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# Tag Archives: algebra

## Let’s Learn Mathematics (Primary Level) 3

Disclaimer: This Mathematics question is purely created for discussion purpose. Any resemblance to actual questions from books or schools is coincidental.

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We have discussed about simple algebra in Let’s Learn Mathematics (Primary Level). Though algebra is formally introduced in secondary schools, the model method is taught in primary schools as an introduction to algebra. Thus, you can see the importance of learning algebra.

The question of this post is considered advanced for primary students, nonetheless it is for primary students. Take your time to understand the question and the solution; they are helpful in learning algebra up to secondary schools.

Q: 3 shirts and a pair of trousers cost \$63.50. A shirt and 2 pairs of trousers cost \$27. Find the cost of a shirt.

A: First you need to draw the diagram to show the relationship of the equations, based on the question. The first sentence (3 shirts and a pair of trousers cost \$63.50) forms an equation and the second equation forms another equation. [Please refer to The Almighty Algebra for notes.]

Diagram to show the relationships

In the above diagram, the blue rectangle represents the shirt and the yellow triangle represents the pair of trousers.

If the student has been solving similar questions, he / she will start by adding or subtracting the two equations.

s = a shirt, t = a pair of trousers

Equation 1, 3s + t = 63.50

Equation 2, s + 2t = 27

Adding equation 1 and equation 2, 4s + 3t = 90.50

Subtracting equation 2 from equation 1, 2s − t = 36.50

This is where students may face the problem to continue solving the problem. They have all the equations but the equations are leading them to nowhere. No matter it is adding or subtracting, you end up with more equations with no solution.

You can apply any Mathematics operation to any equation.

The Mathematics operation includes adding, subtracting, multiplying and dividing. Yes, many students have not thought of multiplying the equations. Nonetheless, we do not encourage students to divide the equations because it will lead to decimals or fractions. It is difficult to solve the question, it is more difficult to solve the question with decimals or fractions.

Back to the question, the question is only asking for the cost of a shirt. So, to simplify the question, we must “eliminate” the cost of a pair of trousers.

Equation 1, 3s + t = 63.50

Equation 2, s + 2t = 27

The fastest way to get rid of “t” is to multiply 2 to equation 1 and then subtract equation 2 from the new equation (equation 3).

Equation 1 x2, 6s + 2t = 127 (Equation 3)

Equation 3 − Equation 2, 5s = 100 [There is no more “t” because 2t − 2t = 0]

s = 100 ÷ 5

s = 20

The cost of a shirt is \$20.

## Primary Mathematics Notes

Some parents have been complaining about Primary Mathematics. They say nowadays Primary Mathematics is not “pure” Mathematics anymore. They say that the Mathematics questions are more like playing with word games, if you do not understand the questions, you cannot solve the questions. Nonetheless, we can still solve the “word games” and score high in Mathematics! In addition, the terms are universal, you can keep this note for Secondary Mathematics too.

Caution: Though the wordings used in this post are commonly found in the Mathematics questions, each question is unique and may vary. The most important thing is to read and understand the questions. Treat each question on a case-by-case basis. If you face any problem on Mathematics questions, please feel free to contact me at wendy@joyfulyue.com

Below are some of the common words used in Primary Mathematics questions:

1. as … as = same

Jenny is as tall as Kenny. That means both Jenny and Kenny have the same height.

Variation version:

(a) Jenny has twice as many candies as Kenny. That means if Jenny has 6 candies, Kenny has only 3 candies. In algebra, Jenny’s candies = 2 x Kenny’s candies.

(b) When x is doubled, find y. That means when x = 2x.

2. -er than, the difference of => use subtraction (-)

Kenny has \$50 more than Jenny. That means the difference of the amount of money between Kenny and Jenny is \$50, most questions can be solved using subtraction.

Kenny is taller than Jenny by 2 cm. That means the difference of the height between Kenny and Jenny is 2 cm and Kenny is taller.

3. altogether, the sum of => use addition (+)

How many flowers are there altogether? The question is asking you to add all the flowers mentioned in the question.

What is the sum of money? The question is asking you to add all the money value mentioned in the question.

4. Mathematics language

(a) Subtract 5 from 9 => 9 – 5

(b) A bag cost \$5.00, how much does it cost if Jenny buys 3 bags? => \$5 x 3

Below are the concepts that must be understood by Primary students so that they can tackle more difficult questions:

2. Ratio

3. Average

5. Algebra (This concept will be taught in more details in Secondary school, thus the understanding of the concept is of utmost importance)

In a nutshell, understanding the concepts is the most important thing to learn in Mathematics. Once you have understood, solving more questions will reinforce the understanding and A* is on the way!

