The worlds first riddle!

"But now when ten and half ten years,"

Ten and half ten is fifteen, but nice job with ten and a half, Rap.

Mark:

GIFT
$2

QUIZ
$25

CARDS
40


CRIME
Brimmers


CRIME:
Let's go through the possibilities one by one...

If Anderson lied, then the statements made by Brimmers and Dodge contradict each other, so Anderson cannot be the liar.

If Carter lied, then, again, the statements made by Brimmers and Dodge contradict each other, so Carter cannot be the liar. At this point you should be able to see that the key to the puzzle is in the contradictory statements made by Brimmers and Dodge.

If Dodge lied, then the statements made by Anderson and Brimmers cannot both be true (assuming only one man was guilty).

Therefore we are left with only one option.

If Brimmers lied when he accused Dodge, then Anderson could truthfully be saying that Brimmers did it, Carter could truthfully deny being involved, and Dodge could truthfully claim that Brimmers wrongfully accused him.

Since Anderson was telling the truth when he accused Brimmers, Brimmers is the guilty man.



Mark:
AGES
45 and 15


Once you figure out what the poem is saying, this is a fairly straight forward algebra problem. The author of the poem states that when he married his wife, his age was 3 times larger than his wife's age. Fifteen years later, his age was only 2 times as large as his wife's age. How old were the couple when they got married?

To solve for two unknown variables, we need two equations:

1) H = 3W Husband is equal to 3 times Wife
2) H+15 = 2(W+15) 15 years later, Husband is equal to 2 times Wife.

Since we know from the first equations that H is the same as 3W, we replace the H in the second equation with 3W...

3) 3W+15 = 2(W+15)

Expanding the right hand side of equation 3 gives...

4) 3W+15 = 2W+30

Now we can subtract 15 from both sides of equation 4...

5) 3W = 2W+15

Subtracting 2W from both sides of equation 5 leaves...

6) W = 15

Since we know from equation 1 that Husband's age is 3 times Wife's age, we can deduce that at the time of the wedding, Wife was 15 years old, and Husband was 45 years old. Fifteen years later, Wife would be 30 years old and Husband would be 60 years old, which fulfills the requirements of equation 2.



Rap wrote, "And the headlines read "31 and a half year old pedophile marries a 10 and a half year old girl."

Well, the news is always right, right!


The sum of two whole numbers is 63. The difference between the numbers is less than 10. How many solutions as you can find.



The local bank charges a monthly service charge of $3.50 plus 8 cents for each check. Last October, Michelle's total charges were $5.58. How many checks did she write



The local library fine schedule for overdue books is as follows: 25 cents per day for each of the first three days; 10 cents per day thereafter Sandra paid a fine of $1.35. How many days overdue was her book



The players on the Hawks hockey team wear the numbers from 1 through 18 on the backs of their jerseys. On opening night, as the players were being introduced, Maura noticed that the players were standing in nine pairs. She also noticed that the sum of the numbers on the jerseys of each pair was a perfect square (an integer times itself). The goalie wore number 1. With what numbered player was he paired

63
5 (largest number ranges from 32 to 36)

BANK
26

LIBRARY
9

GOALIE
15

Well yeah! As I've been trying to convince markr, my forehead is sloped because I slap it with my dohs!

Whole #
5 possibilities
31 32 diff=1
30 33 diff=3
29 34 diff=5
28 35 diff=7
27 36 diff=9


Michele wrote 26 cheques

Sandra kept the book over a week and a day

Hockey Team
1-15 is 4^2
2-14 is 4^2
3-13 is 4^2
4-12 is 4^2
5-11 is 4^2
6-10 is 4^2
7-18 is 5^2
8-17 is 5^2
9-16 is 5^2

The goalie was paired with 15

Rap

oops! A week and two days

Rap

packages of cups and plates
a*54=b*42
a*9*6=b*7*6
a*9=b*7
so a=7 and b=9
7 packages of cups
9 packages of plates

Douglass School
x/3+x/3+x/4+100=x
x=1200
400 to Washington
400 to Annapolis
300 to Williamsburg
100 to Monticello

Band & Orchestra
Boolean Algebra
If B is the Set of elements in the Band
& O is the Set of elements in the Orchestra
Then the elements on the B & O is
BUO=B+O-B∩O
and B=23, O=25, B∩O=7
so BUO=23+25-7=41
41 students are in the band and orchestra

Three Wagons
Sum of 1 to 9 is 9*10/2=45 so each wagon carries 45/3=15 pounds
this is distributed among the wagons as
(1,6,8)=15
(2,4,9)=15
(3,5,7)=15

Rap

There have been complaints that the questions have been getting easier. I say it is due to the smart people who answer the questions. However to show I have done my bit to stamp out smart:



When a man went to visit a friend, he realized he had forgotten the apartment number. He does remember the apartment is on the first floor where the apartments are numbered 101 through 150. He also knows the apartment number is an even number and a multiple of 3 and 11. What is the correct apartment number



Professor Nonsense has a theory about how flowers grow. He believes that every night bugs tug at flowers to get them to grow and then open them carefully just before sunrise. He says that the dewdrops on flowers are actually drops of bug sweat. The professor tells about a small county in which dragonflies each open 6 roses, 4 daisies, and 6 daffodils per night; ladybugs each open 1 petunia, 2 pansies, and 1 tulip; moths each open 3 chrysanthemums and 5 violets; and bees each open 17 tiger lilies and 3 iris.

