The worlds first riddle!

"You can eliminate the nickel/quarter pairing since we know that 1 coin is a dime.
Since one of the dimes (say dime1) is already out, it leaves 3 possible combinations -

dime1/dime2,
dime1/nickel,
dime1/quarter.)

Therefore: 1 in 3.

What say you?"

If you had said, "the first one is a dime," I'd agree. But, as stated, if you run this experiment an infinite number of times, 5/6 will allow the opportunity to say, "one is a dime," and 1/5 of those will include both dimes.

[size=7]SERIES
250.59[/size]

Gotta run. Will do the other later.

Mark wrote, "Gotta run. Will do the other later."

You don't scare me. Well, not much, Ok, I'm outta here!


Oh, WELL.
Fairy nuff, as they say in the boondocks.



You drive to the store at 20 mph and return by the same route at 30 mph. discounting the time spent at the store, what was your average speed






Hey Nick, chew on this:

Ten years from now Raju will be twice as old as Juhi was when Mena was nine times as old as Raju.

Eight years ago, Mena was half as old as Juhi will be when Juhi is one year older than Raju will be at the time when Mena will be five times as old as Raju will be two years from now.

When Raju was one year old, Mena was three years older than Raju will be when Juhi is three times as old as Mena was six years before the time when Juhi was half as old as Raju will be when Mena will be ten years older than Mena was when Juhi was one-third as old as Raju will be when Mena will be three times as old as she was when Juhi was born.

How old are they now


Who the freek cares, is what I say, but I am as you not think I arm.

[size=7]LORD HELPLESS
100km/hr
He's got twice as far to go as the back of the train.

AVERAGE SPEED
24mph[/size]

RAJU, JUHI, & MENA
Are their ages integers?
Combining the second paragraph with the fact that Mena was once nine times as old as Raju, I get that Raju and Mena are currently:
7 and 23
15 and 39
23 and 55
31 and 71
39 and 87 or
47 and 103 (this seems like a reasonable maximum)

However, that makes Raju's age odd. In ten years, it will still be odd, but it's supposed to be twice Juhi's age when yada yada yada. That makes Juhi's age a non-integer.

RAJU, JUHI, & MENA

R=31, J=48, M=71 works with the second and third paragraphs, but not with the first.

Mark:


SERIES
250.59

(Each number in the series is multiplied by 1.5)


LORD HELPLESS
100km/hr
He's got twice as far to go as the back of the train.


AVERAGE SPEED
24mph

Let d be the distance to the store, T be the time it gets to get there, t be the time it takes to get back, and A be the average speed (which is what we want to find out). As we know from elementary mathematics, distance equals rate times time:
d = 20T
T = d/20

d = 30t
t = d/30

Now that we have expressions for T and t, we can come up with an equation that describes the round trip:
2d = A(T + t)
2d = A(d/20 + d/30)
2d = A(3d/60 + 2d/60)
2d = A(5d/60)
A = 120d/5d
A = 24




RAJU, JUHI, & MENA

The solution:
Raju is 3, Juhi is 8, and Mena is 15. A little grumbling is in order here, as clue number 1 leads to the situation a year and a half ago, when Raju was 1 1/2, Juhi was 6 1/2, and Mena was 13 1/2.

This sort of problem is easier to explain as a set of equations.
Let t be the year that Tim was born, j be the year that Juhi was born, m be the year that Mena was born, and y be the current year.
As indefinite years come up, let y1, y2, ... be the indefinite years.
Then the equations are:

y + 10 - t = 2(y1 - j) y1 - m = 9 (y1 - t)
y - 8 - m = 1/2 (y2 - j) y2 - j = 1 + y3 - t y3 - m = 5 (y + 2 - t)
t + 1 - m = 3 + y4 - t y4 - j = 3 (y5 - 6 - m) y5 - j = 1/2 (y6 - t) y6 - m = 10 + y7 - m y7 - j = 1/3 (y8 - t) y8 - m = 3 (j - m)
t = y - 3 j = y - 8 m = y - 15


I am sorry you aint getting no support in this quest. I blame Nick, he said ?'challenge' then took off for the hills.



A group of four people, when asked the time gave these different answers -

Jan said it was "3 minutes past10"
Kit said it was "3 minutes to10"
Nicky said it was "6 minutes to10"
Terri said it was "2 minutes past10"

They were all supposed to be telling the same time! They were all wrong.

Their errors (in no particular order) were 2, 3, 4 and 5 minutes.

What was the correct time



There is a flag:

It consists of a red cross on a white background.
Both arms of the red cross are the same width.
The flag measures 60 cm by 91 cm.
If the area of the red arms is equal to the area of white,
what should be the width of the red arms

[size=7]TIME
1 minute to 10

FLAG
21cm[/size]

There is a place where:

Vicars live in a Vicarage.
Abbots in an Abbey.
Parsons in a Parsonage.
Nuns in a Nunnery.
Priors in a Priory.
Deans in a Deanery.

Where do cannons live

Cannons dont live..

Now, if you mean canons...

cannery

Mark:

TIME
1 minute to 10

FLAG
21cm


Francis, "Cannons dont live.. Now, if you mean canons..."

Non! (good to see you Francis, I hope I find you well) Nein! Nix, Nada. - Not even Canyons.

Mark, "cannery"

That's good. However, cannons live on a carriage.




A clock is observed. The hour hand is exactly at the minute mark, and the minute hand is six minutes ahead of it. Later, the clock is observed again. This time, the hour hand is exactly on a different minute mark, and the minute hand is seven minutes ahead of it. What the heck is going on asks, Chai Tee.

