Solution:

Let the distance between Peter's (hereafter referred as P) and Rhea's (hereafter referred as R) be D.
When P and R meet for the first time, P has traveled 500 miles.
In that case, R has traveled (D - 500) miles.
=> Speed of P / Speed of R = 500 / (D - 500) ...... (1)

When P and R meet for the second time, P has traveled (D - 500) + 300 miles
Now, R has traveled (500 + D - 300)
=> Speed of P / Speed of R = 5 / (D - 5) ..... (2)

Equating (1) and (2)
500 / (D - 500) = (D - 2) / (D + 2)
D = 1200 Miles.

Solution:

2nd

When the cylinder is spun for the first and last time, the following six outcomes are possible:

First:
First chamber is fired first. The first player loses.

Second:
Second chamber is fired first. The first player loses.

Third:
Third chamber is fired first. The first player loses.

Fourth:
Fourth chamber is fired first. The second player loses. (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)

Fifth:
Fifth chamber is fired first. The first player loses. (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)

Sixth:
Sixth chamber is fired first. The second player loses. (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)

Thus, if you analyze, you will find that the second player has a 4/6 chance of winning. Therefore if you want to win the game, going second gives you better shot at it.

Solution:

She has misplaced 119 pens in total.

Solution:

11550

Explanation:
There is just one way to find it out and that is the generic computing method.

The number of squares and rectangles formed = (Summation of column numbers)
= (1 + 2 + 3 +......+ 19 + 20) * (1 + 2 + 3 +......+ 9 + 10)
= 210 * 55
= 11550

Solution:

Suppose that we have B buses at present and S refers to the space before we expanded. This implies that
S = B - 12 ----- [1]

When we expanded, we came up with more space. It can be equated as
S + 0.4 x S = B + 12 ------ [2]

Rewriting the equation we get
1.45 = B + 12
Also, B = 1.45 - 12 ----- [3]

The first equation can also be written as
B = S + 12 ----- [4]

If we equate the third and fourth equation, we will have
1.4 S - 12 = S + 12

Let us subtract S from both the sides
0.4 S - 12 = 12

Adding 12 to both sides
0.4 S = 24

Multiplying by ten and dividing by four on both sides
S = 60

If we insert this value in the fourth equation, we will have
B = 72

Thus there are 72 buses.

Initially we had room for 60 buses after expanding the garage space by 40 percent, we will be having enough room to have 84 buses out of which twelve is for future.

Solution:

Impossible

Explanation:
Let us first denote everything with variables.
d= distance to your friend's place
T = time taken to get there
t = time required to get back
R = Speed while returning back

d = 20T
T = d/20

d = Rt
t = d/R

We have made equations for T and t and now we can derive an equation for the round trip.

2d = 40(T + t)
2d = 40(d/20 + d/R)
2d = 40d(1/20 + 1/R)
1 = 20(R/20R + 20/20R)
20R = 20(R+20)
R = R + 20

Do you see the paradox here? You literally have to travel back at an infinite speed if you want to make the average speed of your trip 40 mph. The faster return speed will have lesser impact of the average speed but consider that the quicker your return trip is, the faster you will make it.

We are talking about travelling double the distance in the same time that was taken for one way trip. Thus, if you travel at infinite speed, you will be able to attain an average of 40 mph.

Solution:

4%

Explanation:
Suppose that the initial volume of the water drop was p.

p - (1/4) p = (3/4) p

(3/4) p - (1/5) [(3/4) p] = (3/4) p * [1 - (1/5)] = (3/4) p * (4/5) = (3/5) p

(3/75) p = (1/25) p

1/25 = 4%

Therefore, the required percentage is four percent.

Solution:

Let us first calculate the total number of matches that will be placed:
9C2 * 2
= 9! / (2!7!) * 2
= 8 * 9
= 72
Therefore, every team will play 16 matches.

Let us now glance over the cases that are possible.

Case 1:
Let us say that Team A, Team B and Team C wins 14 matches each (they lose 1 match each from the other two teams), then the Team D can win only (16 - 3 * 2) = 10 matches.

Case 2:
Suppose that, the top 4 teams win 13 matches each.
Team A will win 13 matches and will lose 3 matches from Team B, Team C and Team D.
Team B will win 13 matches and will lose 3 matches from Team A, Team C and Team D.
Team C will win 13 matches and will lose 3 matches from Team A, Team B and Team D.
Team D will win 13 matches and will lose 3 matches from Team A, Team B and Team C.

If this case actually happens, then the Team E will be able to win 8 matches at maximum as Team A, Team B, Team C and Team D will win all the matches played by E.

Case 3:
Suppose that, the top 5 teams win 12 matches each.
Team A will win 12 matches and will lose 3 matches from Team B, Team C, Team D and Team E.
Team B will win 12 matches and will lose 3 matches from Team A, Team C, Team D and Team E.
Team C will win 12 matches and will lose 3 matches from Team A, Team B, Team D and Team E.
Team D will win 12 matches and will lose 3 matches from Team A, Team B, Team C and Team E.
Team E will win 12 matches and will lose 3 matches from Team A, Team B, Team C and Team D.

If this case actually happens, Team E will win 12 matches but it still won't be able to get into the playoffs assuming that it has the lowest run rate among all.

This implies that from case 2, it is evident that a team must win 13 matches at the least to confirm its position in the playoffs.

Solution:

60%

Explanation:
In all such cases where you have to find out the probability of anything with an already given situation, you have to use the formula for conditional probability.

In this case, you have to find out the probability that the second test was passed by a student given that he passed the first test already.

Therefore, P(Second/First) = P (First and Second)/ P (First)
= 0.25/0.42
=0.60
=60 %

Solution:

2/5

Explanation:
This problem can be solved with different approaches. Let us not go into the complicated generic stuff to avoid confusion and try to solve it with simpler means.

Let us find the results separately. There can be 4 different results that give the sum of 5 and there can be 6 different results that give the sum of 7.

Thus, the probability that we get the sum of 4 before the sum of 7 will be:
4/(4+6) = 4/10 = 2/5.