## Let’s Learn Mathematics (Primary Level)

Disclaimer: This Mathematics question is purely created for discussion purpose. Any resemblance to actual questions from books or schools is coincidental.

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Q: The sum of three numbers is 125. The smallest number is 25. The difference between the largest number and the smallest number is 35. Find the second largest number.

A: Method 1 — This is a common question for Primary students in Singapore. First you need to draw the diagram to show the relationship of the three numbers, based on the question. Once you get the diagram correct, that means you have understood the question and it is easier for you to find the answer.

Diagram for the question

To find the second largest number, you need to subtract the smallest number and the largest number from the sum.

The largest number = 25 + 35 = 60

The second largest number = 125 – 25 – 60

= 40

Method 2 — Use variables to represent the three numbers, a = the smallest number, b = the second largest number and c = the largest number.

a + b + c = 125

a = 25

c – a = 35

To find c, c = 35 + a = 35 + 25 = 60

To find b, b = 125 – a – c

= 125 – 25 – 60

= 40

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Pros and cons of the above two methods

Method 1: Pro — This method is taught in school. As long as you get the diagram correct, you are not far from the correct answer.

Con — If the student cannot understand the question and unable to draw diagram, the student will leave the answer blank, which leads to zero mark. Many parents have not been taught about this method when they were studying. This is when both parents and students are at a loss because parents are unable to teach the children how to solve the question in “school way”.

Method 2: Pro — If you are good at algebra, this is a good method to use. Parents are able to teach the children how to solve the question. Often, the parents are afraid to teach the children this way because it is not the way how teachers teach in school. No worry, the students are actually encouraged to use different ways to solve problems, as long as they can find the correct answer.

Con — Algebra is a more complicated concept than diagram, thus some students may not get used to the idea of variables. If the students learn two methods at the same time, they may get confused.

Note to parents: Algebra will be taught in Secondary schools. Thus, if your children are able to solve the question using the model method, you may want to introduce the algebra method to your children as an alternative way to solve the question or to check whether the answer is correct.

## Let’s Learn Mathematics (Secondary Level)

Disclaimer: This Mathematics question is purely created for discussion purpose. Any resemblance to actual questions from books or schools is coincidental.

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Q: A restaurant manager engages a mover to move 800 imported wine glasses from the port to the restaurant. For wine glasses that have safely reached the restaurant, the manager pays \$0.60 each. But, if the wine glasses are broken, the mover has to pay \$6.80 each to the manager as a compensation.

If the mover receives \$450.40 as the payment, how many wine glasses have safely reached the restaurant?

A: At first glance, many students first find the payment for all 800 wine glasses safely reached the restaurant.

Payment for 800 wine glasses = 800 x \$0.60 = \$480

Subsequently, they will find how many are broken by subtracting and dividing,

\$480 – \$450.40 = \$29.60

Here, they face a problem because \$29.60 ÷ \$6.80 = 4.35. There is a remainder and they start to doubt whether the question is written correctly.

The question is written correctly. Below are two methods to solve the problem. The first is using table, which is easy to understand. The second is using algebra.

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1. From the table below, the answer is 796 wine glasses have safely reached the restaurant.

 Quantity of wine glasses safely reached the restaurant Quantity of broken wine glass(es) Total Payment 800 0 800 x \$0.60 – 0 = \$480 799 1 799 x \$0.60 – 1 x \$6.80 = \$472.60 798 2 798 x \$0.60 – 2 x \$6.80 = \$465.20 797 3 797 x \$0.60 – 3 x \$6.80 = \$457.80 796 4 796 x \$0.60 – 4 x \$6.80 = \$450.40

2. From the table above, we can write the algebraic equation to represent the problem.

a = quantity of wine glasses safely reached the restaurant

p = payment

p = a(0.6)-(800-a)(6.8)

Simplify the equation, p = 0.6a – 800(6.8) + 6.8a

p = 7.4a – 5440

Solve the equation, 450.4 = 7.4a – 5440

and you get a = 796.

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Pros and cons of the above two methods

Method 1: Pro — Simple step to solve a seemingly complicated question.

Con — It is time-consuming if the quantity of wine glasses safely reached the restaurant is smaller, you need to continue the list until you find the answer.

Method 2: Pro — Fast and can find the answer easily even if the quantity of wine glasses safely reached the restaurant changes.

Con — The first step to find the algebraic equation is the key, if the equation is wrong, it will lead to wrong answer. For students who are weak in algebra, they may spend longer time to find the correct algebraic equation.