If a + b 0 and a^b

Question

Competition Mode Question If a + b > 0 and a^b < 0 , which of the following must be true? I. a < 0 II. b > 0 III. |b| > |a| A. I only B. I and II only C. I and III only D. I, II and III E. Noe of these Are You Up For the Challenge: 700 Level Questions

Bunuel · Accepted Answer

No, it's a must be true question. If a + b > 0 and a^b < 0 , which of the following must be true? I. a < 0 II. b > 0 III. |b| > |a| A. I only B. I and II only C. I and III only D. I, II and III E. Noe of these Given: a^b < 0 . This to be true a MUST be negative because (positive)^(any number) > 0. Next, since a + b > 0 and a < 0, then b must be positive AND further from 0 than a . So, b < 0 and |b| < |a|. As you can see all three options must be true. Answer: D.

GMATinsight · Answer

a^b<0 i.e. a<0 because -ve value with any non-even exponent is always Negative a+b>0 i.e. b > 0 (because a is negative) Also |b|>|a| because e.g. a = -2 then b >2 only then a+b will be 0 (as given) i.e. I. a<0 (TRUE) II. b>0 (TRUE) III. |b|>|a| (TRUE) Answer: Option D

freedom128 · Answer

a^b<0 --> means a<0 (i) a+b>0 and a<0 --> it must be true that b>0 and |b|>|a| (ii) and (iii). (i), (ii) and (iii) must be true. FINAL ANSWER IS (D) Posted from my mobile device

Jawad001 · Answer

We are given that, a+b>0 and a^b<0, From a^b<0, we can say that a is negative and b is odd number. a<0, is ok b>0, is ok For being a+b>0, III |b|>|a| is true. Hence: (D)

CrackverbalGMAT · Answer

In this question on inequalities, we can use some basic properties of inequalities and solve this question without too much of difficulty.

Let us try to analyse the data given in the question statement.

1) a+b>0. If the sum of two numbers is more than ZERO, there are not many conclusions we can draw from this situation about the signs of the numbers. The only things we can conclusively say are:

2) The second piece of information given in the question is \(a^b\)<0. This means that a is definitely negative and b is positive.

From the above, it’s clear that statement I and II are definitely true. Based on this, answer options A and C can be eliminated.

a is negative, b is positive and a+b>0 means that the absolute value of b is definitely more than the absolute value of a i.e. |b| > |a|. Statement III is also definitely true. Answer options B and E can be eliminated

The correct answer option is D.

As you see, there weren’t too many advanced concepts we applied here. We stuck to the basics and we were able to solve it in less than 2 minutes. This is probably why there is so much emphasis on being good with your basics, regardless of the topic.

Hope that helps!

sujoykrdatta · Answer

a + b > 0 and a^b < 0 Possibilities for a + b > 0 are: #1. a > 0 and b > 0 => fails the test for a^b < 0 - Hence rejected #2. a > 0 and b < 0 , but |a| > |b| (i.e. a is a larger positive and b is a smaller negative) => fails the test for a^b < 0 (since any positive number a raised to any exponent b remains positive) - Hence rejected #3. a < 0 and b > 0 , but |a| < |b| (i.e. b is a larger positive and a is a smaller negative) => passes the test for a^b < 0 on condition that b is an odd number (since any negative number b raised to any odd exponent a remains negative) - Hence possible Working with the statements: I. a < 0 - True II. b > 0 - True III. |a| < |b| - True Answer D

PrankurD · Answer

Bunuel  Isn't the question more as "which of the following can be true"

Sarkar93 · Answer

Bunuel 
A quick question.

a^b will be negative when a<0 (as stated in option I) and b>0(as stated in option II). But at the same time, b must also be odd power for a^b to remain negative. If b ends up being even, then the term a^b will become positive even if a is negative.

Based on this reasoning, I rejected option II.

Let me know if my understanding is wrong.

Bunuel · Answer

Does II say anything about even or odd nature of b? It simply says: b > 0.

From a^b < 0, we can only say that a < 0 and b is not even. But b can be odd, not an integer, positive, or negative.

IcanDoIt23 · Answer

I tried solving this question by putting in different values of a and b. So, randomly I assumed a = -1 and b = 2 which satisfy all the three conditions which are as a<0, b>0, |a|<|b|. However, subsituting these values doesn't satisy the given two equations.

-1 + 2 = 1 !< 0
-1^2 = 1 !<0

Therefore, I selected option E as the answer. Could you help me understand if I'm missing anything here? Thanks!

Bunuel · Answer

If a = -1 and b = 2 then a^b = 1. However we are given that a^b < 0. Thus, a = -1 and b = 2 do not satisfy given condition.

bishal128 · Answer

I agree with this, for negative a if b is even integer results into positive value which violates a^b<0.

If a + b > 0 and a^b < 0, which of the following must be true?

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