Welcome to Math Playground where you can test your math skills by solving a few fun brain teasers.
No worries, you just need to know the basics and think logically. Even kids can solve these puzzles.
Let's start with an easy one.

One brick is one kilogram and half a brick heavy.
What is the weight of one brick?

There is an easy equation which can help:
1 brick = 1 kg + 1/2 brick
And so 1 brick is 2 kg heavy.

Two trains, 200 km apart, are moving toward each other at the speed of 50 km/hour each. A fly takes off from one train flying straight toward the other at the speed of 75 km/hour. Having reached the other train, the fly bounces off it and flies back to the first train. The fly repeats the trip until the trains collide and the bug is squashed.
What distance has the fly traveled until its death?
There is a complicated and an easy way to calculate this cool math puzzle.
Think outside the box.

There is a complicated way counting a sequence. Or simply knowing that if the fly is flying for 2 hours still at the same speed of 75 km/h then it flies a distance of 150 km.

A passenger train leaves New York for Boston traveling at the speed of 80 km/hr. In half an hour a freight train leaves Boston for New York traveling at the speed of 60 km/hr.
Which train will be further from New York when they meet?
(Kids might know the answer faster than the adults :-)

Of course, when the trains encounter, they will be approximately the same distance away from New York. The New York train will be closer to New York by approximately one train length because they're coming from different directions. That is, unless you take "meet" to mean "perfectly overlap".

Let's play a game. If I went halfway to a town 60 km away at the speed of 30 km/hour, how fast do I have to go the rest of the way to have an average speed of 60 km/hour over the entire trip?

This one has not standard solution. You can't reach the desired average speed under the given circumstances. Easy math will prove that you can't make it at 90km/h or at any at any other speed.

The circumference of the Earth is approximately 40,000 km. If we made a circle of wire around the globe, that is only 10 meters (0.01 km) longer than the circumference of the globe, could a flea, a mouse, or even a man creep under it?

It is easy to compare old and new perimeter - original perimeter is 2xPIxR, length of wire is 2xPIx(new R) and find out that the result is about 1.6 m. So a smaller man can go under it and a bigger man ducks.

We know little about this Greek mathematician from Alexandria, called the father of algebra, except that he lived around 3rd century A.D. Thanks to an admirer of his, who described his life by means of an algebraic riddle, we know at least something about his life.
Diophantus's youth lasted 1/6 of his life. He had his first beard in the next 1/12 of his life. At the end of the following 1/7 of his life Diophantus got married. Five years from then his son was born. His son lived exactly 1/2 of Diophantus's life. Diophantus died 4 years after the death of his son.
How long did Diophantus live?

There is an easy equation to reflect several ages of Diophantus:
1/6x + 1/12x + 1/7x + 5 + 1/2x + 4 = x
So the solution (x) is 84 years.

About 1650 B. C., Egyptian scribe Ahmes, made a transcript of even more ancient mathematical scriptures dating to the reign of the Pharaoh Amenemhat III. In 1858 Scottish antiquarian, Henry Rhind came into possession of Ahmes's papyrus. The papyrus is a scroll 33 cm wide and about 5.25 m long filled with funny math riddles. One of the problems is as follows:
100 measures of corn must be divided among 5 workers, so that the second worker gets as many measures more than the first worker, as the third gets more than the second, as the fourth gets more than the third, and as the fifth gets more than the fourth. The first two workers shall get seven times less measures of corn than the three others.
How many measures of corn shall each worker get? (You can have fractional measures of corn.)

2 equations give a clear answer to the given question:
5w + 10d = 100
7*(2w + d) = 3w + 9d
Where w is amount of corn for the first worker, d is the difference (amount of corn) between two consecutive workers. So this is the solution:
1st worker = 10/6 measures of corn
2nd worker = 65/6 measures of corn
3rd worker = 120/6 (20) measures of corn
4th worker = 175/6 measures of corn
5th worker = 230/6 measures of corn

At noon the hour, minute, and second hands coincide. In about one hour and five minutes the minute and hour hands will coincide again.
What is the exact time (to the millisecond) when this occurs, and what angle will they form with the second hand?
(Assume that the clock hands move continuously.)