Dragonflies deposit 10 sweat drops on each flower they open; ladybugs leave 5 drops, moths deposit 40 drops, and bees leave 15 drops. One morning last summer the professor counted 74,360 sweat drops on the flowers. He noticed four times as many 10-drop flowers as 5-drop flowers, and 3 times as many 15-drop flowers as 40-drop flowers.

How many of each kind of insect had been working on the flowers the night before



During the morning senior assembly, Mr. Carter, the school principal, announced that he had good news for the group. He told the students that he has just been informed that 50 percent of those who applied for admission into State University have been accepted. Also, 50 percent of those who applied to Temple Tech have also been accepted. Carla turned to her friend Marcie and said, "Great! I've been accepted I applied to both schools, and 50 percent +50 percent is 100 percent, so I'm in!"

Analyze the action. Is Carla correct

Apartment# 132

Bug drops I haven't a solution
74360=5*2^3*11*13^2=10d+5l+40m+15b
d=4l & m=3b
so
5*2^3*11*13^2=5(2d+l+8m+3b)
2^3*11*13^2=2*4l+l+8*3b+3b
5*2^3*11*13^2=9l+27b=9(l+3b)
This is a line, and there are an infinite number of solutions

Carlas Admission
some students applied and are accepted to both universities and some students applied to both universitied and are accepted to neither. If Carlas math skills are equal to this logic, somehow I fell she's in the latter group.

Rap

BUG SWEAT
There are two solutions:
262 dragonflies, 262 ladybugs, 40 moths, and 48 bees
92 dragonflies, 92 ladybugs, 85 moths, and 102 bees

160D + 20L + 320M + 300B = 74360
16D = 4 * 4L -> 16D = 16L -> D = L
20B = 3 * 8M -> 20B = 24M -> B = 6M/5

Combine the above to get:
180L + 680M = 74360 -> 9L + 34M = 3718

There are only 12 integer solutions to the previous equation. Only two of them result in B being an integer.

CARLA
Counterexample to prove Carla's logic is false:

Carla and Marcie are the only two students to apply to State or Temple, and they each apply to both universities.

Marcie is accepted by both, Carla by neither.

I vote for tougher math problems, but don't consider this a complaint. I appreciate the effort and humor you put into this.

I rate the most recent batch as:

APARTMENT: trivial
BUG SWEAT: easy, but interesting with some work required
CARLA: trivial

Markr. A forehead sloping moment.

Rap

Amy, Store B and charge it

Record store 20 different heterosexual couples, 72 mixed ones

Taxes--itemize

I think this one is False---You still don't know all vectors of Rn. This is a single transformation.

I got this one down as True. The null space are those vectors [X] s.t. A[X]=0.

I got this one as True. If (A-I)^2=0 then A has a unity eigenvalue, which means |A|=|I|=1. Square matrices with nonzero determinants are invertable (e.g. A^-1 exists).

Rap

couples

interesting problem, the couples out of 9 thing that is

mixing them all together gives 9!/(9-2)!=9*8=72. But this equates to ab as different from ba, and in this case they ain't. So the free for all couples is 72/2=36

# of homosexual couples
Lesbians 5!/(5-2)!/2=5*4/2=10
Girly Boys 4!/(4-2)!/2=4*3/2=6

# of breeders
All-Lesbians-Girly Boys
Breeders 36-10-6=20

There is another way to come up with 20 breeders, but its too easy.

Rap

Try wrote: "715 ladybugs; 1430 dragonflies; 2860 moths; 1287 bees"

What were you smoking in England?

The moths alone would leave 2860*8*40=915,200 sweat drops. That's more than ten times the total.

AMY
Assuming the cash discounts are applied after (as opposed to with) the original discount here's what Amy would pay as a percentage of the original price:
Store A
credit: 30%, cash: 27%
Store B
credit: 40%, cash: 32%
Store C
cash: 33.3%

The lowest price is at store A when paying with cash.

TAXES
Itemizing gives a $911 larger deduction.

TRUE/FALSE
I remember practically nothing from my linear algebra course. I must claim ignorance.

Rap is starting to sound like Ahnold. :wink:

Amy, Amy, Amy, got to start reading the problems closer!

Discount is the key word

Store A Credit 70% Cash 73%
Store B Credit 60% Cash 68%
Store C Credit n/a Cash 66.6%

Go to Store A cash or credit

Rap

TRUE/FALSE
OK, I got out the linear algebra book (might as well be Greek).

The row space and the nullspace are orthogonal. As are the left nullspace and the column space.

(A-I)^2 = A^2 - 2A + I
Since (A-I)^2 = 0
2A - A^2 = I
A(2I-A) = I
Therefore, 2I-A is the inverse of A.

Let A be nxn. Attaching a basis of NS(A) to one of CS(A) gives a basis of R^n.
I'll guess that this is false, but that attaching a basis of NS(A) to one of RS(A) does give a basis of R^n.