# One hundred bushels of corn are to be divided among 100 persons. men get 3 bushels each, women 2 and children 1/2. how many men, women and children are present?

Friday, October 1st, 2010

### Solution:

1. Let m = the number of men (an integer)
2. Let w = the number of women (an integer)
3. Let c = the number of children (an integer)
4. Number of bushels = 3m+2w+.5c = 100
5. Number of people = m+w+c=100
6. 0<m<100
7. 0<w<100
8. 0<c<100

If we assume that there are 10 children, then we can solve for two equations and two unknowns:

1. 3m+2w+5 = 100
2. m+w+10=100

We can solve this by subtracting two times the second equation from the first equation:

1. 3m+2w+5 = 100
2. 2m+2w+20=200 (this is two times the second equation shown above)
3. m          – 15 = -100 (this is the result of subtracting two times the second equation from the first equation)
4. m = -85
5. w=100-10- (-85) = 175

The problem with this result is that the values for m and w are outside the allowable range, so clearly the assumption that there are 10 children is wrong. Let’s try increasing the number of children to 80 and see how that impacts the result from m:

1. 3m+2w+40 = 100
2. 2m+2w+160=200
3. m          -125 = -100
4. m                 =  25
5. w = 100-80-25=-5

Here, the result for m is within the range, but the value for w is outside the range. Let’s try reducing the number of children to 70:

1. 3m+2w+35 = 100
2. 2m+2w+140=200
3. m          -105 = -100
4. m                    =    5
5. w = 100-70-5= 25

Now we have a solution that makes sense (m=5, w=25, c=70). However, there may be other answers that work, so let’s try increasing the number of students to 72 and see what happens:

1. 3m+2w+36 = 100
2. 2m+2w+142=200
3. m          -106 = -100
4. m                    =    6
5. w = 100-72-6= 22

This solution also makes sense (m=6, w=22, c=72). Clearly, there can be several correct solutions to this question.

# What is the square root of 444*111?

Friday, October 1st, 2010

### Solution:

1. Notice that 444 * 111 = 4*111*111
2. The square root of 4 is 2
3. The square root of 111*111 is 111
4. So the square root of 4*111*111 = 2*111
5. The answer is therefore 222

# What is 43 over 100 as a decimal.

Friday, October 1st, 2010

### Solution:

The answer is forty three hundredths, which is written as 0.43.

# Son accepted to college. Tuition will not increase for 4 years he is attending. First tuition pymt of 10,000 due in 6 months. Same payment due every 6 months unitl a total of 8 pymts made.The college offers a bank account with fixed APR of 4%(semiannually) for the next 4 yrs. You can with draw money every 6 months. How much money must you deposit today if you intend on not making any further deposits and will make all tuition pymts from this acct, leaving the acct empty when the last pymt is made?

Friday, October 1st, 2010

### Solution:

The APR is 4%, so the semi-annual interest rate is 2%. If you deposit \$10,000 in the bank after every six months, you will have \$85,829.69 in the bank at the end of 4 years:

1. After six months you will have \$10,000
2. After one year you will have 10,000×1.02+10,000 = \$20,200
3. After 1.5 years you will have \$20,200×1.02+10,000 = \$30,604
4. After 2 years you will have \$30,604×1.02+10,000 = \$41,216.08
5. After 2.5 years you will have \$41,216.08×1.02+10,000 = \$52,040.40
6. After 3 years you will have \$52,040.40×1.02+10,000 = \$63,081.21
7. After 3.5 years you will have \$63,081.21×1.02+10,000 = \$74,342.83
8. After 4 years you will have \$74,342.83×1.02+10,000 = \$85,829.69

The present value of that \$85,829.69, discounting at 2% is \$73,254.81 (i.e. \$85,829.69/(1.02^8)), so if you put this amount in the bank now, and make the payments every six months, there will be nothing left in the account after four years.

# Many academic institutions offer a sabbatical policy. Every seven years a professor is given a year free of teaching. A professor earning 70,000 per year who works for a total of 42 years what is the present value of the amount they will earn while on sabbatical, if the interest rate is 6% EAR?

