Home » Posts tagged 'Secondary 1'
Tag Archives: Secondary 1
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.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
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.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
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.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
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.
Integers
Disclaimer: This article is open for discussion. I have bought this reference book for Secondary 1 Express Mathematics, fourth reprint 2012, 2nd Edition, and found some questionable examples.
In the above extract from Page 12, it is a question for the readers to solve. Let’s do it together:
(a) The difference in the sea level between high tide and low tide = 2 + 4 = 6 m
(b) When the sea level is exactly halfway between high tide and low tide, it is 3 m from either tide.
The sea level is (4 – 3) m = 1 m below the clam at this time.
Below is the extract from Page 123, the answer in the book for question 20.
Answer (b) is different from what I have done. From the calculation of the sea level, it is 3 m, which is the same as my answer. Thus, from the picture above, even without calculation, we can deduce that the sea level is 1 m below the clam when the sea level is exactly halfway between high tide and low tide.
With this example and another example on factors and multiples, I hope parents and students will be careful when referring to any reference books available in the market. If you have any doubt, always find a third opinion. Most importantly, have confidence in yourself, do not feel discouraged if your answers are different. Knowledge is gained through questioning.
I am not sure if they have rectified the error, just want to alert parents and students if you have bought the same book with the same version as mine. I welcome comments and feedback. Thank you.
Factors and Multiples
Disclaimer: This article is open for discussion. I have bought this reference book for Secondary 1 Express Mathematics, fourth reprint 2012, 2nd Edition, and found some questionable examples.
Highest common factor
In the above extract from Page 2, 3 is not a factor of 20. The solution should be as below:
Factors of 18 are 1, 2, 3, 6, 9, 18
Factors of 20 are 1, 2, 4, 5, 10, 20
Common factors of 18 and 20 are 1 and 2.
Hence HCF of 18 and 20 is 2.
HCF = highest common factor.
I am not sure if they have rectified the error, just want to alert parents and students if you have bought the same book with the same version as mine. I welcome comments and feedback. Thank you.
Joyful 阅 