1. Colegio San Patricio
2nd Period Math Review´s ANSWERS
School Year 2009 – 2010
Counting Techniques
5) The school’s cafeteria has new options. Now they have: 3 types of salads, 3 types of
hamburgers, 5 types of fruit beverages, and 5 dessert options. How many different
combinations you to choose from?
MULTIPLICATION Technique (you have 4 groups)
Salads = 3
Hamburgers = 3
Fruit Beverages = 5
Desserts = 5
3*3*5*5 = 225 different combinations
6) How many ways are there for choosing any 4 books from the library’s desk from the 12
books they are on the top of the desk?
COMBINATION Technique (you have 1 group = books & order in which you choose the books
does NOT matters.)
nCr = n! (until r) = 12C4 = 12*11*10*9 = 495 combinations
r! 4*3*2*1
7) In how many ways can a jury choose the best 3 essays from the top 20 essays they
received? (The first chosen person will have the 3rd
. place; the second one will have the 2nd
.
place, and the last person will be the 1st
. place.)
PERMUTATION Technique (you have 1 group = essays & order in which the jury selects the
essays DOES matters.)
nPr = n! (until r) = 20P3 = 20*19*18 = 6840 permutations
Grouping and Analyzing Data
8) The following data shows the grades of different students.
74 95 96 85 77 65
83 88 93 12 81 92
100 20 40 35 98 87
a) Create a class interval table (interval, tally, frequency). The first interval is 0-20
b) Graph the information using a frequency polygon graph.

2. a) Interval TableTable:
Intervals Tallies Frequency
0 - 20 II 2
21 - 40 II 2
41 - 60 0 0
61 - 80 III 3
81 - 100 IIIII IIIII II 11
b)Graph
Fre que nc y Po ly g o n Gra ph
0
2
4
6
8
10
12
0-20 21-40 41-60 61-80 81-100
Inte r v al s
Fr ecuency
Multiplying Polynomials
9) (x – 5) (3x + 7) 10) (3a - 2)(a2
-5 a + 5)
3x2
+ 7x – 15x – 35 3a3
– 15a2
+ 15a – 2a2
+ 10a - 10
3x2
– 8x – 35 3a3
– 17a2
+ 25a - 10
11) (6x – 5y)2
12) (5x + 6y) (5x – 6y)
36x2
– 60xy + 25y2
25x2
– 36y2
13) Find an expression for the area of a trapezoid with short base length x + 3, long base 2x +
9, and height is 10.
* Remember: Area = (B1 + b2)H = (x + 3 + 2x + 9) 10 = (3x + 12)10 = (3x + 12)5 = 15x + 60
of trapezoid 2 2 2
14) What is the area of the rectangle? ___b___
a) 7x² + 5xz + 2xy +yz
b) 10x2
+ 5xz + 2xy + yz
c) 7x² + 10xz + xy + 2yz
d) 10x² + 5xz + 2xy + 2yz²
(Remember: You need to multiply each side, because you have 2 binomials. (5x + y)(2x + z).)
Solid Geometry
5x
y
2x z

3. 15) Draw and answer the following
a) Pentagonal Prism Faces: 5 + 2 = 7
Edges: 5 * 3 = 15
Vertices: 5 * 2 = 10
b) Nonagon Pyramid Faces: 9 + 1 = 10
Edges: 9 * 2 = 18
Vertices: 9 + 1 = 10
16) Draw the different views for the following structure:
Front Left Side Right Side Top
17. Volume
Figure Formula Procedure
a) V = Ab*H
A = Pa
2
V = Ab*8cm = 30cm2
(8cm) = 240cm3
A = (10*3cm)2cm = 30cm2
2
b) V = Ab*H
A = Pa
2
V = Ab*12cm = 72cm2
(12cm) = 864cm3
A = (6*6cm)4cm = 72cm2
2
c) V = Ab*H
3
A = Pa
2
V = Ab*10cm = 37.5cm2
(10cm)= 125cm3
3 3
A = (5*5cm)3cm = 37.5cm2
2
Each side = 3cm
H = 8cm
a = 2cm
a = 4m
Each side = 6m
H = 12 m
a = 3cm
Each side = 5cm
H = 10cm

