word problem plzzz help :(?

Question

yolanda and yoko ran in a 100yd dash. When yolanda crossed the finish line yoko was 10yd behind her. the girls then repeated the race with yolanda starting 10yd behind the starting line.if each girl ran at the same rate as before who won the race? by how man yards?



question #2 ? Assuming the girls run at the same rate as before how far behind the starting line should yolanda be in order for the two to finish in a tie?

Answer 1

distance = rate * time
in the first race, yolanda ran 100 yds with rate r in time t
since yoko was 10 yds behind yolanda when yolanda finished, that means she rand 90 yds with rate s in the same amount of time t

so the equations for the first race are
Yolanda: 100=rt
yoko: 90=st

since t is the same for both, you can substite t=90/s (from the 2nd equation) into the first equation to get
100=r (90/s)
100s = 90r
s=90/100 r
so
s=9/10 r

this means that yoko runs at 9/10 the speed of yolanda

When they repeat the race, yolanda must run 110 yds at rate r to finish, and yoko has to run 100 yds at rate (9/10)r to finish

so the equations are
yolanda: 110 = r * t
so t=110/r

yoko: 100 = (9/10)r * x (yoko's time is diffierent)
x=(1000/9)r

it should make sense that whoever has the least amount of time wins the race, so you need to figure out if 110 / r is less than or greater than (1000/9)r
the easist way is to just pick a value for r, and plug it in to see which one is bigger.

lets say r = 10

then 110/10 =11
and (1000/9) / 10=11 and 1/9

yolanda's time is slightly less so she still wins the race. Yoko
is (1/9) * (9/10) r behind, and since we said r=10,
that means she is 1/9 * (9/10) * 10 behind
1/9 * 9 = 1

she is 1 yard behind


for question #2
yolanda's rate is still r
yoko's rate is still (9/10)r

use d=rt, rewrite it as t=d/r

for yoko and yoland to tie, thier times have to be equal
let x be the distance yolanda starts behind the line

yolanda's time = yoko's time
(100 + x) / r = 100 / (9/10)r

again lets use 10 for r

(100 + x) / 10 = 100 / 9
9(100 + x) = 1000
900 + 9x = 1000
9x = 100

so x = 11.11 or 11 and 1/9

Answer 2

For the first as well as the second, you have the asnwer in your question.. If Yolanda started the race by standing 10yd behind the starting line and Yoko is at the starting line, then running at the same rate which they did before, they will end in a tie.

Answer 3

i will set up the equations. You remedy them. a million) The angles upload as a lot as one hundred eighty: x + y = one hundred eighty One attitude is 5 circumstances the different: x = 5y Use substitution to remedy for x and y. answer: x = 30, y = one hundred fifty 2) allow x be the first (smaller) type, then y = x + a million is the subsequent type, so: y = x + a million x + y = 37 or, using one equation: x + (x+a million) = 37 answer: x = 18, y = 19 3) Draw a photo first. Label both equivalent aspects x, and the bottom y. y is 4 inches decrease than between the aspects, so y = x - 4. the fringe is the sum of all aspects of the triangle, so the equations are: y = x - 4 2x + y = 32 you are able to also set this up as one equation: 2x + (x - 4) = 32 answer: x = 12, y = 8 4. Write the equation precisely the way the problem states it. One type (y) is (=) 5 (5) more effective than (+) yet another (x): y = 5 + x Their sum (x + y) is (=) 3 decrease than (-3 +) 3 circumstances the smaller (3x): x + y = -3 + 3x you presently have 2 equations, 2 unknowns. Simplify the second one equation, and remedy using substitution. answer: x = 8, y = 13

Answer 4

1st race had a winner.
2nd race handicapped and was a tie.

Answer 5

Yep, it's a tie...