Maths Riddles

Quadratic, quad(4 wheeler + rat + i (eye) + C)

The youngest child received $16.

26, (5 * 4) + 6 = 26

John and Michael play all the three sports. Alice plays none.
This is after assuming that girls do not play football. If in case, the school had a girl's football league, we would know only that one of the three play no sports at all and the others play all three.

Only ten percent of students study all the four subjects in school.

P = 20 units
Q = 12 units
R = 15 units

Explaining
If you have started looking for a complex formula, we inform you that this requires just the simple algebra equations.

Let the three sides of the box be P, Q and R.

Now we know the area of rectangle is l* b.

P * Q = 240
P * R = 300
Q * R = 180

Let us multiply these three equations:
(P*Q) (P*R) (R*S) = 240 * 300 * 180
P^2 * Q ^2 * R^2 = 12960000
(P*Q*R)^2 = 12960000
P*Q*R = 3600

Now we know that P*Q = 240

Thus R = 3600/240 = 15.

Similarly putting and solving for other variables, we get
P = 20 units
Q = 12 units
R = 15 units

As per the given data,
6 * number of chocolates + 5 * number of sodas = 32

Now, we know that chocolates and sodas are definitely whole numbers; their respective values will be 2 and 4.

First let us assume that all other condition are fair here.
Now the toss will generate a 52:48 distribution in the favour of heads. Therefore, the toss bias factor for heads is 52/48.

In the same manner, the coin will be generating 49:51 distribution in favour of tails which makes the coin bias factor for heads in this case to be 49/51.

So, we have a combined bias factor of (52 * 49) / (51 * 48) = 2548 / 2448 which will be cancelled by a 2448 / 2548 floor factor.
The floor will be generating a distribution of 2448 / (2548 + 2448) : 2548 / (2548 + 2448) in the favour of tails which amounts to 51.00080064% tails approximately.

Let us assume,
radius of pound is R.
speed of beast is S.
speed of lion is S/4.
Circumference = 2 * Pi * R.

Now, if lion swims R/4 distance from center of the pound and then begins to swim across the pond in center implies both lion and beast can take the round trip in same time.
Explanation :
time by beast : 2 * Pi* R * S
time by lion : 2 * Pi* R/4 * S/4 => 2 * Pi* R * S


Now, the lion can move slowly inward toward the center of pound, and begin swimming around the center in a circle from this distance. It is now going around a very slightly smaller circle than it was a moment ago, and thus will be able to swim around this circle FASTER than the beast can run around the shore.

The lion can keep swimming around this way, pulling further away each second, until finally it is on the opposite side of its inner circle from where the beast is on the shore. At this point, the lion aims directly toward the closest shore and begins swimming that way. At this point, the lion has to swim [0.75R feet + 1 millimeter] to get to shore. Meanwhile, the beast will have to run R*pi feet (half the circumference of the pound) to get to where the lion is headed.
The beast runs four times as fast as the lion, but you can see that it has more than four times as far to run:

[0.75R feet + 1 millimeter] * 4 < R*pi

[This math could actually be incorrect if R were very small, but, in that case, we could just say the lion swam inward even less than a millimeter and make the math work out correctly.]

Because the lion has less than a fourth of the distance to travel as the beast, it will reach the shore before the beast reaches where it is and successfully escape.

Since Nancy is the middle at 6th position, there should be 5 girls on both her sides.
5 + 5 = 10
Counting Nancy as well, the total number of girls in that row = 10 + 1 = 11