Problem Solving

Problem

Two horses begin running on an oval course at the same time. One runs each lap in 9 minutes; the other takes 12 minutes to run each lap. How Many minutes after the start will the faster horse have a one lap lead?

(a) 36
(b) 12
(c) 9
(d) 4
(e) 3

The solution provided by Kaplan is certainly adequate. It states that since a lap is completed every 9 minutes by the faster horse and in that same amount of time the slower horse will have completed 9/12 or 3/4 of a lap, thus making the faster horse 1/4 lap ahead of the slower horse. At this point you can simply figure out that it will take four, 9-minute periods to get one full lap ahead of the slower horse which will have taken a total of 9x4 minutes to achieve.

My question, however, is whether it is feasible to address this problem by using the Rate Pie methodology, which I am generally more comfortable with.


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Sorry I don't know the rate pie methodology, but I have a few pointers.

1) For a Rate problem, if two objects are going in the same direction - subtract the speeds. (think of yourself in a car looking at the car next to you. Although both of you have very high speeds, the relation of your speed to the neighboring car is the difference of your speeds). In this situation, the horses ARE going in the same direction.

So you subtract (12-9 = 3). Thus, it takes 3 minutes for the horse to take over the other horse. For a one lap lead you first divide 12/3 = 4 and then multiply it 4*9 = 36

2) For the Rate problem, if two objects are going in a different direction - add the speeds. In this case it will be 12 + 9 = 21, and then take the calculation forward

Thank you. That definitely helps. I had never thought to subtract speeds.

Ryan Ziemba wrote:Problem

Two horses begin running on an oval course at the same time. One runs each lap in 9 minutes; the other takes 12 minutes to run each lap. How Many minutes after the start will the faster horse have a one lap lead?

(a) 36
(b) 12
(c) 9
(d) 4
(e) 3

The solution provided by Kaplan is certainly adequate. It states that since a lap is completed every 9 minutes by the faster horse and in that same amount of time the slower horse will have completed 9/12 or 3/4 of a lap, thus making the faster horse 1/4 lap ahead of the slower horse. At this point you can simply figure out that it will take four, 9-minute periods to get one full lap ahead of the slower horse which will have taken a total of 9x4 minutes to achieve.

My question, however, is whether it is feasible to address this problem by using the Rate Pie methodology, which I am generally more comfortable with.


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Here is another approach.
draw a table like the one attached. the cells in GREEN are the information we are given. the cells in RED is what I am plugging in.
So lets assume a distance for the track (lap). Lets take a LCM of 9 and 12 = 72. so the track is 72miles. Now lets get the rate for the information we are given by using 72 as the total track distance. remember the formula for rate = d/t so we get the rate to be 8 and 6.
Now, all we are looking for is (what number)*8 will give us a distance double the distance of the other horse.
Lets start plugging in the answers: (a) = 36. So 8*36=288. So in 36 minutes, the first horse will cover a distance of 288 miles. OR 4 LAPS because remember one lap = 72 miles.
Now lets plug it in for the second choice 6*36=216. Or 3 laps.
And THAT'S IT! We got our answer. In 36 mins, horse 1 will run 4 laps and horse 2 will run 3 laps.

hey Showbiz,
But could you please explain me...

For a one lap lead you first divide 12/3 = 4 and then multiply it 4*9 = 36


what is the signigicance of which calculataion...what does 12/3 will give us...and why we shld be multiplying by 9 here..will lead to the result....

can you plz detail...thanks..

i am confused about the answer choices here...
as question clearly says in 1 takes 9 minutes and another take 12
CDE is out of equation straight away
i checked state of horses in 24 minutes...not whole numbers
so left A
another way
36/9=4 (9x4=36)
36/12=3 (or 12x3=36)
other 4 choices even doesnot seem closer to this


well these are wild method and may be worked for this question only....or may be because of answer choices....

Horse A takes 9 minutes to complete 1 lap. Therefore speed of Horse A is 1/9 laps/minute

Horse B takes 12 minutes to complete 1 lap. Therefore speed of Horse B is 1/12 laps/minute

Since both horses are moving in the same direction, the relative speed will be

(1/9 - 1/12) = 1/36 laps/minute

The questions asks us the time it would take for the second horse to be 1 lap ahead

Therefore time required for the horse to be 1 lap ahead = 1 Lap lead / (1/36)[relative speed] = 36 minutes