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.
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