Solutions to Set 4

Problem #1: 3
Problem #2: 13
Problem #3: 5256
Problem #4: 109
Problem #5: 91/216
Problem #6: 36
Problem #7: 20
Problem #8: 12
Problem #9: 1000
Problem #10: 87

Explanations

Problem #1: Because a<b and b<c, you know that a<c. The problem also tells you that c<a. Therefore, a and c must be equal, and a=c=1.
Based on the inequalities a<b and b<c, you now know that 1<b and b<1. Therefore, b=1 as well.
Since a, b, and c each have a value of 1, their sum is 3.
Problem #2: There are a total of 13 ways:
[1] 1 quarter
[2] 2 dimes, 1 nickel
[3] 2 dimes, 5 pennies
[4] 1 dime, 3 nickels
[5] 1 dime, 2 nickels, 5 pennies
[6] 1 dime, 1 nickel, 10 pennies
[7] 1 dime, 15 pennies
[8] 5 nickels
[9] 4 nickels, 5 pennies
[10] 3 nickels, 10 pennies
[11] 2 nickels, 15 pennies
[12] 1 nickel, 20 pennies
[13] 25 pennies
Problem #3: The problem says that the sum of the first N counting numbers is 2628, so
1 + 2 + 3 + ... + N = 2628.
The problem asks for the sum of the first N positive even numbers, which is
2 + 4 + 6 + ... + 2N
= 2(1 + 2 + 3 + ... + N)
= 2(2628)
= 5256.
Problem #4: Using the formula for the circumference of a circle, you will find that each of the two curved sections of the track has a length of
.5(50π)
≈ .5(50 * 3.14)
= 78.5 meters
Therefore, the total perimeter can be expressed as
78.5 + x + 78.5 + x
= 157 + 2x meters
Your desired perimeter is 375 meters, so you can incorporate the information into an equation:
157 + 2x = 375
2x = 218
x = 109.
Problem #5: In order to get at least one Zac Efron bobblehead, Katie can get either 1, 2, or 3 Zac Efron bobbleheads. So you could calculate the probability of getting 1, the probability of getting 2, and the probability of getting 3, then add all of those probabilities together. A quicker way, however, is to find the probability of not getting at least one Zac Efron bobblehead (that is, the probability of getting 0) and subtracting that probability from 1.
The probability that Katie will not get a Zac Efron bobblehead in a particular Yummy Cereal box is 5/6. Therefore, the probability that she will not get one in the first box AND she will not get one in the second box AND she will not get one in the third box is:
(5/6) * (5/6) * (5/6) = 125/216.
This is the probability that Katie will NOT get at least one Zac Efron bobblehead. Therefore, the probability that she WILL get at least one is:
1 - 125/216 = 91/216.
Read more about Combining Probabilities.
Problem #6: Label angles B, C, D, and E as shown:

Angle D and E are interior angles of the regular pentagon, so each has a measure of
D = E = (5-2)(180/5)
= 3(36)
= 108 degrees.
Because B is supplementary to D, it has a measure of
180 - 108 = 72 degrees.
Similarly, C is supplementary to E and has a measure of 72 degrees as well.
Since A, B, and C are angles of a triangle, their measures add up to 180.
A + B + C = 180
A + 72 + 72 = 180
A + 144 = 180
A = 36 degrees.
Read more about Formulas About Polygons and about Adding Angles.
Problem #7: Let x equal the total capacity of the test tube, in mL. From the information given, you know that:
(1/8)x + 15 = (7/8)x
15 = (6/8)x
15 = (3/4)x
20 = x
At the end of the problem, the test tube has been filled to full capacity, so there are 20 mL of hydrochloric acid in it.
Read more about Translating Word Problems.
Problem #8: Examine the formula for counting factors. 6 can be written either as 6 or as 2*3. Therefore, if a number has 6 factors, the exponents of the prime factorization are either {5} or {1 and 2}. In the first case, the smallest number would be 2⁵=32. In the second case, the smallest number would be 2²3¹=12. Since 12 is smaller than 32, the answer is 12.
Read more about Counting Factors.
Problem #9: 10 hews equal 5 futzes. 5 futzes equal 50 erbs. 50 erbs equal 1000 gits.
Problem #10: The total difference between 90 and 9 is 81. This increment of 81 is reached in 27 steps. Therefore, the difference reached in each step (AKA the common difference) is 81/27=3. Since z comes right before 90 in the arithmetic sequence,
z = 90 - 3 = 87.
Read more about Completing Arithmetic Sequences.