Whats wrong with getting 5 as your answer?
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
Here's what I did. We know these two things about arithmetic progressions:
where d is the distance between numbers in the sequence. We also know that:
where is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
Hope you see what I did there. After that, rearrange terms so you get:
For that, I get one positive root of t = 10 (also t = -1 which is obviously not useful). Voila! Thanks for the problem, I had fun with that.
Hello, dugongster!
Both of you forgot about the two-hour headstart.
Alvin set out from a certain point and travelled at the rate of 6 kph.
After Alvin had gone for 2 hours, Ben set on to overtake him and went 4km the first hour,
5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour.
After many hours were they together?
Alvin had a two-hour headstart; he is already 12 km ahead.
. . In the next hours, he travels another km.
Alvin has traveled km in the first hours.
During the same hours, Ben has travelled:
. . km. .**
The two distances are equal: .
. .
. .
Answer: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
We want the sum of the numbers from to
The sum of the numbers from to is: .
The sum of the numbers from to is: .
Hence,the sum of the numbers from to is:
. .
Hi Soroban! With due respect (which is considerable, I might add, you being one of the most helpful and, let's face it, awesome people here), I took into account the 2 hour headstart by using (t - 2) as the time in the arithmetic progression (n = (t - 2)), and starting at 4 ( ).
The answer is indeed 10 hours from when Alvin sets out. I do like the way you solved it, however. You just had to add back the 2 hours head start to get the total time of 10, so your calculations are essentially correct. I think the head start is part of the time it took for them to meet, though, and should be counted.
And dugongster, you're quite welcome.