If |a| = b, is a+b ab (1) a = -b (2) a = -3

Question

If |a| = b, is a+b > ab

(1) a = -b

(2) a = -3

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anirudhgvs · Answer

I think (B) is the right answer.

Let's talk about (A):
 Here we can take the value of 'a' to be '0' too. So, that makes (A) insufficient since we cannot say a definitive 'Yes' or 'No'

yashikaaggarwal · Answer

Given: |a| = b, 
To find: is a+b > ab

(1) a = -b
Case 1 : A = 0
B = 0
0+0 = 0*0 (Wrong)

Case 2: A = -2
B = 2
-2+2 > -2*+2
0>-4 (correct)
Insufficient

(2) a = -3
B = 3 (|-3| = 3)
-3+3 > -3*+3
0 > -9
Sufficient

IMO B

Posted from my mobile device

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem 
 • |a| = b
 o This means b is a non-negative number. 
• We need to find the value of a+b > ab

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1 : a = -b
 • Since, b is a non-negative number, so there can be two cases:
 o Case 1: b = 0
  In that case, a = 0, 
 So, a+b = 0 and a*b = 0 
 Here, a+b = a*b 
o Case 1: b is positive
  In that case, a is negative, and a = -b
 So, a+b = 0 and a*b = - b^2 
 Here, a+b > a*b,   
• We are getting contradictory results. 
Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2:   a = -3
 • So, b = |-3| = 3
 o This means, a+b = 3 – 3 = 0, and a*b = -9
o Therefore, a+b > a*b  
Hence, statement 2 is sufficient.
Thus, the correct answer is  Option B.

TestPrepUnlimited · Answer

Statement 1: 
We can plug in to get "is  0 > -b^2  ?" is the question. Simplify one more step we would be asking "is  b^2 > 0  ?". This is always true UNLESS b = 0, which is possible. Then this is insufficient as we might have that case.

Statement 2: 
If a = -3 then we must have b = |-3| = 3. We know both values of a and b now so it must be sufficient.

Ans: B

KaramveerBakshi · Answer

|a| = b (given)
Is a+b>ab?

Statements:

(1) a=-b
a+b>ab for all cases except b=0
 Insufficient

(2) a = -3
Therefore, b = 3
ab = -9
a+b = 0 > -9
 Sufficient

Answer is Option (B)

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If |a| = b, is a+b > ab (1) a = -b (2) a = -3

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