Posted by shivangi on Tuesday, January 8, 2008 at 8:12am.
Yolanda and Yoko ran in a 100-yd dash. When Yolanda crossed the finish line, Yoko was ten yd behind her. The girls then repeated the race, with Yolanda starting 10 yd behing the starting line. If each girl ran at the same rate as before, who won the race? By how many yards?
I didn't undertand this problem, but someone had explained it to me and showed me what the answer would be. yolanda would win again by 1 yd. because yolanda's rate is 10 yds per 10 sec. and yoko's rate is 9 yds. per 10 sec
Apparently there is a sequel to this problem, which is,
b.Assuming the girls run at the same rate as b4, how behind should yolanda be in order 4 the two 2 finish in a tie. i think yolanda should be at least 2 yds. behind the finish line. Can someone plz check this over 4 me and tell me what the right answer would be???
- Mathematics! - Writeacher, Tuesday, January 8, 2008 at 9:02am
- Mathematics! - drwls, Tuesday, January 8, 2008 at 9:17am
Yolanda's speed is V, and Yoko's speed is 0.9 V, since she travels 90% as far in a given time. They do not tell you the actual speed, and you don't have to know it.
a. If Yolanda starts 10 yards back, Yoko travels 100 yards in time
T = 100/(0.9V), while Yolanda travels
V*100/(0.9V) = 111.11 yards. Yolanda ends up 111.11-110 = 1.11 yards ahead.
b. To finish in a tie, let the distance Yolanda stands behind the starting line be d (yards). She must cover 100 + d yards in the same time Yoko goes 100 yards.
100/(0.9V) = (100+d)/V
Cancel out the V's.
111.11 = 100 + d
d = 11.11 yards
- Mathematics! - Reiny, Tuesday, January 8, 2008 at 9:34am
Continuing with where we left off last night, (Writeacher gave you the link to the posting), recall that
Yolanda's rate was 100/t sec and Yoko's rate was 90/t sec.
let the extra distance that Yolanda should run for them to finish in a tie be x yds.
so the time for Yolanda to finish the race is
(100+x)/(100/t) = (100+x)t/100
and Yoko's time to finish exactly 100 yds is 100/(90/t) = (100/90)t
if they are to finish in a tie, their times are equal, so
(100+x)t/100 = (100/90)t , divide both sides by t, t cannot be zero obviously
(100+x)/100 = 100/90
which gives us
x = 100/9 yds or 11.111.. yds
- Mathematics! - Tanya, Tuesday, October 13, 2009 at 5:58pm
The answer is one yard behind the starting line. This is so because if Yolonda runs 10 miles per 10 seconds and Yoko runs 9 miles per 10 seconds that means Yoko will need Yolanda to be 1 yard backwards.
- Mathematics! - Maria, Friday, October 15, 2010 at 12:55am
Tanya is wrong because in 15 seconds, Yolanda had finished the race and Yoko was at the the 90 yd. mark. This means that even if we were to put Yolanda 10 yards behind the starting line, both Yolanda and Yoko would reach the 90 yard mark at the same time, but since Yolanda can run 10 yards in less time than Yoko, she would win the race. Another thing, you all forgot an important aspect of the problem!!! It also says that Yolanda finished the entire race, or 100 yds, in 15 SECONDS! So there is no way that their rates are 10 yds 10 seconds and 9 yards per 10 seconds, Yolanda only took 15 all together!!
- Mathematics! - Zach, Thursday, February 12, 2015 at 7:20pm
She would have to be 2.22 repeating for them to have a tie. Yoku actually runs 8 yards in ten seconds
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