This is the end of my Math Blog, I will continue to update it in the future if I had the opportunity. Now, I will be talking about the great times I had learning these concepts. This concept when taught by my Maths teacher in primary6, it is very useful in helping me to solve many of the challenging problems in primary 6, hence I hope from this Math Blog, the readers will also have a clear understanding of these concepts. If you wan to have a more detailed on these concepts, you can find the contents from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee. Thank you and farewell.
Saturday, August 7, 2010
Small Test
All right guys, today is the test which is to test your knowledge and skill about the 6 lessons. There will be a total of 6 questions and the total mark for this test is 20. Let begin now.
Question 1:
James had a number of coloured balls in his ball pit. 1/4 of the balls were red, 2/3 of the remaining balls were blue and the rest were green. Given that there were 120 red and green balls altogether, how many balls were there in the ball pit? (3 marks)
Question 2:
The ratio od the number of voters in Town A to that of Town B is 4:5. There are 20% more voters in Town C than in Town B. Given that there are 400 more voters in Town C than in Town A, find the total number of voters in the three towns. (3 marks)
Question 3:
80% of the area of Rectangle A is the same as 75% of the area of Rectangle B. Rectangle A is 5cm2 smaler than Rectangle B. Given that they both have a breadth of 5cm, find the difference in their perimeters. (3 marks)
Question 4:
There were 3/5 as many children as adults on a bus. At the next bus stop, 6 children and 5 adults boarded the bus. Then they were 3/4 as many children as adults on the bus. How many children were there on the bus at first? (3marks)
Question 5:
Ray had 75% as much as money as Paul. After Ray received $200 from his uncle and Paul spent $50. Ray had twice as much money as Paul. Find the sum of money Ray had at first. ( 4 marks)
Question 6:
There were 80 oranges and apples in a basketball at first. 2/5 of the oranges and 2/3 of the apples were eaten. In the end, there were 36 of these fruits left. How many oranges were there in the basket at first? (4 marks)
End of the Small Test. Check your work.
Answers:
Q1: 240
Q2: 3000
Q3: 2cm
Q4: 9
Q5: $180
Q6: 35
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Question 1:
James had a number of coloured balls in his ball pit. 1/4 of the balls were red, 2/3 of the remaining balls were blue and the rest were green. Given that there were 120 red and green balls altogether, how many balls were there in the ball pit? (3 marks)
Question 2:
The ratio od the number of voters in Town A to that of Town B is 4:5. There are 20% more voters in Town C than in Town B. Given that there are 400 more voters in Town C than in Town A, find the total number of voters in the three towns. (3 marks)
Question 3:
80% of the area of Rectangle A is the same as 75% of the area of Rectangle B. Rectangle A is 5cm2 smaler than Rectangle B. Given that they both have a breadth of 5cm, find the difference in their perimeters. (3 marks)
Question 4:
There were 3/5 as many children as adults on a bus. At the next bus stop, 6 children and 5 adults boarded the bus. Then they were 3/4 as many children as adults on the bus. How many children were there on the bus at first? (3marks)
Question 5:
Ray had 75% as much as money as Paul. After Ray received $200 from his uncle and Paul spent $50. Ray had twice as much money as Paul. Find the sum of money Ray had at first. ( 4 marks)
Question 6:
There were 80 oranges and apples in a basketball at first. 2/5 of the oranges and 2/3 of the apples were eaten. In the end, there were 36 of these fruits left. How many oranges were there in the basket at first? (4 marks)
End of the Small Test. Check your work.
Answers:
Q1: 240
Q2: 3000
Q3: 2cm
Q4: 9
Q5: $180
Q6: 35
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Use of the Number x Value Method on Real World
Last Lesson (Number x Value Method)
![](https://eee.uci.edu/wiki/images/e/e5/DinoPlaceValue.jpg)
Today will be the last lesson and today's topics is the easiest of all, it is called the Number x Value Method. Unlike the Guess and Check method, the Number x Value method is used when relationships are involved in the question. People use this method to get the answer more efficiently by multiplying the number of units of each item with value of each item.
Question 1:
Question 1:
There are some 10cent and 20 cent coins in a box. There are thrice as many 10 cents coins as 20 cents coins. If the total amount of money in the box is $18, how many coins are there together?