Friday, October 1st, 2010

### Solution:

1. In his seventh year, the professor receives \$70,000. Since the question does not state when he receives this amount, let’s assume it is received at the end of the seventh year as a lump sum. The present value of this is \$46,554 (i.e. \$70,000/(1.06^7))
2. Similarly, the professor receives \$70,000 at the end of year 14, and the present value of this is \$30,961.07
3. Similarly, the professor receives \$70,000 at the end of year 21, and the present value of this is \$20,590.88
4. Similarly, the professor receives \$70,000 at the end of year 28, and the present value of this is \$13,694.11
5. Similarly, the professor receives \$70,000 at the end of year 35, and the present value of this is \$9,107.37
6. Similarly, the professor receives \$70,000 at the end of year 42, and the present value of this is \$6,056.92

Now we can add up each of the six present values, to arrive at the value of all the amounts received during the sabbaticals: \$126,964.34 (if you add up the numbers above you get \$126,964.35, but that’s due to rounding error).

If the professor receives the money at the middle of each year, you can re-calculate the above (e.g. by discounting the seventh year by 6.5 years rather than by 7 years).  You might also want to recalculate this assuming that the salary is paid monthly, and see how it compares to the above result.

# Find the minimum number of students needed to guarantee that 4 of them were born: hint use pigeonhole principle. 1. on the same day of the week; 2. in the same month. 3. Which counting principle applies to the questions above?

Friday, October 1st, 2010

### Solution:

1. There are seven days in a week
2. If there are 7 students, they could all have been born on different days of the week
3. If there are 8 students, we could guarantee that 2 of them were born on the same day
4. If there are 15 students, we could guarantee that 3 of them were born on the same day
5. F there are 22 students, we could guarantee that 4 of them were born on the same day

# A colony of bats is counted every two months. The first four counts are 1200, 1800, 2700, and 4050. If this growth rate continues, (4 points) 1. What is the recurrence relation of the bat population? (2 points) 2. How many bats are there at the 12th count? Show all work. (Hint: solve the recurrence relation above)

Friday, October 1st, 2010

### Solution:

1. Recurrence relation
• Note that 1800/1200 = 1.5.
• Next, take 1.5* 1800 =2700, and that is the next population.
• Now 1.5*2700 = 4050.
2. To calculate the number of bats at the 12th count, take 1200 * 1.511 = 1200 *86.5 = 103,797

# Suppose a Word is a String of 8 letters of the Alphabet with Repeated Letters Allowed: (5 points, 1 point each) Show all work. Do not Answer with just a Number. 1. How many words are there? 2. How many words end with the letter N? 3. How many words begin with R and end with N? 4. How many words start with A or B? 5. How many words begin with A or end with B

Friday, October 1st, 2010

### Solution:

1. If you have 26 choices for each letter in a word, then there are 26 such one-letter words, 262 two-letter words, 263 three-letter words, etc. There would be 268 eight-letter words.
2. There would be 267 eight-letter words that end with the letter N, which is the same as the number of 7 letter words (just add “N” to the end of each seven-letter word.
3. There would be 266 eight-letter words that begin with R and end with N (just add R to the front and N to the end of all six-letter words.
4. There would be 2*267 eight-letter words that begin with an A or B (add A to the beginning of every seven-letter word, then add B to the beginning of every seven-letter word).
5. There would be 2*267 eight-letter words that begin with A or end with B (add A to the beginning of every seven-letter word, then add B to the end of every 7 letter word).

# Complete These Ordered Pairs: (0, ) (2, ) (-1, ) y=3x-1 (-1, ) (0, ) (1, ) 2x+y=3 (0, ) (-1, ) (1, ) y=2x+1

Friday, October 1st, 2010

### Solution:

Each ordered paid represents x and y: (x,y)

To calculate y, you can plug in x.

If y=3x-1, then the given values of x will result in the pairs (0,-1) (2,5) (-1,-4).

If y=3-2x, then the given values of x will result in the pairs (-1,5) (0,3) (1,1).

If y=2x+1, then the given values of x will result in the pairs (0,1) (-1,-1) (1,3).

# Cos 2x=0.32

Monday, September 6th, 2010

### Solution:

First let’s apply the reverse cosine function to both sides of the equation:

cos-1(cos(2x)) = cos-1 (0.32) = cos-1 ()

Now if you take the inverse cosine the  cosine of 2x, you get 2x…so:

2x = cos-1 ()

2x 71.34°

x 35.67°  (which is approximately the same as 0.622 radians)

This looks like it might be the final answer, but actually it’s only one of the many correct answers. One way to see this is by graphing y=cos(2x) and seeing where it intersects the line y=0.32. If you do this, you will see that the intersections occur at multiple points:

As you can see, the intersections occur at x 0.622 radians, 2.519 radians, 3.764 radians, 5.660 radians, etc. The general form of the solution set is as follows: