Commission and Profit 9-6
Some sales jobs pay an amount based on how much you sell. This amount is called a commission.
Like a discount, the commission can be expressed as a percent or as an amount of money.
amount of commission = percent of commission X total sales.
Using the examples from our textbook,
Maria sold $42,000 word of insurance in January. If her commission is 3% of the total sales, what was the amount of her commission in January?
amount of commission = percent X total sales
0.03 X 42,000 = 1260
Her commission was $1,260.
Profit is the difference between total income and total operating costs.
profit = total income – total costs
The percent of profit is the percent of total income that is profit
percent of profit = profit/total income
A shoe store had an income of $8600 and operating costs of $7310. What percent of the store's income was profit?
profit= income- total costs = 8600 -7310 = 1290
percent of profit = profit/total income = 1290/8600 = 0.15
So the percent of profit was 15%.
Practice finding 10%-- its easy--- just move the decimal over one place.
We practiced finding 20%. Just double what you got for 10%.
MATH AT WORK:
Caterer
A caterer provides food for parties, weddings, bar/bat mitzvahs, and other events. Caterers plan the menu, buy the ingredients, and cook the food. Often they provide seating and music as well. For each event, a caterer determines the cost per guest. The catering business requires a thorough knowledge of rations, proportions, and percents.
Wednesday, May 25, 2011
Tuesday, May 24, 2011
Math 6 Honors (Period 6 and 7)
Discount and Markup 9-5
A discount is a decrease in the price of an item. A markup is an increase in the price of an item. Both of these changes can be expressed as an amount of money or as a percent of the original price of the item. A store may announce a discount of $3 off the original price of $30 basketball, or a discount of 10%
A warm-up suit that sold for $42.50 is on sale at a 12% discount. What is the sale price?
Method 1: Use the formula
amount of change = percent of change X original amount
= 12% X $42.50
therefore the discount is 0.12 X 42.50 or 5.10
The amount of discount is $5.10
The sale price is 42.50 – 5.10 = $37.40
Method 2: Since the discount is 12%, the sale price is 100% - 12% = 88%.
The sale price is 0.88 X 42.50 = $ 37.40
When you know the amount of discount you subtract to find the new price. When dealing with a markup you add to find the new price.
The price of a new car model was marked up 6% over the previous year’s model. If the previous year’s model sold for $7800, what is the cost of the new car? {and what kind of a car could that be?}
Method 1: Use the formula
amount of change = percent of change X original amount
= 6% X 7800
Therefore the markup is 0.06 X7800= $468
The new price is 7800 + 468 = $8268
Method 2: Since the markup is 6% the new price is 100% + 6% or 106% of the original price. so the new price is 1.06 X7800 = $8268
This year a pair of ice skates sells for $46 after a 15% mark up over last year’s price. What was last year’s price?
This year’s price is 100 + 15 or 115% of last year’s price. Let n present last year’s price
46 = (115/100)n
46 = 1.15n
46/.15 = 1.15n/1.115
40 = n
So last year’s price was $40.
A department store advertised eclectic shavers at a sale price of $36.
If this is a 20% discount, what was the original price?
The sale price is 100 - 20 or 80% of the original price. Let n represent the original price.
36 = (80/100)n
36 = .8n
36/.8 = .8n/.8
45 = n
The original price was $45.
Check to see that your answers are logical and reasonable.
Try these: A service station (that’s gas station, now—they no longer provide service!!) give cash customers a 5% discount on the price of gasoline. If gasoline regularly sells for $3.00 a gallon, what is the discounted price?
A store marks up the price of a $5 item to $12. What is the percent of markup?
A discount is a decrease in the price of an item. A markup is an increase in the price of an item. Both of these changes can be expressed as an amount of money or as a percent of the original price of the item. A store may announce a discount of $3 off the original price of $30 basketball, or a discount of 10%
A warm-up suit that sold for $42.50 is on sale at a 12% discount. What is the sale price?
Method 1: Use the formula
amount of change = percent of change X original amount
= 12% X $42.50
therefore the discount is 0.12 X 42.50 or 5.10
The amount of discount is $5.10
The sale price is 42.50 – 5.10 = $37.40
Method 2: Since the discount is 12%, the sale price is 100% - 12% = 88%.
