This scribe is all about paints used in a part of a room.
here is the problem:
if you can paint 2/5 of paint can, in a quarter of a room, how many paint can/s will be used when you will paint a whole room?
to answer this problem, one strategy can be used in solving this problem. it is ratio table. the picture shows how this problem will be solved. but the answer is not yet shown, it only showed how to get the whole room (1/4 x 4). to make 1/4 into a whole number which is one, the numerator and the denominator must be the same so since it was 1/4, we multiply it into 4 to get 4/4 or 1....one thing in ratio table is that when you do it in one side, do it on the other side. so we multiply the number 4 too, to answer the question.
this is to show how you multiply the fraction to 4... and 2 2/3 is the answer to this problem that can paint the whole room
this shows the answer..,.its done...the answer is
2 2/3
Thursday, January 31, 2008
Wednesday, January 30, 2008
Angelo's scribe
Today, Mr Harbeck was away (AWWWWWWWWWWW) so we just did fraction word problems.
I'll show you how I did three of them:
Explanation1: I needed to find out 3/8 of 24, so i first divided 8 to 24. The answer was 3, and that meant 3 is 1/8 of 24. So, to get 3/8 i had to times 1/8 three times. That's why i did 3 times 3.
Explanation2: I needed to find out 3/5 of 125, so i first divided 5 into 125. The answer was 25, and that meant 25 is 1/5 of 125. So, to get 3/5 i had to times 1/5 three times. That's why i did 25 times 3.
Explanation3: I needed to find 7/8 of 160, so i first divided 8 into 160. The answer was 20, and that meant 20 is 1/8 of 160. So, to get 7/8 i had to times 1/8 seven times. That's why i did 20 times 7.
This is how I did my questions. Hope you guys agree!
I'll show you how I did three of them:
Explanation1: I needed to find out 3/8 of 24, so i first divided 8 to 24. The answer was 3, and that meant 3 is 1/8 of 24. So, to get 3/8 i had to times 1/8 three times. That's why i did 3 times 3.
Explanation2: I needed to find out 3/5 of 125, so i first divided 5 into 125. The answer was 25, and that meant 25 is 1/5 of 125. So, to get 3/5 i had to times 1/5 three times. That's why i did 25 times 3.
Explanation3: I needed to find 7/8 of 160, so i first divided 8 into 160. The answer was 20, and that meant 20 is 1/8 of 160. So, to get 7/8 i had to times 1/8 seven times. That's why i did 20 times 7.
This is how I did my questions. Hope you guys agree!
CARL873's scribepost
7)There are 25 pupils in the class, 3/5 of the pupils support Chelsea and the remainder support Arsenal. how many pupils support Arsenal?
8)The local normally sells Mars bars for 40 pence. The shopkeeper says I can buy them for 1/4 less than the normaal price. How much can I buy a Mars bar for?
9) Last year, Mr Murrin weighed 60 kg. This year he weighs 2/5 more. How much does Mr Murrin weigh this year?
8)The local normally sells Mars bars for 40 pence. The shopkeeper says I can buy them for 1/4 less than the normaal price. How much can I buy a Mars bar for?
9) Last year, Mr Murrin weighed 60 kg. This year he weighs 2/5 more. How much does Mr Murrin weigh this year?
jordan's scribe
the thing we did today in class was a question sheet about how do we use the method that Mr harbeck taught as. here is some question we did......
4. The weather fore caster says that it is 20 c in london but only 7/10 as hot in new york. how hot is it in new york?
Jullian's Fraction word PROBLEMS
The thing we did today in class was a question sheet about adding, subtracting, multiplying fractions. Here are some of the questions we did....
Question Number#
10. McDonalds sell milkshakes in two sizes. A small milkshake contains 300ml and a large milkshake contains 2/3 more.
(i) How much does a large milkshake contain? 500ml
Because a 2/3 of a small milkshake is 200 then I added it with 300 because it is more than the small milkshake 200ml+300ml=500ml.
Question Number#
10. McDonalds sell milkshakes in two sizes. A small milkshake contains 300ml and a large milkshake contains 2/3 more.
(i) How much does a large milkshake contain? 500ml
Because a 2/3 of a small milkshake is 200 then I added it with 300 because it is more than the small milkshake 200ml+300ml=500ml.
