Mark, I have at last found the first file to the Math questions you answered. In the interest of clarity, I post the full question along with the answer I have. I think we agree on most points. As for the others, I make no claims to be right, so I give the reasoning.
The search for the others continues.
ONLY WORD THAT ENDS IN 'MT'
It came to me in a dream last night.
"If the puzzle you solved before you solved the puzzle you solved after you solved the puzzle you solved before you solved this one, was harder than the puzzle you solved after you solved the puzzle you solved before you solved this one, was the puzzle you solved before you solved this one harder than this one?"
>>> Yes - Reduced, it becomes "if the puzzle before this was harder than this, was the puzzle before this harder than this?"
What order is denoted by the following prefixes?
>>> Duty watches on a ship.
Aye, aye Sir.
163.013
A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank?
Let x be the input rate in gallons per hour into the tank.
Let y be the output rate in gallons per hour from the tank for each value.
Let w be the number of gallons in the tank.
We know that it takes 2.5 hours to empty the tank with 10 valves open. Over the 2.5 hours the sum of the water in the tank initially and the water going into the tank will equal the amount of water leaving the tank. Let's set that up as an equation.
w + 2.5x = 2.5*10y
We also know that it takes 5.5 hours to empty the tank with 6 valves open. Let's set that up as an equation.
w + 5.5x = 5.5*6y
I don't like decimals so let's multiply both equations by 2:
2w+5x=50y
2w+11x=66y
To solve the problem we need to know the relationship between w and x, so lets solve for y in the first equation and substitute in the second:
y=(2w+5x)/50.
Substituting this in the second equation...
2w+11x=66*(2w+5x)/50
100w+550x=132w+330x
220x=32w
w=220x/32
w=6.875x
Therefore, 6.875 times the input rate equals the capacity of the tank. Thus, it would take 6.875 hours to fill the tank.
165.015
You have a box of matches and six 64-minute fuses. Each fuse burns exactly 64 minutes but the rate of burning is inconsistent. A half a fuse would not necessarily burn in 32 minutes. The fuses are also not equally inconsistent. Two cut fuses of the same length would not necessarily burn in the same time. Without the use of a clock, scissors, or anything other than the matches and the fuses how can you create a 60-minute fuse?
There at least four possible answers to this problem.
Following is my original solution.
1. At time t=0 light fuse 1 at both ends and fuses 2, 3, 4, and 5 at one end. Fuse 1 will have 32 minutes remaining, fuses 2-5 will have 64 minutes remaining.
2. At time t=32 fuse 1 will have burned out. At this moment light fuse 2 at the other end. Fuse 2 will have 16 minutes remaining, fuses 3-5 will have 32 minutes remaining.
3. At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end. Fuse 3 will have 8 minutes remaining, fuses 4,5 will have 16 minutes remaining.
4. At time t=56 fuse 3 will have burned out. At this moment light fuse 4 at the other end. Fuse 4 will have 4 minutes remaining, fuse 5 will have 8 minutes remaining.
5. At time t=60 fuse 4 will have burned out. At this moment start fuse 6. Fuse 5 will have 4 minutes remaining, fuse 6 will have 64 minutes remaining.
6. At time t=64 fuse 5 will have burned out. At this moment snuff out fuse 6. Fuse 6 will have 60 minutes remaining when relit.
Solution 2
1. At time t=0 light fuse 1 at both ends and fuse 2 at one end. Fuse 1 will have 32 minutes remaining and fuse 2 will have 64 minutes remaining.
2. At time t=32 fuse 1 will have burned out. At this moment light the other end of fuse 2 and fuse 3 at one end. Fuse 2 will have 16 minutes remaining and fuse 3 will have 64 minutes remaining.
3. At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end and light fuse 4 at one end. Fuse 3 will have 24 minutes remaining and fuse 4 will have 64 minutes remaining.
4. At time t=72 fuse 3 will have burned out. At this moment light fuse 5 from both ends. Fuse 4 will have 40 minutes remaining and fuse 5 will have 32 minutes remaining.
5. At time t=104 fuse 5 will have burned out. At this moment light fuse 4 from the other end and light fuse 6 from one end. Fuse 4 will have 4 minutes remaining and fuse 6 will have 64 minutes remaining.
