2018 Challenging Math Questions for PSLE

2018 Challenging Math Questions for PSLE

Be better prepared for PSLE Math with these 7 Challenging Math Questions. Specially curated by our team of Math experts, these adapted questions come from a pool of 2018 questions from Top Schools’ Primary 6 Prelim Exam Papers.

Test yourself to see if you can solve these questions. Check out our detailed solutions if you need any help along the way.

Knowing how to solve these questions will definitely help you level up your Math skills in Primary 6 and increase your chances of getting that A* in your Math exam easily.

1. Question on the Topic of Speed

(Adapted from Rosyth 2018 PSLE Prelim Paper P2/Q15)

Crab A and Crab B started walking at the same time from the opposite ends of a 30-cm box. Each crab would turn in the opposite direction and continue walking upon reaching the end of the box. The average speed of Crab A was 1 cm/s and the average speed of Crab B was 0.6 cm/s.

How many times did they meet each other if they walked for 10 min?

(Assuming that their turning time is negligible.)

Find out the answer to this speed question
Speed-question-1

The period given to us is 10 minutes.

Let’s convert 10 minutes into seconds.

1 min = 60 s

10 min = 60 x 10 = 600 s

Important note:

Whenever the faster crab walks from one end of the box to the other end of the box, it overtakes the slower crab ONCE regardless of where the slower crab is.

Hence, to find the number of times both crabs met, we simply need to find the number of 30-cm paths that the faster crab takes.

In this case, Crab A is faster.

Total distance traveled by Crab A in 10 minutes

= 1 x 600

= 600 cm

No. of times both crabs meet

= 600/30

= 20

2. Question on the Topic of Percentage

(Adapted from Rosyth 2018 PSLE Prelim Paper P2/Q11)

Miss Li and Miss Ho had 1674 plastic bags. Miss Li used up 25% of her plastic bags and Miss Ho used up 70% of her plastic bags. After that, Miss Li had twice as many plastic bags left as Miss Ho.

a) How many plastic bags did Miss Li have in the end?

b) What was the percentage decrease in the total number of plastic bags?

Answer to Part a) of this Percentage Question

a) Let’s start by working backwards!

The question ends with Miss Li having twice as many plastic bags as Miss Ho. So here’s the model.

percentage-question-1

We know that Miss Ho used up 70% of her plastic bags.

This means that the number of plastic bags she had in the end must make up of 100% – 70% = 30% of the total number of plastic bags that she had.

percentage-question-2

We can now split 1 big unit into 3 smaller units.

Since Miss Li used up 25% of her plastic bags,the remaining 6 smaller units that Miss Li had must form 100% – 25% which is 75% of the number of plastic bags she had.

75% = 3/4

3/4 is represented by 6 units.

1/4 will be represented by 2 units ( 6/3 ).

percentage-question-3

From the model,

18 units = 1674 plastic bags

6 units = 1674 /3 = 558 plastic bags

Answer to Part b) of this Percentage Question

b) Total no. of units = 18

No. of units that represents the decrease = 9

% decrease in the total number of plastic bags = 9/18 = 50%

3. Question on the Topic of Average

(Adapted from 2018 Tao Nan Prelim Paper P2/Q8)

The total value of the numbers printed on some labels is 520. Each label is printed with a different 3-digit odd number. The average value of all the numbers is 130. The difference between the greatest and smallest number is 6.

Find the smallest number printed on the label.

Answer to this Average Question

Number of numbers printed = 520 / 130 = 4

Now that we know we have 4 numbers,

let’s try to visualise what is given in the question.

Since all the numbers are different and they are odd numbers,

The difference between the numbers must be in 2.

(Think of the odd numbers 1, 3, 5 etc.)

We also know that the difference between the smallest and greatest number is 6.

Hence, we can think of the numbers as shown below:

average-question-1

520 – 2 – 4 – 6 = 508

Smallest number = 508 / 4 = 127

4. Question on the Topic of Area of Triangle

(Adapted from 2018 Maha Bodhi Prelim Paper P2/Q13)

In the rectangle shown below, AB = 28 cm and BC = 21 cm. The ratio of DE : EF : FB = 3 : 2 : 5, GC is ¾ of BG and AF = BF.

What fraction of the rectangle ABCD is shaded?

Area-of-Triangle

Answer to this Question on Area of Triangle

Let’s break the rectangle up into 2 parts – Triangle ABD and Trianlge BCD.

Area-of-Triangle-1

Triangle ADE, Triangle AEF and Triangle AFB share the same height.

(*Indicated in blue)

Area of Triangle ADE = 1/2 x 3 x Height

Area of Triangle AEF = 1/2 x 2 x Height

Area of Triangle AFB = 1/2 x 5 x Height

Area-of-Triangle-2

Triangle DBG and Triangle DBC share the same height DC.

Area of Triangle DBG = 1/2 x 4 x DC

Area of Triangle DBC = 1/2 x 3 x DC

Triangle ABD

Shaded : Unshaded

= 2 : 3 + 5

= 2 : 8

= 1 : 4

Triangle BCD

Shaded : Unshaded

= 4 : 3

Since the area of Triangle ABD and Triangle BCD are equal,

we need to find the common multiple between 5 and 7 which is 35.

Triangle ABD

Shaded : Unshaded

= 1 : 4

= 7 : 28

Triangle BCD

Shaded : Unshaded

= 4 : 3

= 20 : 15

Fraction of rectangle ABCD that is shaded

= (7  + 20) / (7 + 28 + 20 + 15)

= 27/70

5. Question on the Topic of Circles

(Adapted from 2018 Pei Hwa Prelim Paper P2/Q3)

Three circular wheels are places side by side as shown below. AB is 7.5 cm and it cuts through the centres of all the circles.

Find the circumference of the 3 wheels. (Take pi = 3.14)

circle-question

Answer to this Question on Circumference of Circle

Circumference of the wheels

= pi x d

= 3.14 x 7.5

= 23.55 cm

6. Question on the Topic of Geometrical Shapes

(Adapted from 2018 CHIJ St Nicholas Prelim Paper P2/Q17)

Lily drew a figure that is made up of 3 different squares and a circle with a diameter of 10 cm.

What is the total shaded area? (Take pi = 3.14)

geometry-circles-squares

Answer to this Question on Geometrical Shape

Let’s add in 2 lines and move the 2 shaded triangles to the middle to form a square as shown.

geometry-circles-squares-1

From the above, we can tell that the total area of the shaded triangles is 1/4 of the unshaded square.

geometry-circles-squares-2

Radius = 10/2 = 5 cm

Area of circle

= pi x 5 x 5

= 3.14 x 25

= 78.5 cm^2

Area of unshaded square

= 1/2 x 5 x 5 x 4

= 50 cm^2

Area of shaded triangles

= 1/4 x 50

= 12.5 cm^2

Therefore, area of shaded part

= Area of circle – Area of unshaded square + Area of shaded triangles

= 78.5 – 50 + 12.5

= 41 cm^2

7. Question on the Topic of Pattern

(Adapted from 2018 ACS Prelim Paper P2/Q12)

Santa uses some identical butter biscuits and chocolate biscuits to form some figures that follow a pattern as shown.

a. The table shows the number of butter biscuits and chocolate biscuits for the first 3 figures. Complete the table for Figure 4.

Figure Number 1 2 3 4
Number of chocolate biscuits 4 9 16
Number of butter biscuits 3 5 7
Total number of butter biscuits and chocolate biscuits 7 14 23

b. A figure in the pattern has a total of 529 chocolate biscuits. What is the figure number?
c. Another figure in the pattern has a total of 63 butter biscuits. What is the total number of butter biscuits and chocolate biscuits in this figure?

Answer to Part a) of this Question on Pattern

a) Let’s think of how the no. of chocolate biscuits is linked to the figure no.!

Number of chocolate biscuits = (Figure no. + 1)^2

We start off with 3 butter biscuits in Figure 1.

As the figure no. changes, the no. of butter biscuits also changes, but how is the no. of butter biscuits linked to the figure no.?

Number of butter biscuits = 3 + [(Figure no. – 1) x 2]

Hence, for Figure 4,

Number of chocolate biscuits

= (Figure no. + 1)^2

= (4+1)^2

= 5^2

= 25

Number of butter biscuits

= 3 + [(Figure no. – 1) x 2]

= 3 + [(4 – 1) x 2]

= 3 + [3 x 2]

= 3 + 6

= 9

Total no. of chocolate biscuits and butter biscuits = 25 + 9 = 34

Answer to Part b) of this Question on Pattern

b) Number of chocolate biscuits = (Figure no. + 1)^2

Since we want to find the square root of 529, we can use esitmation.

20 x 20 = 400

25 x 25 = 625

Hence, the number we are looking for must be between 20 to 25.

Thinking about the last digit, which similar digit will give you 9 at the back when you multiply them?

3 x 3 = 9

So our guess is 23!

Number of chocolate biscuits = (Figure no. + 1)^2

529 = (Figure no. + 1)^2

 Figure no. + 1 = 23

Figure number

= 23 – 1

= 22

Answer to Part c) of this Question on Pattern

c) Number of butter biscuits = 3 + [(Figure no. – 1) x 2]

63 = 3 + [(Figure no. – 1) x 2]

[(Figure no. – 1) x 2] = 63 – 3 = 60

(Figure no. – 1) = 60/2 = 30

Figure no. = 30 + 1 = 31

Number of chocolate biscuits

= (Figure no. + 1)^2

= (31 + 1)^2

= 32^2

= 1024

Total number of butter biscuits and chocolate biscuits

= 1024 + 63

= 1087

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Comments
  • Krithiga

    This is very useful thank you so much for the information

    Reply
  • ccyxcrystal

    this helped tremendously for my cousin as he is weak in math and thank you for these questions.

    Reply
  • Anonymous

    Good

    Reply
    • Tham Jun Xuan

      this improve my math to A1,thank you

      Reply
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