The sale price is 0.88 X 42.50 = $ 37.40
When you know the amount of discount you subtract to find the new price. When dealing with a markup you add to find the new price.
The price of a new car model was marked up 6% over the previous year’s model. If the previous year’s model sold for $7800, what is the cost of the new car? {and what kind of a car could that be?}
Method 1: Use the formula
amount of change = percent of change X original amount
= 6% X 7800
Therefore the markup is 0.06 X7800= $468
The new price is 7800 + 468 = $8268
Method 2: Since the markup is 6% the new price is 100% + 6% or 106% of the original price. so the new price is 1.06 X7800 = $8268
This year a pair of ice skates sells for $46 after a 15% mark up over last year’s price. What was last year’s price?
This year’s price is 100 + 15 or 115% of last year’s price. Let n present last year’s price
46 = (115/100)n
46 = 1.15n
46/.15 = 1.15n/1.115
40 = n
So last year’s price was $40.
A department store advertised eclectic shavers at a sale price of $36.
If this is a 20% discount, what was the original price?
The sale price is 100 - 20 or 80% of the original price. Let n represent the original price.
36 = (80/100)n
36 = .8n
36/.8 = .8n/.8
45 = n
The original price was $45.
Check to see that your answers are logical and reasonable.
Try these: A service station (that’s gas station, now—they no longer provide service!!) give cash customers a 5% discount on the price of gasoline. If gasoline regularly sells for $3.00 a gallon, what is the discounted price?
A store marks up the price of a $5 item to $12. What is the percent of markup?
Algebra (Period 1)
Motion Word Problems 8-5 & 10-7
rt=d problems
(rate)(time)=distance
Back in Pre-Algebra, these were fairly simple word problems:
1) If you go 60 mph for 3 hours, how far have you gone? (180 miles)
2) You've gone 100 miles in 2 hours. What was your average speed?
(100/2 = 50 mph)
3)You've driven 1000 miles at an average speed of 25 mph. How long did it take you? (1000/25 = 40 hours)
Now the problems become more difficult. Usually they involve 2 cars, trains, planes, etc. One car is the "slow" car and the other is the "fast" car.
Just like the mixture problems, it helps to make a matrix and also draw a picture to help you understand the words.
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, IN THE SAME DIRECTION, AT DIFFERENT TIMES, WHEN WILL THE 2ND CAR CATCH UP WITH THE FIRST CAR?
2 cars leave 3 hours apart. One travels 72 mph. The other travels 120 mph. The slower car leaves first. When with the faster car catch up with the slower car?
Use the set up forms from class... you can find more blank forms online!!
SLOW 72 t 72t
FAST 120 t - 3 120(t - 3)
(fast car left 3 hours later so driving 3 less hours or t - 3)
At the point when the fast car catches and overtakes the slow car, what is true of the distances of the 2 cars at that exact moment???
They are equal!
WHAT IS THE EQUATION?
Set the 2 cars' distances equal:
72t = 120(t - 3)
72t = 120t - 360
-48t = -360
t = 7.5 hours (slow car's time on the road)
t - 3 = 7.5 - 3 = 4.5 hours (fast car's time on the road)
CHECK:
The 2 cars should have traveled the same distance because one car catches up with the other car:
slow: (72)(7.5) = 540
fast: (120)(4.5) = 540
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT DIRECTIONS, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?
2 cars leave going in different directions. One travels 60 mph. The other travels 50 mph. In how many hours will the cars be 605 miles apart?
CAR (RATE) (TIME) = DISTANCE
SLOW 50 t 50t
FAST 60 t 60t
(They travel the same amount of time)
What is true of the distances the 2 cars travel?
Together they travel 605 miles because they are 605 miles apart.
WHAT IS THE EQUATION?
Set the 2 cars' distances as a SUM to 605.
50t + 60t = 605
110t = 605
t = 5.5 hours
So in 5 1/2 hours the 2 cars will be 605 miles apart.