There are a few ways of solving this one. I like the following simple way of thinking. The given situation (when the hour and minute hands overlay) occurs in 12 hours exactly 11 times after the same time. So it's easy to figure out that 1/11 of the clock circle is at the time 1:05:27,273 and so the seconds hand is right on 27,273 seconds. There is no problem proving that the angle between the hours hand and the seconds hand is 131 degrees.

A swimming pool has four faucets. The first can fill the entire pool with water in two days, the second - in three days, the third - in four days, and the last one can fill the pool in 6 hours.
How long will it take to fill the pool using all 4 faucets together?

Because there are 24 hours in one day, in one hour fills the first tap 1/48, the second tap 1/72, the third tap 1/96 and the fourth tap fills 1/6 of the reservoir. That is all together (6+4+3+48) / 288 = 61/288. The reservoir will be full in 288/61 hours, which is 4 hours 43 minutes and about 17 seconds.

A square medieval castle on a square island is under siege. All around the castle there is a square moat 10 meters wide. Due to a regrettable miscalculation the raiders have brought footbridges, which are only 9.5 meters long. The invaders cannot abandon their campaign and return empty-handed.
How can the assailants resolve their predicament?

You can put one foot-bridge over one corner (thus a triangle is created). Then from the middle of this foot-bridge lay another foot-bridge to the edge (corner) of the castle. You can make a few easy equations confirming that this is enough.

Just a little side note. "The Castle" puzzle reminds me of another cool math joke - student's answer on a math test where "x" should have been found.

Crossing the Desert

A military car carrying an important letter must cross a desert. There is no petrol station in the desert, and the car's fuel tank is just enough to take it half way across. There are other cars with the same fuel capacity that can transfer their petrol to one another. There are no canisters or rope to tow the cars.
How can the letter be delivered?
You can play a little game with model cars to simulate this puzzle.

There are 4 cars needed, including the car with the important letter (which travels to the middle of the desert). Its empty tank must be filled to the top to get to the end of desert. The way between the military base (where the cars and petrol is) and the middle of the desert can be divided into 3 thirds. 3 cars will go in their thirds back and forth and overspilling 1/3 of their tanks. This way the tank of the important car will be filled and the letter will be delivered.

A distant planet "X" has only one airport located at the planet's North Pole. There are only 3 airplanes and lots of fuel at the airport. Each airplane has just enough fuel capacity to get to the South Pole. The airplanes can transfer their fuel to one another.
Your mission is to fly around the globe above the South Pole with at least one airplane, and in the end, all the airplanes must return to the airport.

Divide the way from pole to pole to 3 thirds (from the North Pole to the South Pole 3 thirds and from the South Pole to the North Pole 3 thirds).
1st step - 2 aeroplanes to the first third, fuel up one aeroplane which continues to the second third and the first aeroplane goes back to the airport.
2nd step - 2 aeroplanes fly again from the airport to the first third, fuel up one aeroplane which continues to the second third and the first aeroplane goes back to the airport.
3rd step - So there are 2 aeroplanes on the second third, each having 2/3 of fuel. One of them fuels up the second one and goes back to the first third, where it meets the third aeroplane which comes from the airport to support it with 1/3 of fuel so that they both can return to the airport. In the meantime, the aeroplane at the second third having full tank flies as far as it can (so over the South Pole to the last third before the airport).
4th step - The rest is clear - one (of the two) aeroplane from the airport goes to the first third (the opposite direction as before), shares its 1/3 of fuel and both aeroplanes safely land back at the airport.

A magic wish-granting rectangular belt always shrinks to 1/2 its length and 1/3 its width whenever its owner makes a wish. After three wishes, the surface area of the belt's front side was 4 cm^{2}.
What was the original length, if the original width was 9 cm?

