Yes, there will be a point in time when the distance between any two pairs of cars will again be x miles.

Let’s assume that the track has a circumference of C miles. When the leading car catches up with the last car, it will have completed one full lap of the track, and the other two cars will have completed some fraction of a lap. Let’s call this fraction f, where 0 < f < 1.

At this point, the leading car is at the starting point, and the other two cars are x miles apart on the track. The middle car is fC miles ahead of the trailing car, and (1-f)C miles behind the leading car. The trailing car is x + fC miles behind the leading car, and (1-f)C – x miles ahead of the middle car.

As the cars continue to move, the distance between the leading and middle car will decrease at a rate of 55 – v miles per hour, where v is the speed of the middle car. Similarly, the distance between the middle and trailing car will increase at a rate of v – 45 miles per hour.

There will be a point in time when the distance between the leading and middle car, and the distance between the middle and trailing car, are both equal to x. Let’s call this time t. At time t, the distance between the leading car and the middle car will be x, which means that the middle car will be (1-f)x miles behind the leading car. The distance between the middle car and the trailing car will also be x, which means that the trailing car will be x + fC – (1-f)x = fC + x miles behind the leading car.

We can solve for t by setting the distances equal to x and solving for v. We get:

55t – vx = 55x vx – 45t = 45x

Solving for v and equating the two expressions, we get:

v = (55x + 45t) / (t + x)

Substituting this expression for v into the equation for the distance between the middle and trailing car, we get:

fC + x = vt – 45t = (55x + 45t) / (t + x) * t – 45t

Solving for t, we get:

t = C / (10 + 2f)

This is the time at which the distance between any two pairs of cars will again be x miles. Note that this time will be different for different values of f, but there will always be a time when the distance between the cars is x.

To guarantee at least one “four of a kind,” we need to find the worst-case scenario in which we keep taking cards out without getting a four of a kind. In other words, we want to keep removing the cards that are least likely to help us get a four of a kind.

There are 13 ranks in a standard deck of cards (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King), so there are 13 possible four of a kinds. In order to guarantee getting at least one four of a kind, we need to have at least 13 * 4 = 52 cards in our hand (since we need one card of each rank to make a four of a kind).

However, the question asks for the minimum number of cards that must be taken out from the mixed cards to guarantee a four of a kind. So we need to think about the worst-case scenario, where we keep taking out cards without getting a four of a kind.

The worst-case scenario occurs when we take out 48 cards without getting a four of a kind. At this point, we have 4 cards from each rank in our hand, and the next card we draw will complete at least one four of a kind. So the minimum number of cards that must be taken out to guarantee at least one four of a kind is 48.

Answer: 14 minutes

If I cut the 50 feet long of cloth into 10 pieces of 5 feet long it takes 10 minutes. Than I  stack the pieces of cloth on top of each other and cut them all at the same time, making 4 more cuts to split each piece into 5 one-foot-long handkerchiefs so another 4 minutes

Here are some:

1. Finland and Andorra > finlandorra
2. Serbia and Biafra > Serbiafra
3. Ireland and Andora > Irelandora
4. Belarus and Russia > Belarussia
5. Sweden and Denmark > Swedenmark
6. Vietnam and Namibia > Vietnamibia
7. Nauru and Uruguay > Nauruguay
8. Nicaragua and Guatemala > Nicaraguatemala
9. Niger and Germany > Nigermany

I will  put one blue gem in one sack and the other 99 gems into the sack with the red gems so now the probability will we 0.5 * (1/1 + 99/199) = 0.7487 which is almost 3/4

Anwer:  2 Hours
Explanation:
Jacob can eat 2 chocolates in 10 minutes so in one hour he can eat 60 * 2 / 10 = 12
Jolly can eat 7 chocolates in 20 minutes o in one hour  she can eat 60* 7 /20 =21
So in one hour they can eat 27+12+21 =60
so it takes 2 hours to eat 120 chocolates.

The problem is because of vapor barrier. The chocolate ice cream is probably located at the entrance to the store and because its more popular , 
in a separate container ant the time to buy it is short, However the strawberry ice cream is located at the back of the store probably with a small queue,
so shopping time is longer.
So when it takes more time to buy the ice cream, the engine was enough to cool down, and the vapor plug would open, while when buying the chocolate ice cream the steam
is not released to the engine chokes.
 

If a material floats on another then obviously the first material is lighter. So the answer is: False