CHECK:
If you plug in 5.5 hours for each car to find each cars distances, they should add to 605 miles.
slow: (50)(5.5) = 275 miles traveled
fast: (60)(5.5) = 330 miles traveled
275 + 330 = 605 miles
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN SAME DIRECTION, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?
2 cars leave going in the same direction. One travels 45 mph. The other travels 35 mph. In how many hours will the cars be 15 miles apart?
CAR (RATE) (TIME) = DISTANCE
SLOW 35 t 35t
FAST 40 t 40t
(They travel the same amount of time)
What is true of the distances the 2 cars travel?
They are getting further and further apart as the minutes go by.
The DIFFERENCE of the 2 cars is 15 miles after a certain amount of time.
WHAT IS THE EQUATION?
40t - 35t = 15
5t = 15
t = 3 hours
So in 3 hours the 2 cars will be 15 miles apart.
CHECK:
If you plug in 3 hours for each car to find each cars distances, their distances should subtract to 15 miles.
slow: (35)(3) = 105 miles traveled
fast: (40)(3) = 120 miles traveled
120 - 105 = 15 miles
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT OR SAME DIRECTION, IN THE SAME AMOUNT OF TIME, EACH TRAVELS A DIFFERENT DISTANCE, WHAT IS THEIR SPEED?
One car travels 20 mph faster than the other car. One car travels 240 miles while the other travels 180 miles. Find their average speeds.
CAR (RATE) (TIME) = DISTANCE
SLOW r 180/r 180
FAST r + 20 240/(r + 20) 240
This time you have the distance and know that the faster car is 20 mph faster than the slower car. To find the time for each car, use the fact that d/r = t so divided each car's distance by their rates.
WHAT IS THE EQUATION?
The times of these 2 cars is equal (left at same time and stopped at same time) so set their times equal:
180/r = 240/(r + 20)
Multiply both sides equally by the LCM:
(r)(r + 20)(180/r) = (r)(r + 20)240/(r + 20)
Cross cancel:
180(r + 20) = 240(r)
180r + 3600 = 240r
3600 = 60r
r = 60 mph (the slower car)
r + 20 = 60 + 20 = 80 mph (the faster car)
CHECK:
If you plug in the speeds, you should find that both cars traveled the same amount of time:
slow: (60)t = 180 t = 3 hours
fast: (80)t = 240 t = 3 hours
A BOAT/PLANE TRAVELS WITH THE CURRENT ON THE DEPARTING LEG OF THE JOURNEY AND TRAVELS AGAINST THE CURRENT ON THE RETURN LEG. WHAT IS THE SPEED OF THE BOAT/PLANE IN STILL WATER/AIR?
A boat travels with a current of 6 mph for 3 hours and then returns home against the same current. The trip home takes 5 hours. What is the speed of the boat in still water?
BOAT/PLANE (RATE) (TIME) = DISTANCE
WITH CURRENT r + 6 3 3(r + 6)
AGAINST CURRENT r - 6 5 5(r - 6)
r is the speed in still water and 6 is the speed of the current. The boat with need less time to go the same distance with the current than against it.
WHAT IS THE EQUATION?
The distance to the boat's location and the distance home must be equal.
Set the distances equal:
3(r + 6) = 5(r - 6)
3r + 18 = 5r - 30
48 = 2r
r = 24 mph (speed in still water)
r + 6 = 24 + 6 = 30 mph (speed with the current)
r - 6 = 24 - 6 = 18 mph (speed against the current)
CHECK:
If you plug in the speed with and against the current with the hours traveled, you get the distances to and home. Those distances should be equal:
3(30) = 5(18)
90 = 90
rt=d problems
(rate)(time)=distance
Back in Pre-Algebra, these were fairly simple word problems:
1) If you go 60 mph for 3 hours, how far have you gone? (180 miles)
2) You've gone 100 miles in 2 hours. What was your average speed?
(100/2 = 50 mph)
3)You've driven 1000 miles at an average speed of 25 mph. How long did it take you? (1000/25 = 40 hours)
Now the problems become more difficult. Usually they involve 2 cars, trains, planes, etc. One car is the "slow" car and the other is the "fast" car.
