The worlds first riddle!

hi try
back again

w'aleykum Aas-salam .elhamdolallah.

A happy and healthy new year to all!




any way thanks



fishes: 25

"Zsuo eodj upis aasj sqqu mreu rst/
Vsmu eptl piyy vpfr "


It looks like some sort of code.

Vsm upi eptl oy piy

Usamashaker chalks up the first points.
"Fishes: 25"


The answer is 25 fish. Each man throws away a fish before he takes his third of the total. Therefore, the amount of fish in the cooler must be divisible by 3 after one fish is thrown away.
First man sees 25 fish. He throws away one fish leaving 24 fish. He then takes 8 fish (24/3) and that leaves 16 fish.
Second man sees 16 fish. He throws away one fish leaving 15 fish. He then takes 5 fish (15/3) and that leaves 10 fish.
Third man sees 10 fish. He throws away one fish leaving 9 fish. He then takes 3 fish (9/3).
If you try the other numbers smaller than 25 (22, 19, 16, 13, 10, 7, and 4) that are divisible by 3 after throwing away one fish you will see they do not solve the problem.




Can you find two consecutive numbers whose squares when subtracted equal 43


If a steamship weighs 10,000 tons
and sails 5,000 miles,
with a cargo large of overshoes
and carving knives and files,
if the mates were almost 6 feet tall
and the bos'n near the same,
would you subtract or multiply
to find the captain's name

Hi, Tryagain, glad to have you back!




21
22

Steam-ship , sails
captain, commander
something of that kind

Fowl:
8 chickens, 12 ducks.

3, 7, 10, 11, 12

If you're going to rerun items, at least get them right the second time around :wink:

17 is missing from the list.

Welcome back. With your daily dose of puzzles gone, and those damn neopets problems clogging up the works, it's been rough.

what does this mean? Multiply the sum times 3

n*3? so the number is 7

n^3? so the number is -41.55

DrewDad
Fowl:
8 chickens, 12 ducks.

Some may say it was a lucky guess. But, I won't call it a foul.



8 chickens and 12 ducks.

Let c represent the number of chickens. Let d represent the number of ducks.

The total number of animals bought was 20. So c + d = 20.

The man received a $2 discount per chicken and a $0.50 discount per duck. The total discount he received was $22. So 2c + 0.50d = 22.

Now we have two equations with two varaibles. First we will rewrite the first equation to find the value of c.
c + d = 20
c = 20 - d

Now we can substitue the vale of c from above into the second equation and solve for d.
2c + 0.50d = 22
2(20 - d) + 0.50d = 22
40 - 2d + 0.50d = 22
40 - 1.50d = 22
-1.50d = -18
d = 12

Now solve for c in the first equation.
c + d = 20
c + 12 = 20
c = 8


Mark, the only member to have a thread named after him writes,

"3, 7, 10, 11, 12
If you're going to rerun items, at least get them right the second time around . 17 is missing from the list. "

Doh!
What a good start to the New Year!

"…those damn neopets problems clogging up the works, it's been rough."

Your story has touched my heart. Never before have I come across a person who has more problems than you. Please accept these few words as a sign of my deepest sympathy. (- Deleted -).


O drr yjru jsbr yjrot pem yjtrsf mpw. :wink:




This statement has been passed by the CA Medical Board as statistically possibly correct.

I have never, ever, posted the following -


The population of Dale is 4,800 more than Ville. If 3,100 people move from Ville to Dale, the population of Dale will be eleven times the population of Ville.

What was the original population of Dale and Ville



A freight train leaves Ville at 6:00 pm and is travelling towards Boro at 15 mph. A passenger train leaves Boro at 6:00 pm and is travelling towards Ville at 85 mph. Ville and Boro are 200 miles apart.

When and where will the two trains meet

Sorry Usamashaker. I was interrupted by a video conference with Merry Olde Europe.

You said, "n*3? so the number is 7"



I am 7.

Let n represent the number.

Add 47 to me.
47 + n

Multiply the sum by 3.
(47 + n)3

Divide the product by 2.
((47 + n)3) / 2

You end up with 81.
((47 + n)3) / 2 = 81
(47 + n)3 = 162
47 + n = 54
n = 7

at 8.00 pm at 30 miles from ville


population: 4200.......9000


never mind

What was the original population of Dale and Ville.

Usamashaker, who is on a roll says,

"population: 4200.......9000"



Let x = original population of Ville.
Let x + 4800 = original population of Dale.

Now 3100 people move from Ville to Dale.

The new population of Ville = x - 3100
The new population of Dale = x + 4800 + 3100

Now the new population of Dale is eleven times the new population of Ville.

So now we have:
x + 4800 + 3100 = 11(x - 3100)
x + 7900 = 11x - 34100
7900 = 10x - 34100
42000 = 10x
4200 = x

Original population of Ville was 4200.
Original population of Dale was 9000.



When and where will the two trains meet?

Usamashaker, "at 8.00 pm at 30 miles from ville"



There are two trains travelling towards each other. The freight train is going 15 mph and the passenger train is going 85 mph. Their total speed, relative to each other, is 15 mph + 85 mph = 100 mph.
Use the formula Distance = Rate x Time to solve the problem. The distance is 200 miles and the rate is 100 mph. Let t be the time the two trains travel.
Now we have 200 miles = (100 mph)t.
So t = 200 miles / 100 mph. t=2 hours.

The trains meet at 8:00 pm.

Using the same formula (Distance = Rate X Time) we can now determine how far each train traveled.
Freight train. d = (15 mph)(2 hours). d=30 miles.
Passenger train. d = (85 mph)(2 hours). d=170 miles.

