Abba Casabra

[Nishant.Chandra]

If a and b are distinct integers and their product is not equal to zero, is a > b?

(1) (a^(3)b - b^(3)a)/(a^(3)b + b^(3)a - 2a^(2)b^(2)) < 0

(2) b < 0

Source: Question # 3 for Advanced Quant 750.

Here is the Explanation:

(1) INSUFFICIENT: To analyze statement (1), we must first simplify. Let's start with the top of the fraction:

a^(3)b - b^(3)a
ab(a^(2) - b^(2))

Recognizing the quadratic template x^(2) - y^(2) = (x + y)(x - y) allows us to simplify further:
ab(a + b)(a - b)

Next, let's tackle the bottom of the fraction:

a^(3)b + b^(3)a - 2a^(2)b^(2)
ab(a^(2) + b^(2) - 2ab)
ab(a^(2) - 2ab + b^(2))

Recognizing the quadratic template x^(2) - 2xy + y^(2) = (x - y)(x - y) allows us to simplify further:
ab(a - b)(a - b)

Putting the top and bottom of the fraction back together yields:

ab(a + b)(a - b)/ab(a - b)(a - b)
(a + b)/(a - b)

So statement (1) simplifies to (a + b)/(a - b) < 0. If a - b is positive (in other words, if a > b), we can simplify even further:

IF a > b:
a + b < a - b
b < -b
2b < 0
b < 0

Question 1: How did we get "a + b < a - b" If we multiply by 0 (because <0), then we should get a+b<0. So why did we assume that '<0' is '<1'

This tells us that b < 0 whenever a > b. The converse of this statement is also true. Whenever b < 0, a must be greater than b for the inequality to hold (try plugging sample values into (a + b)/(a - b) < 0 to verify that when b is negative, an a value smaller than b fails to satisfy the inequality).

This tells us that b < 0 whenever a > b. The converse of this statement is also true. Whenever b < 0, a must be greater than b for the inequality to hold (try plugging sample values into (a + b)/(a - b) < 0 to verify that when b is negative, an a value smaller than b fails to satisfy the inequality).

However, if a - b is negative (in other words, if a < b), we must flip the sign when multiplying through by a - b:
IF a < b:
a + b > a - b
b > -b
2b > 0
b > 0

Question 2: How did we get "a + b > a - b" If we multiply by 0 (because <0), then we should get a+b>0. So why did we assume that '<0' is '<1'

This tells us that b > 0 whenever a < b. The converse of this statement is also true. Whenever b > 0, a must be less than b for the inequality to hold (try plugging sample values into (a + b)/(a - b) < 0 to verify that when b is positive, a values greater than b fail to satisfy the inequality).

Since we have no information about the sign of b, we don't know if a > b, and we can conclude that statement (1) is insufficient.

(2) INSUFFICIENT: Simply knowing that b < 0 is not enough to determine if a > b. This statement gives no information about a.

(1) AND (2) SUFFICIENT: Since we know from statement (1) that a > b whenever b < 0, and since we know from statement (2) that b < 0, we can conclude that a > b.

I know that when we multiply by a negetive constant the sign becomes opposite. But what you have done above needs clarification.

I shall be awaiting your answers.

[JonathanSchneider]

Hm, I'm in agreement with you, Nishant. We would need that 0 to be a 1 in order to cross-multiply and have the "a - b" term remain. Currently, the "a - b" term would multiply by the 0 to yield 0. This is not particularly helpful, as we would then know only that a + b > 0 when a - b < 0, and vice versa. This, of course, is already known, as only such an arrangement would create the overall negative fraction. It seems we need to update this one.

