Alex left his house to walk to the local Mall at a steady 3 mph. His brother, Ben, left the house exactly seven minutes later, walking at a steady pace of 3.5mph, taking exactly the same route. How long would it take Ben to catch up with his bother? And when they do meet, what was the distance travelled from the house?. Thanks for any help.
Here is another way of looking at it:
Since the ratio of their walking is 3:3.5=6:7. But since the LCM of 6 and 7 is 42, therefore they will meet in 42 minutes after Ben left the thouse, or 42+7=49 minutes after Alex the house. The distance travelled will be:42/60 x 3.5 =49/60 x 3=2.45 miles from the house when they meet.
+37084 Let's find Alex' head start: 7 min = 7/60 hr
7/60 x 3 = 21/60 mile head start.
Now Alex distance will equal Ben's distance when they meet.
Rate x time = distance
Alex' distance = 21/60 + 3 t = 3.5 t Solve for t = 42/60 = 42 minutes
In 42 minutes, Ben will have walked 42/60 x 3.5 = 2.45 miles
Here is another way of looking at it:
Since the ratio of their walking is 3:3.5=6:7. But since the LCM of 6 and 7 is 42, therefore they will meet in 42 minutes after Ben left the thouse, or 42+7=49 minutes after Alex the house. The distance travelled will be:42/60 x 3.5 =49/60 x 3=2.45 miles from the house when they meet.
+129839 Here's yet another solution:
Alex walks at 3 mph...which equals 3 *5280/60 = 264 ft. per minute.. ....so .....in 7 minutes....he is........7*264 = 1848 feet ahead of Ben
And Ben walks at 3.5 mph = 3.5 * 5280 / 60 = 308 ft per minute
So....the distance that Ben closes on Alex every minute = 308 - 264 = 44 ft
So....to close the 1848 ft....it will take Ben 1848/44 = 42 minutes after he starts.......and he will be
3.5 (42/60) = 2.45 miles from home