(ii)If Mr.Murrin drinks 2/3 of a small milkshake and Miss Hoyne 1/2 of a large milkshake who drinks the most? Miss Hoyne drinks the most because, 2/3 of Mr.Murrins 2/3 of 300 is 200ml and Miss Hoyne's 1/2 of the 500ml milkshake is 250ml.
11. Half a can of paint covers 1/3 of a room. How much paint is needed to cover the whole room? 1 and 1/2 of paint is needed to paint the whole room.....
11. Half a can of paint covers 1/3 of a room. How much paint is needed to cover the whole room? 1 and 1/2 of paint is needed to paint the whole room.....
Wednesday, January 23, 2008
Ivanne's Scribe Post
Today, we learned how to multiply mixed numbers. Here are the steps in multiplying mixed numbers.
Example:
1 1/4 x 1 1/2
First, you have to convert the mixed numbers to improper fractions.
Denominator x Whole number + Numerator
1 1/4 = 4 x 1 + 1 = 5/4
(4 multiplied by 1 equals 4 plus 1 equals 5. The denominator stays the same.)
1 1/2 = 2 x 1 + 1 = 3/2
(2 multiplied by 1 equals 2 plus 1 equals 3. The denominator stays the same.)
Now, we have 5/4 x 3/2.
Second, you have to multiply the numerators.
5 x 3 = 15
Third, you have to multiply the denominators.
4 x 2 = 8
Now, we have 15/8.
Fourth, we have to simplify the answer. To convert an improper number to mixed number, we have to divide them.
The answer will be the whole number, the remainder will be the numerator, and the divisor will be the denominator. So, we now have 1 7/8.
1 1/4 x 1 1/2 = 1 7/8
We can also use the numerator swap...
Example:
2 1/5 x 1 2/3
We will do the first step...convert the mixed numbers to improper fractions.
2 1/5 = 5 x 2 + 1 =11/5
1 2/3 = 3 x 1 + 2 = 5/3
Now, we have 11/5 x 5/3.
To multiply the fractions easier, we will do the numerator swap.
11/5 x 5/3 will become 5/5 x 11/3.
Now, we have to simplify these fractions.
5/5 = 1
11/3 = 11/3
Our fractions now are 1 x 11/3.
To multiply a whole number and a fraction, you just need to put 1 as the denominator of the whole number.
Now, we have to multiply the numerators..
1 x 11 = 11
And the denominators..
1 x 3 = 3
**Or you can just multiply them like 1 x 1/3 and the answer is 1/3 because any number you multiply by 1, the answer is always the number.
We now have 11/3. We need to simplify them.
So, 2 1/5 x 1 2/3 = 3 2/3
Example:
1 1/4 x 1 1/2
First, you have to convert the mixed numbers to improper fractions.
Denominator x Whole number + Numerator
1 1/4 = 4 x 1 + 1 = 5/4
(4 multiplied by 1 equals 4 plus 1 equals 5. The denominator stays the same.)
1 1/2 = 2 x 1 + 1 = 3/2
(2 multiplied by 1 equals 2 plus 1 equals 3. The denominator stays the same.)
Now, we have 5/4 x 3/2.
Second, you have to multiply the numerators.
5 x 3 = 15
Third, you have to multiply the denominators.
4 x 2 = 8
Now, we have 15/8.
Fourth, we have to simplify the answer. To convert an improper number to mixed number, we have to divide them.
The answer will be the whole number, the remainder will be the numerator, and the divisor will be the denominator. So, we now have 1 7/8.
1 1/4 x 1 1/2 = 1 7/8
We can also use the numerator swap...
Example:
2 1/5 x 1 2/3
We will do the first step...convert the mixed numbers to improper fractions.
2 1/5 = 5 x 2 + 1 =11/5
1 2/3 = 3 x 1 + 2 = 5/3
Now, we have 11/5 x 5/3.
To multiply the fractions easier, we will do the numerator swap.
11/5 x 5/3 will become 5/5 x 11/3.
Now, we have to simplify these fractions.
5/5 = 1
11/3 = 11/3
Our fractions now are 1 x 11/3.
To multiply a whole number and a fraction, you just need to put 1 as the denominator of the whole number.
Now, we have to multiply the numerators..
1 x 11 = 11
And the denominators..
1 x 3 = 3
**Or you can just multiply them like 1 x 1/3 and the answer is 1/3 because any number you multiply by 1, the answer is always the number.
