If ab ≠ 0, is ab > a/b ?

Hi Bunell,

Can you please explain this question through plugging in. For (2) , ab>-(a/b) .. So for what values this will give yes and no and for which values after combining 1 and 2 we will get the req answer

Also wanted to add. From (2) a/b and ab has to be of the same signs . i tried plugging in values in 2. a=8,b=4 so ab=32 and a/b=2 . Similarly a=-4,b=-2 , ab=8, a/b=2 and also with a=1/8,b=8 so ab=1, a/b=1/72. So for all values ab>a/b.

Please let me knw what i am doin wrong

For (2):

If a=b=1 (ab + a/b > 0), then ab = a/b. Hence in this case the answer to the question whether ab > a/b is NO.

If a=b=2 (ab + a/b > 0), then ab > a/b. Hence in this case the answer to the question whether ab > a/b is YES.

Hope it's clear.

Wow! Great approach Buneul. It never occurred to me to simplify the equation in the question.

If ab ≠ 0, is ab > a/b ?

Is ab > \frac{a}{b}? --> is ab - \frac{a}{b}>0? --> \frac{a(b^2-1)}{b}>0? --> \frac{a}{b}*(b^2-1)>0?

(1) |b| > 1 --> both sides are non-negative, thus we can square: b^2>1 --> b^2-1>0. We need to know the sign of a/b. Not sufficient.

(2) ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0. We need to know the sign of b^2-1. Not sufficient.

(1)+(2) b^2-1>0 and \frac{a}{b} > 0, thus their product \frac{a}{b}*(b^2-1)>0. Sufficient.

Hi bunuel,

I did not get the highlighted part of the solution.
Can you please explain why both a and b cannot be negative. I think if we consider values of a and b = -2 and -3 respectively. ab + a/b will remain positive overall.

Thanks in advance for your help.

Both ab and a/b cannot be negative, not a and b.

Why cant I solve the equation as ab2 > a hence b2>1 hence (a) the ans ???

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.

So, we can neither cross-multiply ab > a/b by b nor reduce it by a.

If b is positive, then yes we'll get ab^2 > a but if b is negative, then when multiplying by negative value we should flip the sign of the inequality and in this case we would get ab^2 < a.

The same way if a is positive, then b > 1/b but if it's negative, then b < 1/b.

Hi Bunnel, Considering the above quote, is the below approach correct?

Given that ab>a/b

Case 1: if a > 0 then b > 1/b ; true if b>1
Case 2: if a < 0 then b< 1/b ; true is b <1

State 1 says lbl >1 cant conclude anything from this

State 2 says ab + a/b > 0 so from this we know ab & a/b are both positive which means ab is positive, meaning a & b have same sign.
Cant conclude anything further

From State 1 & 2;
lbl > 1 and ab have same sign so only case 1 valid.

Thanks in advance,
Aj

Hi AjChakravarthy,

Few suggestions on your analysis:

Case 1: if a > 0 then b > 1/b ; true if b>1--> b > 1/b is also true when -1 < b < 0. Consider b = -0.5, in this case -0.5 > -2. Hence we can't say that b > 1/b is true for only b > 1.

Case 2: if a < 0 then b< 1/b ; true is b <1 --> b < 1/b is not true for cases when -1 < b < 0. Hence we can't say b < 1/b is true for all values of b < 1.

From State 1 & 2;
lbl > 1 and ab have same sign so only case 1 valid-- > |b| > 1 can be written as b > 1 if b is +ve or -b > 1 if b is -ve. This can be simplified to b > 1 or b < -1. Hence |b| > 1 can't be inferred as b being always positive.

Hope its clear!

Regards
Harsh

Despite reading the question and explanation again and again I still can't understand " ab + a/b > 0. Notice that ab and a/b have the same sign. Both of them cannot be negative, because in this case their sum would also be negative, therefore both are positive: a/b > 0."

Bunuel could you please help?

Let me try to explain.

When you are given that ab+a/b > 0 ---> \(a* \frac{b^2+1}{b} >0\) , now realise that \(b^2+1\) is ALWAYS > 0 for all values of b and thus for making \(a* \frac{b^2+1}{b} >0\) you only need to worry about a/b > 0

Once you see that the given statement 2 is reduced to a/b > 0 ---> this can only mean that both a and b have the SAME signs. Either both are <0 or both are >0. You can also see it this way:

when a/b >0--> consider 4 cases:

a>0 and b>0 ---> YES for a/b >0
a>0 and b<0 ---> NO for a/b >0
a<0 and b>0 ---> NO for a/b >0
a<0 and b<0 ---> YES for a/b >0

Thus, you get "YES" only when both a and b are of the SAME sign.

You do not have to worry about the actual sign of a or b as shown below:

Once you see that statements need to be combined, you see that the original equation is ab > a/b ? ---> ab-a/b >0 ---> a(b-1/b) > 0 ---> \(\frac{a*(b^2-1)}{b} > 0\)

---> \(\frac{a*(b+1)(b-1)}{b} > 0\), this is ONLY dependent on the sign of (b+1)(b-1) as a/b will ALWAYS be positive for all values of a and b (as both a and b are of the SAME sign).

Statement 1 mentions that |b| >1 --> b>1 or b<-1 , for these ranges, (b+1)(b-1) is > 0 ALWAYS and hence you get a definite "YES", making C as the correct answer.

Hope this helps.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If ab ≠ 0, is ab > a/b ?

(1) |b| > 1

(2) ab + a/b > 0

Modify the original condition and the question and multiply b^2 on the both equations, which becomes ab^3>ab? --> ab^3-ab>0? -->ab(b^2-1)>0?. There are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) &2), |b|>1 -> b^2>1, ab and a/b get to have the same sign in 2). That is, ab>0 -> a/b>0(this is divided by b^2 and as b^2 is a positive number, direction of the sign doesn’t change.) So, 2) becomes ab>0, which is yes and sufficient. Therefore, the answer is C.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Solution from "Thursdays With Ron"

Bunuel
Can you please explain the highlighted part in detail. From my understanding, even if both a and b are -ve, the answer option would be C. THanks!

The point is that ab and a/b will have the same sign: either both are positive or both are negative. But if ab and a/b are negative, then their sum will also be negative, thus ab + a/b > 0 means that ab and a/b are both positive.

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