i could not undrstand thsi question

[defeated_soldier]

The winning relay team in a high school in a high school sports competition clocked 48 min for a distance of 13.2 km. The runners A,B,C .......

i could not understand what does it mean by 'clocked' .....does it wants to mean simply that the race 13.2 km ended in 48 min.

is this alright ?

can you plz explain.

Regards

 

[pka]

“…to mean simply that the race 13.2 km ended in 48 min.” I really do not know the answer. If I were doing the question, I would assume that is the meaning. Perhaps the full text of the question give you some more clues.

 

[defeated_soldier]

The winning relay team in a high school sports competition clocked 48 min for a distance of 13.2 km. The runners A,B,C and D maintained the speeds of 15km/hr,16 km/hr,17 km/hr and 18km/hr respectively. What is the ratio of the time taken by B to the time taken by D ?

a) 5:16
b) 5:17
c) 9:8
d)8:9

How do i solve it ?



lets take B took T1 time.
D take T2 time.

so the ratio is T1/T2.

now how do i calculate T1 and T2 now ?

we know T1=S1/V1 and T2=S2/V2

but we dont know S1 and S2 ...so i cant calculate T1 and T2 ......and thats why i cant not get the ratio.


can you help please ?

thnks

 

[pka]

I am hopeless when sports are involved in a question.
But I think that each member of a relay team runs the same distance.
If that is true then you can solve this:
16T<SUB>B</SUB>=18T<SUB>D</SUB>, where T<SUB>X</SUB> is time for X.

 

[defeated_soldier]

thanks for the response.

 

[Gene]

From the numbers given clocked must mean the four runners averaged 48 munutes per lap of 13.2km but is is not relevant. In a relay race they each run the same distance so the ratio doesn't depend on what the distance is. You have it as long as S1=S2=S.
T1/T2 = (S/V1)/(S/V2) = V2/V1

 

[defeated_soldier]

Excellent tips!

thanks

 

[soroban]

Hello, defeated_soldier!

A strange problem *

The winning relay team in a high school sports competition clocked 48 min for a distance of 13.2 km.
The runners A, B, C and D maintained speeds of 15 kph,16 kph, 17 kph and 18kph, respectively.
What is the ratio of the time taken by B to the time taken by D ?

$\displaystyle a)\;5:16\;\;\;\;b)\;5:17\;\;\;\;c)\;9:8\;\;\;\;d)\;8:9$

In a relay race, each runner runs the same distance . . . call it $\displaystyle d.$

Runner $\displaystyle B$ ran $\displaystyle d$ km at 16 kph . . . His time: $\displaystyle \,t_{_B}\,=\,\frac{d}{16}$ hours.

Runner $\displaystyle D$ ran $\displaystyle d$km at 18 kph . . . His time: $\displaystyle \,t_{_D}\,=\,\frac{d}{18}$ hours.

The ratio of their times is: \(\displaystyle \L\,\frac{t_{_B}}{t_{_D}}\;=\;\frac{\frac{d}{16}}{\frac{d}{18}}\;=\;\frac{18}{16}\;=\;\frac{9}{8}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*

Neither the total time nor the length of the track was relevant.


In fact, the "48 minutes" is very sloppy time-keeping.

Each runner ran 3.3 km at his respective speed.

The total time is: \(\displaystyle \,\frac{3.3}{15}\,+\,\frac{3.3}{16}\,+\,\frac{3.3}{17}{\,+\,\frac{3.3}{18} \:=\:\frac{9837.3}{12240}\:=\:0.80370098\) hours

$\displaystyle \;\;$which equals 48 minutes, 13.32 seconds.

And track events are measured to the nearest hundredth of a second.