ddm wrote:A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10 th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30 th of a minute. What is B’s speed in m/s?

(A) 12

(B) 14

(C) 16

(D) 18

(E) 20

Its A 12

1 race:
A runs 480 m in t seconds
B 0.9 x 480 m in t+6 seconds ...............................................1

2nd race:
A runs 480 m in t seconds
B runs 0.7 x 480 m in t-2 seconds .........................................2

Speeds of the runners are constant in both the races

Equate 1 and 2

0.9*480 / (t+6) = 0.7*480 / (t-2)

solve for T

t = 30

substitute the value of t in equation 1

0.9* 480/( 30+6) = 12.

Hope its clear

Can you please explain how you got the 0.9 and 0.7

thanks!!

I figured it out....don't worry

I'm sorry i don't get it, how did you get 0.9 and 0.7?

chykie wrote:I'm sorry i don't get it, how did you get 0.9 and 0.7?

in the first race A gives B a head start of 48 m which is 10% ( 48/480)
thus b has to cover the remaining distance only which is 90% of 480m.

similarly 0.7 . B is 144 mts ahead which is 30% ( 144/480) . thus have 2 cover only 70% of the total distance hence 0.7

hope its clear now...

do let me know if u have any doubts..

sudhir3127 wrote: 1 race:
A runs 480 m in t seconds
B 0.9 x 480 m in t+6 seconds ...............................................1

2nd race:
A runs 480 m in t seconds
B runs 0.7 x 480 m in t-2 seconds .........................................2

Guys, please, help me to understand why it was concluded that in each race time was equal?

I got following:
1 race:
A runs 480 m in t(1) seconds
B 0.9 x 480 m in t(1)+6 seconds ...............................................1

2nd race:
A runs 480 m in t(2) seconds
B runs 0.7 x 480 m in t(2)-2 seconds .........................................2

A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10 th of a minute. In the second heat,
A gives B a head start of 144 m and is beaten by 1/30 th of a minute. What is B's speed in m/s?

(A) 12

(B) 14

(C) 16

(D) 18

(E) 20


equation 1: 480-48 / Sb - 480/ Sa = 6
480/Sa - 480-144/Sb = 2


solve both to get Sb as 12

I solved this problem in my way. Very long but clear (i think)

A and B ran a race of 480 m.
S=480

In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute.
B starts 48 m ahead A. But A will be the firast at the finish. B needs additional 1/10th of a minute (6 sec) to get to the finish. Time of A and B are equal here.
So, we got 2 equations using the distance formula:
Racer A: 480 m in t1 seconds => speed of racer A = 480/t1 m/s
Racer B: (480-48) m in t1+6 seconds => speed of racer B = (480-48)/t1+6 = 432/t1+6 m/s


In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
So, again we got 2 equations using the distance formula:
Racer A: 480 m in t2+2 seconds => speed of racer A = 480/t2+2 m/s
Racer B: (480-144) m in t2 seconds => speed of racer B = 336/t2 m/s

Speed of each racer are constant, thus:
speed of racer A = 480/t1 = 480/t2+2
speed of racer B = 432/t1+6 = 336/t2

From 480/t1 = 480/t2+2 we have:
1/t1 = 1/t2+2 (divided by 480)
t1=t2+2
substitute this equation to 432/t1+6 = 336/t2
432/t2+8 = 336/t2
solve to get t2 = 28

substitute t2 = 28 in
speed of racer B = 336/t2 m/s
speed of racer B = 12 m/s

Answer A

ronniecoleman wrote:
equation 1: 480-48 / Sb - 480/ Sa = 6
480/Sa - 480-144/Sb = 2


solve both to get Sb as 12

Yes, this reasoning is more clear than my

4meonly wrote:

ronniecoleman wrote:
equation 1: 480-48 / Sb - 480/ Sa = 6
480/Sa - 480-144/Sb = 2


solve both to get Sb as 12

Yes, this reasoning is more clear than my

Great work !

I'm trying to solve it another way, but I don't know if it works. Can someone else tell me if this is valid...

1st heat:

delta time = 6 sec

delta distance = 48 m

T = D/(Sb-Sa)

6 = 48/(Sb-Sa) valid equation?