Data Sufficiency

Brent@GMATPrepNow wrote:Ann and Bea leave X-ville at the same time and travel towards Y-ville, which is 70 kilometers away. Their individual speeds are constant, but Ann's speed is greater than Bea's speed. Upon reaching Y-ville, Ann immediately turns around and drives toward X-ville until she meets Bea. When they meet, how far has Bea traveled?

1) Ann's speed is 30 kilometers per hour greater than Bea's speed
2) Ann's speed is twice Bea's speed

Alternate approach for Statement 2:
Bea travels half as fast as Ann.
Thus, when Ann has traveled the WHOLE distance to Y-ville, Bea will be only HALFWAY to Y-ville.
As a result, half the total distance -- 35 miles -- will then separate Bea and Ann.
For Bea and Ann to meet, they must now WORK TOGETHER to cover the 35 remaining miles between them.
Since Bea travels 1km for every 2km that Ann travels, Bea will travel 1 of every 3 miles that are traveled when she and Ann work together.
Implication:
Bea will travel 1/3 of the 35 remaining miles between her and Ann.
Thus:
Total distance traveled by Bea = (half the total distance) + (1/3)(35 remaining miles between Bea and Ann) = 35 + (1/3)(35) = 140/3 miles.
SUFFICIENT.

GMATGuruNY wrote:
Alternate approach for Statement 2:
Bea travels half as fast as Ann.
Thus, when Ann has traveled the WHOLE distance to Y-ville, Bea will be only HALFWAY to Y-ville.
As a result, half the total distance -- 35 miles -- will then separate Bea and Ann.
For Bea and Ann to meet, they must now WORK TOGETHER to cover the 35 remaining miles between them.
Since Bea travels 1km for every 2km that Ann travels, Bea will travel 1 of every 3 miles that are traveled when she and Ann work together.
Implication:
Bea will travel 1/3 of the 35 remaining miles between her and Ann.
Thus:
Total distance traveled by Bea = (half the total distance) + (1/3)(35 remaining miles between Bea and Ann) = 35 + (1/3)(35) = 140/3 miles.
SUFFICIENT.

Dear GMATGuru,

How you do you interpret the same data in Red in case that two cars are chasing each other?

Can you please show with an example or direct to solved example in case of chasing?

Thanks

Mo2men wrote:

GMATGuruNY wrote:
Alternate approach for Statement 2:
Bea travels half as fast as Ann.
Thus, when Ann has traveled the WHOLE distance to Y-ville, Bea will be only HALFWAY to Y-ville.
As a result, half the total distance -- 35 miles -- will then separate Bea and Ann.
For Bea and Ann to meet, they must now WORK TOGETHER to cover the 35 remaining miles between them.
Since Bea travels 1km for every 2km that Ann travels, Bea will travel 1 of every 3 miles that are traveled when she and Ann work together.
Implication:
Bea will travel 1/3 of the 35 remaining miles between her and Ann.
Thus:
Total distance traveled by Bea = (half the total distance) + (1/3)(35 remaining miles between Bea and Ann) = 35 + (1/3)(35) = 140/3 miles.
SUFFICIENT.

Dear GMATGuru,

How you do you interpret the same data in Red in case that two cars are chasing each other?

Can you please show with an example or direct to solved example in case of chasing?

Thanks

After Bea arrives at the halfway point and Ann arrives at Y-ville, Ann TURNS AROUND and travels back TOWARD BEA, while Bea continues her journey and travels TOWARD ANN.
Since Bea and Ann are traveling TOWARD EACH OTHER, they are NOT chasing each other.
Rather, they are WORKING TOGETHER to cover the 35 miles between them.
Thus, the reasoning in the red portion applies.

If Ann is chasing Bea, and we want to determine the distance that Ann and Bea will each travel before Ann catches up to Bea, we could apply the following line of reasoning:
Let d = the distance between Ann and Bea.
Distance traveled by Ann = (Ann's rate)/(Difference between Ann's rate and Bea's rate) * d.
Distance traveled by Bea = (Bea's rate)/(Difference between Ann's rate and Bea's rate) * d.

