A challenging area problem

Question

A figure is shaded as shown below :-

The horizontal boundary line and vertical boundary line of the shaded area are away from the centre of the circle with the dimension shown.  Find the difference of the shaded area and unshaded area.

Answer :

For a primary school maths problem, it seems to be impossible to calculate the difference as complex formula for such calculation is not covered in primary school.  Moreover, the radius or the diameter of the circle was not given.  However, if you use some visual analysis, you will simplify the area of the shaded part as below :-

For a primary school maths problem, it seems to be impossible to calculate the difference as complex formula for such calculation is not covered in primary school. Moreover, the radius or the diameter of the circle was not given. However, if you use some visual analysis, you will simplify the area of the shaded part as below :-

By dropping 2 mirror image vertical line and horizontal line and flip the 2 portion, the shaded portion will cover the semi-circle plus 2 rectangles with the dimension shown.

The unshaded area will cover the semi-circle minus the 2 rectangles. So the difference will be 4 rectangles

Difference of the shaded area and unshaded area = 4 x 4cm x 6cm

                                                                           = 96 cm²

A common word problem

Today I will cover a simple concept question which involve 2 scenarios yet this type of question frequently appears in test & exam. 

Question

Mr Lee has a bag of sweets to distribute to his pupils.

If he gives each pupil 8 sweets, he will have 8 sweets left.

If he gives each pupil 6 sweets, he will have 86 sweets left. How many pupils are there in his class? How many sweets does he has in his bag?

Method 1

Dfference in number of sweets per pupils is 8 - 6 = 2

Difference in number of sweets left 86 - 8 = 78

Number of pupils = 78 / 2 = 39

Total number of sweets 8 x 39 + 8 = 320

( or 6 x 39 + 86 = 320 )

Method 2

Let the number of pupils be 1 unit

Total sweets = 8 units + 8 = 6 units + 86

                  8U - 6U  86 - 8

                          2U  78

                          1U  39

Total sweets = 8 x 39 + 8 = 6 x 39 + 86 = 320

Ans : 320 sweets and 39 pupils

Challenging Question - Think Out of the Box

Question :

The figure below, not drawn to scale, is made up of 3 squares.  Find the area of the shaded parts

Ans :

Area of Big Rectangle = 18cm x 14cm
                                      252 cm²

Area of unshaded Big triangle = ¹/₂ * 14cm * 16cm
                                             = 112 cm²

Area of unshaded small triangle = ¹/₂ * 8cm * 8cm
                                                = 32 cm²

Area of Small unshaded Rectangle = 2cm x 6cm
                                                     = 12 cm²

Area of shaded parts = 252cm² - 112cm² - 32cm² - 12cm²
                                 = 96 cm²



                                                                                          Ans : 96 cm²

Explanation :

This question require a little thinking.  As the slope of the shaded parts extended from one square to the other, it do not cut perfectly at the corner of the 10cm square.  Therefore you will need to extend the boundary into a bigger rectangle and minus off the unshaded portion.  This type of question will train the students to think unconventionally. The diagram is illustrated separately at the bottom.

More Word Problem - 2 scenerios

Question :

Mr Tan bought a bag of sweets for his class.  If he gives 13 sweets to each of his pupils equally, he has 4 sweets left.  If he gives 8 sweets to each of his pupils equally, he has 214 sweets left. How many pupils are there in his class?

Method 1 :

Differences in sweets to each pupils        13 - 8 = 5
Difference in total of sweets left              214 - 4 = 210

Nos. of pupils  =  Difference in total of sweets left / Differences in sweets to each pupils
                      =  210 / 5 = 42


                                                                                    Ans : 42 pupils


Methods 2 :

Let the Nos. of pupils in the class be 1 unit
Total sweets = sweet gave + sweets left
        13 units + 4  =  8 units + 214
        13 units - 8 unit 214 - 4
                       5 units210
                       1 unit 210 / 5 = 42


                                                                                    Ans : 42 pupils

Remarks :

Many students approach to solve this type of problem with guess & check method.  Although it is one of the approve ways of solving the problem, personally, I feel it take more time with guess & check method.  With the above 2 methods, it will give student more time to check their test papers by solving the problem faster.  It is also my advice to student that whenever problem need to be solve by guess and check.  They should reserve it to the last and spend their time firstly on question that they can solve quickly.          

