Is a |a|

Question

Is a < b?

(1) a^3 = ab

(2) |b| > |a|

Source: 4Gmat

pr3mk5 · Answer

OA - E

Statement 1: a^3 = ab

If a = +ve then b should be +ve. And many values will satisfy this condition. For ex a = 2 and b = 4. In this case a < b.

If a = -ve then b should be positive. Again many values would satisfy this condition. For ex a = -2 and b = 4. In this case too a < b.

But if a = 0, then B can be any value. b = -1 or 1. In one case a>b and in other case a < b. No clear solution.

Statement 2: |b| > |a|

Lets take two values. a = -2 and b = -3. Since in mod b would be greater than a; however a is greater than b.
if a = 2 and b = 3, then b is greater than a. No clear solution.

Combining both.

If a = 0. Then b could be any value. Could be 2 or -2.
For positive value of b it'd be greater than a
but for negative value of b it'd be lesser than a.
Hence no clear solution even after combining.

EgmatQuantExpert · Answer

Solution

To find:

• Whether a < b?
Analysing Statement 1

Hence, statement 1 is not sufficient to answer

Analysing Statement 2

Hence, statement 2 is not sufficient to answer

Combining Both Statements
Considering the combined data available in both the statements, we can write

Hence, the correct answer is option E.

Answer: E

Bob2018 · Answer

it should be e

from 1st st. a can be 0 or a square can be equal to b
from 2nd, b can either be less than a or more than a

chetan2u · Answer

1) a^3=ab... a(a^2-b)=0 So either a=0 or a^2=b or both are 0 Insufficient 2) |b|>|a| So if b>0, yes a

soumya170293 · Answer

Answer would be E , I think.
a^3 = ab 
a^2 = b

as bcz square can't be negative so b must be positive.
but a can be positve or negative so we can't decide. ( for taking 1 a can be equal to b or by taking any number b is always greater than a . )

| b | > | a |

b > a ( when taking + ve)

or

-b > -a ( when taking -ve )
b < a so we can't decide.

Archit3110 · Answer

#1
a^3=ab
or say
a^2=b
now a= +/-2 and b=4
in sufficient
#2
|b| > |a| 
we can have +/- values of b so in sufficient to conclude that b>a

from 1& 2
we cannot derive anything in common to conclude b>a
IMO E

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Is a < b? (1) a^3 = ab (2) |b| > |a|

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