Solution 1 :
Value (cents) --> 10 Number of units--> 3
Value (cents) --> 20 Number of units --> 1
Total amount:
3 x 10 =30
1 x 20 = 20
Total: 50units
Hence,
50units--> 18 x 1000= 1800
1unit--> 1800 divide 50 = 36
4units --> 4 x 36 = 144#
Question of the day:
A goldfish costs $12. An angelfish costs $3 less. Mr Ho paid $195 for some goldfish and angelfish. he bought thrice as many angelfish as goldfish.
a) How many fish did he buy altogether?
b) How much more did he spend on goldfish than angelfish?
Answer: a)20 b) $15
This is the end and there will be a small test two weeks later, please prepare well.
Solution 1 :
Value (cents) --> 10 Number of units--> 3
Value (cents) --> 20 Number of units --> 1
Total amount:
3 x 10 =30
1 x 20 = 20
Total: 50units
Hence,
50units--> 18 x 1000= 1800
1unit--> 1800 divide 50 = 36
4units --> 4 x 36 = 144#
Question of the day:
A goldfish costs $12. An angelfish costs $3 less. Mr Ho paid $195 for some goldfish and angelfish. he bought thrice as many angelfish as goldfish.
a) How many fish did he buy altogether?
b) How much more did he spend on goldfish than angelfish?
Answer: a)20 b) $15
This is the end and there will be a small test two weeks later, please prepare well.
Use of the Simultaneous Concepts on Real World
Simultaneous Concept is very useful in many ways, firstly salesman who always had problems solving the number of products can always use this concept to accurately find out the number of products needed. Also, housewives can also use this method to go to the market to buy the right amount of food etc. And also this concept is very simple to comprehend, hence it is very useful even in the real world.
Lesson 5 (Simultaneous Concept)
This will be the second last of out lesson. So, listen carefully to today lesson and remember there will be a test after Lesson 6 which is the last lesson. Topics taught today will be called the Simultaneous Concept. This concept when used in the right way, can assist us to solve non-rountine problems efficiently.
Question1 :
There are 45 pupils in a class. 60% of the boys and 20% of the girls wear spectacles. There are 19 pupils in the class who wear spectacles. How many girls wear spectacles?
Solution1 :
60% of the boys + 20% of the girls --> 19pupils
300% of the boys + 100% of the girls --> 95pupils
100% of the boys + 100% of the girls --> 45 pupils
200% of the boys --> 95 - 45= 50 pupils
20% of the boys --> 50 divide 10= 5 pupils
100% of the boys --> 5 x 5= 25 pupils
Number of girls in the class = 45 - 25
= 20
Number of girls who wear spectacles = 20% of 20
= 20/100 x 20
= 4#
Question2 :
5 apples and 3 oranges cost $2.70 altogether. 2 apples and 4 oranges cost $2.20 altogether. How much does a dozen apple cost?
Solution2 :
5 Apples + 3 Oranges --> $2.70
2Apples + 4 Oranges --> $2.20
20apples + 12 oranges --> $10.80
6 apples + 12 oranges --> $ 6.60
14 apples --> $4.20
12apples --> $3.60#
Question of the day:
There were 80 oranges and apples in a basketball at first. 2/5 of the oranges and 2/3 of the apples were eaten. In the end, there were 36 of these fruits left. How many oranges were there in the basketball at first?
Answer: 35#
This is the end of Lesson 5.
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Question1 :
There are 45 pupils in a class. 60% of the boys and 20% of the girls wear spectacles. There are 19 pupils in the class who wear spectacles. How many girls wear spectacles?
Solution1 :
60% of the boys + 20% of the girls --> 19pupils
300% of the boys + 100% of the girls --> 95pupils
100% of the boys + 100% of the girls --> 45 pupils
200% of the boys --> 95 - 45= 50 pupils
20% of the boys --> 50 divide 10= 5 pupils
100% of the boys --> 5 x 5= 25 pupils
Number of girls in the class = 45 - 25
= 20
Number of girls who wear spectacles = 20% of 20
= 20/100 x 20
= 4#
Question2 :
5 apples and 3 oranges cost $2.70 altogether. 2 apples and 4 oranges cost $2.20 altogether. How much does a dozen apple cost?
