Story line reference :

Math concept reference :

__Sarin learns Average Number, Percentage, Tables and Charts (Math Concept)__

__Sarin learns concept of ratio in school__

*Challenge yourself with the question before look out for the given solution!*

**Upper primary school mathematics question UPQ588**

The pie chart below shows the type of buns sold at a shop last week. The shop sold 900 buns in all. Half of the buns sold were chocolate. The shop sold 20% more cheese buns than vanilla buns.

a. How many banana buns did the shop sell?

b. What percentage of the buns sold last week were cheese buns?

*Solution:*

a. 100% ==> 900

1% ==> 9 buns

17% ==> 9 × 17 = __153__ buns

The number of Banana buns = 153

b. Total percentage of Cheese buns and Vanilla buns

= 50% – 17%

= 33%

33% ==> 33 × 9 = 297

The ratio of cheese buns to vanilla buns = 120 : 100 = 6 : 5

11 units = 297

1 unit = 27 buns

The number of cheese buns sold

= 6 × 27

= 162

The percentage of cheese buns sold

= 162 ÷ 900 × 100%

= __18%__

Reference Reading:

**Sarin participates in school anniversary event part 1**

__Sarin learns concept of Fraction (Mathematics concept)__

__Sarin learns Average Number, Percentage and Charting in school (Math Concept)__

__Sarin learns concept of ratio in school__

*Challenge yourself with the question before look out for the given solution!!!*

**Upper primary school mathematics question UPQ587**

There were 352 people attended the anniversary dinner in the school hall. 75% of them were invited guests and the rest were students. Later in the night, some invited guests decided to leave and the number of invited guests was 3/5 that of the total number of people in the hall.

How many invited guests decided to leave the dinner?

Solution:

The number of invited guests at first = 352 × 75% = 264

The number of students at first = 352 – 264 = 88

No student had left the hall,

Hence,

At final, 2/5 of the people in the hall were students,

2 units = 88

1 unit = 44

The number of invited guests finally = 3 × 44 = 132

The number of invited guests left the club = 264 – 132 = __132__

__Alternative solution(1):__

Per the model,

8 units = 352

1 unit = 44

The number of guests left the dinner

= 3 × 44

= __132__

__Alternative solution(2):__

75% people in the dinner were invited guests,

Guests : Students : Total = 75 : 25 : 100 = 3 : 1 : 4 = 6 : 2 : 8

Finally,

Guests : Students : Final Total = 3 : 2 : 5

There were no students left the hall,

8 units = 352

1 unit = 44

Guests left the dinner in unit = 6 – 3 = 3 units

The number of guests left the dinner

= 3 × 44

= __132__

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Read references:

**Sarin participates in school anniversary event part 1**

__Sarin learns concept of Fraction (Mathematics concept)__

*Challenge yourself with the question before look out for the given solution!!!*

**Upper primary school mathematics question UPQ586**

Sarin used some red and yellow balloons to decorate the school hall for the school anniversary event. 2/3 of the red balloons and 6/7 of the yellow balloons were small balloons. If he had a total of 65 large balloons and 4/5 of them were red balloons, how many balloons did he has altogether?

Solution:

4/5 of the big balloons were red balloons,

5 units = 65

1 unit = 13

The number of big red balloons = 4 × 13 = 52

The number of big yellow balloons = 13

2/3 of the red balloons are small balloons,

Then

1/3 of the red balloons are small balloons

Total number of red balloons = 3 × 52 = 156

6/7 of the yellow balloons were small balloons,

Then,

1/7 of the yellow balloons were small balloons,

Total number of yellow balloons = 7 × 13 = 91

Total number of balloons = 156 + 91 = __247__

__Alternative solution(1):__

By math model,

Per the model,

5 units = 65

1 unit = 13

Total number of balloons = 19 × 13 = __247__

Alternative solution(2) :

By ratio method,

2/3 of the red balloons are small balloons,

R(big) : R(small) : R = 1 : 2 : 3 = 4 : 8 : 12

** ** 6/7 of the yellow balloons were small balloons,

Y(big) : Y(small) : Y = 1 : 6 : 7

Since,

4/5 of the big balloons were red balloons,

R(big) : Y(big) : Big = 4 : 1 : 5

By matching R(big) to 4,

R(big) : Y(big) : R(small) : Y(small) : Big : Total = 4 : 1 : 8 : 6 : 5 : 19

Therefore,

5 units = 65

1 unit = 13

Total number of balloons = 19 × 13 = __247__

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Story reference :** **

Math concept reference:

