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by finance » Fri Jul 22, 2011 4:05 am
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
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by knight247 » Fri Jul 22, 2011 4:29 am
Answer is A

in the 1st heat
A
Distance=x
T=t1
B
Distance=x-48
T=t2

t2-t1=1(60)/10=6 sec
t2-t1=6
can be rewritten as

(x-48)/s2-x/s1=6
x/s1=(x-48)/s2-6.....(1)

in the 2nd heat
A
Distance=x
T=t3
B
Distance=x-144
T=t4

t3-t4=1(60)/30=2
t3-t4=2
Can we rewritten as

x/s1-(x-144)/s2=2 (both ppl's speed are the same in heat 1 and 2)

x/s1=2+(x-144)/s2.....(2)

combining 1 and 2 we get

2+(x-144)/s2=(x-48)/s2-6
(2s2+x-144)/s2=(x-48-6s2)/s2
2s2+x-144=x-48-6s2
8s2=96
s2=12

Hence A
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by pemdas » Fri Jul 22, 2011 4:33 am
>>> assumption that the rates of A and B in the first and the second heats are constant.

distance = 480, A's rate=x, B's rate=y
the first heat parameters, (480-48)/y - 480/x = 1/10
the second heat parameters, 480/x - (480-144)/y = 1/30

By solving the system of equations we get the answer

{ 432x-480y=xy/10
{ 480y-336x=xy/30

432x-480y=3(480y-336x), 1440x=1920y, 3x=4y, x=4y/3

432(4y/3)-480y=(4y^2)/30, 576y-480y=2y^2/15, 96y=(2y^2)/15, 2y=96*15, y=720
and x=720*4/3=960

x is A's rate (speed), y is B's rate (speed) --> 720/60 or 720 meters per minutes makes 12 m/s


a


finance 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
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by HeintzC2 » Fri Jul 22, 2011 11:51 am
finance 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
Without requiring solving simultaneous equations:
We know that B lost by 6 seconds when given a 48m head start, but won by 2 seconds when given a 144m head start.

This means that B made up 8 seconds in the race between the 48m and 144m intervals.
Since we're assuming B is running at a constant pace the entire race, (144m-48m)/8s = 12m/s

Correct answer is A
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by finance » Fri Jul 22, 2011 1:53 pm
Yes..OA is A..can someone pls.solve according to Manhattan table approach?
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