4. 18) What is the volume of a cube if the length is 6cm, width is 4cm, and it’s height is 11cm?
19) Find the volume of the following figure.
20) A cereal company decided to increase the height of its boxes by 30% and reduce the
width in order to maintain the same volume. What will be the new height and width if the
length stays the same?
Initially: Length = 20 cm
Width = 30cm
Height = 40cm
Answer:
Actual cereal’s box volume= (20cm)(30cm)(40cm) = 24,000cm3
.
NEW MEASURES =
L= 20cm (stays the same)
H= 52cm (it increased a 30% = 40 +30% = 40 + 12 = 52cm)
W= ?
* They what the SAME volume, so:
24,000cm3
=(20cm)(52cm)W =
24,000cm3
= 1040cm2
* W =
24,000cm3
/ 1040cm2
= W
W = 23.08cm.
So this is the new size of the width in order to keep the same volume of the boxes.
V = l*w*h = 8cm(5cm)(9cm) = 264cm3
V= (4cm*5cm*6cm) – (2cm*2cm*6cm)
= (120cm3
) – (24cm3
) = 96cm3
Remember: You need to “take away” the volume of the
empty space (the hole, which has the same deepness of the
complete figure). That’s why you subtract the volume of the
hole FROM the volume of the COMPLETE figure.

5. 21)Express the perimeter of the figure in simplest form.
Remember: Perimeter is the addition of ALL the sides of the figure. So, you just combine “like
terms”.
P= (2n-1)+(2n-1)+(-3n2
+n+3)+(-6n+8)+(5n2
+3n-4) = 2n2
+2n+5
22) The figure below shows two parallel lines cut by a transversal. In each exercise find the
measures of all eight angles under the given conditions.
a) m ‹ 3 = 2x + 40 and m ‹ 7 = 3x + 27
• ‹ 3 and ‹ 7 are CORRESPONDING angles, so they are EQUAL.
2x+40 = 3x+27
-27 + 40 = 3x – 2x
13 = x
So, if x= 13, the value of the angle is= 2(13) +40 = 26+40 = 66°
If ‹ 3 = 66°, then ‹ 4 = 114° (because they’re supplementary angles). Now you can fill the
rest of the angles.
b) m ‹ 4 = 3x + 40 and m ‹ 5 = 2x
• ‹ 4 and ‹ 5 are CO-INTERIOR or ALLIED angles, so they are SUPPLEMENTARY ANGLES.
3x+40 + 2x = 180°
5x + 40 = 180
5x = 140
X = 28
So, if x= 28, the values of the angles are= ‹ 4 = 3(28) +40 = 84+40 = 124° and ‹ 5= 2(28) = 56°
Now you can fill the rest of the angles.
2n-1
-6n+8
5n2
+3n-4
2n-1
-3n2
+n+3
1
2
4
3
5 8
6 7

6. 21)Express the perimeter of the figure in simplest form.
Remember: Perimeter is the addition of ALL the sides of the figure. So, you just combine “like
terms”.
P= (2n-1)+(2n-1)+(-3n2
+n+3)+(-6n+8)+(5n2
+3n-4) = 2n2
+2n+5
22) The figure below shows two parallel lines cut by a transversal. In each exercise find the
measures of all eight angles under the given conditions.
a) m ‹ 3 = 2x + 40 and m ‹ 7 = 3x + 27
• ‹ 3 and ‹ 7 are CORRESPONDING angles, so they are EQUAL.
2x+40 = 3x+27
-27 + 40 = 3x – 2x
13 = x
So, if x= 13, the value of the angle is= 2(13) +40 = 26+40 = 66°
If ‹ 3 = 66°, then ‹ 4 = 114° (because they’re supplementary angles). Now you can fill the
rest of the angles.
b) m ‹ 4 = 3x + 40 and m ‹ 5 = 2x
• ‹ 4 and ‹ 5 are CO-INTERIOR or ALLIED angles, so they are SUPPLEMENTARY ANGLES.
3x+40 + 2x = 180°
5x + 40 = 180
5x = 140
X = 28
So, if x= 28, the values of the angles are= ‹ 4 = 3(28) +40 = 84+40 = 124° and ‹ 5= 2(28) = 56°
Now you can fill the rest of the angles.
2n-1
-6n+8
5n2
+3n-4
2n-1
-3n2
+n+3
1
2
4
3
5 8
6 7