karthikpandian19 wrote:Dannetta has a jar containing red pens and black pens. If she selects one pen at random from the jar, what is the probability that the pen will be red?

If Dannetta removes 9 black pens from the jar, there will be an equal number of red and black pens remaining in the jar.
If Dannetta removes 1/6 of the black pens, 3/7 of the pens remaining in the jar will be red.

On the contrary, this question should take very little time.
To determine P(red), we need to know what fraction of the pens are red (or, alternatively, what fraction are black).

Statement 1: If Dannetta removes 9 black pens from the jar, there will be an equal number of red and black pens remaining in the jar.
Different fractions are possible.
If 1 black pen and 1 red pen remain, then prior to the removal of the 9 black pens, B=10 and R=1, implying that P(R) = 1/11.
If 10 black pens and 10 red pens remain, then prior to the removal of the 9 black pens, B=19 and R=10, implying that P(R) = 10/29.
INSUFFICIENT.

Statement 2: If Dannetta removes 1/6 of the black pens, 3/7 of the pens remaining in the jar will be red.
In other words, the remaining 5/6 of the black pens will constitute 4/7 of the remaining pens in the jar:
This information is sufficient to determine what fraction of the pens are black (and thus what fraction are red).
SUFFICIENT.

The correct answer is B.

Here's the math for statement 2:
The remaining 5/6 of the black pens will constitute 4/7 of the remaining pens in the jar.
Let B = black and T = total.
(5/6)B = (4/7)(T - (1/6)B)
(5/6)B = (4/7)T - (4/42)B
(39/42)B = (4/7)T
B = (24/39)T.

Thus, if T=39, then B=24 and R=15.
P(R) = 15/39 = 5/13.

Please note that WE SHOULDN'T DO THIS MATH WHEN WE TAKE THE ACTUAL TEST.
As soon as we realize that statement 2 is sufficient to determine what fraction of the marbles are black, we should pick answer choice B and move on.

OA is B

GMATGuruNY wrote:

karthikpandian19 wrote:Dannetta has a jar containing red pens and black pens. If she selects one pen at random from the jar, what is the probability that the pen will be red?

If Dannetta removes 9 black pens from the jar, there will be an equal number of red and black pens remaining in the jar.
If Dannetta removes 1/6 of the black pens, 3/7 of the pens remaining in the jar will be red.

On the contrary, this question should take very little time.
To determine P(red), we need to know what fraction of the pens are red (or, alternatively, what fraction are black).

Statement 1: If Dannetta removes 9 black pens from the jar, there will be an equal number of red and black pens remaining in the jar.
Different fractions are possible.
If 1 black pen and 1 red pen remain, then prior to the removal of the 9 black pens, B=10 and R=1, implying that P(R) = 1/11.
If 10 black pens and 10 red pens remain, then prior to the removal of the 9 black pens, B=19 and R=10, implying that P(R) = 10/29.
INSUFFICIENT.

Statement 2: If Dannetta removes 1/6 of the black pens, 3/7 of the pens remaining in the jar will be red.
In other words, the remaining 5/6 of the black pens will constitute 4/7 of the remaining pens in the jar:
This information is sufficient to determine what fraction of the pens are black (and thus what fraction are red).
SUFFICIENT.

The correct answer is B.

Here's the math for statement 2:
The remaining 5/6 of the black pens will constitute 4/7 of the remaining pens in the jar.
Let B = black and T = total.
(5/6)B = (4/7)(T - (1/6)B)
(5/6)B = (4/7)T - (4/42)B
(39/42)B = (4/7)T
B = (24/39)T.

Thus, if T=39, then B=24 and R=15.
P(R) = 15/39 = 5/13.

Please note that WE SHOULDN'T DO THIS MATH WHEN WE TAKE THE ACTUAL TEST.
As soon as we realize that statement 2 is sufficient to determine what fraction of the marbles are black, we should pick answer choice B and move on.