We now have 11/3. We need to simplify them.
So, 2 1/5 x 1 2/3 = 3 2/3
Tuesday, January 22, 2008
Anne's Scribe
Today we learned about multiplying fractions. We made a pamphlet yesterday and filled in boxes. In one particular box we talked about Standard Algorithim and Array Models to help us multiply fractions easier and with better understanding.
In that box we learned about making ugly fractions into friendly fractions. How to do that was to take an ugly fraction (for example) 20/33 x 11/30. Those are what you can call ugly fractions! To make it easier for you and other people, you can just do a numerator swap (as Mr. Harbeck says) with the 20 and 11. So now your fraction is 11/33 x 20/30.
Now you can simplify both fractions. 11/33 can be simplified into 1/3 because 33 can be divided by 11, and 11 can be divided by 11. Or, 11 can fit into 33, 3 times. Or another way to show that it can be changed into 1/3 is because 11 and 33 is divisble by 11. 20/30 can be simplified into 2/3 because they are both divisible by 10. 10 fits into 20, 2 times. 10 fits into 30, 3 times. How hard was that?!
Now the final step is to get both fractions and multiply them. Now you are ended up with a much more easier and pretty fraction. Get ready for extreme multiply makeover! 1/3 (11/30) x 2/3 (20/30). Since we already talked about multiplying fractions by using cross multiplying we can now get a fraction without having to multiply large, ugly numbers. Wasn't that easy?!
click for a picture http://www.freewebs.com/annev/math.JPG
In that box we learned about making ugly fractions into friendly fractions. How to do that was to take an ugly fraction (for example) 20/33 x 11/30. Those are what you can call ugly fractions! To make it easier for you and other people, you can just do a numerator swap (as Mr. Harbeck says) with the 20 and 11. So now your fraction is 11/33 x 20/30.
Now you can simplify both fractions. 11/33 can be simplified into 1/3 because 33 can be divided by 11, and 11 can be divided by 11. Or, 11 can fit into 33, 3 times. Or another way to show that it can be changed into 1/3 is because 11 and 33 is divisble by 11. 20/30 can be simplified into 2/3 because they are both divisible by 10. 10 fits into 20, 2 times. 10 fits into 30, 3 times. How hard was that?!
Now the final step is to get both fractions and multiply them. Now you are ended up with a much more easier and pretty fraction. Get ready for extreme multiply makeover! 1/3 (11/30) x 2/3 (20/30). Since we already talked about multiplying fractions by using cross multiplying we can now get a fraction without having to multiply large, ugly numbers. Wasn't that easy?!
click for a picture http://www.freewebs.com/annev/math.JPG
Monday, January 21, 2008
Jomarie's Scribe
Today in class we learned about Multiplying & Dividing - Standard Algorithm and Array Models.
Standard Algorithm is basicly multiplying the numerators, multiplying the denominators and simplifying your answer if possible. The example we did together was 2/3 x 1/2 = __. Which the answer is 1/3. How we got it is....
Array Model is a picture (rectangle shaped) showing your answer, in picture form. It is basicly multiplying the denominator to make a large rectangle and multiplying the numerator to make a small rectangle (most likely inside the large rectangle). The example was the same thing, 2/3 x 1/2 = __, but you have to show it in picture form.
Wednesday, January 16, 2008
Mariya's Scribe
today, we learned about adding fractions.
mr harbeck showed us how to add fractions with an example. our example was 4/5 + 8/15. we knew that 4/5 + 8/15 didnt equal to 12/20. so, what we did was find a common denominator. so, 4/5 became 12/15. what we did to get 12/15 was that we multiplied 4/5 by 3. now, we can add the two fractions. so, 12/15 + 8/15 added up to 20/15. 20/15 is an unpropper fraction so we changed it to 1 5/15 or simplified it to 1 1/3.
mr harbeck showed us how to add fractions with an example. our example was 4/5 + 8/15. we knew that 4/5 + 8/15 didnt equal to 12/20. so, what we did was find a common denominator. so, 4/5 became 12/15. what we did to get 12/15 was that we multiplied 4/5 by 3. now, we can add the two fractions. so, 12/15 + 8/15 added up to 20/15. 20/15 is an unpropper fraction so we changed it to 1 5/15 or simplified it to 1 1/3.
today, we also did another investigation. in today's investigation, we are doing a problem in two places in Brooklyn, Carrol Garden and Flatbush. what we know about the two places is that the community wants to build a park for their neighbourhood and both of their space for the park is 50 m by 100 m.