Case 1: d=100, Ann's rate is twice Bea's rate
Since Ann is 100 miles BEHIND Bea, she must travel 100 more miles than Bea to CATCH UP to Bea.

Let Ann's rate = 2mph and Bea's rate = 1mph
Distance traveled by Ann = 2/(2-1) * 100 = 200 miles.
Distance traveled by Bea = 1/(2-1) * 100 = 100 miles.

Let Ann's rate = 4mph and Bea's rate = 2mph
Distance traveled by Ann = 4/(4-2) * 100 = 200 miles.
Distance traveled by Bea = 2/(4-2) * 100 = 100 miles.

Let Ann's rate = 20mph and Bea's rate = 10mph
Distance traveled by Ann = 20/(20-10) * 100 = 200 miles.
Distance traveled by Bea = 10/(20-10) * 100 = 100 miles.

In every case, Ann travels 100 more miles than Bea, with the result that she catches up to Bea.
Thus, as the three cases above illustrate, the distance traveled by Ann = 200 miles, while the distance traveled by Bea = 100 miles.

Case 2: d=100, Ann's rate is three times Bea's rate
Since Ann is 100 miles BEHIND Bea, she must travel 100 more miles than Bea to CATCH UP to Bea.

Let Ann's rate = 3mph and Bea's rate = 1mph
Distance traveled by Ann = 3/(3-1) * 100 = 150 miles.
Distance traveled by Bea = 1/(3-1) * 100 = 50 miles.

Let Ann's rate = 6mph and Bea's rate = 2mph
Distance traveled by Ann = 6/(6-2) * 100 = 150 miles.
Distance traveled by Bea = 2/(6-2) * 100 = 50 miles.

Let Ann's rate = 30mph and Bea's rate = 10mph
Distance traveled by Ann = 30/(30-10) * 100 = 150 miles.
Distance traveled by Bea = 10/(30-10) * 100 = 50 miles.

In every case, Ann travels 100 more miles than Bea, with the result that she catches up to Bea.
Thus, as the three cases above illustrate, the distance traveled by Ann = 150 miles, while the distance traveled by Bea = 50 miles.

GMATGuruNY wrote:

Mo2men wrote:

GMATGuruNY wrote:
Alternate approach for Statement 2:
Bea travels half as fast as Ann.
Thus, when Ann has traveled the WHOLE distance to Y-ville, Bea will be only HALFWAY to Y-ville.
As a result, half the total distance -- 35 miles -- will then separate Bea and Ann.
For Bea and Ann to meet, they must now WORK TOGETHER to cover the 35 remaining miles between them.
Since Bea travels 1km for every 2km that Ann travels, Bea will travel 1 of every 3 miles that are traveled when she and Ann work together.
Implication:
Bea will travel 1/3 of the 35 remaining miles between her and Ann.
Thus:
Total distance traveled by Bea = (half the total distance) + (1/3)(35 remaining miles between Bea and Ann) = 35 + (1/3)(35) = 140/3 miles.
SUFFICIENT.

Dear GMATGuru,

How you do you interpret the same data in Red in case that two cars are chasing each other?

Can you please show with an example or direct to solved example in case of chasing?

Thanks

After Bea arrives at the halfway point and Ann arrives at Y-ville, Ann TURNS AROUND and travels back TOWARD BEA, while Bea continues her journey and travels TOWARD ANN.
Since Bea and Ann are traveling TOWARD EACH OTHER, they are NOT chasing each other.
Rather, they are WORKING TOGETHER to cover the 35 miles between them.
Thus, the reasoning in the red portion applies.

If Ann were chasing Bea, I would not apply this line of reasoning.

Dear Mitch,

Thanks for your help.However, you missed my point as I was not clear enough. My question above was a general question not related to the original problem. I want to see how to apply the same reasoning in CHASING situation.

Mo2men wrote:Thanks for your help.However, you missed my point as I was not clear enough. My question above was a general question not related to the original problem. I want to see how to apply the same reasoning in CHASING situation.

I've amended my second post above to answer your question.
Please check the revised post.