# Why do i solve the sum in this way (mathematics)?

The sum is as given below:

"There are two cars A and B.They start from the same point at the same time and travel in a circular path .Car A can complete one round in 18 minutes whereas car B can complete the round in 16 minutes. After how much time will they meet?"

my maths teacher told me to take the LCM(least common multiple) of their respective completion times of a round. Can anybody please explain to me as to why i should take the LCM. Also i would be greatful if you could explain it simply.Cheerio.

"There are two cars A and B.They start from the same point at the same time and travel in a circular path .Car A can complete one round in 18 minutes whereas car B can complete the round in 16 minutes. After how much time will they meet?"

my maths teacher told me to take the LCM(least common multiple) of their respective completion times of a round. Can anybody please explain to me as to why i should take the LCM. Also i would be greatful if you could explain it simply.Cheerio.

active| newest | oldestI may not be seeing the same picture as everyone else!

I assume that both cars are travelling in the same direction. the 16 minutes car is red the 18 minutes car is green.

After 16 minutes the red car will have finished one lap - The green car will not finish until a further 2 minutes has passed.

Therefore the red car lapped the green car 2 minutes earlier in the lap - ie after 16 - 2 minutes or 14 minutes? isn't this right?

The question as stated does not say the cars must meet at the start point.

Car A starting point 8 laps, car B starting point 9 laps is the only place they meet assuming they are traveling in the same direction. The least common number is 2.

If they are traveling in opposite directions then it is much less.

If that is the question as the teacher gave it to the class I would tell him to take an English class he uses assumptive English.

The question is phrased badly leaving readers to make assumptions.

On a clock face the hands line up every hour just not in the same place hence my initial answer the car lapping at 16 mins passes the other before the car lapping at 18 mins gets back to the start line.

Traveling in the same direction car B only gains two minutes per lap on car A, so other than at the start they only meet on the ninth lap of car B and the eight lap of car A at the start. it is the only times they meet no others.

Joe

Nope, the way I read the question the cars pass each other on every lap every 16 minutes just not in the same place. Just like the hands on a clock pass every hour.

The bit MISSING from the question is "when will the cars meet back at the start line."

If the cars are traveling in the same direction then car B gains a two minute lead every lap until it makes 1 full lap more than car A

A clock is a ratio of 60 to 1 the cars are 1.1 to 1 ratios

1 lap car A

1 lap 2 minutes car B

2 lap car A

2 lap 4 minutes car B

and so on.

Because the most common multiple is 1 and it won't help you.

Regarding this method of finding the

LeastCommonMultiple of the completion time for one round, doing that will give you the smallest time needed for either car to do anintegernumber of rounds.Coincidentally, that method gives you the right answer for the numbers given, since it turned out that 144 minutes was exactly 8*18 minute turns for car A, and 9*16 minute turns for car B. So both cars travelled an integer number of turns, and they met exactly where they started.

However, in general the cars might NOT meet after an exact integer number of turns.

Supposing car A completes one turn in 17 minutes, and car B completes one turn in 14 minutes. The least common multiple of 17 and 14 is 17*14 = 238. So we can expect the cars to meet at 17*14=14*17 = 238 minutes, when car A has done exactly 14 turns, and and car B has done exactly 17 turns.

However the cars will also meet at time

beforethen, e.g. at t = 79+1/3 minutes, when car A has moved:a = (C/17)*(79+1/3) = (4 +2/3)*C = (4 +2/3) turns

and car B has moved:

b= (C/14)*(79+1/3) = (5 +2/3)*C = (5 +2/3) turns

I'll start by calling the length of the circular path C.

Car A moves a distance a = a(t) = (C/18)*t.

Car B moves a distance b = b(t) = (C/16)*t.

For both of those expressions, time t is measured in minutes.

When the cars meet again on this circular road, their respective distances traveled will be the same, plus or minus some integer number of the whole circle C. The condition for where-when the cars meet can be written:

a +n*C = b, [where n is some positive integer]

I wrote it that way because in positive time, a is always lagging behind b because car A is travelling more slowly than car B.

Next I substitute in those expressions for a and b from above. Then the Cs divide out, and I am left with an expression relating n and t.

If I did the math right, the

firsttime the cars meet, occurs at n=1 and t=144 minutes.(and the zeroth time they meet is when they start, at n=0 and t=0 minutes)

Weeeelll, unless I am off by some chance, the LCM is not for the first time they meet, but the first time they'll meet in the exact same spot.

You could think of the cars as the long and the short had of the clock, which travel at different speeds. For instance, the long hand take 60 minutes to complete a circle, while the short one needs 12*60 = 720 minutes. If you take the LCM, which in this case is 720 minutes, you'll get the time needed to meet at the exact same spot they started from (12:00 for instance). Which makes sense, as the only time those two hands will meet there again, or be in the exact same positions, is after 12 hours have passed.

However, the first time those two will actually meet will be, by logic, sometime between 1:00 and 2:00 (again, starting from 12:00/00:00), so after something between 60 and 120 minutes. Because the long hand will have to pass the short one every something-hour.

If you ponder over that problem for a bit (or, you know, cheat by using google) you should be able to find the correct formulas to calculate the actual time those two cars will meet.

Hint:

You'll need to break it down to how many degrees (of a circle) each car will travel per time unit and search for a common degree X at the same time point t.

Ask yourself when the first time both cars have travelled equal distances must be, and it becomes obvious.