Solution2 :
5 Apples + 3 Oranges --> $2.70
2Apples + 4 Oranges --> $2.20
20apples + 12 oranges --> $10.80
6 apples + 12 oranges --> $ 6.60
14 apples --> $4.20
12apples --> $3.60#
Question of the day:
There were 80 oranges and apples in a basketball at first. 2/5 of the oranges and 2/3 of the apples were eaten. In the end, there were 36 of these fruits left. How many oranges were there in the basketball at first?
Answer: 35#
This is the end of Lesson 5.
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Use of the External Transfer with Changed Quantities Concept on Real World
Lesson 4 External Transfer with Changed Quantities Concept
Today we will be learning about my favourite concept of all, called the External Transfer with Changed Quantities Concept. It can be quite difficult if u do not understand completely about them, so pay attention. External Transfer with Changed Quantities Concept involves changes to both quantities. The quantities at the beginning and the end are different, thus making it very difficult to solve such word problems using the traditional model approach.
Question1 :
The number of ten-cent coins in a box was 1/2 the number of fifty-cent coins. Syed took out 5 fifty-cent coins and exchanged them for ten-cent coins. Then he put the money back into the box. The number of fifty-cent coins became 5/8 the number of ten-cent coins. How much money was there in the box?
Solution 1:
Ten-cent Coins Fifty-cent Coins
1unit 2 units Initial
+25 - 5 Change
8 parts 5 parts Final
x5 x8
Ten-cent Coins Fifty-cent Coins
5unit 16 units Initial
+125 -40 Change
40 parts 40 parts Final
5unit 16 units Initial
+125 -40 Change
40 parts 40 parts Final
The final quantities are equal (40parts).
Hence:
Ten-cent coins : 5units + 125 + 40
Equal
Fifty-cent coins : 16 units
11units: 125 +40 = 165
1unit: 165 divide 11 =15
2units : 2x15= 30
Total amount of money= (15x10) + (30x 50)
= 1650 (cents)
= $16.50#
Question 2:
Joyce had 1/5 as many erasers as Ethan at first. Then their monther gave Ethan 12 more erasers and Joyce 5 more erasers. The ratio of the number of Ethan's erasers to the number of Joyce's erasers became 4:1. How many erasers did Ethan have at first?
Solutions2 :
Joyce Ethan
1unit 5 units Initial
+5 +12 Change
1 parts 4 parts Final
x4
1unit 5 units Initial
+5 +12 Change
1 parts 4 parts Final
x4
Joyce Ethan
4unit 5 units Initial
+20 +12 Change
4 parts 4 parts Final
Their final quantities are equal (4parts).
Joyce: 4units + 20
Equal
Ethan: 5units + 12
1unit: 20-12 =8
Number of erasers Ethan had at first= 5units
= 40#
Question of the day:
Ray had 40% as many pencils as Rajon at first. Ray gave 8 pencils away and Rajon gave 40 pencils away. Then Ray had 0.5 as many pencils as Rajon. How many pencils did Ray have at first?
Answer: 48#
Note: This lesson is very important.
End of Lesson 4
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Use of the Equal Concepts on Real World
Third Lesson (Equal Concept)
Today, we will be learning about the Equal Concept. The word problems in this category explicitly compare two fractions that represent equal quantities. Many a time, these fractions take the form of percentages, decimals or ratios.
Question1 :
James and Jessie had a total of $850. After Jessie spent 1/4 of her money and James spent 1/2 of his money, both had an equal amount of money left. Find the total amount of money each had at first.
Solution1 :
Jessie 1- 1/4= 3/4
James1 - 1/2 = 1/2
3/4 Jessie = 1/2 James
3/4 Jessie = 3/6 James
10 units : $850
1 unit: $85
$85 x 6 =$510#
$85 x 4 = $340#
Question 2 :
Andrew and Ray shared a sum of money. Andrew spent 0.4 of his money and Ray spent 55% of his money. In the end, Ray had thrice as much money as Andrew. Given that Ray had $300 more than Andrew at first, find the amount of money Andrew had at first.