Here's a variation on a famous puzzle by Lewis Carroll, who wrote Alice's Adventures in Wonderland.
A group of 100 soldiers suffered the following injuries in a battle: 70 soldiers lost an eye, 75 lost an ear, 85 lost a leg, and 80 lost an arm.
What is the minimum number of soldiers who must have lost all 4?

Add up all the injuries, and you find that 100 soldiers suffered a total of 310 injuries. That total means that, at a minimum, 100 soldiers (it is, of course, not 100 soldiers, but 100 as calculation from 310 person-injuries out of 400 possible) lost 3 body parts, and 10 (the remainder when dividing 310 by 100) must have lost all 4 body parts. (In reality, as many as 70 may have lost all 4 body parts.)
Another way to solve it is to draw a line of 100 parts and compare injuries from opposite ends of the line, finding the intersection part of all 4 injuries. If left side of line (LS), right side (RS) and intersection (I), then:
70 (LS) and 75 (RS), then 45 (I)
45 (LS) and 85 (RS), then 30 (I)
30 (LS) and 80 (RS), then 10 (I) soldiers must have lost all 4 parts

One glass has 10 cl of tonic water and another 10 cl of fernet. Pour 3 cl of tonic into the glass with fernet and after mixing thoroughly, pour 3 cl of the mixture back into the glass with tonic water.
Is there more tonic in the glass of fernet or more fernet in the glass of tonic?
(Ignore the chemical composition!)

There is exactly as much tonic in the glass of fernet as there is fernet in the glass of tonic.

I have two US coins totaling 55 cents. One is not a nickel.
Figure out what the coins are.

This was just a catch question. One of the coins is really not a nickel because nickel is the other coin. So it is a fifty cent piece and a nickel - it said ONE isn't a nickel, but the OTHER ONE IS!

No two inhabitants have the same number of hairs on their head.

No inhabitant has exactly 518 hairs.

There are more inhabitants in town than hairs on any individual inhabitant's head.

What is the highest possible number of inhabitants?

There can live maximum of 518 people in the town. By the way, it is clear that one inhabitant must be baldy, otherwise there wouldn't be a single man in the town.

An anthropologist studying a primitive tribe in a remote location in the Amazon basin, had uncovered a strange tribal custom. Whereby, if a husband found out his wife was unfaithful to him, he must execute her in a public ceremony in front of the whole tribe on the same day at midnight. It so happened that every man in the tribe knew about every cheating wife except his own, since telling a man about his cheating wife was against the tribal honor code. On the day of his departure, the anthropologist held a tribal meeting and made the announcement: "I know there are unfaithful wives in this tribe." On the ninth day thereafter all cheating wives were executed at the same time.
How many unfaithful wives were there?

If there are n unfaithful husbands (UHs), every wife of an UH knows of n-1 UH's while every wife of a faithful husband knows of n UHs. [this because everyone has perfect information about everything except the fidelity of their own husband]. Now we do a simple induction: Assume that there is only one UH. Then all the wives but one know that there is just one UH, but the wife of the UH thinks that everyone is faithful. Upon hearing that "there is at least one UH", the wife realizes that the only husband it can be is her own, and so shoots him. Now, imagine that there are just two UH's. Each wife of an UH assumes that the situation is "only one UH in town" and so waits to hear the other wife (she knows who it is, of course) shoot her husband on the first night. When no one is shot, that can only be because her OWN husband was a second UH. The wife of the second UH makes the same deduction when no shot is fired the first night (she was waiting, and expecting the other to shoot, too). So they both figure it out after the first night, and shoot their husbands the second night. It is easy to tidy up the induction to show that the n UHs will all be shot just on the n'th midnight.

Find the mistake in these mathematical equations.
x = 2
x(x-1) = 2(x-1)
x^{2}-x = 2x-2
x^{2}-2x = x-2
x(x-2) = x-2
x = 1

The equation is solved the right way, apart from one little detail. There must be stated that x does not equal y, because there would be dividing by zero, which is not defined in maths.