Zoser wrote:Statement 1: Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour.
Notice that we could easily duplicate this scenario in real life.
Start with Car X 1 mile ahead of car Y (given info)
Have Car X drive at 50 mph and car Y at 40mph.
Use a stopwatch to time how long it takes for Car X to be 2 miles ahead of Y.
As you can see, we have enough information to answer the target question
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Hi,
Can you explain how to solve this algebraically using statement 1 info?
Thanks
You bet!
At time = 0, car X is 1 mile ahead of car Y.
We want to know the time it takes for car X to be
2 miles ahead of car Y
In other words,
we want to know the time it takes for the GAP between the two cars to increase by 1 mile.
Here's one approach:
Statement 1: Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour.
So,
in 1 hour, car X travels 50 miles, and car Y travels 40 miles
In other words,
in 1 hour, car X travels 10 miles FARTHER than car Y travels
In other words,
in 1 hour, the GAP between the cars increases by 10 MILES
From this, we can make many conclusions.
For example,
in 2 hours, the GAP between the cars increases by 20 MILES
Likewise,
in 3 hours, the GAP between the cars increases by 30 MILES
In 1/2 hour, the GAP between the cars increases by 5 MILES
In 1/5 hours, the GAP between the cars increases by 2 MILES
In 1/10 hours (i.e., 6 minutes), the GAP between the cars increases by 1 MILE
DONE!
Brent Hanneson - Creator of GMATPrepNow.com
