Post on 17-Jan-2018
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Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
Every one of these problems can be set up with the same format.
100%
ofis
Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
Every one of these problems can be set up with the same format :
100%
ofis
To set up the problem, go thru the given information and insert known values into the format…
100%
ofis
Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
Every one of these problems can be set up with the same format :
100%
ofis
To set up the problem, go thru the given information and insert known values into the format…
In our example above…23 is what percent of 50
100%
of23
Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
Every one of these problems can be set up with the same format :
100%
ofis
To set up the problem, go thru the given information and insert known values into the format…
In our example above…23 is what percent of 50
100%
5023
Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
Every one of these problems can be set up with the same format :
100%
ofis
To set up the problem, go thru the given information and insert known values into the format…
In our example above…23 is what percent of 50
100%
5023
It’s now a basic proportion and you find what is missing…in this case %
Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
100%
5023
A short cut for solving proportions, one that gets all the Algebra OUT is this :1. Begin with the partner of the unknown ( red box )2. Move around the outside the proportion ( green box )3. Divide, then multiply to get the unknown
Divide
Multiply
Percent – Base and rate Problems
You might know the “base / rate” problems as “is / of” problems. For example, 23 is what percent of 50 ?
100%
5023
A short cut for solving proportions, one that gets all the Algebra OUT is this :1. Begin with the partner of the unknown ( red box )2. Move around the outside the proportion ( green box )3. Divide, then multiply to get the unknown
Divide
Multiply
%462322350100
Percent – Base and rate Problems
EXAMPLE # 2 : What is 35 % of 80
Percent – Base and rate Problems
EXAMPLE # 2 : What is 35 % of 80
100%
ofis
Percent – Base and rate Problems
EXAMPLE # 2 : What is 35 % of 80
10035
80is
100%
ofis
Percent – Base and rate Problems
EXAMPLE # 2 : What is 35 % of 80
10035
80is
100%
ofis
Divide
Multiply
28358.03510080
Percent – Base and rate Problems
EXAMPLE # 3 : 15 is 20 % of what number
100%
ofis
Percent – Base and rate Problems
EXAMPLE # 3 : 15 is 20 % of what number
10020
of15
100%
ofis
Divide
Multiply
7510075.01002015
APPLICATION PROBLEMS
In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.
APPLICATION PROBLEMS
In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.
EXAMPLE : And inspector rejects 12 out of 250 parts due to being out of tolerance. What percentage of parts were rejected ?
APPLICATION PROBLEMS
In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.
EXAMPLE : And inspector rejects 12 out of 250 parts due to being out of tolerance. What percentage of parts were rejected ?SOLUTION :
What is the problem asking us to find ?
APPLICATION PROBLEMS
In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.
EXAMPLE : And inspector rejects 12 out of 250 parts due to being out of tolerance. What percentage of parts were rejected ?SOLUTION :
What is the problem asking us to find ?The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere.
APPLICATION PROBLEMS
In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.
EXAMPLE : And inspector rejects 12 out of 250 parts due to being out of tolerance. What percentage of parts were rejected ?SOLUTION :
What is the problem asking us to find ?The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere.
I can see the “of” easily…
APPLICATION PROBLEMS
In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.
EXAMPLE : And inspector rejects 12 out of 250 parts due to being out of tolerance. What percentage of parts were rejected ?SOLUTION :
What is the problem asking us to find ?The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere.
I can see the “of” easily…Which leaves the 12 as the “is”…
APPLICATION PROBLEMS
In most percent application problems, you have to determine what the problem is asking for and then use the is / of proportion to solve.
EXAMPLE : And inspector rejects 12 out of 250 parts due to being out of tolerance. What percentage of parts were rejected ?SOLUTION :
What is the problem asking us to find ?The problem is asking for the % of rejects, so the “is” and “of” part is in the wording somewhere.
I can see the “of” easily…Which leaves the 12 as the “is”…
Now just solve like we did before…
Divide
Multiply
100÷250×12=4.8 %
APPLICATION PROBLEMS
EXAMPLE #2 :A motor is said to be 80% efficient if the output power delivered is 80% of the input power received. How many horsepower does a motor have if it is 80% efficient with a 6.20 horsepower output.
APPLICATION PROBLEMS
EXAMPLE #2 :A motor is said to be 80% efficient if the output power delivered is 80% of the input power received. How many horsepower does a motor have if it is 80% efficient with a 6.20 horsepower output.
SOLUTION :
Again we can quickly identify one part of the proportion… 80 % goes in the “%” spot.
APPLICATION PROBLEMS
EXAMPLE #2 :A motor is said to be 80% efficient if the output power delivered is 80% of the input power received. How many horsepower does a motor have if it is 80% efficient with a 6.20 horsepower output.
SOLUTION :
Again we can quickly identify one part of the proportion… 80 % goes in the “%” spot.
Since the motor IS 80% efficient, the known output ( 6.20 hp ) goes in the “is” spot…
APPLICATION PROBLEMS
EXAMPLE #2 :A motor is said to be 80% efficient if the output power delivered is 80% of the input power received. How many horsepower does a motor have if it is 80% efficient with a 6.20 horsepower output.
SOLUTION :
Again we can quickly identify one part of the proportion… 80 % goes in the “%” spot.
Since the motor IS 80% efficient, the known output ( 6.20 hp ) goes in the “is” spot…
Once again follow the rules for solving the unknown …
APPLICATION PROBLEMS
EXAMPLE #3 :By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ?
APPLICATION PROBLEMS
EXAMPLE #3 :By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ?
SOLUTION : 𝑖𝑠𝑜𝑓 =
%100
APPLICATION PROBLEMS
EXAMPLE #3 :By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ?
SOLUTION : 𝑖𝑠𝑜𝑓 =
135100
When production increases or decreases, the beginning percent is always 100%. An increase will add to the 100%, a decrease will subtract from the 100%. Since production increased by 35%, our new production rate is
100% + 35% = 135%.
APPLICATION PROBLEMS
EXAMPLE #3 :By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ?
SOLUTION : 270𝑜𝑓 =
135100
When production increases or decreases, the beginning percent is always 100%. An increase will add to the 100%, a decrease will subtract from the 100%. Since production increased by 35%, our new production rate is
100% + 35% = 135%.
The new number of pieces is a result of the increase or decrease in production and must be placed in the proportion relative to the increase or decrease.
APPLICATION PROBLEMS
EXAMPLE #3 :By replacing high – speed cutters with carbide cutters, a machinist increases productivity by 35%. Using carbide cutters, 270 pieces per day are produced. How many pieces per day were produced by the steel cutters ?
SOLUTION : 270𝑜𝑓 =
135100
When production increases or decreases, the beginning percent is always 100%. An increase will add to the 100%, a decrease will subtract from the 100%. Since production increased by 35%, our new production rate is
100% + 35% = 135%.
The new number of pieces is a result of the increase or decrease in production and must be placed in the proportion relative to the increase or decrease.
Solve for unknown :