[ammarjavaid007]

Hello,

Great explanation Nishant but theres something I'm not getting. I tried this question today and was able to simplify the first statement to (a+b)/(a-b)<0 and from here I proceeded to plug in numbers to test the inequality. It was my understanding that whenever a>b, b<0 but this doesn't apply for all numbers.

let a=-2 & b=-6

(-2+-6)/(-2--6) = -2 (satisfies)

let a=4 & b=-6

(4+-6)/(4--6) = -1/5 (satisfies)

but when a=10 & b=-6

(10+-6)/(10--6) = 1/4 (doesn't satisfy)

I dont understand how then that whenever b<0, the inequality will be satisfied. Maybe I'm looking too much into it idk. Plug in is usually how i proceed with these types of questions and it didn't work here.

Unless i've misread something or have overlooked a hidden constraint, I'm not sure what I'm doing wrong.

Can someone please help?

Thank you.

[RonPurewal]

Yeah, that derivation is weird.

Basically, we know that (a + b)/(a - b) is negative"”meaning that (a + b) and (a - b) have opposite signs.

Here's how I interpret this:

* If a - b is positive, then a + b must be negative (that's the only way you'll get a negative quotient). So ...
Negative < positive;
a + b < a - b.
I don't see a way to get there algebraically"”at least not any way that's halfway simple.

* If a - b is negative, then, using signs again, a + b is positive.
So...
Positive > negative;
a + b > a - b.
Again, I don't see an easy algebra way.

[RonPurewal]

Does the ANSWER KEY actually mention "multiplying through"? I suppose that's the real issue here.

[RonPurewal]

Final insight:
Anything that's less than 0 is also, of course, less than 1. (Anyone who's under 40 years old is also under 50 years old, and so on.)

So, if I KNOW that (a + b)/(a - b) < 0, then I also know that (a + b)/(a - b) < 1. So, if (a + b)/(a - b) < 1 is sufficient, then, a fortiori, (a + b)/(a - b) < 0 is also sufficient.

(This is dangerous ground to tread, though, unless you know exactly what you're doing. For instance, if (a + b)/(a - b) < 1 is NOT sufficient, then (a + b)/(a - b) < 0 might still be sufficient!)

[StaceyKoprince]

Hi, guys

(Ron, thanks for bringing this to my attention.)

The solution is correct, but it's not explained very well. I've just re-written it in the system so that the explanation actually makes sense now! I've pasted it below.
* * *

(1) INSUFFICIENT: To analyze statement (1), first simplify. Start with the top of the fraction:

(a^3)b - (b^3)a = ab(a^2 - b^2)

Use the quadratic template x^2 - y^2 = (x + y)(x - y) to simplify further:

ab(a + b)(a - b)

Next, tackle the bottom of the fraction:

(a^3)b + (b^3)a - 2(a^2)(b^2) = ab(a^2 + b^2 - 2ab) = ab(a^2 - 2ab + b^2)

Use the quadratic template x^2 - 2xy + y^2 = (x - y)(x - y) to simplify further:
ab(a - b)(a - b)

Put the top and bottom of the fraction back together:

ab(a + b)(a - b) / ab(a - b)(a - b) = (a + b)/(a - b)

So statement (1) simplifies to (a + b)/(a - b) < 0. Because the quotient is negative, one of the terms has to be positive and one has to be negative. If a - b is positive, then a + b has to be negative. In that case:

a + b < a - b
b < -b
2b < 0
b < 0

In addition, if b < 0 and a - b is positive, then a must be greater than b. (If you're not sure, try some real numbers to verify.) Therefore, if a - b is positive, then b < 0 and a > b.

However, if a - b is negative, then a - b has to be negative. In that case:

a + b > a - b
b > -b
2b > 0
b > 0

In addition, if b > 0 and a - b is negative, then a must be smaller than b. (If you're not sure, try some real numbers to verify.) Therefore, if a - b is negative, then b > 0 and a < b.

The problem provides no information about the sign of b, so statement (1) is insufficient.

(2) INSUFFICIENT: Simply knowing that b < 0 is not enough to determine if a > b. This statement gives no information about a.

(1) AND (2) SUFFICIENT: From statement (1), a > b whenever b < 0. From statement (2) b < 0, so a > b.

The correct answer is (C).