Is a > 0 ?
(1) a^3  a < 0
(2) 1  aÂ² > 0
My answer :
(1)
a^3  a = a(a+1)(a1) < 0
Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0. SUFFICIENT
It makes sens doesn't it ?
(2) is useless,
My answer is A.
In the mgmat book on inequalities, they do it another way and their answer is C. What's wrong with mine ?
Thanks
Inequalities advanced question MGMAT
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St 1 :
a ( a^2 1) < 0
so a <0 or ( a^2 1) <0 . Insufficient.
St 2: 1  aÂ² > 0
which is nothing but (a^2 1 )<0 ( Just multiply 1 in the prompt of st 2)
Still we are not sure of a as just a^2 <1, where in which a can be 1<0<1. Insufficent.
Combining St1 & st 2, we have (a^2 1 )<0, so a must be greater than 0 for the st 1 to hold true.
Pick C.
a ( a^2 1) < 0
so a <0 or ( a^2 1) <0 . Insufficient.
St 2: 1  aÂ² > 0
which is nothing but (a^2 1 )<0 ( Just multiply 1 in the prompt of st 2)
Still we are not sure of a as just a^2 <1, where in which a can be 1<0<1. Insufficent.
Combining St1 & st 2, we have (a^2 1 )<0, so a must be greater than 0 for the st 1 to hold true.
Pick C.
 kmittal82
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What if a = 0.1? Then you have 0.1 x 1.1 x 0.9 < 0 . This means this statement has a solution for a > 0 as well as a < 0.Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0.
If the question had stated a is an integer, then your explanation would have been good.
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One way to evaluate the statements is to determine the critical points. I posted an explanation of this approach here:waggrave wrote:Is a > 0 ?
(1) a^3  a < 0
(2) 1  aÂ² > 0
My answer :
(1)
a^3  a = a(a+1)(a1) < 0
Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0. SUFFICIENT
It makes sens doesn't it ?
(2) is useless,
My answer is A.
In the mgmat book on inequalities, they do it another way and their answer is C. What's wrong with mine ?
Thanks
https://www.beatthegmat.com/inequalityc ... 89518.html
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HI Since it is not given that X is an integer, you need to keep in mind that the number can be a fraction as well. Let's take an example  X=1/2waggrave wrote:Is a > 0 ?
(1) a^3  a < 0
(2) 1  aÂ² > 0
My answer :
(1)
a^3  a = a(a+1)(a1) < 0
Hence a < 0 cause none of them can equal to 0 since the product is inferior to 0, and the 3 of them have to be inferior to 0. SUFFICIENT
It makes sens doesn't it ?
(2) is useless,
My answer is A.
In the mgmat book on inequalities, they do it another way and their answer is C. What's wrong with mine ?
Thanks
in this case, although your number is positive the answer will be less than zero..
(1/8)  (1/2) <0
Also if we take a negative value  say for example (2) then also we will get the resultant less than 0. hence the first condition is also NOT sufficient by itself.
(1) a^3  a < 0
(2) 1  aÂ² > 0
Lets consider the first case (1) a^3  a < 0
a^3 < a >(A)
here a could be +ve/ ve Integer fraction or zero
now lets assume a is +ve Integer. Equation (A) will not be satisfied.
now lets assume a is ve Integer. Equation (A) will be satisfied.
now lets assume a is +ve fraction. Equation (A) will be satisfied.
now lets assume a is ve fraction. Equation (A) will not be satisfied.
only combining option (2) we can come out with a ans
(2) 1  aÂ² > 0
Lets consider the first case (1) a^3  a < 0
a^3 < a >(A)
here a could be +ve/ ve Integer fraction or zero
now lets assume a is +ve Integer. Equation (A) will not be satisfied.
now lets assume a is ve Integer. Equation (A) will be satisfied.
now lets assume a is +ve fraction. Equation (A) will be satisfied.
now lets assume a is ve fraction. Equation (A) will not be satisfied.
only combining option (2) we can come out with a ans
This time no looking back!!!
Navami
Navami
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I received a PM requesting that I show how this DS can be solved by determining the critical points.waggrave wrote:Is a > 0 ?
(1) a^3  a < 0
(2) 1  aÂ² > 0
Statement 1: aÂ³  a < 0.
a(a+1)(a1) < 0.
The critical points are a=0. a=1, a=1.
These are the only values where a(a+1)(a1) = 0.
When a is any other value, a(a+1)(a1) < 0 or a(a+1)(a1) > 0.
To determine the range of a, test one value to the left and right of each critical point.
Plug a < 1 into aÂ³  a < 0:
Let a = 2.
(2)Â³  (2) <0.
6 < 0.
This works.
Plug 1 < a < 0 into aÂ³  a < 0:
Let a = 1/2.
(1/2)Â³ (1/2) < 0.
3/8 < 0.
Doesn't work.
Plug 0 < a < 1 into aÂ³  a < 0:
Let a = 1/2.
(1/2)Â³  1/2 < 0.
3/8 < 0.
This works.
Plug a > 1 into aÂ³  a < 0:
Let a = 2
(2)Â³  2 < 0.
6 < 0.
Doesn't work.
Two ranges work in statement 1:
a < 1.
0 < a < 1.
Since a can be negative or positive, insufficient.
Statement 2: 1  aÂ² > 0
1  aÂ² > 0
(1+a)(1a) > 0.
The critical points are a = 1 and a = 1.
These are the only values where 1  aÂ² = 0.
When a is any other value, 1  aÂ² < 0 or 1  aÂ² > 0.
To determine the range of a, test one value to the left and right of each critical point.
Plug a < 1 into 1  aÂ² > 0:
Let a = 2.
1  (2)Â² > 0.
3 > 0.
Doesn't work.
Plug 1 < a < 1 into 1  aÂ² > 0:
Let a = 0.
1  0Â² > 0.
1 > 0.
This works.
Plug a > 1 into 1  aÂ² > 0:
Let a = 2.
1  2Â² > 0.
3 > 0.
Doesn't work.
The only range that works is 1 < a < 1.
Since a can be negative or positive, insufficient.
Statements 1 and 2 combined:
Ranges that satisfy statement 1: 0 < a < 1 or a < 1.
Range that satisfies statement 2: 1 < a < 1.
The only range that satisfies both statements is 0 < a < 1.
Since a must be positive, sufficient.
The correct answer is C.
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