__Sarin learns Average Number, Percentage, Tables and Charts (Math Concept)__

*Challenge yourself with the question before look out for the given solution!!!*

**Upper primary school mathematics question UPQ585**

Sarin’s mother bought some number of liquid soaps at an average price of $11 from the Kluang Mall. If she were to buy another liquid soap which cost $37, the average price would then become $13. How many numbers of liquid soaps did she buy?

*Solution:*

The math model,

The original average price = $11

The final average price = $13

The difference = 13 – 11 = $2

The added liquid soap to be $13 (The average number)

The remaining dollars = 37 – 13 = $24

The number of liquid soaps she buy = 24 ÷ 2 = __12__

Read the posting on **Sarin participates in school anniversary event part 1 **for story on school anniversary.

Read also the posting on ** Sarin learns the concept of Fraction (mathematics concept)** to understand Fraction.

*Challenge yourself with the question before look out for the given solution!!!*

**Upper primary school mathematics question UPQ584**

In the school anniversary event, Class 6A and Class 6B had to sell 480 dining tickets altogether. After Class 6A sold 3/4 of their share and Class 6B sold 5/8 of their share, each of the class had the same number of tickets left. How many tickets did Class 6B have to sell?

Solution:

Construction the model based on both classes have same number of tickets left finally.

From the model,

20U = 480

1U = 24 tickets

The number of tickets Class 6B need to sell = 8 × 24 = __192__

__Alternative solution:__

The unsold tickets are equal,

1/4 = 3/12 of Class 6A is equal to 3/8 of Class 6B

Total tickets = 480

Total in units = 12 + 8 = 20

20 units = 480

1 unit = 24 tickers

The number of tickets Class 6B need to sell = 8 × 24 = __192__

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The blog postings are about the Singapore Math. The readers can learn from the postings about Solving Singapore Primary School Mathematics. The blog presents the Math Concept, the Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can acquire the skill of dealing with the Math Modeling, the Math Problem Solving and the Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on __Shapes__, __aerimeter__ and Area.

You can read posting **Sarin learns Perimeter, Area and Volume in school (Math Concept) **to learn the concept of volume.

Read also the posting on **Sarin learns Shapes and Solids in school (Math Concept)** to know about solid.

*Challenge yourself with the question before look out for the given solution!!!*

**Upper primary school mathematics question UPQ583**

The figure is made of 2 identical circles of diameter 14 cm. Find the area of the yellow region. (take π = 3.14)

Solution:

Area of the rectangle = 14 × 28 = 392 cm^{2}

Area of the two circles = 2 × π × 7 × 7 = 307.72 cm^{2}

The differences = 392 – 307.72 = 84.28 cm^{2}

The yellow region is of the differences

= 84.28 ÷ 4

= __21.07__ cm^{2}

Alternative solution:

The area of the blue square = 7 × 7 = 49 cm^{2}

The area of the quadrant = × × 7 × 7 = 38.465 cm^{2}

The difference = 49 – 38.465 = 10.535 cm^{2}

The area of yellow region = 2 × 10.535 = __21.07__ cm^{2}

Read the posting on **Sarin learns Shapes and Solids in school (Math Concept)** to know about shapes.

Read the posting on ** Sarin learns Lines and Angles (Math concept)** to understand the concept of lines and angles.

*Challenge yourself with the question before look for the given solution!!!*

**Upper primary school mathematics question UPQ582**

In the figure, not drawn to scale, ABC is an equilateral triangle, ABD is an isosceles triangle, and ABE is a straight line. Find ∠ DBE.

*Solution:*

Since ABC is an equilateral triangle,

∠ CAB = ∠ ABC = ∠ BCA = 60^{0}

∠ BAD = 60^{0} – 33^{0} = 27^{0}

ABD is an isosceles triangle,

AB = BD,

∠ BAD = ∠ BDA = 27^{0}

ABE is a straight line and ABD is a triangle,

∠ DBE = 27^{0} + 27^{0} = 54^{0}^{ }(Exterior angle of a triangle)

Read the posting on **Sarin learns Shapes and Solids in school (Math Concept)** to know about shapes.

