Solutions to Set 2

Problem #1: 100%
Problem #2: 5/18
Problem #3: 3780
Problem #4: 8
Problem #5: 35/36
Problem #6: 90
Problem #7: 120
Problem #8: 140
Problem #9: 1120
Problem #10: 6

Explanations

Problem #1: The product will be 0 if and only if at least one of the three cards has 0 written on it. There are only two cards without 0 on them. Therefore, no matter how you choose three cards, at least one of them must have 0. The product will always be 0, so the the probability is 100%.
Read more about The Zero-Product Property.
Problem #2: For each individual roll, the probability of obtaining a 3 is 1/6. The probability of obtaining a 3 on the first roll OR obtaining a 3 on the second roll is:
1/6 + 1/6 - (1/6)*(1/6)
= 1/6 + 1/6 - 1/36
= 11/36.
Within this probability, we have counted the outcome in which Chuck rolls a 3 both times. However, we do not want to count that outcome, since the problem states "exactly one 3." Therefore, we need to calculate the probability of rolling a 3 both times, then subtract that from 11/36. The probability of obtaining a 3 on the first roll AND obtaining a 3 on the second roll is:
(1/6)*(1/6)
= 1/36.
Our final answer, therefore, is:
11/36 - 1/36
= 10/36
= 5/18.
Read more about Combining Probabilities.
Problem #3: Use Heron's Formula.
s = (11 + 12 + 13)/2
s = 18.
A = √ (18(18-11)(18-12)(18-13))
A = √ (18(7)(6)(5))
A = √ (3780)
A² = 3780.
Read more about Heron's Formula.
Problem #4: Let p and c represent the number of pigs and cows, respectively. You can then write the following equations based on the ratios given:
(p + 7)/(c + 20) = 1/2
(p + 7)/c = 3/2
The first equation can be re-written as
2(p + 7) = c + 20.
The second equation can be re-written as
2(p + 7) = 3c.
Substituting 2(p + 7) in the two equations, you have
c + 20 = 3c
20 = 2c
10 = c.
Substitute 10 for c into one of your previous equations:
2(p + 7) = 3c
2(p + 7) = 3*10
2(p + 7) = 30
p + 7 = 15
p = 8.
Read more about going From Ratios to Algebraic Equations.
Problem #5: The only way to roll a sum of 12 is by rolling two 6s. The probability of rolling a 6 on each individual die is 1/6. Therefore, the the probability of rolling 6s on both dice is
(1/6) * (1/6) = 1/36.
This is the probability of rolling a sum of 12. Therefore, the probability of NOT rolling a sum of 12 is
1 - 1/36 = 35/36.
Read more about Combining Probabilities
Problem #6: The least common multiple of 10 and 45 is 90. Since N is a multiple of both 10 and 45, it must also be a multiple of 90. The number 90 itself has 12 factors, as desired, so N=90.
Read more about Least Common Multiple and Counting Factors.
Problem #7: The prime numbers between 20 and 40 are 23, 29, 31, and 37. The sum of these is 120.
Read more about Prime Numbers.
Problem #8: Based on the ratio given, Woozy is to receive 7/15 of the total land.
(7/15) * 300 acres
= 140 acres.
Read more about going From Ratios to Algebraic Equations.
Problem #9: The number of combinations of 8 hats chosen 3 at a time is 56. The number of combinations of 6 hats chosen 3 at a time is 20. Since any of the 3-hat combinations can be put together with any of the 3-scarf combinations, there are total of 56*20=1120 ways to choose 3 hats and 3 scarves.
Read more about Calculating Combinations.
Problem #10: If x, y, and z represent the hundreds, tens, and ones digits of A, respectively, then A can be written as:
100x + 10y + z.
B can be written as:
100z + 10y + x.
Since the sum of A and B is 625,
(100x + 10y + z) + (100z + 10y + x) = 625
101x + 20y + 101z = 625
101(x + z) + 20y = 625
Because the sum of the hundreds digit and ones digit of A is 5,
x + z = 5.
So, substitute 5 for x + z in the other equation:
101(x + z) + 20y = 625
101(5) + 20y = 625
505 + 20y = 625
20y = 120
y = 6
Since the tens digit of B is y, this gives us the answer we are looking for, 6.
Read more about going From Digits to Algebraic Expressions.