X----------------Y
d = 9 miles
Jan starts from X at a speed of 2 miles per hour and Terry starts from Y at a speed of 3 miles per hour.
Time taken when they will meet \(\frac{9}{2+3}\) = 1.8 hours = 108 min.
Total distance covered by Jan = 2*1.8 = 3.6 miles
Total distance covered by Terry = 3*1.8 = 5.4 miles
Now each person travels for 108 min so we need to find the distance between them 42 minutes before they meet.
It means they each will travel for 66 minutes.
Now, Jan covers a distance of 2.2 miles in 66 minutes and Terry covers a distance of 3.3 miles in 66 minutes.
Distance remaining between them is \((3.6 - 2.2) + (5.4-3.3)\) = 1.4 + 2.1 = 3.5
OA, DCarcass wrote:
If Jan starts walking from Town X to Town Y, a distance of 9 miles, at 2 miles per hour at the same time as Terry starts walking from Town Y to Town X at 3 miles per hour along the same path as Jan, how many miles apart will Jan and Terry be exactly 42 minutes before they meet?
A. 2.0
B. 2.5
C. 3.0
D. 3.5
E. 4.0
Kudos for the right answer and explanation