Solution 2:
3x 3/5 Andrew = 9/20 Ray
9/5 Andrew = 9/20 Ray
Difference: 15 units : $300
1 unit : 300 divide 15 = $20
Original amount : 5x 20 =$100#
Question of the day:
A bag contained some red and green marbles. 20% of the red marbles and 0.6 of the green marbles were removed. In the end, there were half as many red marbles as green marbles. Given that there were 30 more green marbles than red marbles at first, find the number of red marbles in the bag at first.
Answer : 10#
This is the end of lesson 3.
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Question1 :
James and Jessie had a total of $850. After Jessie spent 1/4 of her money and James spent 1/2 of his money, both had an equal amount of money left. Find the total amount of money each had at first.
Solution1 :
Jessie 1- 1/4= 3/4
James1 - 1/2 = 1/2
3/4 Jessie = 1/2 James
3/4 Jessie = 3/6 James
10 units : $850
1 unit: $85
$85 x 6 =$510#
$85 x 4 = $340#
Question 2 :
Andrew and Ray shared a sum of money. Andrew spent 0.4 of his money and Ray spent 55% of his money. In the end, Ray had thrice as much money as Andrew. Given that Ray had $300 more than Andrew at first, find the amount of money Andrew had at first.
Solution 2:
3x 3/5 Andrew = 9/20 Ray
9/5 Andrew = 9/20 Ray
Difference: 15 units : $300
1 unit : 300 divide 15 = $20
Original amount : 5x 20 =$100#
Question of the day:
A bag contained some red and green marbles. 20% of the red marbles and 0.6 of the green marbles were removed. In the end, there were half as many red marbles as green marbles. Given that there were 30 more green marbles than red marbles at first, find the number of red marbles in the bag at first.
Answer : 10#
This is the end of lesson 3.
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Use of the Repeated Identity Concepts on Real World
Second Lesson (Repeated Identity Concept)
Today we will be learning about the Repeated Identity Concept. This convept usually involves around an object or person that is repeated in the question. We can solve the solve the question either by the model or the units method. Just like the Remainder Concept which we learnt earlier, the Repeated Identity Concept can exist in many other forms such as fractions, ratios and the percentage.
Question 1:
Jennifer has 3/4 as many biscuits as Melissa. Melissa has 1/4 as many bisuits as Tom. If the three of them have a total of 92 bicuits, how many biscuits does Jennifer have?
Solution 1:
Jennifer: Melissa
3 : 4
Melissa : Tom
1 : 4
Jennifer : Melissa : Tom
3 : 4 : 16
total units: 23units
23units: 92
1unit: 4
3units: 12#
Question2 :
The number of spectators at a basketball match on Wednesday was 10% more than on Tuesday. On Thursday, there were 20% less spectators than on Wednesday. Given that there were 4800 more spectators on Tuesday than on Thursday, find the number of spectators on Tuesday.
Solution 2:
Wed: Tue: Thur
11: 10 : 8.8
10 - 8.8 = 1.2
1.2units: 4800
10units : 40000#
Question of the day:
Ray allen, Paul Pierce, Rajon Rondo and Kevin Garnett shared $168. Kevin Garnett received 1/7 of the total amount of money received by Ray Allen. Paul Pierce and Rajon Rondo. Ray Allen received 3/4 of the total amount of money received by Paul Pierce and Rajon Rondo. Paul Pierce received 2/5 as much as Rajon Rondo. How much did Paul Pierce receive?
Answer: $24#
This is the end of the second lesson, please revise as there will be a small test at the end of the 5 lessons.
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Question 1:
Jennifer has 3/4 as many biscuits as Melissa. Melissa has 1/4 as many bisuits as Tom. If the three of them have a total of 92 bicuits, how many biscuits does Jennifer have?
Solution 1:
Jennifer: Melissa
3 : 4
Melissa : Tom
1 : 4
Jennifer : Melissa : Tom
3 : 4 : 16
total units: 23units
23units: 92
1unit: 4
3units: 12#
Question2 :
The number of spectators at a basketball match on Wednesday was 10% more than on Tuesday. On Thursday, there were 20% less spectators than on Wednesday. Given that there were 4800 more spectators on Tuesday than on Thursday, find the number of spectators on Tuesday.
