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A list contains 11 consecutive integers. What is the greatest integer

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A list contains 11 consecutive integers. What is the greatest integer [#permalink]  24 Aug 2015, 22:44
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67% (01:55) correct 33% (01:01) wrong based on 78 sessions
A list contains 11 consecutive integers. What is the greatest integer on the list?

(1) If x is the smallest integer on the list, then (x + 72)^(1/3) = 4
(2) If x is the smallest integer on the list, then 1/64 = x^(-2)

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[Reveal] Spoiler: OA

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A list contains 11 consecutive integers. What is the greatest integer [#permalink]  24 Aug 2015, 22:51
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Statement 1 will have unique value of x ie -8
Statement 2 will have 2 values of x ie +8 & -8
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Last edited by SOURH7WK on 31 Aug 2015, 21:16, edited 1 time in total.
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Re: A list contains 11 consecutive integers. What is the greatest integer [#permalink]  25 Aug 2015, 00:03
Bunuel wrote:
A list contains 11 consecutive integers. What is the greatest integer on the list?

(1) If x is the smallest integer on the list, then (x + 72)^(1/3) = 4
(2) If x is the smallest integer on the list, then 1/64 = x^(-2)

Ans: A

as it is a list of consecutive int we can find the greatest if we can find the any number with its position on the list
1) Solution: (x+72) = 4^3
x+72= 64
x= -8 and x is the smallest so greatest is 2 [Sufficient]
2) 1/64 = x^-2
x^2 = 64
x= +8 and -8
two lists possible [Not Sufficient]
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Re: A list contains 11 consecutive integers. What is the greatest integer [#permalink]  26 Aug 2015, 03:38
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As answered above, In order to find the greatest integer, we need to find the value of x.

(1) provides single value of x. Enough
(2) provides two values of x. Not enough

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Re: A list contains 11 consecutive integers. What is the greatest integer [#permalink]  26 Aug 2015, 09:40
Bunuel wrote:
A list contains 11 consecutive integers. What is the greatest integer on the list?

(1) If x is the smallest integer on the list, then (x + 72)^(1/3) = 4
(2) If x is the smallest integer on the list, then 1/64 = x^(-2)

Kudos for a correct solution.

IMO : A

St 1:
$$(x + 72)^(1/3) = 4$$
Cubing on both sides
$$(x + 72)= 4^3$$
x=-8
Suff

St 2:
1/64 = x^(-2)
x = 8 or -8
Not suff
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Re: A list contains 11 consecutive integers. What is the greatest integer [#permalink]  27 Aug 2015, 19:20
1. (x+72) ^1/3 = 4 ; cubing both the side; (x+72)=64 => x=-8; so we will be able to get answer.

2. 1/64 = 1/x^2 ; x = -8, +8; therefore not sufficient to answer the question.

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Re: A list contains 11 consecutive integers. What is the greatest integer [#permalink]  27 Aug 2015, 22:09
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A - X can be obtained by Simple math as a fixed value -8
Therefore, the series will be unique (-8,-7,-6,...0,1,2)
Hence it's Sufficient.

B - The Value of X obtained is +/- 8.
Hence, the series will not be unique (Two different series 8, 9, 10,......18 or -8,-7,-6,.....2)
Not sufficient.
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A list contains 11 consecutive integers. What is the greatest integer [#permalink]  27 Aug 2015, 22:50
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Given that the list contains consecutive integers, it's enough if we find one of the numbers and its position in the list to have all numbers in the list. So, if we happen to know the smallest element in the list, we also know the largest element in the list.

Let's consider statement(1)

$$(x+72)^{1/3} = 4$$
$$=> \sqrt[3]{(x+72)} = 4$$
Cubing both sides,
$$=> x + 72 = 64$$
$$=> x = -8$$
With this, we can also find the largest number in the list. So, A/D stands.

Consider (2)

$$1/64 = x^{-2}$$
$$=> 1/64 = 1/x^2$$
$$=> x^2 = 64$$
$$=> x = \pm8$$

We have two values for x here and hence the greatest element in the list can vary based on this.
Hence, A
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Re: A list contains 11 consecutive integers. What is the greatest integer [#permalink]  27 Aug 2015, 23:55
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1) solving 1 x=-8
2) solving 2 x=+8 and -8
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Re: A list contains 11 consecutive integers. What is the greatest integer [#permalink]  30 Aug 2015, 08:43
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Expert's post
Bunuel wrote:
A list contains 11 consecutive integers. What is the greatest integer on the list?

(1) If x is the smallest integer on the list, then (x + 72)^(1/3) = 4
(2) If x is the smallest integer on the list, then 1/64 = x^(-2)

Kudos for a correct solution.

GROCKIT OFFICIAL SOLUTION:

If we can determine the smallest integer on the list or a specific integer on the list when the list is written in increasing order, we can determine the greatest integer on the list.

1) Sufficient: We’re given one variable and one equation for the smallest integer on the list. That means we could solve for smallest integer and add 10 to find the greatest integer. If you don’t see this, consider:

$$(x+72)^{\frac{1}{3}} = 4$$

Cubing both sides, x + 72 = 64. Then x + 72 = 64 and x = -8. Adding 10 to -8, the greatest integer is 2. Eliminate choices B, C and E.

2) Insufficient: If $$\frac{1}{64}=x^{-2}$$ then $$\frac{1}{64}=\frac{1}{{x^2}}$$ and $$x^2=64$$, so x could be -8 or 8.

There are two different possibilities for the smallest integer on the list, so there must be two different possibilities for the greatest integer on the list. Statement 2) is insufficient, leaving the correct answer choice as A.
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A list contains 11 consecutive integers. What is the greatest integer [#permalink]  01 Oct 2015, 02:58
1) (x+72)^1/3 =4
or, x+72 = 4^3 = 64
or, x=- 8...so sufficient
2) 1/64 = x^-2
x= 8 or -8 ....insufficient
ans: A
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A list contains 11 consecutive integers. What is the greatest integer   [#permalink] 01 Oct 2015, 02:58
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