How much time elapsed between the first and second observations

[size=7]CLOCK
2:12[/size]

Clock


I mark time dost thou?

The hour hand is exactly on a minute mark five times per hour -- on the hour, twelve minutes past the hour, twenty four minutes past, thirty six minutes past, and forty eight minutes past.

Let X be the number of hours, and Y be the number of minutes past the hour. When the hour hand is on a minute mark, the position of the hour hand is 5X + Y/12, and the position of the minute hand is Y. On the first occasion, Y = 5X + Y/12 + 6. This is equivalent to 60X = 11Y - 72.

Since Y can only take one of the values in the set { 0, 12, 24, 36, 48 }, it can be determined that the only legal values for the equation are X = 1 and Y = 12. So the time is 1:12.

Similarly, the second occasion's equation is 60X = 11Y - 84. The only legal values here are X = 3 and Y = 24.

Therefore, the time is [size=7]3:24.

Between 1:12 and 3:24, two hours and twelve minutes have elapsed.[/size]


On Try's tombstone , it is said that; one sixth of his life was spent in childhood and one twelfth as a teenager . One seventh of his life passed between the time he became an adult and the time he married ; five years later, his son was born . Alas, the son died four years before he did . He lived to be twice as old as his son did .

How old did Try live to be

[size=7]OLD TRY
84[/size]

The Classic Greek Algebra Problem
x/6+x/12+x/7+5+x/2+4=x
& markr beats me to it again.

Rap

Back from your leave of absence?

Mark:

OLD TRY
84


How did he know that?

One way

Let x be the age of the man at the time of his death. His age consists of the number of years he spent as a child plus the number of years he spent as a teenager plus the number of years between his coming of age and his marriage plus five plus his son's age (half his own) plus four. In mathematical terms, this is:

x = x/6 + x/12 + x/7 + 5 + x/2 + 4
x = 14x/84 + 7x/84 + 12x/84 + 42x/84 + 9
x = 75x/84 + 9
84x/84 - 75x/84 = 9
9x/84 = 9
x/84 = 1
x = 84


Wait! What is this?


Raprap:

The Classic Greek Algebra Problem
x/6+x/12+x/7+5+x/2+4=x


I echo, echo, echo Marks comment. Good to see you beat the rap, rap.

Whilst one can admire the beauty and simplicity of classic Greek algebra, a Try Angle would be:

Try lived 1/12 of his life as a teenager. There are seven teen years.

7 x 12 = 84 :wink:






Christmas Conundrums.


i. How many days is it from Wednesday the 1st August to the first Saturday in September


ii. There are two clocks. One of them gains 6 seconds in every hour, while the other loses 9 seconds in every hour. If they are both set to show the same time, and then set going, how long will it be before they are exactly 1 hour different


iii. The number FIVE as written using block capitals contains exactly 10 strokes or segments of a straight line. Find a number which when written out in words (using no tricks) contains as many strokes as the number says.


iv. Can you find four consecutive ODD numbers, which add up to a total of 80


v. What is the opposite of "NOT OUT"

[size=7]i) inclusive 32 days, exclusive 30. If August 1st is a Wednesday, September 1st is a Saturday
ii)Clock a gains 15 seconds an hour on B (6+9=15). An hour is 3600 seconds, 3600/15=240 hours. So in 240 hours (10 days) clock A will be one hour ahead of clock B.
iii) dunt no.
iv) 17+19+21+23=80
v) opposite of Not out=out[/size]

Rap

[size=7]i. 31
ii. 10 days
iii. 29
iv. 17+19+21+23
v. OUT[/size]

Rap:


i) Inclusive 32 days, exclusive 30. If August 1st is a Wednesday, September 1st is a Saturday

ii) Clock a gains 15 seconds an hour on B (6+9=15). An hour is 3600 seconds, 3600/15=240 hours. So in 240 hours (10 days) clock A will be one hour ahead of clock B.

iii) dunt no.

iv) 17+19+21+23=80

v) opposite of Not out=out


Mark:

i. 31
ii. 10 days
iii. 29
iv. 17+19+21+23
v. OUT


The score after one round:

Try - 100%

Rap - 80%

Mark - 80%


Do try to keep up dudes. :wink:



Christmas Conundrums. Part II


vi. When one particular number, written in figures, is turned upside down it increases in value by 21. Which number is that

vii. A car has travelled 24,000 km and, in that distance, has worn out 6 tyres. Each tyre travelled the same distance. How far did each separate tyre travel

viii. If 6 cats can catch 6 rats in 6 minutes, how many cats are needed to catch 10 rats in 10 minutes

ix. Find two whole numbers which, when multiplied together give an answer of 41

x. A man bought 50 metres of rope in a shop. He did not know it, but the metre-rule used to measure the rope was 1cm short. What length of rope did he actually get





OMG - Gus is dead!


The accident happened at the weekend as Gus returned to his stables. He tripped and gashed skin already badly injured in the fire.

Mr McWilliam said: "I'm in no doubt that it was as a result of the fire; he was never the same. I would just like anyone who knows who did this to go to the police. It is tragic, Gus was always frightened after this sick attack on him."

On a vet's advice, the 26-year-old horse, named Gus, was put down at stables near the home of owners Geoffrey and Kath McWilliam, in Cedar…

Mr McWilliam added: "We've had an amazing response from the community, he will be missed by a large number of people."

A police investigation into the arson attack is continuing. R.I.P. Gus.


Not part of the riddle set - but harrowing if you knew Gus.