16, 12,  8 years old

27 days

$100

Michel = secretary,
Lisette = president,
Louis = treasurer

If there are 20 physics students who all take mathematics, and half of the mathematics students study physics, there must be 40 students in the mathematics class. If half of the 30 literature students take mathematics, then 15 of them take mathematics. Since none of the literature students study physics, only five students in the mathematics class study neither physics nor literature.

1. Fill the 3-liter bottle and pour it into the empty 5-liter bottle.
2. Fill the 3-liter bottle again, and pour enough to fill 5-liter bottle. This leaves exactly 1 liter in the 3-liter bottle.
3. Empty the 5-liter bottle; pour the remaining 1 liter from the 3-liter bottle into the 5-liter bottle.
4. Fill the 3-liter bottle and pour it into the 5-liter bottle. The 5-liter bottle now has exactly 4 liters.

1. Fill the 5-liter bottle and pour water from it into the 3-liter bottle until it is full.
   This leaves 2 liters in the 5-liter bottle.
2. Empty the 3-liter bottle and pour the 2 liters of water from the 5-liter bottle into the 3-liter bottle.
3. Fill the 5-liter bottle again.
4. Fill the 3-liter bottle from the 5-liter bottle. Since the 3-liter bottle had 2 liters of water, only one liter is transferred leaving exactly 4 liters of water in the 5-liter jug.

Hydrogen chloride combines with sodium hydroxide to form common salt (sodium chloride) and water.

HCl + NaOH  NaCl + H₂O

If you said $10.00 Dollars, you are wrong! It is not that simple.

This is a problem that requires setting up two equations with two unknowns to find the answer.

The box (B) plus the ring (R) cost $200.00 Dollars:

B + R = 200

The jewelry box is valued at $190.00 Dollars more than the ring, i.e., the price of the box (B) is equal to the price of the ring (R) plus $190.00 Dollars:

B = R + 190

Now we can substitute the right side of the second equation in the first equation and solve for R.

R + 190 + R = 200
2 R = 200 – 190
2 R = 10
R = 10/2
R = 5

The ring is worth $5.00 Dollars.
The price of the box is $190.00 plus $5.00, which is $195.00 Dollars, and the box plus the ring add up to $200.00. So, everything checks out.

he words “Numerology” and “special significance” in the statement of the problem should have given you a hint that this was not a simple mathematical problem.

The sequence (2, 6, 10, 15, 19) corresponds to every fourth floor starting from 2 in the building where I live. As you can see in the picture of the elevator panel, the number 13 is skipped to avoid “bad luck”.

• 4 is considered unlucky by the Chinese and Japanese, because it sounds like the word “death”.
• 7 is considered lucky by most cultures.
• 13 is considered unlucky by most Western cultures.
• 666 is considered evil by some Christians because it is the number of the “beast” according to the Bible (Revelation 13:18).

Even if we don’t believe in the superstitions of numerology, we have to deal with the numerical traditions of our culture.

The problem gives us the hint that water will freeze in space. Looking at the phase diagram for water, we can see that below a pressure of 0.006 atmospheres (611.7 Pascals) water cannot exist in the liquid state. In the vacuum of space, water can only be gaseous or solid. Any liquid water ejected into space will boil, and the evaporation will cool some of the remaining water below the freezing point.

Phase diagram for water

The heat for melting or freezing water, also called the Heat of Fusion, is 80 calories per gram at 0°C. It is necessary to remove 80 calories of energy to freeze one gram of water at 0°C. The heat for boiling or condensing water at 100°C, called the Heat of Vaporization, is 539 calories per gram. The following graph summarizes the effect of energy on the temperature of water in its solid, liquid, and gaseous phases.

Since the problem does not state the temperature of the water ejected into space, we can do the calculations for 0°C and for 100°C, which are the minimum and maximum temperatures for water at normal pressure within the space capsule. We can solve the problem as follows:

Let X be the amount of water that evaporates, and 100-X the amount of water that turns into ice.

For water at 0°C, we can set up the following equation, where (100-X) 80 is the number of calories required to freeze 100-X grams of water at 0°C, and 539 X is the number of calories removed by the evaporation of X grams of water.

(100 – X) 80 = 539 X
X = 8000/619 = 12.9 grams

If the ejected water has a temperature of 0°C, only 12.9 grams need to evaporate to freeze the remaining water (87.1 grams).

For water at 100°C, it is first necessary to reduce the temperature to 0°C. This requires 100 calories per gram. In addition, 80 calories are required to freeze one gram of water. Thus, we need to remove 180 calories to freeze 1 gram of water at 100°C. We can set up the following equation, where (100-X) 180 is the number of calories required to freeze 100-X grams of water at 100°C, and 539 X is the number of calories removed by the evaporation of X grams of water.

(100 – X) 180 = 539 X
X = 18000/719 = 25.0 grams

For water ejected into space, 87.1 grams turn into ice if the water is initially at 0°C, and 75.0 grams turn into ice if the water is at 100°C. Thus, at least 75 grams out of the original 100 grams or 75% will turn into ice.