1. Colegio San Patricio
1st Period Review WITH ANSWERS
School Year 2010 – 2011
Name ___________________________________________________ List No._________
Date_____________________________________________________ 8th_________
I. Working with real numbers
3(4 − 8)
1) 8 − (−10) + 5 − 20 − 10 2) (−5) + + 10 ÷ 5
2
2
8 + 10 + 5 – 20 -10= 25 + 3 (-4) + 2 = 25 -12 + 2 = 25 -6+2= 21
23-20-10= 3 – 10 = -7 2 2
3) 15 ÷ 3 • 3 + 2 • (5 − 2) + 8 ÷ 4 3) (20 − 15) • [ 20 ÷ 2 − ( 2 • 2 + 2)]
5 *3+2(3)+2= 15+6+2= 23 (5)(10-(4+2))= 5 (10-6) = 5 (4) = 20
II. Adding and Subtracting Polynomials
4) (15 x + 3 x + 2 x − 1) + (12 x − 3x + 2 x − 4 x + 3)
6 4 6 3 2
5) 3x5 + 6x3 - 2x2 + x - 2
27x6+3x4-3x3+2x2-2x+2 + 2x5 - 3x4 - 6x3 + 4x2 - 3x + 5
5x5 – 3x4+2x2 -2x+3
6) (7 x − x + x − 1) − (2 x + 3x + 6) 7) (3x − x − 7) − (5 x + 5 − x + 2 x)
3 2 2 2 3 2
7x3-x2+x-1-2x2-3x-6= 3x2 – x - 7- 5x3 – 5 + x2 - 2x=
7x3 - 3x2 - 2x – 7 -5x3 + 4x2 - 3x - 12
8) Subtract (5 x + 3 x − 7) from (−1 + 7 x − 2 x)
2 2
-1+7x2 - 2x – (5x + 3x2 -7) =
4x2 - 7x +6
9) Express the perimeter of the figure in simplest form.
2n-1
5n2+3n-4 Perimeter = addition of all sides =
2n-1 P = 2n2 + 2n + 5
-6n+8
-3n2+n+3
III. Drawing and classifying angles
10) Draw the fallowing angles and classify.
a) 65° b) 135° c) 260°
Type: _acute_ Type _obtuse_ Type_reflex_

2. 11) Measure the following angles
288° (reflex) 144° (obtuse) 83° (acute)
12) What is the complement of an angle that measures 65° ?
65° + 25° = 90°
13) What is the supplement of an angle that measures 120° ?
120° + 60° = 180°
14) Two angles are supplementary. The measure of one angle is twice as large as the
measure of the other. Find the measure in degrees of each angle.
X + 2x =180°
2X= 120° X= 60°
3x = 180
X= 180/3 = 60
15) Line MN and line RS intersect in point t (vertex)
a) If m ‹ RTM = 5x and m ‹ NTS = 3x + 10 , find ‹ RTM
M 5x = 3x +10 m < RTM = 5(5) = 25°
S 5x-3x=10
5x
t 3x + 10 2x=10
R
X= 5
N
b) If m ‹ MTS = 4x – 60 and m ‹ NTR = 2x , find ‹ MTS
M
4x - 60 S
4x-60 =2x m<MTS = 4(30) - 60
t 4x-2x= 60 = 60°
R 2x 2x = 60
N X= 30
c) If m ‹ RTM = 7x + 16 and m ‹ NTS = 3x + 48 , find ‹ NTS
M 7x+16= 3x+48 m<NTS= 3(8)+48
S
7x + 16 7x-3x= 48 -16 = 24 +48 = 72°
t 3x + 48 4x= 32
R X= 8
N

3. 16) 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
Remember: Angles 3 & 7 are CORRESPONDING, 66° 114°
1 4 66° 114°
5 8
so they are EQUAL, or have the SAME measure. 114° 66° 66°
2 3 6
114° 7
2x +40 = 3x + 27
40 -27 = 3x – 2x 2(13) +40= 26 + 40= 66°
13 = x
b) m ‹ 4 = 3x + 40 and m ‹ 5 = 2x
Remember: Angles 4 & 5 are ALLIED Interior,
56° 124°
So they are SUPPLEMENTARY, or both make up a 180°. 56° 1 124°
4 5 8
56°
3x + 40 + 2x = 180° m<4= 3(28)+40= 84+40= 124° 2
124° 3 6
124° 7
56°
5x= 180 – 40 m<5= 2(28) = 56°
5x = 140
X= 28
IV. Ratios, Rate, Proportions and Scale Factors
17) Find the amount of tax in a Television that costs $650 with a tax rate of 8%.
$650 - 100% 650 (8) = 5200 /100 = $52
X 8% or 650 (.08) = $52
18) During a sale a store offers a discount of 25% off any purchase. What is the regular
price of a dress that a customer purchased for $73.50?
x 100% 73.50(100) / 75 = $98
$73.50 75% or x – (0.25x) = $73.50
0.75x = 73.50
X= 73.50/0.75 = $98
19) Last year Maria’s rent was $600 per month. This year, her rent increased to $630 per
month. What was the percent of increase in her rent?
$600 - 100% 630*100=63000/600= 105%
$630 x so the rent increase a 5%.
20) If the sales tax on $150 is $7.50 what is the percent of the sales tax?
$150 100% 7.50*100/150 = 5%
$7.50 x
21) Can an 8 inch by 12 inch photograph be reduced to 3 inch by 5inch photograph?
Explain why or why not?
No, because they are NOT using the same scale factor to reduce the picture.
8 ≠ 3
12 5

4. 22) The ratio of two numbers is 1:4, and the sum of these numbers is 40. Find the numbers.
1: 4 = 1+4 = 5, so 40 ÷5 = 8
So, 8: 32 (1= 8, then 4*8= 32)