Word problem

Doomtown is 200 miles due west of Sagebrush , and Joshua is due west of Doomtown. At 9am Mr Archer leaves Sagebrush for Joshua. At 1 pm. Mr Sassoon leaves Doomtown for Joshua. If Mr. Sassoon travels at an average speed 20 mph faster than Mr. Archer and they each reach Joshua at 4 pm how fast is each traveling?

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Jun 11, 2010
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Word Problem
by: Karin

This problem is dealing with distance, time and rate; therefore, we will be using the distance formula. D = rt.

Let's first think of all the information that we know and don't know.

Mr. Sassoon:
Distance = D (We don't know the distance between Doomtown and Joshua)

Rate = r + 20 (he travels 20mph faster than Mr. Archer)

time = 3 hours (1:00pm - 4:00 pm)

Mr. Archer:
Time: 7 hours (travels from 9am - 4pm)

Rate: r (we don't know his rate)

Distance: D + 200 We labeled d as the distance between Doomstown and Joshua and we know that Sagebrush is 200 more miles from Doomtown.

So, we now know that we have 2 unknown variables: r and d.

Let's use the distance formula to write 2 equations; 1 for Mr. Sassoon, and 1 for Mr. Archer.

Mr. Sassoon:
d= rt
d = (r+20)3
d = 3r + 60 (Distribute the 3)

Mr. Archer:
d = rt
d+200 = r7

d + 200 = 7r (written properly)

Now we can use our knowledge of solving a system of equations word problem. Using the substitution method is best because we know what d is equal to.

So, let's substitute 3r + 60 into the equation: d + 200 = 7r

3r + 60 + 200 = 7r
3r + 260 = 7r

Now let's solve for r.
3r + 260 = 7r
3r - 3r + 260 = 7r - 3r
260 = 4r
260/4 = 4r/4
65 = r

Since r = 65, we know that Mr. Archer travels 65 mph and Mr.Sassoon travels 85 mph. (r+20)

I hope this helps!
Karin





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