<p>These two seem easy but I can't seem to figure out a specific way to solve it. I know d=r x t but I get two unknowns D:
Here are the problems:
A boat can reach it's destination in 3 hours. If it goes twice as fast it can reach 10 miles more than it's destination, in 2 hours. How fast does it originally go?</p>
<p>A man can drive a certain distance in 5 hours. If he increased his speed by 10miles per hour, he could travel the same distance in 4 1/6 hours. What is the distance he travels?</p>
<p>The first one is 10mph.
(10mph x 3 hrs = 30 miles
(20mph x 2 hrs = 40 miles)</p>
<p>the second is 250 miles
(If you divide 5 hrs by 4 1/6 hrs, you get 1.2. Then you must think, 10 mph is 20% of what number? 50! So add the 10 mph to the 50 mph and u get 60 mph.
50 mph x 5 hrs = 250 miles
60 mph x 4 1/6 hrs = 250 miles)</p>
<p>don’t know if my explainations helped in any way… I’m good at math, but I can’t really explain it. :p</p>
<p>1) r*t = d.
we don’t know the distance, but we don’t need it, either. original is 3(r) = d and the faster one is 2(2r) = d + 10.
Then this becomes 3r = d and 4r = d +10. Subtracting these two equations gives -r = -10, or r = 10, so 10mph.
EDIT: or just plug in d = 3r into the other equation, giving 4r = 3r +10, so r = 10.</p>
<p>2) d = rt. let his one of his speed equation be 5r = d. Let 4 1/6 be 25/6. When the speed is increased, it is 25/6(r+10) = d.
so that’s (25r + 250)/6. Let 5r from the original be 30r/6. When you subtract the two equations 30r/6 = d and (25r+250)/6 = d, you get (30r - 25r - 250)/6 = 0. So that’s 5r -250 = 0, or 5r = 250. so r is 50. Plug it into the original, and that’s 5*50 = 250.
EDIT: you plug it back into the equation because you want the distance.</p>