In Carrol Garden's, 3/4 of the lot is devoted to a playground and 2/5 of the playground is blacktop.
In Flatbush, 2/5 of the lot is devoted to a playground and 3/4 of the playground is blacktop.
these are our questions.
- is one neighbourhood getting more blacktop than the other?
- is there more blacktop space in one lot than in the other lot ?
- how would you convince the people in the two neighbourhoods that your conclusion is correct ?
Sunday, January 13, 2008
Karen's Scribe
Last Friday in class we did a test about adding and subtracting fractions.
I will be explaining two questions from the test. The following questions will be #'s 3&4.
Question #3
To find the answer I used $ strategy. First I ignored the 5 from 5 9/10 then I have to figure out the answer for 9/10 using 100, so then it would be 100÷10(denominator)=10 multiply by 9(numerator) equals 9o it would be 90/100. Next is 3 4/5 again I ignored 3 so then for now we have 4/5, 100 divided by 5 is 20 multiply by 4 is equals 80/100.
Now we have 10/100, we still have 5 and 3 that we ignored previously. I just did 5-3 = 2 then included 2 to the answer
Question #4
Using the $ money strategy. I did kind of the same thing for the question #3 that I just answered previously. I also ignored the whole for each fraction so then for now I have 2/5 and 13/20
But since you can't subtract 40/100 to 65/100. Now I have to take the whole from 5 9/10 to subtract it to 1 ( 5-1) = 4 then I add 40 + 100 since we have 40/100 I added them together so it became 140/100for 3 65/100 I just leave it the same way then now we have 4 140/100 subtract by 3 65/100 = 1 75/100 although we can still simplify it.
I simplified 75/100 divided by 5 = 15/20 then again divided by 5 = 3/4 then that's where I added 1 to get 1 3/4
I will be explaining two questions from the test. The following questions will be #'s 3&4.
Question #3
To find the answer I used $ strategy. First I ignored the 5 from 5 9/10 then I have to figure out the answer for 9/10 using 100, so then it would be 100÷10(denominator)=10 multiply by 9(numerator) equals 9o it would be 90/100. Next is 3 4/5 again I ignored 3 so then for now we have 4/5, 100 divided by 5 is 20 multiply by 4 is equals 80/100.
Now we have 10/100, we still have 5 and 3 that we ignored previously. I just did 5-3 = 2 then included 2 to the answer
Question #4
Using the $ money strategy. I did kind of the same thing for the question #3 that I just answered previously. I also ignored the whole for each fraction so then for now I have 2/5 and 13/20
But since you can't subtract 40/100 to 65/100. Now I have to take the whole from 5 9/10 to subtract it to 1 ( 5-1) = 4 then I add 40 + 100 since we have 40/100 I added them together so it became 140/100for 3 65/100 I just leave it the same way then now we have 4 140/100 subtract by 3 65/100 = 1 75/100 although we can still simplify it.
I simplified 75/100 divided by 5 = 15/20 then again divided by 5 = 3/4 then that's where I added 1 to get 1 3/4
marc's scribe
The strategy i used to answer the questions was common denominator.
To get the common denominator, you have to find a number that both of the denominators go into.
To get the common denominator, you have to find a number that both of the denominators go into.
Question 1:
4/5 + 9/15
Question 2:
4 3/8 - 1 1/2
Wednesday, January 9, 2008
Kadille Dave's Scribe Post (Subtracting Fractions)
in subtracting fractions, steps must be memorized and strategies must be used in different kinds of fractions. different kinds of fractions can be mixed number minus mixed number, mixed number minus proper fraction, even with different denominators, proper fractions with the same denominators, proper fractions with different denominators.*
*the denominators depend on the complication or ugliness of the two numbers. so different strategy can be used or shortcuts can be used in the ugliness of the denominators.
some examples or equations with different varieties will be shown here with explanations
1. subtracting proper numbers (answer can be simplified)
*common denominator
in this picture, the equation have the same denominators and the minuend is larger than the subtrahend meaning that the numerator in the first fraction is higher than the second one.
now, the answered is shown, the answer is right but the answer can be simplified. but how? the next illustration will show it.
to simplify the unsimplified answer such as 2/4, we need to get the factors of the numerator and the denominator, circle the same two factors that each number have (like shown in the picture). align the same factors, not in different factors to be organized and easy to understand (as shown in the picture). but...
if there are more factors that are the same, circle them, align them then multiply the common factors(shown in the picture) to get the GCF...if the numbers are too big and hard to identify, try to use prime factorization...it is used to find the prime factors of a big number
now we have the simplified answer...
so the solution of this equation looks like this
*clock method
i didn't draw a clock but this shows the minutes in each fraction.
