SEA Maths 2023

SEA YEAR 2023
Section 1
1. Write the numeral 91 005 in words.
Tens of Thousands Thousands Hundreds Tens Ones

10 000 1 000 100 10 1
9 1 0 0 5
Ninety + thousand five units
Answer: Ninety-one thousand and five
2. Arrange the numbers below in descending order.
3 162 3 612 3 261 3 126
We can place the numbers on a Place Value Chart
Thousands Hundreds Tens Ones

3 1 6 2
3 6 1 2
3 2 6 1
3 1 2 6
Since all digits in the thousands column have the same value, we start by observing
the next column which is the hundreds column.
The digit with the largest value is 6. So, 3 612 is the largest of the four numbers.
The next largest digit is 2, so 3 261 is the second largest number.
We now move to the tens column and examine the tens digit in the two remaining
numbers which are 3 162 and 3 126. Their tens digits are 6 and 2.
Since 6 is larger than 2, then 3162 is larger than 3126.
So, 3 162 is the third largest number and 3 126 is the smallest of the given four
numbers.
Answer: 3 612, 3 261, 3 162, 3 126
Copyright ©2023.Some Rights Reserved. faspassmaths.com 1

3
3. Write 2 as an improper fraction.
8
8 Alternatively:
1=
8 3 ( 2 ´ 8) + 3
2 =
8 16 8 8
2 = 2´ =
8 8 16 + 3
=
3 16 3 8
2 = +
8 8 8 19
=
19 8
=
8
19
Answer:
8
4. Divide 628 by 12.
052
12 6 2 8
-6 0 ¯
28
-2 4
4 (Remainder)
4 1
Answer: 52 with remainder 4 or 52 or 52 in its lowest form
12 3
5. 82 + 62 =
82 + 6 2 = ( 8 ´ 8 ) + ( 6 ´ 6 )
= 64 + 36
= 100
Answer: 100

6. Round 15 296 to the nearest thousand.
In rounding to the nearest thousand, we need to express this number in thousands or larger.
The digits to the right of the thousands digit will be replaced by zero.
The value of the digit to the immediate right of the thousands digit is critical in deciding if
the number of thousands must be increased or not. We refer to the hundreds digit, in this
case, as the deciding digit. If this digit is less than 5, we round down and if it is 5 or more,
we round up by adding one thousand to the thousands digit.
Tens of Thousands Thousands Hundreds Tens Units

10 000 1 000 100 10 1
1 5 2 9 6

Remains the same This digit is less than 5.
So, we round down by replacing it and all digits to
its right by 0.
Answer: 15 000
1
7. Max ate 5 cherries which were of the total number of cherries he picked. How many
9
cherries did Max pick?
1
If 5 cherries were of the number picked, then, the number of cherries picked is
9
5 × 9 = 45
Alternatively:
1
of the number of cherries picked is 5.
9
The whole has 9 equal parts, and one part is 5.
The whole is 5 × 9 = 45.

Answer: 45

8. Write the total value of the bills and the coin shown below.
Total in $ and cents is:
$20 + $20 + $5 + $1 + $0.05

$20.00
$20.00
$ 5.00
$ 1.00
$ 0.05
$46.05
Answer: $46.05
9. Calculate 15% of 120.
15
15% =
100
15
15% of 120 = ´ 12 0
10 0
3
15
= ´12
2 10
6
3 ´ 12
=
21
= 3´ 6
= 18
Answer: 18

10. Write the digits 1 2 3 4 in the squares below to create an addition
problem with the largest sum.
The two largest digits are 3 and 4. These two digits are placed in the tens column. It
doesn’t matter where the last two digits, 1 and 2 are placed. Either with the 3 or the 4
will make no difference in the final sum.
Answer:
Tens Units Tens Units
(10) (1) OR. (10) (1)
4 1 4 2
+ 3 2 + 3 1
7 3 7 3
11. A part of the calendar for the month of April is shown below.
April
Sun Mon Tue Wed Thu Fri Sat
10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25
On what day of the week is the 2nd of the April?
By counting backwards and forwards the calendar is completed to show:
April
Sun Mon Tue Wed Thu Fri Sat
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
We observe that the 2nd of April is on a Sunday.
Answer: Sunday

12. The total mass of a pineapple and a pawpaw is shown on the scale below. The mass of
the pineapple is 2.75 kg.
What is the mass of the pawpaw, in grams?
The pointer on the scale is midway between 3 and 4.