More Interesting question

A figure is made up of a square and a rectangle overlapped each other as shown below. The ratio of the area of the square to the ratio of the are of the rectangle is 3:2. The shaded area of the rectangle is 3/8 of the area of the rectangle. The total unshaded area is 42cm² . What is the area of the rectangle?

It is relative common for MOE to set a typical question like this. Firstly ratio is provided between the area of the square & area of the rectangle. Then the fraction of shaded area of the rectangle is provided. Finally, the area of the unshaded is provided so that you can determine the area of rectangle.

In order to compare the item properly, we can compare the items in units as below :-

Since the shaded area of the rectangle is 3/8. We can represent the shaded portion of the rectangle as 3 units and the unshaded portion as 5 units. As the area of the rectangle take up 2 parts, which is 8 units in total, we can simply said that 1 parts is equal to 4 units. Since the rectangle is 2 parts then the square is 3 parts, which will take up 12 units. The interlapped took up 3 units, hence the unshaded portion of the square will be 9 units (12units - 3 units). The unshaded portion make up by unshaded rectangle portion (5 units) and the unshaded square portion ( 9 units), so the total unshaded area is 14 units.

14 units     42cm²

1 units       42/ 14 = 3cm²

         

Since the area of rectangle is made up of 8 units

8 units      3cm² X 8 = 24cm²

Interesting word problem - Remainder Concept

I saw this interesting word problem posted in one of the thread in Kiasu parents forum (http://www.kiasuparents.com) the question look something like this :-

Question :

Kumar spent $1729 on a set of encyclopedia. He spend 1/4 of the remainder on a a camera and still had 2/5 of his money left. Find the total amount of money he had at first.

This is a typical question based on remainder concept. While it is relatively easily for experienced teacher to solve the question. However, in teaching, it is not how much you know, but rather how easy you make it for the children to understand. You have to approach it from the mentality of your students. There are a few points which you have to specially take note as below :-

1. The amount given $1729 is a relatively odd number and some students will perceive it as tough to calculate.

2. 3/4 of the remainder which makeup the Final fraction of 2/5 is not very clear cut to some students who are not fast learner. (as you need to further divide the 2/5 to 6/15 so that the 3/4 of the remainder fit nicely into the subunit. You could split the 1/4 & 3/4 of the remainder properly into the subunits nicely.)

3. After further divide the final fraction into 15 equal parts, it will not be easy for your students or your kids to work on as they have to be careful visually as well as when they carry out the computing.

4. For students that you have not introduce the remainder concept, you will need to guide and train them to identify why this question belong to the remainder concept. This question also give you to opportunity to train them how to think.

Let approach it from the model method

From the question we can draw the model below

We know that Kumar had spent the money on the encyclopedia and camera are left with 2/5 of the total amount. We know the amount spent on the encyclopedia but not the final fraction (We could immediately calculate the total amount if we know the final fraction of the encyclopedia. We know the final fraction left but not the actual money left. The great clue is the money left is 2/5 of the total and it is also 3/4 of the remainder after buying the encyclopedia and camera. So trying to relate the amount left which is 3/4 of the remainder to the final fraction we have to compare as below

We had to split the each 1/5 to 3 equal units so each 1/4 will make up 2/15 of the final fraction and 3/4 will make up 6/15 of the final fraction as indicated below

Hence we can find out the final fraction of the encyclopedia by substract 2 units (camera) and 6 units (left) from the total 15 units. Hence the encyclopedia is 7 units as shown below:-

Therefore the total amount is as work out as below

                            7 units  1729

                            15 units1729 / 7 * 15 = 3705  

Word Problem - Fraction

Question :

XiuLi has $459 and YuLi has $1560. After both of them spent equal amount of money, XiuLi has 1/4 as many money as YuLi. So how much money has YuLi spent ?

Model Methods

Different at beginning = Different after spent

So 3 units $1560 - $459 = $1101

1 unit $1101 /3 = $367

YuLi spent = XiuLi spent = $459 - $367 = $92

                                                                                              ANS : $92