*money method $
this shows the value in each fraction using money as a strategy
note: you will know that the answer is simplified if the factors of two numbers of the answer, the numerator and the denominator are prime or composite with prime. every prime number has a factor of 1 and the number itself so 1 is useless in finding the prime factors of a number because 1 is a prime nor composite number (1 x 1, 1 x 2, 1 x 3, 1 x ...)
2. subtraction of mixed number and proper fraction or a mixed number (the same denominator)
*money
the solution is shown here using money
*clock
the solutions shown in this post is clearly shown and what i did here is clock method
3. subtraction of mixed number and proper fraction or a mixed number (different denominator)
*least common denominator (LCD)
common denominator must be use when solving an equation with different denominators so in getting the common denominator of the two denominators, you can multiply them together, in prime numbers or prime numbers with composite numbers, you must do this method. but for more ease, the picture will be showed next...
this picture showed how to get the LCD of the two denominators.
the lcd we found is 20 so i changed the fraction with the same denominators. and the pictures shows the similar changed of the numerators and the denominators.
the solution looks like this
note: if the numerator of the minuend, is smaller than the subtrahend, borrow a whole and add it to the numerator, the whole must be shown as a fraction (1= a whole=??/??) the numerator and the denominator must be the same because it represents a whole number and the number must be the same as the common denominator of the equation so the answer would be
*the denominators depend on the complication or ugliness of the two numbers. so different strategy can be used or shortcuts can be used in the ugliness of the denominators.
some examples or equations with different varieties will be shown here with explanations
1. subtracting proper numbers (answer can be simplified)
*common denominator
in this picture, the equation have the same denominators and the minuend is larger than the subtrahend meaning that the numerator in the first fraction is higher than the second one.
now, the answered is shown, the answer is right but the answer can be simplified. but how? the next illustration will show it.
to simplify the unsimplified answer such as 2/4, we need to get the factors of the numerator and the denominator, circle the same two factors that each number have (like shown in the picture). align the same factors, not in different factors to be organized and easy to understand (as shown in the picture). but...
if there are more factors that are the same, circle them, align them then multiply the common factors(shown in the picture) to get the GCF...if the numbers are too big and hard to identify, try to use prime factorization...it is used to find the prime factors of a big number
now we have the simplified answer...
so the solution of this equation looks like this
*clock method
i didn't draw a clock but this shows the minutes in each fraction.
*money method $
this shows the value in each fraction using money as a strategy
note: you will know that the answer is simplified if the factors of two numbers of the answer, the numerator and the denominator are prime or composite with prime. every prime number has a factor of 1 and the number itself so 1 is useless in finding the prime factors of a number because 1 is a prime nor composite number (1 x 1, 1 x 2, 1 x 3, 1 x ...)
2. subtraction of mixed number and proper fraction or a mixed number (the same denominator)
*money
the solution is shown here using money
*clock
the solutions shown in this post is clearly shown and what i did here is clock method
3. subtraction of mixed number and proper fraction or a mixed number (different denominator)
*least common denominator (LCD)
common denominator must be use when solving an equation with different denominators so in getting the common denominator of the two denominators, you can multiply them together, in prime numbers or prime numbers with composite numbers, you must do this method. but for more ease, the picture will be showed next...
this picture showed how to get the LCD of the two denominators.
the lcd we found is 20 so i changed the fraction with the same denominators. and the pictures shows the similar changed of the numerators and the denominators.
the solution looks like this
note: if the numerator of the minuend, is smaller than the subtrahend, borrow a whole and add it to the numerator, the whole must be shown as a fraction (1= a whole=??/??) the numerator and the denominator must be the same because it represents a whole number and the number must be the same as the common denominator of the equation so the answer would be
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