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Difficulty:
35%
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67% (01:24) correct
33%
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wrong
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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 \((x + 72)^{(\frac{1}{3})} = 4\)
(2) If x is the smallest integer on the list, then \(\frac{1}{64} = x^{(-2)}\)
Kudos for a correct solution.
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(1) If x is the smallest integer on the list, then \((x + 72)^{(\frac{1}{3})} = 4\)
(2) If x is the smallest integer on the list, then \(\frac{1}{64} = x^{(-2)}\)
Kudos for a correct solution.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
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Answer A)
Statement 1 will have unique value of x ie -8
Statement 2 will have 2 values of x ie +8 & -8
Statement 1 will have unique value of x ie -8
Statement 2 will have 2 values of x ie +8 & -8
Bunuel
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]
Posts: 15
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
Hence, A is the answer.
(1) provides single value of x. Enough
(2) provides two values of x. Not enough
Hence, A is the answer.
Bunuel
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
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.
Hence answer is A
2. 1/64 = 1/x^2 ; x = -8, +8; therefore not sufficient to answer the question.
Hence answer is A
hiteshahire22
Joined: 15 Feb 2015
Last visit: 16 Sep 2016
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Location: United States
Concentration: General Management, Technology
GPA: 3
WE:Business Development (Manufacturing)
Posts: 19
Correct Answer A.
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.
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.
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
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
AvengerGmat
Joined: 04 Aug 2013
Last visit: 03 Jul 2017
Posts: 9
Location: United States
Concentration: Operations, Marketing
WE:Engineering (Manufacturing)
Posts: 9
Answer is A
1) solving 1 x=-8
2) solving 2 x=+8 and -8
1) solving 1 x=-8
2) solving 2 x=+8 and -8
Bunuel
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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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
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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