Math Is Fun Forum

It is not an equation. An equation must have an equals sign = in it.

Yes!

The puzzle was posted in this old threaad in this forum that I’ve been browsing. It didn’t appear that anyone had got it. So I re-posted it here to make sure it was resolved.

BTW, I didn’t get it either. I had to search the Internet for the solution.

The answer to Q1 is the same as the answer to Q2.

Or, if you want to analyse Q1 just for the fun of it …

when the day after tomorrow is yesterday = in three days’ time
when the day before yesterday was tomorrow = four days ago

But in three days’ time it will be exactly one week after four days ago. Therefore the “today” in Q1 can be any day of the week – and so the only way to answer the question is with reference to “today” not as a particular day of the week but as any day in general.

The month numbers you have there are for the beginning of the month. Thus your month 10 would mean beginning of that month, meaning only 9 full months have passed. Since the question wants 10 months, I think you should add another month’s calculation … which should bring the total number of rabbits to 288.

Okay, assume we start with a male–female pair and all babies are born as male-female pairs.

AA = Adult (mature) pair, JJ = Juvenile (one-month old) pair, BB = Baby (newborn) pair

Starting with AA:

Total rabbits at end of 10 months = 2×(28 + 13 + 19) = 120.

Neither does omy. I’m just going by lightning’s strange logic.

McCague? I’ve heard of a bowler called Martin McCague who used to play for Kent.

One possible answer is 1 hour 50 minutes 600⁄13 seconds (i.e. he waited from 480⁄13 seconds past 11:04 to 300⁄13 seconds past 12:55).

There are other possible answers (e.g. he might have waited from shortly before 10:10 to shortly after 13:50).

You can try log[sub]10[/sub] instead of ln if that helps a little.

5:30.

You only need the first two clues. Then just look at which two times differ by 2 hours 5 minutes.

The puzzle follows the conventional format for such logic puzzles: you are given several sets of n elements each, and the aim is to find the correct bijections between the sets. So if you have matched (n−1) members between two sets, the last remaining member of one set must automatically match that of the other set.

Anyway, I’m out of puzzles for the moment – you brainy people have solved the ones I have too quickly. I’ll try and think of another puzzle or two now.

1. Substitute x = −2 and y = 5 into the two equations. Then you get two simultaneous equations in a and b instead of in x and y. So just solve for a and b.

2. Squaring both sides of the first equation, x[sup]2[/sup] − 2xy + y[sup]2[/sup] = 64. From the second equation, xy = 6. Substitute this into the squared equation and find x[sup]2[/sup]+y[sup]2[/sup]. (Note: You do not need to find x and y, just x[sup]2[/sup]+y[sup]2[/sup].)

3. Let the wind speed be v km/h.
When flying with the wind, the speed of the plane was 300+v.
The time taken to fly 900 km at this speed was t[sub]1[/sub] = 900⁄(300+v). (Remember: time = distance ∕ speed.)
When flying against the wind, the speed of the plane was 300−v.
The time taken to fly 600 km at this speed was t[sub]2[/sub] = 600⁄(300−v).
Now put t[sub]1[/sub] = t[sub]2[/sub] and you can solve for v.