If a and b are non-zero integers, and a/b 1, then which of

Question

If a and b are non-zero integers, and a/b > 1, then which of the following must be true? A. a > b B. 2a > b C. a^2< b^2 D. ab > b E. a^3 < b^3

Bunuel · Accepted Answer

If a and b are non-zero integers, and a/b > 1, then which of the following must be true?

A. a > b
B. 2a > b
C. a^2< b^2
D. ab > b
E. a^3 < b^3

If b is positive, then we have that a>b>0.
If b is negative, then we have that a<b<0.

A. a > b. Not necessarily true.

B. 2a > b. Not necessarily true.

C. a^2< b^2 --> |a|<|b|. Not necessarily true.

D. ab > b. If a>b>0, then ab>b (the product of two positive integers is obviously greater than either one of them) and if a<b<0, then ab=positive>negative=b. So, this statement is always true.

E. a^3 < b^3 --> a<b. Not necessarily true.

Answer: D.

Hope it's clear.

CrackverbalGMAT · Answer

Lets plug in here

a = -3, b = -2
A - Eliminated
B - Eliminated
C - Eliminated
D - True
E - Eliminated

navigator123 · Answer

A & B are non zero integers then, a>b .. Assume a = 10 & b = 20. Not true2a>b.. assume a = 10 and b =40 20>40?? Not true. B^2 > a^2. assume a = 5 and b = 2. 4 > 25?? Not true ab > b assume a = -3 and b =7 -21>7?? Not true a^3 < b^3.. assume a = 3 and b = 2 9 < 8?? Not true. I guess question is not correctly worded or i am making a mistake somewhere.

fozzzy · Answer

Sorry I have updated the question!

TGC · Answer

I Aspire to learn even 10% of the approaches you use to solve a math question.

Rgds,
TGC !!

jlgdr · Answer

Yeah very good question a bit complicated, maybe I took the long way home but let's see We get a-b/b > 0 One has to scenarios 1) If a-b>0, then b>0, therefore a>0. So all in all a>b>0 2) If a-b<0, then b<0, therefore a<0. So all in all 0>b>a Let's check out the answer choices A. a > b Not always true, could be scenario 2 B. 2a > b Not true at all C. a^2< b^2 Too many different possibilities D. ab > b This is equal to b(a-1)>0 Now we have two scenarios as well If b>0, then a>1. Note that we are told that a.b are integers. Therefore b>0 means that b has to be 1 at least and therefore a>1. If b<0, then a<-1. Same as above So it works for both of our scenarios E. a^3 < b^3 This is the same as to say b>a, which is not always true Hence answer is D Hope it helps Kudos rain!!!

Enael · Answer

If a = 0.5 and b = 0.1

Then a/b = 0.5/0.1 = 5

When I tried to plug in the values into the d) option I got a*b = (0.5)(0.1) = 0.05 which isn't greater than b = 0.1

So option D also wouldn't hold true.

Am I doing something wrong?

Bunuel · Answer

Yes. We are told that a and b are non-zero  integers .

bhatiavai · Answer

I guess the best approach here would be to Plug in values...
Try a=-3 and b = -2

All get eliminated except for D

whoisjon2 · Answer

Hi Brunel,

I am still confused. I can't see the connection between the question asking : a/b>1 , and the answers below

Could you or anyone kindly explaain this

Thank you

Bunuel · Answer

The question does not asks whether a/b > 1. It says that a/b > 1. So, the question is: given that a/b > 1, which of the following options must be true?

JeffTargetTestPrep · Answer

If a/b > 1, then a and b must have the same sign, with |a| > |b|. For example, a = 3 and b = 2 OR a = -3 and b = -2. Let’s analyze each given answer choice. A. a > b Let a = -3 and b = -2; we can see that A is not true. B. 2a > b Let a = -3 and b = -2; we can see that B is not true. C. a^2 < b^2 Let a = 3 and b = 2; we can see that C is not true. D. ab > b Let a = 3 and b = 2; we have ab > b. Let a = -3 and b = -2; we have ab > b. We see that D can be true, but let’s also analyze choice E before we conclude that choice D is definitely true. E. a^3 < b^3 Let a = 3 and b = 2; we can see that E is not true. Answer: D

LakerFan24 · Answer

my .02:

* MUST BE TRUE questions tend to be D or E. so even if you're running out of time or trying to burn the Q, you have a 50/50 here.

we're told that:
1) a & b = integers AND ARE NOT 0 (aka NO FRACTIONS)
2) \(\frac{a}{b}\)>1

Let's make sense of (2): for a fraction to be GREATER THAN 1, what needs to be true?
a) The numerator must be greater than the denominator
b) The numerator AND denominator must share the same sign (i.e. both must be POSITIVE or NEGATIVE)

KEYS TO THE PROBLEM:
> if b = positive (let's say b=2), then we can say a=4. that way, 2>1.
> if b = negative (let's say b=-2), then we can say a=-4. that way, 2>1.
** Both of these conditions match the above proofs.
** Very helpful to PICK #s to solve

Let's plug in these sets of #s for A/C A->E.
A) a>b --> what if a=-4, b=-2? Incorrect.
B) 2a>b --> what if a=-4, b=-2? Incorrect.
C) \(a^{2}\)<\(b^{2}\) --> what if a=-4, b=-2? Incorrect.
D) ab>b --> try: a=-4, b=-2. then, 8>-2. or: a=4, b=2. then, 8>2. Correct.
E) \(a^{3}\)<\(b^{3}\) -->what if a=4, b=2? then 64<8. Incorrect.

sarthak1701 · Answer

a and b are non zero and a/b > 1, therefore a and b are of the same sign and the ABSOLUTE value of a is greater than b.

Let's take the following values of a and b =

a = 3, b = 2

a = -3, b = -2

Only D satisfies both sets of these numbers.

bumpbot · Answer

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