Solution 2:
Wed: Tue: Thur
11: 10 : 8.8
10 - 8.8 = 1.2
1.2units: 4800
10units : 40000#
Question of the day:
Ray allen, Paul Pierce, Rajon Rondo and Kevin Garnett shared $168. Kevin Garnett received 1/7 of the total amount of money received by Ray Allen. Paul Pierce and Rajon Rondo. Ray Allen received 3/4 of the total amount of money received by Paul Pierce and Rajon Rondo. Paul Pierce received 2/5 as much as Rajon Rondo. How much did Paul Pierce receive?
Answer: $24#
This is the end of the second lesson, please revise as there will be a small test at the end of the 5 lessons.
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
Use of the Remainder Concepts on Real World
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhSEteUoTxl7DIojjD1oEz4vHaxv7SZxiW1qogL7OWsoDSLwB9HCHm9knTJSM-RVgm-NB60Qey89TyhudKBC7kKeyqN9c6_Yr9ibbTEbR2QhtHgDN4JohMJbzaPg9LCT4ILvk6X-4h8t-g/s200/W020090331530584034578.jpg)
The use of the Remainder Concept in real life is for those housewifes who frequently and students. The two people mentioned before always carried money to supermarkets and the school. Hence, with the introduction of the Remaineder Concept, they will be able to count the amount of money or how much they had spent in a day and the amount of money saved with effiency As a result, the use of the Remainder Concept plays a large role in peoples' real daily life.
Friday, August 6, 2010
First Lesson (Remainder Concept)
All right guys, this will be our very first lesson. There will be a total of 10 weeks for us to master our Math Skills. So, today we will be learning about the Remainder Concept. Remainder Concept deals with fractions and percentages of the remainders. The recommended method to solve this type of word problems is usually by using the model method. However, model method has its constraints. Often, it depends largely on the numbers used in the questions to determine the complexity of the models.
Firstly, you have to learn about what is fractions and percentages. Go online to have a basic understanding of them. Now, I believe you have a clear understanding of fractions and percentages. Now I will be giving some simple questions for you to answer.
Question1 : Mr Tan used 1/2 of his salary to pay his bills and give 2/3 of the remainder to his wife. He saved the rest of the money. What is the fractions of the money received his wife and the rest of the money?
Solution1 : 1/2 = 3/6 (Bills) 2/3 x 1/2 (Amount of money left) = 2/6 = 1/3 (Wife)#
1/3 x 1/2 ( Rest of the savings) = 1/6 ( Savings) #
Statement: The fractions of the money received his wife and the rest of the money is 1/3 and 1/6.
Question 2 : Mr Tan spent $1280 of his salary on a television set and 1/3 of the remainder on a VCD player. If he had 2/5 of his salary left, how much was his salary?
Solution2 :
Total Salary :
$1280 (Television Set)
2/5 divide 2/3 = 3/5 (Remainder)
Remainder:
1/3 (VCD Player)
2/3 ( Rest of the Salary)
1/3 x 3/5 = 1/5 (Total)
2/3 x 3/5 = 2/5 (Total)
Hence: 2/5 of Total Money = $1280
2/5 : $1280
5/5 : $3200#
Question of the day:
Mrs Goh went to Nike Shopping Mall with $200. She spent 45% of her money on a shirt and 60% of the remainder on a pair of shoes.
a) How much did she spend on the pair of shoes? b) How much has she left?
Answer:
a) $66 b) $44
End of Lesson 1
Reference from the book "Challenging Maths Problems Made Easy" by Ammiel Wan Chee Hong
About Me
New Maths Post
Dear readers, this is going to be my very first Math Blog, please leave any comments for me to improve it. Anyway, the purpose for this Blog Post is to share some of my Math Concepts which I had learnt in my previous year from a book called "Challenging Maths Problems Made Easy" and it is still very useful till now. The target audience is for those who has the same interest as me. Some of the Math Formula are Remainder Concept, Repeated Identity Concept, Equal Concept, and a lot more. So, I will be sharing all my experiences learning this concepts and I will uploading it once per two week for my Maths Ace assignment. So, please wait for my newest post next week. :D
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