1. ## Distance problems

Hello,

These are some SAT prep problems that I'm not sure about...

How long would a 1000 mile trip take if I was traveling at 50 mph for the first 500 miles and 60 mph for the rest of the trip?

A boat can reach its destination in 3 hours. If it goes twice as fast as it can reach 10 miles more than its destination, in 2 hours. How fast does it originally go?

A man can drive a certain distance in 5 hours. If he increased his speed by 10 mph, he could travel the same distance in 25/6 hours. What is the distance he travels?

2. Originally Posted by m58
How long would a 1000 mile trip take if I was traveling at 50 mph for the first 500 miles and 60 mph for the rest of the trip?
For the first 500 miles at 50 mph implies h = 500/50 = 10 hours, for the second 500 miles at 60 mph implies h = 500/60 = 8.33 hours.

Do you follow?

Similar logic applies to the 2nd and 3rd questions, have a go!

3. Ah, that's far easier than how I originally tried it! Thanks!

For the second problem, I did:
D = RT
3R = (R)(3)
D + 10 = (2R)(2)
D = 4R - 10
Input (4R - 10) into the first equation to get: 3 = 4R - 10/R -> 3R = 4R - 10 -> 10 = R

For the third problem, how's this answer: 500/4 = 125, so his distance will be 125 more mph than before

4. ## distance is the same

Originally Posted by m58
Hello,
A man can drive a certain distance in 5 hours. If he increased his speed by 10 mph, he could travel the same distance in 25/6 hours. What is the distance he travels?
since the distance is the same and we are just adusting the rate you have

$\displaystyle R_{1}5 = D = R_{2}t$

or

$\displaystyle R(5) = (R+10)(\frac{25}{6})$

thus $\displaystyle R=50$

so initially if the trip took $\displaystyle 5$ hours at $\displaystyle 50mph$ then the distance was $\displaystyle 250$miles
so if you add $\displaystyle 50+10= 60$

this matches the $\displaystyle \frac{250}{60} or \frac{25}{6}$hours