1
Hence, the combined mass of the pineapple and the pawpaw is 3 kg or 3.5 kg.
2
3.5 kg = 3.5 ´1000
= 3 500 g 3500g
-2 7 5 0 g
The mass of the pineapple is 2.75 kg.
750g
2.75 kg = 2.75 ´1000
= 2 750 g
\Mass of the pawpaw is (3 500 – 2 750) g = 750 g
Alternatively:
Mass of the pawpaw is
1 kg = 1000 g
3 . 5 0 kg \ 0.75 kg = 1000 ´ 0.75
- 2 . 7 5 kg = 750 g
0 . 7 5 kg
Answer: 750 grams

13. Identical cubes are packed into a box as shown below.
How many cubes are in the box?
The length of the box fits 5 cubes.

The width of the box fits 4 cubes.
The height of the box fits 2 cubes.
\The number of cubes in the box = 5 ´ 4 ´ 2 = 40
Answer: 40 cubes
14. Calculate the difference in length between the eraser and the pen.
1 cm 4 cm 6 cm 17 cm
\Length of eraser = (4 – 1) cm
= 3cm
\Length of pen = (17 – 6) cm

= 11cm
The difference in length between the pen and the eraser = (11 – 3) cm = 8 cm
Answer: 8 cm

15. The pyramid shown below has a square base.
How many edges of the pyramid have a length of 6 cm?
The slant edges of the pyramid are 6 cm in length.
There are 4 slant edges
The four slant edges of the pyramid are therefore each of length 6 cm.
Answer: 4 edges

16. The arrow below is pointing to B. The arrow moves three quarter-turns in an anti-
clockwise direction.
To which letter is the arrow now pointing?
After three quarter-turns in an anti-clockwise direction, the arrow will point to C.
Answer: C

17. Name the type of triangle shown below.
In the above triangle all three sides are unequal in length. The triangle is called scalene.
Answer: Scalene
18. The table below shows the show sizes of some students.
Shoe Size 9 8 7 6 5
Number of Students 11 3 12 9 3
What shoe size represents the mode?
The shoe size with the highest frequency represents the mode.
The shoe size 7 occurs 12 times.
Since 12 is highest frequency, the modal shoe size is 7.
Answer: 7
19. Calculate the mean of the numbers below.
32 14 24 0 5 32
14
Total of all numbers: 32 + 14 + 24 + 0 + 5 = 75 24
0
There are 5 numbers. + 5
75
Total of the numbers !"
Mean = = " = 15
Number of numbers
Answer: 15

20. The incomplete tally chart below shows the pets owned by students.
Pets Owned by Students

Pet Number Tally
Birds 8
|||| |||
Fishes 10
|||| ||||
Dogs 12
Complete the tally chart to show the number of students who own dogs.
The completed tally chart now looks like:
Pets Owned by Students

Pet Number Tally
Birds 8
|||| |||
Fishes 10
|||| ||||
Dogs 12
|||| |||| ||

Section 2
21. Write the missing digit in each box below.
9 2 1
- 2 6 3 7
6 5 8
Consider the ones column, we cannot subtract 7 ones from 1 one, so we re-group:
2 tens and 1 ten and 11 ones as shown below:
Th H T O
1 11
9 2 1
–2 6 3 7
6 5 8 4
We know the difference between the two numbers is 6 584.

