# race and time problem...pls help

• Jan 22nd 2013, 08:03 PM
hisajesh
race and time problem...pls help
In a 1000m race, if A gives B a 40 m start, A wins by 19 seconds, but if A gives B 30 seconds start, B wins by 40 m. Find the time each takes to run a 1000 m?

My attempt:
1. A gives B a 40 m start, A wins by 19 seconds: so B runs 960 m in x+19 seconds and A runs 1000 m in x seconds.
Speed A =960/x+19
Speed B=1000/x

2. A gives B 30 seconds start, B wins by 40 m: so B runs 1000 m in y-30 seconds and A runs 960 m in y seconds.
Speed A = 960/y
Speed B=1000/y-30

When I try to equaute, it does not work.
The book answer is A takes 125 seconds and B takes 150 seconds. Appreciate your help.
• Jan 24th 2013, 03:27 AM
kylehk
Re: race and time problem...pls help
Quote:

Originally Posted by hisajesh
In a 1000m race, if A gives B a 40 m start, A wins by 19 seconds, but if A gives B 30 seconds start, B wins by 40 m. Find the time each takes to run a 1000 m?

My attempt:
1. A gives B a 40 m start, A wins by 19 seconds: so B runs 960 m in x+19 seconds and A runs 1000 m in x seconds.
Speed A =960/x+19
Speed B=1000/x

2. A gives B 30 seconds start, B wins by 40 m: so B runs 1000 m in y-30 seconds and A runs 960 m in y seconds.
Speed A = 960/y
Speed B=1000/y-30

When I try to equaute, it does not work.
The book answer is A takes 125 seconds and B takes 150 seconds. Appreciate your help.

Hi, your solution logic is good. But I've spot some places where it went wrong.

In Step 1, you correctly state that "A gives B a 40 m start, A wins by 19 seconds: so B runs 960 m in x+19 seconds and A runs 1000 m in x seconds."

Then Speed A should be defined by the ratio of the distance A runs (refer to your statement, A runs 1000m) to the time A takes (still refer to your statement, it's x seconds). So Speed A = 1000/x. Similarly, Speed B=960/(19+x).

In Step 2, you define that "A runs 960m in y secons" and correctly define Speed A=960/y. The mistakes comes from the time you think it takes for B. Since B first runs 30 seconds while A stands still, and B then runs another y seconds together with A: So B totally runs 1000m in (y+30) seconds (instead of y-30 as you suggested).

Double check your definitions of Speed A and Speed B and you can get the sample answer 125 and 150.
• Jan 24th 2013, 03:39 AM
kylehk
Re: race and time problem...pls help
Just find that the above equations is complicated to solve.

Below is another direction inducing less computation.

Let's now directly define variables required to solve, i.e., time $t_{A}$ for A to finish the race and time $t_{B}$ for B to finish the race.

Then Speed A: $s_{A}=\frac{1000}{t_{A}}$
Speed B: $s_{B}=\frac{1000}{t_{B}}$

In Step 1: in the time for A to finish plus another 19 seconds, B runs only 960m. We thus can find the following equation:
$t_{A}+19 =\frac{960}{s_{B}}=\frac{960}{\frac{1000}{t_{B}}}$; (1)

In Step 2: in the time for A to finish 960m plus another 30seconds, B runs 1000m. We thus have the following equation:
$\frac{960}{s_{A}}+30=t_{B}$
Substituting $s_{A}$ by $s_{A}=\frac{1000}{t_{A}}$, we have
$\frac{960}{\frac{1000}{t_{A}}}+30=t_{B}$; (2)

Now using equation (1) and (2), you could directly solve $t_{A}$ and $t_{B}$.
• Jan 25th 2013, 09:28 PM
kylehk
Re: race and time problem...pls help
Hi hisajesh, thanks for your thanks:)
• Jan 29th 2013, 11:36 AM
hisajesh
Re: race and time problem...pls help
@Kylekh: You are welcome :-)

I further played with this problem...

I tried to find x and y by equating the speed of A in both the case and equating the speed of B in both the cases.

Equating speed A on both the cases, I get
1000/x=960/y.....

Equating speed B on both the cases, I get
960/x+19=1000/y+30

• Jan 29th 2013, 11:40 AM
hisajesh
Re: race and time problem...pls help
But...If I tweak the equations without changing the meaning, I could get the answer like:
1000/x=960/y-30......
960/x+19 =1000/y.....

I could get x =125 seconds and y =150 seconds.
• Jan 29th 2013, 07:08 PM
kylehk
Re: race and time problem...pls help
Quote:

Originally Posted by hisajesh
@Kylekh: You are welcome :-)

I further played with this problem...

I tried to find x and y by equating the speed of A in both the case and equating the speed of B in both the cases.

Equating speed A on both the cases, I get
1000/x=960/y.....

Equating speed B on both the cases, I get
960/x+19=1000/y+30

Hi, thanks for your question. I think this confusion comes from the indistinction of x, y you defined and the time you want to find. Double check Posts #1 and #2. You'll find realize that the x you defined is the time for A to run 1000m, while the y you defined is the time for A to run 960m, not the time for B to run 1000m.

So by solving the equations as in Posts 1, 2 and 5, I'm sure you can find x=125 (i.e., A runs 1000m in 125 seconds) and $y=125\times\frac{960}{1000}$ (i.e., A runs 960m in y seconds).

Then it's normal that you didn't get y=150 seconds. Let me know if any further discussion I might provide.
• Feb 1st 2013, 07:55 PM
hisajesh
Re: race and time problem...pls help
Got it...
Here is my revised question....

1. A gives B a 40 m start, A wins by 19 seconds: so B runs 960 m in x+19 seconds and A runs 1000 m in x seconds.
Speed A =1000/x
Speed B=960/x+19

2. A gives B 30 seconds start, B wins by 40 m: so B runs 1000 m in y+30 seconds and A runs 960 m in y seconds.
Speed A = 960/y
Speed B=1000/y+30

Here I consider Speed A is equal in both the cases. so I have 1000/x=960/y, I get x=25 and y=24..can you tell me what is wrong in my approach.
• Feb 1st 2013, 09:04 PM
kylehk
Re: race and time problem...pls help
Quote:

Originally Posted by hisajesh
Got it...
Here is my revised question....

1. A gives B a 40 m start, A wins by 19 seconds: so B runs 960 m in x+19 seconds and A runs 1000 m in x seconds.
Speed A =1000/x
Speed B=960/x+19

2. A gives B 30 seconds start, B wins by 40 m: so B runs 1000 m in y+30 seconds and A runs 960 m in y seconds.
Speed A = 960/y
Speed B=1000/y+30

Here I consider Speed A is equal in both the cases. so I have 1000/x=960/y, I get x=25 and y=24..can you tell me what is wrong in my approach.

There are two variables x, y. So you need a system of two equations to solve them.

More specifically, using only 1000/x = 960/y, you could not specify the values of x and y.

By equating Speed A and Speed B, both the following two equations should be solved together to get x and y:

1000/x = 960/y
960/(x+19) = 1000/(y+30)