Read the posting on ** Sarin learns Lines and Angles (Math concept)** to understand the concept of lines and angles.

*Challenge yourself with the question before look for the given solution !!!*

**Upper primary school mathematics question UPQ581**

In the figure, ABCD is a square and EFD is an isosceles triangle. ∠AED = 72º and ∠ EFB = 51º. Find ∠ EDF.

*Solution:*

ABCD is a square,

∠ EBF = 90^{0} or right angle

Therefore, EBF is a right-angled triangle,

∠ BEF = 180^{0} – 90^{0} – 51^{0} = 39^{0}

AEB is a straight line,

∠ DEF = 180^{0} – 72^{0} – 39^{0} = 69^{0}

EDF is an isosceles triangle with DF = DE,

∠ DFE = ∠ DEF = 69^{0}

And,

∠ EDF = 180^{0} – 69^{0} – 69^{0} = __42 ^{0}__

You can read the story of Sarin and the siblings making greeting cards for their mother by reading the posting on **Sarin, Hairu and Fatimah make greeting cards for Mother’s Day** in Welcome page.

You can read the posting on **Sarin learns Perimeter, Area and Volume in school(Math Concept) **to learn the concept of area.

*Challenge yourself with the question before look for the given solution!*

**Upper primary school mathematics question UPQ580**

Sarin, Hairu and Fatimah made a greeting card for her mother. They made the greeting card as like a photo frame. They used 4 identical coloured rectangular shape papers and arrange them as the figure shown to make the frame of the greeting card.

The perimeter of each of the coloured rectangular shape paper is 30 cm and the inner square formed by the frame is 23 cm^{2} more than the total area of the coloured rectangular papers. Find the area of the inner square.

*Solution:*

Based on the arrangement of the rectangular papers shown in the figure

The shape of the greeting card is a square

The outer length of the frame = The sum of the length and breadth one rectangular paper

Given the perimeter of one rectangular paper = 30 cm

The outer length = 30 ÷ 2 = 15 cm

The area of the greeting card = 15 × 15 = 225 cm^{2}

From the model,

The area of 2 inner squares = 225 + 23 = 248 cm^{2}

The area of an inner square = 248 ÷ 2 = __124__ cm^{2}

Or,

The area of 2 frames = 225 – 23 = 202 cm^{2}

The area of the frame = 202 ÷ 2 = 101 cm^{2}

The area of the inner square = 225 – 101 = __124__ cm^{2}

Read the posting on **Sarin participates in school anniversary event part 1 **for story on school anniversary.

Read the posting on ** Sarin learns concept of ratio in school** to understand Ratio.

*Challenge yourself with the question before look out for the given solution*

**Upper primary school mathematics question UPQ579**

Fatimah were given equal number funfair coupons to sell. They did not manage to completely sell all their share of coupons. Sarin sold 7 times as many coupons as Hairu and was left with 15 unsold coupons . 1/2 of the coupons sold by Fatimah was 5 more than those by Hairu. The 3 of them sold a total of 260 funfair coupons.

a. How many coupons did Fatimah sell?

b. How many coupons did each of them receive?

Solution:

a. The total number of coupons sold = 260

From the model,

10 units + 10 = 260

10 units = 250

1 units = 25 coupons

The number that Fatimah sold = 2 × 25 + 10 = __60__ coupons

b. The number of coupons each of them has = 7 × 25 + 15 = __190__ coupons

__Alternative solution:__

Half of Fatimah sold was 5 more than Hairu sold,

Fatimah : Hairu = (2U+10) : 1U

Sarin sold 7 times as many as Hairu sold,

Sarin : Hairu = 7U : 1U

The ratio of number of coupons sold,

Sarin : Hairu : Fatimah = 7U : 1U : (2U+10)

In total, 7U + 1U + 2U + 10 = 10U + 10

Given the total number of ticket sold = 260

Hence,

10U = 250

1U = 25

a. The number of coupons sold by Fatimah = 2 × 25 + 10 = __60__

b. The number of coupons each of them has at first

= 7 × 25 + 15 = __190__

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