We know that we subtracted 2 637 from a number to get this difference.
– 2 637 = 6 584
= 6 584 + 2 637
= 9 221
The completed problem looks like:

9 2 2 1
- 2 6 3 7
6 5 8 4

5
22. 6 - 2 =
14
5 5 Alternatively:
6-2 = 4- 5 84 33
14 14 6-2 = -
5 14 14 14
= 3 +1-
14 51
=
14 5 14
= 3+ -
14 14 9
=3
14 - 5 14
= 3+
14
9
= 3+
14
9
=3
14
9
Answer: 3
14
23. Karen uses 0.65 metres of cloth to make 1 skirt.
How many metres of cloth would she need to make 12 similar skirts?
1 skirt uses 0.65 metres of cloth

12 similar skirts will require 0.65 m × 12 Use 65 (0.65× 100) to
avoid a decimal
multiplicand
65× 12
=(60 × 12) + (5 × 12)
= 720 + 60 = 780
The correct answer is

780 cm ÷ 100 = 7.8 m
Answer: 7.8 metres

24. Kim and Ben estimated the solution to the following problem.
60 ´18
Kim’s Estimate Ben’s Estimate

600 1200
Explain who had the better estimate and how it was calculated.
Both rounded the multiplier, 18 to a multiple of 10 to make the calculations easier.
Kim rounded down 18 to 10 and got a lower estimate of 60 × 10 = 600

Ben rounded up 18 to 20 and got a higher estimate of 60 × 20 = 1 200
10 18 20
Since 18 is closer to 20 than 10, rounding 18 to 20 will give a better estimate than
rounding 18 to 10.
Ben’s estimate is better than Kim’s estimate because he chose 20.
Alternatively, we obtain the exact value of 60 × 18 = 1080
• The difference between Kim’s estimate and the exact value is 1080 – 600 = 480
• The difference between Ben’s estimate and the exact value is 1200-1080 =120
600 1080 1200

Kim’s Exact Ben’s
estimate Value estimate
Ben’s estimate is closer to the exact value than Kim’s estimate since 120 is less than
480.
Ben, therefore, has the better estimate.

25. Shop A and Shop B sell the same packet of milk. Shop A sells 4 packets of milk for
$26.00. Shop B sells 3 packets of milk and a container for $26.00. The price of the
container is $3.20.
What is the difference in the price of a packet of milk at Shop A and Shop B?
At Shop A, 4 packets of milk cost $26.00.

\Cost of 1 packet of milk = $26.00 ÷ 4
= $6.50
At Shop B, 3 packets of a milk and a container for $3.20 cost $26.00.

Hence, the 3 packets of milk at shop B will cost:
26.00
- 3.2 0
2 2.80
\Cost of 1 packet of milk at B = $22.80 ÷ 3

= $7.60
The difference in cost of a packet of milk at Shop B and Shop A = $7.60 - $6.50
= $1.10
Answer: At shop B one packet of milk is $1.10 more than it is at shop A.

26. Andy, Tom and Brad shared $85 among themselves. Andy received $10 more than
Tom while Brad received $20 more than Andy.
How much money did each boy receive?
Let us represent each boy’s share in a diagram.
Tom Tom’s share
Tom $10 Andy’s share is $10 more than Tom
Tom $10 $20 Brad’s share is Andy’s share plus $20.
We know that the whole is $85. So, we can combine all the shares to make $85.
If we add the total in the known boxes we get: $10+$10+$20 = $40
This $40 is part of the $85, so we subtract it from $85 to get: $85 – $40 = $45
What remains is 3 equal units where each unit represents Tom’s share.
So, Tom’s share × 3 = $45
Tom’s share $45 ÷ 3 = $15
Tom got $15
Andy got $15 + $10 = $25
Brad got $15 + $10 + $20 = $45
Answer: Tom got $15

Andy got $25
Brad got $45

27. The cost of a bag is three times the cost of a cap. The total cost of 2 bags and 3 caps is
$180.
What is the cost of 1 cap?
Representing the cost of a cap as one unit, we have:
1 cap
1 bag
2 bags and 3 caps
In all, there are 9 equal units.

1 unit represents the cost of a cap.
Cost of 9 caps = $180
Cost of 1 cap = $180 ÷ 9
= $20
Answer: $20

28. The first 4 elements of a pattern are shown below.
How many squares will form the 11th element?
The number of squares in the 1st element = 1 + 2 + 3

The number of squares in the 2nd element = 2 + 3 + 4
The number of squares in the 3rd element = 3 + 4 + 5
The number of squares in the 4th element = 4 + 5 + 6
Hence, following the pattern, the number of squares in the 11th element = 11 + 12 + 13
= 36
Answer: 36 squares
State the pattern rule.
The rule for the number of squares is always the sum of three numbers.
The first number is the number of the pattern. The second number is the number of
pattern + 1. The third number of the pattern is the number of pattern + 2.
So, if the number of the pattern is n, then the 1st number is n, the 2nd number will be
n + 1 and the third number will be n + 2 .
The number of squares in the nth pattern = n + n + 1 + n + 2 which simplifies to 3n + 3

29. Draw a rectangle with an area of 18 cm2 on the grid below.
Our rectangle is 9 cm long and 2 cm wide.
We could have drawn a rectangle 6 cm ´ 3 cm or any other rectangle where length ´

width = 18 cm2.
For whole number dimensions there is a third possibility, 18 cm ´ 1 cm, but clearly this
will not fit on the given grid.

30. A clock shows that the time in Trinidad is 6:00 a.m. The clock is 15 minutes slow. The
time in Trinidad is 2 hours ahead of the time in Canada.
What is the correct time in Canada?
Time on the clock is 6:00 a.m.

Clock is 15 minutes slow.
\Correct time in Trinidad when the clock reads 6:00 am is = 6 : 00 + 0 :15 a.m.
= 6 :15 a.m.
The correct time in Canada will be 6 :15 a.m.

-2 : 00
4 :15 a.m.
Answer: 4:15 a.m.

31. Four identical squares are placed side by side to form the rectangle ABCD.
Calculate the difference between the perimeter of one square and the perimeter of the
rectangle ABCD.
Perimeter of one square = 12 cm× 4

= 48 cm
Perimeter of rectangle = 2 ( 48 + 12) cm

= 120 cm
Difference between the perimeter of the rectangle and the perimeter of the square
= (120–48) cm
= 72 cm
Answer: 72 cm

32. Keisha’s luggage was over limit by 5 kg. When she removed her hairdryer and
shampoo, her luggage was below the limit by 500 g. When she put back the hairdryer,
the luggage was over the limit by 2 kg.
What was the mass of the hairdryer, in grams?
We are given the following statements
1. Luggage with hairdryer and shampoo is over the limit by 5 kg.
2. Luggage without hairdryer and shampoo is below the limit by 500 g
3. Luggage with hairdryer and no shampoo is over by 2 kg.
Luggage with hairdryer and shampoo is 5 kg over.

Luggage with hairdryer is 2 kg over.
Hence, the mass of the shampoo is (5 – 2) kg = 3 kg
Since the luggage was above by 5 kg when both items were present and it was below
by 500 g after removing both items, then the hairdryer and the shampoo together weigh
5 kg + 500g = 5.5 kg.
The mass of the hairdryer

= Mass of both hairdryer and shampoo – Mass of shampoo
= (5.5 – 3) kg
= 2.5 kg
= (2.5 × 1 000) g
= 2 500 grams
Alternatively
Since the luggage was below the limit by 500g without the hairdryer and shampoo, and
2 kg over the limit after putting in the hairdryer, the hairdryer must weigh:
2 kg + 0.5 kg = 2.5 kg = 2 500g
under Limit over
0.5 kg 2 kg 5 kg
No items HD only HD + S
Answer: 2 500 grams

33. Five quadrilaterals are shown below.
Write the name of the quadrilateral that matches the properties given.
Properties Name
No lines of symmetry, one pair of parallel
sides, one angle greater than a right angle
Two lines of symmetry, two pairs of parallel

sides, not right angles
The figure with no lines of The figure with two lines of symmetry, two pairs of
symmetry, one pair of parallel parallel sides, not right angles is the given rhombus.
sides, one angle greater than a right
angle is the given trapezium.
The completed table now looks like:
Properties Name
No lines of symmetry, one pair of parallel sides,
one angle greater than a right angle Trapezium
Two lines of symmetry, two pairs of parallel
sides, not right angles Rhombus

34. Complete the shape on the grid below using AB as the line of symmetry.
We reflect each vertex of the given figure in the line AB. Each reflected vertex would
lie on the opposite side of AB and the same distance from AB. This can be located by
counting the blocks. The new vertices are now connected by straight lines to complete
the symmetrical figure.
The completed shape looks like:

35. The incomplete graph below shows the runs scored by a cricket team in 3 matches.
The total runs scored in Match 1 and Match 2 was 100.

How many runs were scored in Match 3?
The bar for Match 1 is 3 ‘blocks’ high.

The bar for Match 2 is 2 ‘blocks’ high.
Hence, 5 blocks represent a total of 100 runs.

\Each block on the vertical axis represents 100 runs. ÷ 5 = 20 runs
Hence, in Match 3 the number of runs scored is four blocks high and will represent
20 ´ 4 = 80 runs
Answer: 80 runs

36. The incomplete table below shows Paul’s marks in 5 subjects. Paul’s mean mark was
68.
Paul’s Marks in Five Subjects
English
Subject Science Mathematics Art Music Language
Arts
Mark 80 55 47 73
Calculate Paul’s mark in Mathematics.
Paul’s mean mark in 5 subjects is 68.

\Paul’s total marks in all 5 subjects = 68 ´ 5
= 340
Paul’s total marks in four of the subjects = 80 + 55 + 47 + 73

= 255
Hence, Paul’s mark in Mathematics = 340 - 255

= 85
Answer: 85 marks

Section 3
37. Tara bought an equal number of pies and doughnuts. She spent a total of $70.
Pies Doughnuts
2 for $5 3 for $10
How many pies did Tara buy?
Tara can buy pies in sets of two and she can buy doughnuts in sets of three.
The number of pies can be 2, 4, 6, 8, 10, 12, …

The number of doughnuts can be 3, 6, 9, 12, …
For equal numbers of pies and doughnuts, the options are 6, 12, … or multiples of 6.
Assume she bought 6 of each, we calculate her total cost as follows:
2 pies are sold for $5

\6 pies will cost $5× 3 = $15
3 doughnuts for $10

\6 doughnuts will cost $10× 3 = $20
6 pies and 6 doughnuts cost $ (15 + 20) = $35
If Tara spent $70 then she must have bought twice as many as 6 of each of the items
since $70 = $35 × 2.
12 pies and 12 doughnuts will cost $35 × 2 = $70
Answer: 12 pies

38. Dave made a decoration by sticking rectangular strips of Bristol board together. When
two strips were stuck together, there was an overlap of a length of 4 cm from each
strip.
Dave stuck 13 strips of Bristol board, each of length 15 cm to make the decoration.
Calculate the length of Dave’s decoration.
Assuming Dave overlaps the strips by placing 4 cm of the second strip over the first
strip and then continues in this manner. Starting from the first strip, each strip will have
a 4 cm overlap at its right end, so that only 15 – 4 = 11 cm will be exposed. Assuming
he had only 5 strips, then the 5th strip will have no overlap on its right end and so15 cm
will be exposed.
11 cm 11 cm 11 cm 11 cm 15 cm
1 2 3 4 5
4 cm 4 cm 4 cm 4 cm
The completed length of the decoration will comprise:

12 strips of length 11 cm and the 13th strip of length 15 cm
Length of Dave’s decoration (11 × 12) + 15 𝑐𝑚

= (132 + 15) 𝑐𝑚
= 147 𝑐𝑚

39. Jenna drew a five-sided shape with one right angle. Two sides of the shape are shown
on the grid below.
a) Complete the shape by joining some of the dots on the grid.

b) Describe all the angles in the shape.
This shape has
Acute angles at vertices A, D and E.
Right angle at C.
Reflex angle at B.
Alternatively:
F G
This shape has
Acute angles at vertices K

and H.
Right angle at G.
Reflex angles at F and L L
K H

40. Brandy and Clara played 3 rounds in a video game. The incomplete table shows their
scores.
Scores in Video Game
Student Round 1 Round 2 Round 3

Brandy 90 82
Clara 48
Medals were given based on the average score of 3 rounds
Medals Based on Average Score

Bronze Silver Gold
61 – 90 91 – 120 121 – 150
a) What is the lowest score that Brandy should obtain in Round 3 to qualify for a
silver medal?
The lowest average to obtain a silver medal is 91.

So, the total that must be scored in all three games = 91´ 3 = 273
Brandy scored, in the first two games, a total of 90 + 82 = 172
Hence, the lowest score that Brandy needs in Round 3 is 273 - 172 = 101
Answer: 101 points
b) Clara’s total score was 140. Her score in Round 2 was three times her score in
Round 3.
What was Clara’s score in Round 3?
Clara’s score in both Rounds 2 and 3 would total 140 - 48 = 92

Score in Round 2 is three times the score in Round 3.
Round 2 Round 3 4 parts = 92

1 part = 92 ÷ 4
=23
Hence, the score in Round 3 is 23
Answer: 23 points

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SEA YEAR 2023

1. Write the numeral 91 005 in words.

Tens of Thousands Thousands Hundreds Tens Ones

Ninety + thousand five units

Answer: Ninety-one thousand and five

2. Arrange the numbers below in descending order.

3 162 3 612 3 261 3 126

We can place the numbers on a Place Value Chart

Thousands Hundreds Tens Ones

Answer: 3 612, 3 261, 3 162, 3 126

Copyright ©2023.Some Rights Reserved. faspassmaths.com 1

4. Divide 628 by 12.

Copyright ©2023.Some Rights Reserved. faspassmaths.com 2

Tens of Thousands Thousands Hundreds Tens Units

The whole is 5 × 9 = 45.

Copyright ©2023.Some Rights Reserved. faspassmaths.com 3

Total in $ and cents is:

$20 + $20 + $5 + $1 + $0.05

9. Calculate 15% of 120.

Copyright ©2023.Some Rights Reserved. faspassmaths.com 4

On what day of the week is the 2nd of the April?

By counting backwards and forwards the calendar is completed to show:

We observe that the 2nd of April is on a Sunday.

Copyright ©2023.Some Rights Reserved. faspassmaths.com 5

What is the mass of the pawpaw, in grams?

The pointer on the scale is midway between 3 and 4.

Answer: 750 grams

Copyright ©2023.Some Rights Reserved. faspassmaths.com 6

How many cubes are in the box?

The length of the box fits 5 cubes.

\Length of pen = (17 – 6) cm

Copyright ©2023.Some Rights Reserved. faspassmaths.com 7

How many edges of the pyramid have a length of 6 cm?

The slant edges of the pyramid are 6 cm in length.

There are 4 slant edges

Copyright ©2023.Some Rights Reserved. faspassmaths.com 8

To which letter is the arrow now pointing?

After three quarter-turns in an anti-clockwise direction, the arrow will point to C.

Copyright ©2023.Some Rights Reserved. faspassmaths.com 9

What shoe size represents the mode?

19. Calculate the mean of the numbers below.

Copyright ©2023.Some Rights Reserved. faspassmaths.com 10

Pets Owned by Students

The completed tally chart now looks like:

Pets Owned by Students

Copyright ©2023.Some Rights Reserved. faspassmaths.com 11

21. Write the missing digit in each box below.

We know the difference between the two numbers is 6 584.

The completed problem looks like:

Copyright ©2023.Some Rights Reserved. faspassmaths.com 12

23. Karen uses 0.65 metres of cloth to make 1 skirt.

1 skirt uses 0.65 metres of cloth

The correct answer is

Answer: 7.8 metres

Copyright ©2023.Some Rights Reserved. faspassmaths.com 13

Kim’s Estimate Ben’s Estimate

Both rounded the multiplier, 18 to a multiple of 10 to make the calculations easier.

Kim rounded down 18 to 10 and got a lower estimate of 60 × 10 = 600

Ben’s estimate is better than Kim’s estimate because he chose 20.

Alternatively, we obtain the exact value of 60 × 18 = 1080