Just like the mixture problems, it helps to make a matrix and also draw a picture to help you understand the words.
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, IN THE SAME DIRECTION, AT DIFFERENT TIMES, WHEN WILL THE 2ND CAR CATCH UP WITH THE FIRST CAR?
2 cars leave 3 hours apart. One travels 72 mph. The other travels 120 mph. The slower car leaves first. When with the faster car catch up with the slower car?
Use the set up forms from class... you can find more blank forms online!!
SLOW 72 t 72t
FAST 120 t - 3 120(t - 3)
(fast car left 3 hours later so driving 3 less hours or t - 3)
At the point when the fast car catches and overtakes the slow car, what is true of the distances of the 2 cars at that exact moment???
They are equal!
WHAT IS THE EQUATION?
Set the 2 cars' distances equal:
72t = 120(t - 3)
72t = 120t - 360
-48t = -360
t = 7.5 hours (slow car's time on the road)
t - 3 = 7.5 - 3 = 4.5 hours (fast car's time on the road)
CHECK:
The 2 cars should have traveled the same distance because one car catches up with the other car:
slow: (72)(7.5) = 540
fast: (120)(4.5) = 540
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT DIRECTIONS, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?
2 cars leave going in different directions. One travels 60 mph. The other travels 50 mph. In how many hours will the cars be 605 miles apart?
CAR (RATE) (TIME) = DISTANCE
SLOW 50 t 50t
FAST 60 t 60t
(They travel the same amount of time)
What is true of the distances the 2 cars travel?
Together they travel 605 miles because they are 605 miles apart.
WHAT IS THE EQUATION?
Set the 2 cars' distances as a SUM to 605.
50t + 60t = 605
110t = 605
t = 5.5 hours
So in 5 1/2 hours the 2 cars will be 605 miles apart.
CHECK:
If you plug in 5.5 hours for each car to find each cars distances, they should add to 605 miles.
slow: (50)(5.5) = 275 miles traveled
fast: (60)(5.5) = 330 miles traveled
275 + 330 = 605 miles
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN SAME DIRECTION, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?
2 cars leave going in the same direction. One travels 45 mph. The other travels 35 mph. In how many hours will the cars be 15 miles apart?
CAR (RATE) (TIME) = DISTANCE
SLOW 35 t 35t
FAST 40 t 40t
(They travel the same amount of time)
What is true of the distances the 2 cars travel?
They are getting further and further apart as the minutes go by.
The DIFFERENCE of the 2 cars is 15 miles after a certain amount of time.
WHAT IS THE EQUATION?
40t - 35t = 15
5t = 15
t = 3 hours
So in 3 hours the 2 cars will be 15 miles apart.
CHECK:
If you plug in 3 hours for each car to find each cars distances, their distances should subtract to 15 miles.
slow: (35)(3) = 105 miles traveled
fast: (40)(3) = 120 miles traveled
120 - 105 = 15 miles
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT OR SAME DIRECTION, IN THE SAME AMOUNT OF TIME, EACH TRAVELS A DIFFERENT DISTANCE, WHAT IS THEIR SPEED?
One car travels 20 mph faster than the other car. One car travels 240 miles while the other travels 180 miles. Find their average speeds.
CAR (RATE) (TIME) = DISTANCE
SLOW r 180/r 180
FAST r + 20 240/(r + 20) 240
This time you have the distance and know that the faster car is 20 mph faster than the slower car. To find the time for each car, use the fact that d/r = t so divided each car's distance by their rates.
WHAT IS THE EQUATION?
The times of these 2 cars is equal (left at same time and stopped at same time) so set their times equal:
180/r = 240/(r + 20)
Multiply both sides equally by the LCM:
(r)(r + 20)(180/r) = (r)(r + 20)240/(r + 20)
Cross cancel:
180(r + 20) = 240(r)
180r + 3600 = 240r
3600 = 60r
r = 60 mph (the slower car)
r + 20 = 60 + 20 = 80 mph (the faster car)
CHECK:
If you plug in the speeds, you should find that both cars traveled the same amount of time:
slow: (60)t = 180 t = 3 hours
fast: (80)t = 240 t = 3 hours
A BOAT/PLANE TRAVELS WITH THE CURRENT ON THE DEPARTING LEG OF THE JOURNEY AND TRAVELS AGAINST THE CURRENT ON THE RETURN LEG. WHAT IS THE SPEED OF THE BOAT/PLANE IN STILL WATER/AIR?
