a
Case i) 0< a < 1 , which you pointed out; In this case a^5 will be even lesser fraction of the fraction a
Case ii) a < -1 ; in this case , a^5 and a both are negative numbers but a^5 is of higher magnitude than a

So,you see the statement 2 alone does not suffice to find out the range for the set which would differ in above both cases.

So, now checking each statement :
1) a is negative i.e a<0, for which range would be different for the cases
(i) if -1 <= a < 0
(ii) if a< -1 ; So,not sufficient

2) a^5<a for which range would be different for the cases
(i) if 0 < a <1
(ii) if a < -1 ; So,not sufficient

However,combining these two statements yield
a < -1, in which case we can find the range. Range would be a^2 - a^5

Kudos is the best form of appreciation

Can anyone else give it a try? To be honest I still don't really understand it, even with the already posted solutions...

the question is asking whether a^2 is the largest number in the set and whether a^5 is the smallest.

from 1

since a is -ve we know a^2 is largest but we cant be sure whether a^5 is smallest ( true if a is integer for example and false if a is a -ve fraction).. insuff

from 2

a^5< a .... a^5 - a < 0 , i.e. a(a^4 - 1) < 0 this is true if a is -ve and larger than 1 or a is a +ve fraction... insuff

both together

a is a -ve fraction larger than 1 thus surely range is a^2 - a^5
C

A set contains the following 5 numbers: a2,a5,a,a2,a5a2,a5,a,a2,a5. Is the range of the set equal to a2−a5a2−a5?

(1) a is negative
(2) a5<a

For statement 1 . the set range is differenent when we select value less than - 1 and when we select value between -1 and 0
statement 2 a(a^4-1)< 0
a(a^2+1) (a+1)(a-1)<0
this will hold when a< -1 or a is between 0 an 1

combining statement 1 and 2
answer is C

Unable to understand how 0<a<1 was conceived.

If a(a^4-1)< 0
a(a^2+1) (a+1)(a-1)<0

correct me if i am wrong ; a<0 else a<-1 else a<1
How do you get a is between 0 and 1 ??? ( i understand the portion a<-1)

IMO its C
statement 1 says A is negative , but we dont now if a is in fraction form or integer form - insufficient
statement 2- a^5 < a , which means that a could be fraction or negative but not both-insufficient
stat 1 + stat 2 - a is negative and is not in fraction form

Let me give it a go.

(1) a is negative, then a can be negative integer or negative fraction. Lets take 2 examples, a=-2 and a=-1/2
For a=-2, the sets can be arranged as a^5, a, a/2, a/5, a^2.. So range is given by a^2 - a^5. hence answer to the question is YES
But for a=-1/2, the sets can be arranged as a, a/2, a/5, a^5, a^2.. So range is given by a^2 - a. Hence answer to the question is NO
INSUFFICIENT

(2) a^5 < a; this means a is either negative integer or a positive fraction is possible
We have already solved for negative integer above, and the answer was YES
Positive fraction, lets say a = 1/2, the sets can be arranged as a^5, a/5, a/2, a^2, a.. So range is given by a - a^5. Hence answer to the question is NO
INSUFFICIENT

Combining (1) & (2)
a MUST be a negative integer, and we have already solved this before which give the answer YES

Hence, the answer is C

is it really "95% hard" category question?

Cannot understand this part..........But for a=-1/2, the sets can be arranged as a, a/2, a/5, a^5, a^2.. So range is given by a^2 - a. Hence answer to the question is NO
. wouldn't this give us a^5 -1/32, which is the smallest of all?

Harps
See the highlighted part again. For a = -1/2,
a^2 = 1/4 and a^ 5 = -1/32(is this smallest or is this 2nd largest )
Thus,
a^2 - a = 1/4 - (-1/2) = 1/4 + 1/2 = 3/4 * 1
a^2 - a^5 = 1/4 - (-1/32) = 3/4 * 3/8

Which one is largest.?
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Good numbers to test: \(-2, -1/2, 0, 1/2, 2\)

(1) If \(a < 0,\) then \(a^2\) is the largest out of all the numbers since its raised to an even power.

What's the smallest number? If \(a = \frac{-1}{2}\), the smallest number is a.

If \(a = -2\), the smallest number is \(a^5\).

We could get two different ranges; INSUFFICIENT.

(2) \(a^5 < a\)

Either \(a < -1\) or \(0 < a < 1\). INSUFFICIENT.

(1&2) Combined, \(a < -1\).

We can conclude the range is \(a^2 - a^5\)
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Bunuel Can you help me with this? Doesn't GMAT Considers definition of range to be difference between largest positive and smallest positive . There's infact a GMATPrep Question on this where the set has some negative and positive values and you're asked about range. Is something wrong with Veritas or am I thinking too much about it

The range of a set is the difference between the largest and smallest elements of the set. For example, the range of {-5, 0, 7, 11} is 11 - (-5) = 16.
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Bunuel Thanks! I confused this question with https://gmatclub.com/forum/list-k-consi ... 47855.html.
Not realizing that this has modified something

Given \(a^2, a^5, a, \frac{a}{2}, \frac{a}{5}\), do the first two values yield the greatest difference between any two values in the set?

Statement 1: \(a\) is negative
Case 1: \(a\) is a negative fraction between -1 and 0
Plugging \(a=-\frac{1}{2}\) into \(a^2, a^5, a, \frac{a}{2}, \frac{a}{5}\), we get:
1/4, -1/32, -1/2, -1/4, -1/10
Here, the greatest difference is NOT yielded by the first two values, so the answer to the question stem is NO.

Case 2: \(a\) is less than -1
Plugging \(a=-2\) into \(a^2, a^5, a, \frac{a}{2}, \frac{a}{5}\), we get:
4, -32, -2, -1, -2/5
Here, the greatest difference IS yielded by the first two values, so the answer to the question stem is YES.

Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2: \(a^5 < a\)
Case 2 also satisfies statement 2.
In Case 2, the answer to the question stem is YES.

Case 3: \(a\) is a positive fraction between 0 and 1
Plugging \(a=\frac{1}{2}\) into \(a^2, a^5, a, \frac{a}{2}, \frac{a}{5}\), we get:
1/4, 1/32, 1/2, 1/4, 1/10
Here, the greatest difference is NOT yielded by the first two values, so the answer to the question stem is NO.

Since the answer is YES in Case 2 but NO in Case 3, INSUFFICIENT.

Statements combined:
Only Case 2 satisfies both statements.
In Case 2, the answer to the question stem is YES.
SUFFICIENT.


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