# Math Help - word problem.

1. ## word problem.

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 !!

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.

3. Nope. Ben started 2 hours after

4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.

5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:

$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ 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:

$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$

Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$

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.

6. Hello, dugongster!

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?

. . In the next $\,h$ hours, he travels another $6h$ km.
Alvin has traveled $6h + 12$ km in the first $\,h$ hours.

During the same $\,h$ hours, Ben has travelled:
. . $4 + 5 + 6 + \hdots + (h+3) \:=\:\dfrac{h(h+7)}{2}$ km. .**

The two distances are equal: . $\dfrac{h(h+7)}{2} \:=\:6h + 12$

. . $h^2 + 7h \:=\:12h + 24 \quad\Rightarrow\quad h^2 - 5h - 24 \:=\:0$

. . $(h - 8)(h + 3) \:=\:0 \quad\Rightarrow\quad h \:=\:8,\:-3$

Answer: . $\text{8 hours}$

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**

We want the sum of the numbers from $\,4$ to $h\!+\!3.$

The sum of the numbers from $\,1$ to $h\!+\!3$ is: . $\dfrac{(h+3)(h+4)}{2}$

The sum of the numbers from $\,1$ to $\,3$ is: . $1+2+3 \:=\:6$

Hence,the sum of the numbers from $\,4$ to $h\!+\!3$ is:

. . $\dfrac{(h+3)(h+4)}{2} - 6 \;\;=\;\; \dfrac{h^2 + 7x + 12 - 12}{2} \;\;=\;\;\dfrac{h(h+7)}{2}$

7. great. thanks to both of you. grep, you rock !!!

8. Originally Posted by Soroban
Hello, dugongster!

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 ( $a_1 = 4$).