# Posts Tagged with ‘Homework Answer’

Friday, October 1st, 2010

### Solution:

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

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

- 3m+2w+5 = 100
- m+w+10=100

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

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

- 3m+2w+40 = 100
- 2m+2w+160=200
- m -125 = -100
- m = 25
- 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:

- 3m+2w+35 = 100
- 2m+2w+140=200
- m -105 = -100
- m = 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:

- 3m+2w+36 = 100
- 2m+2w+142=200
- m -106 = -100
- m = 6
- 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.

Friday, October 1st, 2010

### Solution:

- Notice that 444 * 111 = 4*111*111
- The square root of 4 is 2
- The square root of 111*111 is 111
- So the square root of 4*111*111 = 2*111
- The answer is therefore 222

Friday, October 1st, 2010

### Solution:

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

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:

- After six months you will have $10,000
- After one year you will have 10,000×1.02+10,000 = $20,200
- After 1.5 years you will have $20,200×1.02+10,000 = $30,604
- After 2 years you will have $30,604×1.02+10,000 = $41,216.08
- After 2.5 years you will have $41,216.08×1.02+10,000 = $52,040.40
- After 3 years you will have $52,040.40×1.02+10,000 = $63,081.21
- After 3.5 years you will have $63,081.21×1.02+10,000 = $74,342.83
- 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.

Friday, October 1st, 2010

### Solution:

- 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))
- Similarly, the professor receives $70,000 at the end of year 14, and the present value of this is $30,961.07
- Similarly, the professor receives $70,000 at the end of year 21, and the present value of this is $20,590.88
- Similarly, the professor receives $70,000 at the end of year 28, and the present value of this is $13,694.11
- Similarly, the professor receives $70,000 at the end of year 35, and the present value of this is $9,107.37
- 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.

Friday, October 1st, 2010

### Solution:

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

Friday, October 1st, 2010

### Solution:

- 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.

- To calculate the number of bats at the 12
^{th} count, take 1200 * 1.5^{11} = 1200 *86.5 = 103,797

Friday, October 1st, 2010

### Solution:

- If you have 26 choices for each letter in a word, then there are 26 such one-letter words, 26
^{2} two-letter words, 26^{3} three-letter words, etc. There would be 26^{8} eight-letter words.
- There would be 26
^{7} 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.
- There would be 26
^{6} 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.
- There would be 2*26
^{7} 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).
- There would be 2*26
^{7} 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).

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).

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: