SEA MATHEMATICS 2024
Section 1
1. Write the numeral for two hundred and thirty thousand, five hundred and sixty-three.
Two hundred and thirty thousand
Five hundred
Sixty-three
230000
+
500
63
230563
m
Answer: 230 560
.co
2. Write the value of the underlined digit in the numeral below.
95 367
1 000
5
­
Tens
Units
hs
Thousands Hundreds
100
3
10
6
1
7
at
Tens of
Thousands
10 000
9
m
5 ´1000 = 5000
Answer: 5 000
as
s
3. A common factor of 15 and 18 is 1. What other factor is common to 15 and 18?
sp
3
5
1 5
5
1
2
3
3
1 8
9
3
1
fa
15 = 1´ 3 ´ 5
18 = 1´ 2 ´ 3 ´ 3
Besides 1, we observe that 3 is the only other factor common of 15 and 18.
Answer: 3
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Page 1 of 30
4. Write ONE of the following symbols in the box below to make the number sentence
correct.
>
=
<
>
8639
.co
Answer: 8693
m
8693
8639
The thousands digits (8) and the hundreds digits (6) are the same in both numbers.
However, the tens digit (9) of 8693 is bigger than the tens digit (3) of 8639. This makes
8639 the larger number.
The symbol > means greater than.
5. 9 9 2 7
Alternatively:
1 0 3
6. 7 -
as
s
Answer: 103
1 0 3
m
0
hs
0 2 7
- 2 7
9 9 2 27
at
9 9 2 7
-9
2
=
3
fa
sp
7 2
1 3
3 ( 7 ) - 1( 2 ) 21 - 2
=
3
3
19
=
3
1
=6
3
Answer: 6
Alternatively:
2
2
7 - = 6 +13
3
3 2
= 6+ 3 3
1
= 6+
3
1
=6
3
1
3
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Page 2 of 30
7. Write 0.40 as a fraction in its lowest terms.
40
100
20
=
50
2
= (in its lowest terms)
5
0.40 =
.co
m
2
5
Answer:
8. 3.12 ´ 4 =
hs
3 .1 2
´
4
12.48
m
9. 15% of 300 =
at
Answer: 12.48
15 3
´ 300
100
= 15 ´ 3
= 45
as
s
15% of 300 =
fa
sp
Answer: 45
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Page 3 of 30
10. An incomplete pattern is shown below.
36,
28,
21,
15,
______,
6
What is the missing element in the pattern?
36 - 8 = 28
28 - 7 = 21
21 - 6 = 15
15 - 5 = 10
m
\
.co
And 10 - 4 = 6
Answer: 10
sp
as
s
m
at
hs
11. What is the length of the pencil shown below?
fa
The left end of the pencil is at the 1.5 cm mark.
The right end of the pencil is at the 9.5 cm mark.
Hence, the length of the pencil is 9.5 - 1.5 = 8 cm
Answer: 8 cm
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Page 4 of 30
12. Aidan left home at 6:45 a.m. and arrived at school at 7:25 a.m.
How long was his journey?
Hr
6
Arrival time
7 :
Departure time 6 :
0 :
Min
85
25
45
40
[Recall 1 hour = 60 minutes]
fa
sp
as
s
m
at
hs
.co
m
Answer: 40 minutes
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Page 5 of 30
hs
The opposite sides of the four-sided figure
are equal and each internal angle is a right
angle. The shape is a rectangle.
1
Length of the rectangle = 4 cm
2
Width of the rectangle = 4 cm
1
\Area of the rectangular shape = 4 ´ 4 cm 2
2
æ9 ö
= ç ´ 4 ÷ cm 2
è2 ø
= 18 cm 2
fa
sp
as
s
m
at
What is the area of the shape?
.co
m
13. A shaded shape is shown on the 1 cm grid below.
Alternative Method
There are 16 whole squares and 4 half-squares. The 4 halves are equal to 2 wholes.
Area of rectangle = 16 cm2 + 2 cm2 = 18 cm2
Answer: 18 cm2
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Page 6 of 30
.co
What is the difference between their masses?
1 kg = 1000 g
\ 2.5 kg = 2.5 ´1000
hs
= 2500 g
= 2 5 0 0
-1 6 0 0
at
Difference in masses
m
14. Two packs of sugar are shown below.
fa
sp
as
s
Answer: 900 g
m
9 0 0
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Page 7 of 30
.co
m
15. Gillian is standing at O facing NW. She makes quarter turns and is now facing SE.
as
s
m
at
hs
What is the least number of quarter turns made by Gillian?
sp
To face SE, Gillian can make 2 quarter turns anti-clockwise, first to SW and then to
SE.
She could also have made 2 quarter turns clockwise to be in the same position. first to
NE and then to SE.
fa
Answer: 2 quarter turns
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Page 8 of 30
hs
.co
m
16. How many vertices are there in the solid shown below?
at
The vertices number 6 are named A to F, as shown.
fa
sp
as
s
m
Answer: 6 vertices
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Page 9 of 30
.co
m
17. Draw ALL the lines of symmetry on the shape below.
The figure has two lines of reflective symmetry. These are drawn ‘dotted’. The figure
can be folded along either line without any overlapping.
fa
sp
as
s
m
at
hs
The completed figure with the lines of symmetry, shown dotted, looks like:
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Page 10 of 30
18. The mean of three numbers is 12. Two of the numbers are 10 and 11. What is the third
number?
Mean of the three numbers is 12.
Hence, the sum of the three numbers is 12 ´ 3 = 36
Two numbers are 10 and 11.
Their total = 10 + 11
= 21
.co
m
Hence, the third number is 36-21
= 15
Answer: 15
19. The tally chart below shows the types of gifts students received.
hs
Gifts Students Received
Gift
Tally
Bicycles
at
|||| ||||
Books
|||| |
m
Cellphones
as
s
Tablets
|||| |||| |||| ||
|||| |||| ||||
Which gift represents the mode?
sp
Gift
Bicycles
fa
Books
Cellphones
Tablets
Tally
|||| ||||
|||| |
|||| |||| |||| ||
|||| |||| ||||
Total
5+ 4 = 9
5 +1 = 6
5 + 5 + 5 + 2 = 17
5 + 5 + 5 = 15
The mode is the item that occurs most often. Cellphones are received as gifts more
than any other mentioned item.
\The mode is cellphones.
Answer: Cellphones
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Page 11 of 30
20. The pictograph below shows the colours of students' bookbags.
Colours of Bookbags
Red
.co
m
No. of
bookbags
Black
Blue
Colours
Brown
hs
If 32 of the bookbags are black, how many are blue?
m
at
4 faces represent 32 bags.
32
\1 face will represent
= 8 bags
4
The number of blue bags is represented by 3 faces
as
s
3 faces represent: 8 bags × 3 = 24 bags
fa
sp
Answer: 24 bags
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Page 12 of 30
Section 2
3
´ 32 =
4
2
-1
Left-hand side:
3
´ 32 = 24
4
Equating both sides:
2
–1 = 24
\
2
= 24 + 1
2
= 25
m
21.
.co
= 25
=5
hs
=5
Answer:
at
22. A packet of sweets was shared among 4 students. Each student received 15 sweets and
there were 5 sweets remaining.
m
What was the total number of sweets in the packet?
as
s
Each student received 15 sweets.
Therefore, the four students received a total of 15 ´ 4 = 60 sweets
The number of sweets remaining = 5
sp
\The total number of sweets that were in the packet = 60 + 5
= 65
fa
Answer: 65 sweets
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Page 13 of 30
.co
m
23. Two fraction models, A and B, are shown on the grid below.
Explain why the fraction models represent equivalent fractions.
hs
Answer: In both models, the whole is the same size.
m
at
In diagram A, the shape is divided into two equal parts with one part shaded. Hence,
!
the fraction shaded represents " of the square A.
In diagram B, the shape is divided into four equal parts with the shaded region
"
consisting of two parts. Hence, the fraction shaded is of square B.
#
sp
as
s
To show that both fractions are equivalent, we can divide shape A into four quarters.
fa
In both shapes, the shaded area comprising two whole squares are shaded.
Hence,
1 2
= are equivalent fractions since they represent the same part of the whole.
2 4
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Page 14 of 30
24. Two fruit stalls sell mangoes at the prices shown below.
Stall A
Stall B
$9.00 for 6 mangoes
$5.00 for 4 mangoes
at
hs
Stall B:
4 mangoes cost $5.00
$5.00
\1 mango will cost
4
$5.00
36 mangoes will cost
´ 36 = $45.00
4
.co
Stall A:
6 mangoes cost $9.00
$9.00
\1 mango costs
6
$9.00
36 mangoes will cost
´ 36 = $54.00
6
m
What is the cheaper price of 36 mangoes between Stall A and Stall B?
m
Since $45 < $54 , the mangoes are cheaper in Stall B.
$5 4 . 0 0
as
s
- $4 5 . 0 0
$ 9 . 0 0
sp
Hence, the 36 mangoes will cost $9.00 less in Stall B than in Stall A.
fa
Answer: $9.00 cheaper in Stall B
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Page 15 of 30
25. The number line below shows the values of A and C.
sp
as
s
m
at
hs
.co
m
The length of AC is 4 times the length of BC. What is the value of B?
2
1
Length of AC = 5 - 4
9
3
47 39
=
9
9
8
=
9
AC is 4 times the length of BC, so BC is one quarter the length of AC
8
\Length of BC = ÷ 4
9
2
=
9
+
"
$
"
= 5$
fa
B
2 2
\B = 5 9 9
=5
Answer: 5
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Page 16 of 30
26. An arch was made using 280 balloons. For every 4 red balloons, 3 blue and 7 green
balloons were used.
How many blue balloons were used to make the arch?
A set comprising 4 red + 3 blue + 7 green balloons form a repeated pattern.
The set has a total of 14 balloon
If 280 balloons were used in all then the number of sets is 280 ÷ 14 = 20
m
There are 3 blue balloons in each set of 14 balloons.
Hence, the number of blue balloons in 20 sets will be 20 × 3 = 60
.co
Answer: 60 blue balloons
at
How many rings were in each set?
hs
27. Kai bought sets of jewelry containing rings and bracelets. Each set cost $25 and
contained 3 more rings than bracelets. Kai spent a total of $300 and received 24
bracelets.
$300
$25
= 12 sets
as
s
m
Number of sets in $300 at $25 per set =
There are 12 sets containing 24 bracelets
So 1 set will contain 24 ÷ 12 = 2 bracelets
sp
Each set has 3 more rings than bracelets
\The number of rings per set = 2 + 3
= 5 rings
fa
Answer: 5 rings
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Page 17 of 30
28. David has $150.00 to buy pencils and rulers.
Pencils
Rulers
$1.84 each
$8.13 each
Explain how estimation can be used to determine whether or not David has enough
money for 15 pencils and 15 rulers.
.co
m
Answer: Pencils cost 16¢ less than $2.00.
Rulers cost 13¢ more than $8.00.
Hence, one pencil and one ruler costs a little less than $8.00 + $2.00 = $10.00.
So, 15 pencils and 15 rulers will cost less than $10 ´15 = $150 .
David has $150.
\David will have sufficient money to buy 15 pencils and 15 rulers.
at
What is the length of the wire?
hs
29. A piece of wire is bent for a rectangle of width 8 cm. The length of the rectangle is 6
cm longer than the width.
m
Length of rectangle = Width + 6
= 8+6
sp
as
s
= 14 cm
fa
Length of wire = Perimeter of the rectangle = (14 + 8) ´ 2 cm
= 22 cm × 2
= 44 cm
Answer: 44 cm
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Page 18 of 30
30. Phillip plays football every 3 days and cricket every 4 days. He played football and
cricket on 5th February.
Sun
Mon
FEBRUARY
Tue Wed Thurs
1
4
5
6
11
12
13
8
Sat
3
9
10
.co
m
7
Fri
2
What will be the next date on which Phillip will play both football and cricket?
hs
Cricket is played on the following dates in February: 5, 9, 13, 17, 21, …
at
Football is played on the following dates in February: 5, 8, 11, 14, 17, 20…
The common day is date is 17th February.
m
\Phillip will play both games again on 17th February.
fa
sp
as
s
Answer: 17th February
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Page 19 of 30
.co
m
31. A box is packed with identical cubes, as shown below.
Width of the box : 6 cubes
Length of the box : 8 cubes
Height of the box : 4 cubes
at
Volume of the box = 6×8×4 cubes
= 192 cubes
hs
How many more cubes are needed to fill the box completely?
m
The number of cubes in the box = 19
as
s
Hence, the number of additional cubes required to fill the box = 192 - 19
= 173 cubes
fa
sp
Answer: 173 cubes
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Page 20 of 30
32. Mervyn started to tile a room at 8:40 a.m. He took 4 minutes to lay each tile. After he
laid each set of 30 tiles, he took a 45-minute break. Mervyn laid a total of 90 tiles.
At what time did he finish laying all the tiles?
Time taken to lay one tile = 4 minutes
.co
Break time = 45 minutes
m
Time taken to lay 30 tiles = 30 ´ 4 minutes
= 120 minutes
= 2 hours
hs
\Time taken to lay the remaining 90 – 30 = 60 tiles = 60 ´ 4 minutes
= 240 minutes
= 4 hours
at
\Time taken to lay all 90 tiles + break time = 2 hours + 45 minutes + 4 hours
= 6 hours 45 minutes
m
Hr Min
1
= 8: 4 0
85 minutes = 1 hour and 25 minutes
as
s
Start time
Time taken to finish = 6 : 4 5 +
15 : 2 5 h
sp
15 : 25 - 12 = 3: 25 p.m.
fa
Answer: 3:25 p.m.
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Page 21 of 30
m
at
hs
.co
m
33. In the diagram below, AB is a line of symmetry. Shade 2 squares to complete the
symmetrical shape.
fa
sp
as
s
The completed diagram looks like:
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Page 22 of 30
at
hs
.co
m
34. An incomplete hexagon ABCDEF, is shown on the grid below. Insert the point C on
the grid such that the hexagon has two right angles, and draw lines to complete the
hexagon.
fa
sp
as
s
m
The completed diagram looks like:
The right angles are at A and D.
(C can be any point along the straight line marked DX (not including D).
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Page 23 of 30
35. Olivia scored 86, 90 and 70 on three tests. She can earn a Grade A if her mean score is
at least 80.
What is the lowest score she can obtain on the fourth test to earn a Grade A?
Total in the three previous tests = 86 + 90 + 70
= 246
m
To earn a Grade A, Olivia requires a minimum total of 80 ´ 4 = 320
.co
\Olivia requires a minimum score of 320 - 246 = 74 in the 4th test.
fa
sp
as
s
m
at
hs
Answer: 74
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Page 24 of 30
36. The block graph below shows the number of tickets sold by students at a school.
as
s
m
at
hs
.co
m
Tickets Sold by Students
How many students sold at least 3 tickets?
sp
Number who sold 5 tickets = 7
Number who sold 4 tickets = 5
Number who sold 3 tickets = 10
fa
Hence, the number who sold at least 3 tickets, which is 3 or more tickets = 7 + 5 + 10
= 22
Answer: 22 tickets
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Page 25 of 30
Section 3
37. The cost of 1 bag, 1 book and 1 pen is $45. Alex bought 1 bag, 1 book and 2 pens and
paid a total of $51. The cost of 1 bag is twice the cost of 1 book.
What is the cost of 1 bag?
Cost of 1 bag + 1 book + 1 pen = $45
Cost of 1 bag + 1 book + 2 pens = $51
m
The equation below shows the difference in cost of the two sets of items:
.co
(1 bag+1 book +2 pens) – (1 bag+1 book +1 pen) = $51– $45.
\Cost of 1 pen = $51 - $45
= $6
at
hs
Cost of 1 bag + 1 book + 1 pen = $45
Cost of 1 bag + 1 book + $6 = $45
Cost of 1 bag + 1 book
= $45 – $6 = $39
as
s
m
Cost of 1 bag = 2 ´ Cost of 1 book. We can represent this information as follows:
Cost of 1 book
Cost of 1 bag
1 bag and 1 book cost $39
sp
\Cost of 1 book = $39 ÷ 3
= $13
fa
Hence, the cost of 1 bag = $13 ´ 2
= $26
Answer: $26
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Page 26 of 30
hs
.co
m
38. One bottle of liquid soap has the same capacity as 6 bottles of hand sanitizer. Four
bottles of shampoo have the same capacity as 12 bottles of hand sanitizer.
The capacity of 1 bottle of liquid soap is 1.5 litres.
at
What is the capacity of 1 bottle of shampoo, in millilitres?
as
s
m
Capacity of 1 bottle of soap = 1.5 litres
= 1.5 × 1000 ml
= 1 500 ml
1 bottle of liquid soap holds the same as 6 bottles of hand sanitizer
1500
Hence, 1 bottle of hand sanitizer holds
ml = 250 ml
6
sp
12 bottles of hand sanitizer will hold 250 ´12 = 3000 ml
fa
12 bottles of hand sanitizer hold the same as 4 bottles of shampoo
Hence, 3000 ml = Capacity of 4 bottles of shampoo
3000
ml
4
= 750 ml
\Capacity of 1 bottle of shampoo =
Answer: 750 ml
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Page 27 of 30
m
39. Lollipop sticks are used to form a geometrical pattern, as shown below.
1
2
3
4
6
9
Figure
No. of
lollipop
sticks
1
at
7
8
___
9
10
___
26
4
5
6
7
8
9
10
6
9
11
14
16
19
21
24
26
m
3
+3
+2
The completed table looks like:
Figure
1
2
3
No. of
lollipop
4
6
9
sticks
sp
6
2
as
s
+2
5
hs
Figure
No. of
lollipop
sticks
4
4
.co
a) Complete the table below by writing the number of lollipop sticks that will
form Figure 4 and Figure 9.
+3
+3
+2
+2
+2
+3
4
5
6
7
8
9
10
11
14
16
19
21
24
26
fa
b) Describe the pattern rule.
Answer: from the 1st figure which has 4 sticks, we add 2 (4+2) to get 6 sticks in
figure 2, then add 3 (6+3) to get 9 sticks in figure 3. The pattern is repeated,
that is, add 2 to get the number of sticks in figure 4 (9+2) and add 3 to get the
number of sticks in figure 5 (11+3) and so on.
So, starting with figure 1 we +2 then +3, repeating this pattern for all the
subsequent figures.
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Page 28 of 30
40. The incomplete bar graph below shows the time, in seconds, taken by 5 girls to run a
race. The average time taken by the 5 girls to run the race was 25 seconds.
as
s
m
at
hs
.co
m
Time Taken to Run a Race
a) Calculate the time taken by Bella to run the race.
sp
Total time taken by all 5 girls = 25 seconds × 5
= 125 seconds
fa
Time taken by Jemma
Ariel
Ella
Laura
Total
=
35 s
10 s
20 s
45 s
110 s
\Bella took (125 – 110) seconds = 15 seconds
Answer: 15 seconds
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Page 29 of 30
b) What was the time taken by the fastest runner?
Time taken by:
Jemma
35
Ariel
10
Ella
20
Laura
45
Bella
15
Ariel was the fastest since she took the lowest time of 10 seconds.
fa
sp
as
s
m
at
hs
.co
m
Answer: 10 seconds
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Page 30 of 30