The trains meet 30 miles from Ville and 170 miles from Boro.




John, Matt, and Richard can build a garage in 6 days.
If Matt and Richard work together, they can build the garage in 9 days.
If John and Richard work together, they can build the garage in 8 days.

How many days would each man take to build the garage if they worked alone

GARAGE
John: 18 days
Matt: 24 days
Richard: 14.4 days

Mark puts his score on the door with:

GARAGE
John: 18 days
Matt: 24 days
Richard: 14.4 days


John would take 18 days, Matt would take 24 days, and Richard would take 14.4 days to build the garage.

Let J (John) build 1/J part of the garage in 1 day.

Let M (Matt) build 1/M part of the garage in 1 day.

Let R (Richard) build 1/R part of the garage in 1 day.

Since all three working together can build the garage in 6 days, then in one day they can build 1/6 of the garage.

So 1/J + 1/M + 1/R = 1/6

Since Matt and Richard can build the garage in 9 days, then in one day they can build 1/9 of the garage.

So 1/M + 1/R = 1/9

Since John and Richard can build the garage in 8 days, then in one day they can build 1/8 of the garage.

So 1/J + 1/R = 1/8

Now we substitute values into the first equation and solve.
1/J + 1/M + 1/R = 1/6
1/J + 1/9 = 1/6
1/J = 1/6 - 1/9
1/J = 3/18 - 2/18
1/J = 1/18
J=18 John can build the garage in 18 days.

1/J + 1/M + 1/R = 1/6
1/M + 1/8 = 1/6
1/M = 1/6 - 1/8
1/M = 4/24 - 3/24
1/M = 1/24
M=24 Matt can build the garage in 24 days.

1/J + 1/M + 1/R = 1/6
1/18 + 1/24 +1/R = 1/6
1/R = 1/6 - 1/18 - 1/24
1/R = 12/72 - 4/72 -3/72
1/R = 5/72
R=72/5
R=14.4 Richard can build the garage in 14.4 days.




Four men were shipwrecked on an island. They didn't have any food with them, so they searched the island for food and found some pineapples. After they had gathered some pineapples, they decided to rest and they all fell asleep.

The first man woke up and he was extremely hungry so he ate 1/3 of the pineapples. He then went back to sleep.

The second man woke up and he was extremely hungry so he ate 1/3 of the pineapples. He then went back to sleep.

The third man woke up and he was extremely hungry so he ate 1/3 of the pineapples. He then went back to sleep.

The fourth man woke up and he was extremely hungry, but he ate only his rightful share of the remaining pineapples. There were 6 pineapples left.

How many pineapples did the men gather

pineapples:27

Usamashaker writes, "pineapples:27"

The answer is 27 pineapples.
To determine the answer you need to determine how many pineapples were available to each man.

The fourth man took his rightful share (1/4 of the total pineapples left when he woke up). There were 6 pineapples left after he took his share. Therefore 6 is 3/4 of the total pineapples that were available to him. Let x represent the total number of pineapples.
(3/4) x =6 So x = 8

The third man took 1/3 of the total pineapples. We know now that there were 8 pineapples left after he took his share. Therefore 8 is 2/3 of the total pineapples that were available to him. Let x represent the total number of pineapples.
(2/3) x =8 So x = 12

The second man took 1/3 of the total pineapples. We know now that there were 12 pineapples left after he took his share. Therefore 12 is 2/3 of the total pineapples that were available to him. Let x represent the total number of pineapples.
(2/3) x =12 So x = 18

The first man took 1/3 of the total pineapples. We know now that there were 18 pineapples left after he took his share. Therefore 18 is 2/3 of the total pineapples that were available to him. Let x represent the total number of pineapples. (2/3) x =18 So x = 27




Five clever newsboys formed a partnership and disposed of their papers in the following manner. Tom Smith sold one paper more than one quarter of the whole lot. Billy Jones disposed of one paper more than a quarter of the remainder. Ned Smith sold one paper more than a quarter of what was left, and Charley Jones disposed of one paper more than a quarter of the remainder.

At this stage the Smith boys had together sold just one hundred papers more than the Jones boys had sold. Little Jimmy Jones, the youngest kid in the bunch, now sold all the papers that were left. The three Jones boys sold more papers than the two Smith boys, but how many more

NEWSBOYS
220 (total was 1020)

HARE
to tie, he has to go 5 times as far in 2/7 the time. Therefore, he has to go more than 17.5 times as fast, or more than 16.5 times faster.

Hair today, gone tomorrow. It would appear Mark is a good sport

The diameter of the circular track has no bearing on the problem. When they meet, the hare has gone 1/6 of the way while the tortoise travelled 17/24.

The tortoise has therefore been moving 17/4 times as fast as the hare. The hare has 5/6 of the distance yet to go as compared to 1/6 for the tortoise, so the hare must go five times faster than the tortoise, or 85/4 times faster than he went before.


Now, while he is training for the world duvet wrestling championship, something a little bit different


Lay out 16 cards. Ace, 2, 3 and 4.

Two players alternately turn down a card and add it to the common score, and the player who makes the score of 22, or forces his opponent to go beyond that number, wins.

For example, A turns down a 4, B turns down a 3 (counting 7), A turns down a 4 (counting 11), B plays a 2 (counting 13), A plays 1 (14), B plays 3 (17), and whatever A does, B scores the winning 22 next play.

Again, supposing the play was 3-1, 1-2, 3-3, 1-2, 1-4, scoring 21, the second player would win again, because there is no 1 left and his opponent must go beyond 22.

Which player should always win, and how

HARE
Oops. I blew the hare question. My calcs were based on the tortoise getting a 1/4 head start.