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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.

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

Solution:

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

Solution:

1. 1L = 1.057 fl qt
2. 1L = 1000 ml
3. 1.079 fl qt = 1.079 x 1.057 L = 1.140503 L
4. 1.140503 L = 1140.503 ml

Solution:

Base ten blocks are a mathematical manipulative used to learn basic mathematical concepts including addition, subtraction, number sense, place value and counting. You can manipulate the blocks in different ways to express numbers and patterns. Generally, the 3-dimensional blocks are made of a solid material such as plastic or wood and come in four sizes to indicate their individual place value: Units (one’s place), Longs (ten’s place), Flats (hundred’s place) and Big Blocks (thousand’s place). There are also computer programs available that simulate base ten blocks.

Since the number that is being modeled has a zero in the tens place and a zero in the ones place, there are no Longs or Units. Since the number is less than 999, there are no Big Blocks. That leaves only Flats, and there must be five Flats. Since each Flat represents one hundred, the number must be 500.

Solution:

1. A quantitative variable is a phenomenon measured in amounts, that is, numerical units. For example, a numerical score is a quantitative variable.
2. The measurement is based on a scale of 1 to 4.

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.

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.

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

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 12^th count, take 1200 * 1.5¹¹ = 1200 *86.5 = 103,797