6. At time t=108 fuse 4 will have burned out. At this moment snuff out fuse 6. Fuse 6 will have 60 minutes remaining.
Solution 3
1. Light #1 at both ends, and #2 and #3 at one end. Unlight #2 and #3 when #1 burns out. This makes #2 and #3 both 32-minute fuses.
2. Light #2 at both ends, and #4 at one end. Unlight #4 when #2 burns out. This makes #4 a 48-minute fuse.
3. Light #4 at both ends, and #3 at one end. Unlight #3 when #4 burns out. This makes #3 an 8-minute fuse.
4. Light #3 at both ends, and #5 at one end. Unlight #5 when #3 burns out. This makes #5 a 60-minute fuse.
Solution 4
1. At time t=0 light fuse 1 at both ends and fuses 2, 3, and 4 at both ends. Fuse 1 will have 32 minutes remaining, fuses 2-4 will have 64 minutes remaining.
2. At time t=32 fuse 1 will have burned out. At this moment light fuse 2 at the other end. Fuse 2 will have 16 minutes remaining, fuses 3,4 will have 32 minutes remaining.
3. At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end. Fuse 3 will have 8 minutes remaining, fuses 4 will have 16 minutes remaining.
4. At time t=56 fuse 3 will have burned out. At this moment light fuse 4 at the other end and fuse 5 at one end. Fuse 4 will have 4 minutes remaining and fuse 5 will have 64 minutes remaining.
5. At time t=60 fuse 4 will have burned out. At this moment snuff out fuse 5. Fuse 5 will have 60 minutes remaining when relit.
166.016
Fifty contestants for the Miss America pageant were arranged in ten rows and five columns on the Boardwalk in Atlantic City. One viewer of this display noticed that of the tallest women in each column that Miss Maryland was the shortest of the five. Another viewer noticed that of the shortest women in each row that Miss Nevada was the tallest of the ten. There are no ties, all fifty women have different heights. Who is taller, Miss Maryland or Miss Nevada?
There are three possible scenarios:
1. If Miss Maryland and Miss Nevada are in the same column then Miss Maryland would be taller, because she is the tallest in that column.
2. If Miss Maryland and Miss Nevada are in the same row then Miss Maryland would be taller, because Miss Nevada is the shortest in that row.
3. If Miss Maryland and Miss Nevada share neither a row nor column then there must be some other contestant who is both in Miss Maryland's column and Miss Nevada's row. Call this woman Miss Texas. To illustrate her position consider the following diagram.
xxxxx
xmxxx
xxxxx
xxxxx
xxxxx
xxxxx
xtxxn
xxxxx
xxxxx
xxxxx
In this digram the m stands for Miss Maryland, the n for Miss Nevada, and the t for Miss Texas.
Miss Maryland must be taller than miss Texas, because miss Maryland is the tallest in her column. Miss Texas must be taller than Miss Nevada, because Miss Nevada is the shortest in her row. Thus if M>T>N then M>N, or Miss Maryland is taller than Miss Nevada.
167.017
The front wheels of a wagon measure 3.5 feet in diameter. The rear wheels measure 4.25 feet in diameter. While the wagon is stopped somebody makes a chalk mark on both a front and rear wheel. How far must the wagon travel before both chalks marks return to their initial position at the same time?
The circumference of the front wheel is 3.5*pi and the circumference of the rear wheel is 4.25*pi. We must find the least common multiple of both these numbers. Both these numbers are divisible by .25*pi, so the least common multiple is .25*pi*LCM(14,17) = .25*pi*14*17 = 59.5*pi =~ 186.92.
168.018
5 cows can eat 2 acres of grass in 10 days. 7 cows can eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?
The problem again is, "5 cows can eat 2 acres of grass in 10 days. 7 cows can eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?"
By the time the cows have eaten all the grass total grass consumed will equal total initial grass plus total grass re-growth. Next, let's define some terms.
x=initial amount of grass in an acre.
y=amount of grass grown in one acre in one day.
Putting the given information in the form of equations we get:
50=2x+20y
210=3x+90y
To put the first equation in more simple English, 5 cows eating for 10 days results in a consumption of 50 units of grass. This is equal to the sum of 2x initial units of grass and 20y units of grass growth (10 days times 2 acres).
Next we must solve for x and y. Rewriting the above equations we get:
150=6x+60y (multiplying by 3)
420=6x+180y (multiplying by 2)
Subtracting the first equation from the second we get 270=120y, so y=9/4. Plugging this into either equation we get x=5/2.
The question to be answered is "How many days will it take 16 cows to eat 7 acres of grass?" Let's let d be the number of days. So setting this up as an equation we get.
16d=7x + 7dy
16d=35/2 + 63d/4
d/4=35/2
d=70.
So it will take 70 days for 16 cows to eat 7 acres of grass.
169.019
A spider eats 3 flies a day. Until the spider fills his quota a fly has a 50% chance of survival if he attempts to pass the web. Assuming 5 flies have already made the attempt to pass, what is the probability that the 6th fly will survive the attempt?
Using the binomial distribution, the probability that n flies out of 5 have been eaten is combin(5,n)*(1/2)5, where combin(5,2)=5!/(n!*(5-n)!).
The probability that 0 flies were eaten is combin(5,0)*(1/2)5=1/32.
The probability that 1 fly was eaten is combin(5,1)*(1/2)5=5/32.
The probability that 2 flies were eaten is combin(5,2)*(1/2)5=10/32.
So the probability that the spider is still hungry is 1/32 + 5/32 + 10/32 = 16/32 = 1/2. The probability the spider is full is 1-1/2=1/2. Thus the probability of a successful attempt to pass is (1/2)*1 + (1/2)*0.5 = 0.75 .
170.010
I am as old as John will be when I am twice as old as John was when my age was half the sum of our present ages. John is as old as I was when John was half the age he will be 10 years from now. How old am I?
90% of this problem is working through the wording into the form of an equation. Let's let i be my age and j be the age of John. The first piece of information tells us...
i = 2*((i+j)/2 + (j-i)) + (j-i)
= 2*(3j-i)/2 + (j-i)
= 4j - 2i
= (4/3)*j
The second piece of information tells us...
j = (j+10)/2 + (i-j)
= i - j/2 + 5 = (2i+10)/3
Combining the two equations we get...
i = (4/3)*(2i+10)/3
= (4/9)*(2i+10)
= (8/9)*i + 40/9
= 40
So I am 40 (and John is 30).
171.011
A, B, and C are three hills, each connected by a network of roads. There are 82 ways to get from A to B, including those routes that pass through C. There are 62 ways to get from B to C, including those routes that pass through A. The number of ways to get from A to C, including those ways passing through B, is less than 300. How many ways are there to get from A to C?
Let's define the number of direct routes from A to B as x.
Let's define the number of direct routes from B to C as y.
Let's define the number of direct routes from A to C as z.
From the information given we know:
x + yz = 82
y + xz = 62
Adding them we get:
x + xz + y + yz = 144
(x + y) * (z + 1) = 144
and subtract them to get:
x - xz - y + yz = 20
(y - x) * (z - 1) = 20
The factors of 20 are {1, 2, 4, 5, 10, 20}. So (z - 1) is one of {1, 2, 4, 5, 10, 20}. then (z + 1) is one of {3, 4, 6, 7, 12, 22}.
The factors of 144 are: {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. So (z + 1) must be one of these factors.
Common possible (z + 1) values are {1, 2, 3, 4, 6, 12}. Given (x + y) * (z + 1) = 144, possible values of (x + y) are: {144, 64, 48, 36, 24, 12}.
Possible values of (z - 1) are: {-1, 0, 1, 2, 4, 10}.
Concurrently since (y - x) * (z - 1) = 20, then possible values of (y - x) are: {20, 10, 5, 2, -20}.
This leads to five possible solutions. Note that x+y=64 is not listed because then z would be 1 and (y-x)*(z-1) could not equal 20 for any values of x and y.
1. If x+y=144 then z=0, y-x=-20, y=62, x=82
2. If x+y=48 then z=2, y-x=20, y=34, x=14.
3. If x+y=36 then z=3, y-x=10, y=23, x=13.
4. If x+y=24 then z=5, y-x=5, y=14.5, x=9.5.
5. if x+y=12 then z=11, y-x=2, y=7, x=5
The fourth solutions is not possible because there can not be half a path.
The problem limits us to z+xy<300, so that eliminates the first three.
Thus the fifth solution is the correct one. If x=5, y=7, and z=11 then the final answer is z+xy=11+5*7=46.
