- Motion problems
- Coin problems
- Consecutive integer problems
- Solving and graphing inequalities
- Solving e quations
- Ratio Problems
- Percent problems
- Percent change problems
- Properties of numbers
- Adding and subtracting polynomials
Thursday, November 19, 2009
QUARTERLY TEST NOTICE
NOTICE: Quarterly test will be administered next Tuesday, November 24. Topics to be included but not limited to are as follows;
Aim: Review all motion problems




Example: Two cars start from the same point at the same time and travel in opposite directions. The slow car travels at 28 mph,and the fast car travels at 35 mph.
In how many hours will the cars be 252 miles apart?
Given is R for each car and the total distance travelled by both cars going in opposite directions.
Find T, the time, in hours travelled that each car travelled.
Solution: Start by defining a Let Statement:
Let x = the number of hours traveled by each car. Both cars are travelling for the same time T.
Let Slow car distance be D1 and fast car distance be D2
For Car 1 D1 = R1T For Car 2 D2 = R2T
D1 + D2 = Total Distance Now substituting RT for distances D1 and D2
R1T + R2T = Total Distance
28mph x X + 35mph x X = 252 miles
28X + 35X = 252 miles.
63X = 252
X= 4 hrs
Practice Exercise:
http://kutasoftware.com/FreeWorksheets/Distance%20Rate%20Time%20Word%20Problems.
Lesson Review:
Example problem: How far can a man drive out into the country at the average rate of 40 mph and return over the same road at the average rate 30 mph if he travels a total of 7 hours?
Need to find T then substitute in RT to find distance in D=RT
Tout + T back = 7 hrs and Distance out is same as the Distance back
Tout + Tback = 7hrs.
Let Tout = X
X +Tback = 7hrs
subtract X from both sides:
Tback = 7hrs - X
So: T out = X and T back = 7-X
Dout = RT = 40mph x X
Dback = RT 30mph x (7-X)
If Dout = Dback
Then 40X = 30 (7-X)
40X = 210 -30X
40X + 30X =210
70X = 210
X=3 hrs
Dout = RT = 40(3) = 120 miles
as a check Dout should = Dback
Dback = RT = 30 (7-x) = 30(7-3) = 30(4) = 120 miles
Practice Exercise:http://kutasoftware.com/FreeWorksheets/Distance%20Rate%20Time%20Word%20Problems.pdf
Example problem: How far can a man drive out into the country at the average rate of 40 mph and return over the same road at the average rate 30 mph if he travels a total of 7 hours?
Need to find T then substitute in RT to find distance in D=RT
Tout + T back = 7 hrs and Distance out is same as the Distance back
Tout + Tback = 7hrs.
Let Tout = X
X +Tback = 7hrs
subtract X from both sides:
Tback = 7hrs - X
So: T out = X and T back = 7-X
Dout = RT = 40mph x X
Dback = RT 30mph x (7-X)
If Dout = Dback
Then 40X = 30 (7-X)
40X = 210 -30X
40X + 30X =210
70X = 210
X=3 hrs
Dout = RT = 40(3) = 120 miles
as a check Dout should = Dback
Dback = RT = 30 (7-x) = 30(7-3) = 30(4) = 120 miles
Practice Exercise:http://kutasoftware.com/FreeWorksheets/Distance%20Rate%20Time%20Word%20Problems.pdf
Monday, November 16, 2009

Homework: Handout sheet side 2
Lesson Review:
Opposite motion problems: Remember that Distance is = to the rate or speed times the time D = RT in all of the following problems D is replace by the values of RT when known or by one of the values and a variable either R or T.
Example: Two cars start from the same point at the same time and travel in opposite directions. The slow car travels at 28 mph,and the fast car travels at 35 mph.
In how many hours will the cars be 252 miles apart?
Given is R for each car and the total distance travelled by both cars going in opposite directions.
Find T, the time, in hours travelled that each car travelled.
Solution: Start by defining a Let Statement:
Let x = the number of hours traveled by each car. Both cars are travelling for the same time T.
Let Slow car distance be D1 and fast car distance be D2
For Car 1 D1 = R1T For Car 2 D2 = R2T
D1 + D2 = Total Distance Now substituting RT for distances D1 and D2
R1T + R2T = Total Distance
28mph x X + 35mph x X = 252 miles
28X + 35X = 252 miles.
63X = 252
X= 4 hrs
Practice Exercise:
http://kutasoftware.com/FreeWorksheets/Distance%20Rate%20Time%20Word%20Problems.pdf

