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cd?

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Source: — Data Sufficiency |
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by everything's eventual » Mon Sep 10, 2012 8:36 pm
Prepare a table for 1) and check for all signs for c and d

c 3c d -8d^3

- - - +

+ + + -

+ + - +

- - + -


You can see that 3c = -8d^3 only if the signs of c and d are different. So cd will not be positive for 1).

Consider 2)

d > c +4

Let d = 6 and c = 1

6 > 1 + 4

Let d = 6 and c = -1

6 > -1 + 4

So c and d can have same sign or different signs. Therefore we do not know cd is +ve or -ve.

Choose A
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by everything's eventual » Mon Sep 10, 2012 8:39 pm
Sorry I always go wrong in aligning. Please prepare the table in your notebook and you will understand
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by alex.gellatly » Mon Sep 10, 2012 9:09 pm
grandh01 wrote:Is the product of cd positive?

1) 3c= -8d^3
2) d> c + 4
For questions like this, the easiest thing to do (in my opinion) is just plug in simple numbers.

QUESTION: is cd positive?
For cd to be positive that means they have to have the same sign (++ or --)

STATEMENT ONE: 3c=-8d^3
Solve for either c or d^3. we get: -(8/3)*c = d^3.
Now say c=1, plug and go. If c=1 than d^3= -(8/3) You don't need to solve for D. Just notice when c is +, D is -. That means CD = a - number. Remember any number x raised to an odd number (ie x^3) preserves the sign of x. This statement is sufficient.

STATEMENT TWO: d> c + 4
This doesn't really help if c is -3, then d is +, but if c is -100, then d is -. We don't know for sure, insufficient.

The correct answer is thus A.

Hope this helps
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
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by Javoni » Fri Sep 14, 2012 11:38 am
Here we go, lets start fith the 2nd option that seems prettier:-)

2. d>c+4, let c=-1 then d>3, hence cd<0
let c=0 then d>4, hence cd=0
let c=2 then d>6, hence cd>0
Thus, Cross B and D out as the 2nd option is Insufficient

1. 3C=-8D^3, here we get -0.375*C = D^3

Obviously the only way for the equation set forth above to be veracious C and D must have different signs.
For example, let C=1000---->>>>>> D^3 = -375, hence cd<0
let C=-100 -->>>>>>> D^3 = 37.5, hence cd<0
From above we might conclude that the answer can be (A)

Please, correct me if I went awry
Life begins at the End of your Comfort Zone...
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by Brent@GMATPrepNow » Thu Jun 13, 2019 6:15 am
grandh01 wrote:Is the product of cd positive?

1) 3c = -8d^3
2) d > c + 4

OA is A
Target question: Is the product cd positive?

Statement 1: 3c = -8d³
Divide both sides by d to get: 3c/d = -8d²
Divide both sides by 3 to get: c/d = -8d²/3
Rewrite as: c/d = (-8/3)(d²)
Since d² is greater than or equal to zero for all values of d, and since -8/3 is NEGATIVE, we can rewrite our equation as: c/d = (NEGATIVE)(some number greater than or equal to zero)
(NEGATIVE)(some number greater than or equal to zero) = some number that is less than or equal to zero
So, c/d = some number less than or equal to zero
This means the quotient c/d CANNOT be positive
It also mean the product cd CANNOT be positive
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

ASIDE: the important concept here is that, if c/d is positive, then cd is also positive. Likewise, if c/d is negative, then cd is also negative.

Statement 2: d > c + 4
There are several values of c and d that satisfy statement 2. Here are two:
Case a: c = 1 and d = 10. Here, cd = (1)(10) = 10, so the answer to the target question is YES, cd IS positive
Case b: c = -1 and d = 5. Here, cd = (-1)(5) = -5, so the answer to the target question is NO, cd is NOT positive
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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