A boat travels with a current of 6 mph for 3 hours and then returns home against the same current. The trip home takes 5 hours. What is the speed of the boat in still water?
BOAT/PLANE (RATE) (TIME) = DISTANCE
WITH CURRENT r + 6 3 3(r + 6)
AGAINST CURRENT r - 6 5 5(r - 6)
r is the speed in still water and 6 is the speed of the current. The boat with need less time to go the same distance with the current than against it.
WHAT IS THE EQUATION?
The distance to the boat's location and the distance home must be equal.
Set the distances equal:
3(r + 6) = 5(r - 6)
3r + 18 = 5r - 30
48 = 2r
r = 24 mph (speed in still water)
r + 6 = 24 + 6 = 30 mph (speed with the current)
r - 6 = 24 - 6 = 18 mph (speed against the current)
CHECK:
If you plug in the speed with and against the current with the hours traveled, you get the distances to and home. Those distances should be equal:
3(30) = 5(18)
90 = 90
Monday, May 23, 2011
Pre Algebra (Period 2 & 4)
Volume of Prisms & Cylinders 10-7
Remember when I said that the basic formula for area is A = bh
Well, the basic formula for volume of a prism or cylinder is:
V = BH
Where capital B = the area of the base of the prism/cylinder
capital H = the height of the prism/cylinder
THE LABEL IS ALWAYS CUBED!!!
Volume of a rectangular, square, or parallelogram prism:
V = BH
Volume = area of the base times the height of the prism
V = (bh)(H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of a triangular prism:
V = BH
Volume = area of the base times the height of the prism
V = (1/2 bh)(H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of a trapezoidal prism:
V = BH
Volume = area of the base times the height of the prism
V = [(average of the 2 bases)(h)](H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of a cylinder:
V = BH
Volume = area of the base times the height of the prism
V = (π r2)(H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of Pyramids, Cones & Spheres 10-9
To find the volume of a pyramid or cone is just as easy!!!
It's just V= 1/3 BH
That means just find the volume as if it was a prism or cone then just divide it by 3!!!
VOLUME OF A SPHERE (ball) (this one is different) V = 4/3 π r3
Remember when I said that the basic formula for area is A = bh
Well, the basic formula for volume of a prism or cylinder is:
V = BH
Where capital B = the area of the base of the prism/cylinder
capital H = the height of the prism/cylinder
THE LABEL IS ALWAYS CUBED!!!
Volume of a rectangular, square, or parallelogram prism:
V = BH
Volume = area of the base times the height of the prism
V = (bh)(H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of a triangular prism:
V = BH
Volume = area of the base times the height of the prism
V = (1/2 bh)(H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of a trapezoidal prism:
V = BH
Volume = area of the base times the height of the prism
V = [(average of the 2 bases)(h)](H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of a cylinder:
V = BH
Volume = area of the base times the height of the prism
V = (π r2)(H)
The lower case b and h are the base and height of the base (this is plane geometry!)
Volume of Pyramids, Cones & Spheres 10-9
To find the volume of a pyramid or cone is just as easy!!!
It's just V= 1/3 BH
That means just find the volume as if it was a prism or cone then just divide it by 3!!!
VOLUME OF A SPHERE (ball) (this one is different) V = 4/3 π r3
Sunday, May 22, 2011
Algebra (Period 1)
Khan Academy has some excellent You Tube videos to help you review basic Algebra I concepts. I suggest you watch at least some of these: :
How to :
Solve Equations using the Quadratic Equations Part 1
Solving Equations Using the Quadratic Equation Part 2
Solve Quadratic Equations by Completing the Square
How to :
Solve Equations using the Quadratic Equations Part 1
Solving Equations Using the Quadratic Equation Part 2
Solve Quadratic Equations by Completing the Square
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