Homework: Handout sheet all problems
Lesson Review:
Opposite motion problems: Remember that Distance is = to the rate or speed times the time D = RT in all of the following problems D is replace by the values of RT when known or by one of the values and a variable either R or T.
Example: Two cars start from the same point at the same time and travel in opposite directions. The slow car travels at 28 mph,and the fast car travels at 35 mph.
In how many hours will the cars be 252 miles apart?
Given is R for each car and the total distance travelled by both cars going in opposite directions.
Find T, the time, in hours travelled that each car travelled.
Solution: Start by defining a Let Statement:
Let x = the number of hours traveled by each car. Both cars are travelling for the same time T.
Let Slow car distance be D1 and fast car distance be D2
For Car 1 D1 = R1T For Car 2 D2 = R2T
D1 + D2 = Total Distance Now substituting RT for distances D1 and D2
R1T + R2T = Total Distance
28mph x X + 35mph x X = 252 miles
28X + 35X = 252 miles.
63X = 252
X= 4 hrs
Practice Exercise:
http://kutasoftware.com/FreeWorksheets/Distance%20Rate%20Time%20Word%20Problems.pdf
Thursday, November 12, 2009

post test review handout above.
Lesson Review:
Remember that the number of coins times the value of each coin in cents, is = total value of the coins in cents.
Example:
Bill has 4 times as many quarters as dimes. In all he has $2.20. How many coins of each type does she have?
Solution:
Let x= number of dimes (x is always = to the coin in the problems that follows the word "than" or "as") then, 4x is = to the number of quarters
The value of a dime is 10 cents and the value of a quarter is 25 cents
The total value of all the coins is $2.20 cents which is the same as 220 cents after we multiply $2.20 times 100 cents.
Now we write the equation: Dimes + Quarters = $2.20
x + 4x = $2.20 Now I can't add a dime and a quarter without converting then to cents, so:
(10)x + 25(4x) = $2.20 (100)
1 0 x + 100x = 220
110x= 220
x= 2 , so there are 2 dimes.
Therefore 4x quarters is 4(2) which = 8 quarters.
Practice Exercises:
http://mathforum.org/dr.math/faq/faq.coins.html
http://www.purplemath.com/modules/coinprob.htm
http://www.onlinemathlearning.com/coin-problems.html
Lesson Review:
Remember that the number of coins times the value of each coin in cents, is = total value of the coins in cents.
Example:
Bill has 4 times as many quarters as dimes. In all he has $2.20. How many coins of each type does she have?
Solution:
Let x= number of dimes (x is always = to the coin in the problems that follows the word "than" or "as") then, 4x is = to the number of quarters
The value of a dime is 10 cents and the value of a quarter is 25 cents
The total value of all the coins is $2.20 cents which is the same as 220 cents after we multiply $2.20 times 100 cents.
Now we write the equation: Dimes + Quarters = $2.20
x + 4x = $2.20 Now I can't add a dime and a quarter without converting then to cents, so:
(10)x + 25(4x) = $2.20 (100)
1 0 x + 100x = 220
110x= 220
x= 2 , so there are 2 dimes.
Therefore 4x quarters is 4(2) which = 8 quarters.
Practice Exercises:
http://mathforum.org/dr.math/faq/faq.coins.html
http://www.purplemath.com/modules/coinprob.htm
http://www.onlinemathlearning.com/coin-problems.html
Wednesday, November 4, 2009
Aim: Test # 3 Review Sheet 2 ANSWER KEY
.
Homework; Complete Test Review Sheet # 2 (Answers on sheets above)
Test # 3 Tomorrow
Topics to be included are:
Percent problems
Percent change problems
Solving and graphing one step, two step inequalities
Solving inequalties with fractional terms
Solving ratio and proportion problems
Solving consecutive, even and odd integers problems
Equation solving
Properties of real numbers
Adding and subtracting polynomials
Monday, November 2, 2009
AIM: TEST 3 REVIEW


NOTICE: Test #3 this Thursday, November 5.
Topics to be included are:
Percent problems
Percent change problems
Solving and graphing one step, two step inequalities
Solving inequalties with fractional terms
Solving ratio and proportion problems
Solving consecutive, even and odd integers problems
Equation solving
Properties of real numbers
Adding and subtracting polynomials
Test reviews are above (click on sheets to enlarge)
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