I always get this type of question wrong. It's like when I see this kind of question, I forget how I should solve! What's a good approach?
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours
Thank you.
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I think the answer is B this is how i solve it though
D = RT
After 3 hours there would be a distance of 18 miles between the two cyclist
Distance covered by first cyclist
D = 6*3 = 18
The difference in speeds at which the two are cycling is 4 mph
Therefore we assume Second cyclist has to cover a distance of 18 miles at a speed of 4 mph to catch up with the first cyclist.
Hence D = RT
18 = 4T
18 / 4 = T
4/2/4 = T
4 / 1/2 =T
D = RT
After 3 hours there would be a distance of 18 miles between the two cyclist
Distance covered by first cyclist
D = 6*3 = 18
The difference in speeds at which the two are cycling is 4 mph
Therefore we assume Second cyclist has to cover a distance of 18 miles at a speed of 4 mph to catch up with the first cyclist.
Hence D = RT
18 = 4T
18 / 4 = T
4/2/4 = T
4 / 1/2 =T

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At 6mph, in 3 hrs, the first cyclist would have travelled 3*6 = 18 miles.
The second cyclist will take 18/(106) = 4.5 hrs to catchup with the first cyclist.
The procedure to solve the "catchup" type problems is below.
“Catch up” Problem: Two objects (A: r1 , B: r2) moving in same direction; A faster than B (r1 > r2 ) but B ahead of A.
Procedure:
• Find the catch up rate: ie difference of the two rates: r1  r2
• Find catch up distance: ie the distance by which A is behind B.
• Divide the distance by catchup rate : ie d / (r1r2) to get time (t) taken for A to catch up with B (catchup time)
• If the question asks how long it would take to overtake by x miles. Then the prev cal will be modified to have total distance = (catchup distance + overtake distance).
• If the question asks how far had A travelled when A and B meet and if A started x hrs earlier than B, then A’s distance = (x*r1)+catchup distance.
The second cyclist will take 18/(106) = 4.5 hrs to catchup with the first cyclist.
The procedure to solve the "catchup" type problems is below.
“Catch up” Problem: Two objects (A: r1 , B: r2) moving in same direction; A faster than B (r1 > r2 ) but B ahead of A.
Procedure:
• Find the catch up rate: ie difference of the two rates: r1  r2
• Find catch up distance: ie the distance by which A is behind B.
• Divide the distance by catchup rate : ie d / (r1r2) to get time (t) taken for A to catch up with B (catchup time)
• If the question asks how long it would take to overtake by x miles. Then the prev cal will be modified to have total distance = (catchup distance + overtake distance).
• If the question asks how far had A travelled when A and B meet and if A started x hrs earlier than B, then A’s distance = (x*r1)+catchup distance.
One thing clear is they would have travelled the same distance by the time second cyclist catches up. So equate both distance but remember the second guy starts 3 hours behind.kanha81 wrote:I always get this type of question wrong. It's like when I see this kind of question, I forget how I should solve! What's a good approach?
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours
Thank you.
Let t be time the first cyclist starts, then t3 be time of 2nd cyclist. So
10(t3)=6t
t=7.5. So from the time the second guy starts 7.5 hrs would have elasped, but remember he starts 3 hours later. So 7.53= 4.5hrs.
The second probably better or easier way in which you don't have to worry about subtracting elasped time is to let t be the time of the 2nd cyclist. Then t+3 woud be time of first.
10t=6(t+3)
t=4.5
Hope this helps.