Solutions to Set 6

Problem #1: 12
Problem #2: 72
Problem #3: 200
Problem #4: 30
Problem #5: 115
Problem #6: 1/16
Problem #7: 54
Problem #8: 63
Problem #9: 10000
Problem #10: 9

Explanations

Problem #1: Based on a property of right triangles, right triangles ABC, PBA, and PAC are similar.
Since triangles PBA and PAC are similar, the ratio PA/PB is equal to the ratio PC/PA. So,
PA/PB = PC/PA
PA/18 = 8/PA
(PA)² = 8*18
(PA)² = 144
PA = 12.
Problem #2: Since you can shovel 1 driveway in 2 hours, your speed is 1/2 driveways per hour. Since your brother can shovel 1 driveway in 3 hours, his speed is 1/3 driveways per hour. The combined speed is:
1/2 + 1/3 = 5/6 driveways per hour.
Therefore, it takes you 6/5 hours to shovel 1 driveway. This is equivalent to
(6/5) * 60 = 72 minutes.
Problem #3: The base of triangle PCD is CD, which has length 20. The height of triangle PCD is the same as the height of the square, which is also 20. Therefore, the area of PCD is:
(1/2)(base)(height)
= (1/2)(20)(20)
= 200 square feet.
Problem #4: Each of the triangle's sides is the distance between the centers of two tangent circles. Therefore, the length of each side is the sum of two radii. You can thus calculate that the triangle's side lengths are:
2 + 3 = 5
2 + 10 = 12
3 + 10 = 13.
The side lengths 5, 12, and 13 are members of a Pythagorean triple, so it is a right triangle. The base and height are therefore 5 and 12, so the area of the triangle is
(1/2)(base)(height)
= (1/2)(5)(12)
= 30.
Read more about Circles Tangent to Each Other and The Pythagorean Theorem.
Problem #5: Since AD=AE, you can consider the figure symmetrical, so the measure of AEB is equal to the measure of ADC, which is 20 degrees.
ABE is a triangle, so the measures of angles ABE, BAE, and AEB add up to 180 degrees:
ABE + BAE + AEB = 180
ABE + 25 + 20 = 180
ABE = 135 degrees.
Therefore, angle DBF, being a supplement of angle ABE, has a measure of
180 - 135 = 45 degrees.
BDF is a triangle, so the measures of angles BDF, DBF, and BFD add up to 180 degrees:
BDF + DBF + BFD = 180
20 + 45 + BFD = 180
BFD = 115 degrees.
Read more about Adding Angles and Formulas About Polygons.
Problem #6: The possible pairs of integers are (1, 2), (2, 3), (3, 4), etc., all the way up to (224, 225). They are in the form (n, n+1), where n ranges from 1 to 224. Therefore, there are 224 possible pairs.
Of these pairs, how many have a product less than or equal to 225? Notice that 225 is the square of 15. In other words, 15*15=225. Therefore, 14*15 will give a product less than 225, while 15*16 will give a product greater than 225.
You can see, therefore, that a pair (n, n+1) will work if and only if n is 14 or less. With n ranging from 1 to 14, there are 14 such pairs.
The probability of choosing such a pair is:
14/224
= 1/16.
Read more about Perfect Squares.
Problem #7: On each face of the Rubik's cube, there are
3 * 3 = 9 stickers.
Since a cube has 6 faces, there are a total of
9 * 6 = 54 stickers.
Problem #8: The problem states that the lengths of the swords can be side lengths of a triangle. By the Triangle Inequality, the sum of 25 and 39 must be greater than x.
25 + 39 > x
64 > x.
The greatest value of x satisfying this inequality is 63.
Problem #9: First, find the prime factorization of 6⁹⁹. Since the prime factorization of 6 is 2*3,
6⁹⁹ = (2 * 3)⁹⁹
2⁹⁹3⁹⁹.
Therefore, using the formula for counting factors, this number has
(99 + 1)(99 + 1)
= (100)(100)
= 10000 factors.
Read more about Counting Factors.
Problem #10: By definition, the arithmetic mean before including x is
(1 + 2 + 8 + 10)/4
and the arithmetic mean after including x is
(1 + 2 + 8 + 10 + x)/5.
According to the problem, the old arithmetic mean increased by .75 is the new arithmetic mean, so you can write and solve the following equation:
(1 + 2 + 8 + 10)/4 + .75= (1 + 2 + 8 + 10 + x)/5
21/4 + .75 = (21 + x)/5
5.25 + .75 = (21 + x)/5
6 = (21 + x)/5
30 = 21 + x
9 = x.
Read more about The Arithmetic Mean in Equations.