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Primary 3 Mathematics Mock Examinations
New Edition 2018
© Singapore Asia Publishers Pte Ltd
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Mock Examination 1Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S1
13 (3)
a quarter past 414 (4) Number of squares 6 Number of triangles 4 4 triangles 2 squares Total number of squares 6 + 2 = 8 Length of each side of a square 3 m Area of a square 3 m × 3 m = 9 m² Area of 8 squares 9 m² × 8 = 72 m²15 (2)
AH
G
B GH is perpendicular to AB.
16 (1) Cost of a tennis racket $237 Change received $63 Amount of money Jordan gave the cashier $237 + $63 = $30017 (4) Number of oranges Mr Tan bought 590 Number of oranges in a box 9 Number of boxes needed 590 ÷ 9 = 65 R 5 65 boxes with a remainder of 5 oranges. Therefore, another box is needed to pack the remaining
5 oranges. Total number of boxes to pack all the oranges 65 + 1 = 6618 (2) Total mass of Andy and Eileen 112 kg Mass of Andy thrice (3 times) the mass of Eileen
Andy
Eileen 1 unit
1 unit 1 unit 1 unit
?
112 kg
4 units 112 kg 1 unit 112 kg ÷ 4 = 28 kg Mass of Eileen 28 kg19 (2) Since the number of chicken pies and curry puffs sold
are the same, we can use ‘group method’ to solve the question.
Total amount earned $100 Cost of a chicken pie $2 Cost of a curry puff $3 Total cost of a chicken pie and a curry puff $2 + $3 = $5 (1 group) Total number of groups $100 ÷ $5 = 20 20 groups represent 20 chicken pies and 20 curry puffs. Therefore, he sold 20 chicken pies.
SOLUTIONS MOCK EXAMINATION 1Section A1 (3) 4068 + 189 = 42572 (3) 38 less than 40 tens 40 tens = 400 400 – 38 = 3623 (1)
137× 7959
42
4 (2) 383, 633, , 1133, 1383
+ 250 + 250
+ 250 + 250 633 + 250 = 8835 (1)
3 (× 3)
_______ 7 (× 3) = 9 ___ 21
6 (2) Height of your classroom door 2 m = 200 cm (Note: The standard height of the door at home and in
school is around 2 m.)7 (4)
bigger number
smaller number
?
13123594
?
smaller number 3594 bigger number 3594 + 1312 = 4906 Sum of 2 numbers 3594 + 4906 = 85008 (1) $50 – $26.90 = $50.00 – $26.90 = $23.109 (2)
Number of angles smaller than a right angle 210 (3) Letter F
11 (2) The unit for the weighing scale is in kg. Each unit is 200 g. Mass of the durian 1600 g = 1 kg 600 g12 (4)
Total number of parts 6 Number of shaded parts 2
Fraction of shaded parts 2 (÷ 2)
______ 6 (÷ 2) = 1 __ 3
Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S2Mock Examination 1
20 (3) Since Pattern 1 has 6 circles, Pattern 2 has 10 circles
and Pattern 3 has 14 circles, every next pattern will increase by 4 circles.
Pattern 1 6 circles Pattern 2 6 circles + 4 circles = 10 circles Pattern 3 10 circles + 4 circles = 14 circles Pattern 4 14 circles + 4 circles = 18 circles Pattern 5 18 circles + 4 circles = 22 circles Pattern 6 22 circles + 4 circles = 26 circles
Section B21 Six thousand and sixty-nine = 6069
22 3847 is less than 5775. 5775 – = 3847 5775 – 3847 = 1928
23 125 × 8 = tens 125 × 8 = 1000 1000 = 100 tens
24 Smallest 4-digit odd number 3049 (Note: The answer is not ‘0349’ as it will become a
3-digit number ‘349’.)
25 Find a common multiple of 3, 5, 2 (numbers from the denominators).
Common multiple of 3, 5, 2 is 30. (Convert all denominators to the same number.)
2 (× 10)
________ 3 (× 10) = 20 ___ 30
4 (× 6)
_______ 5 (× 6) = 24 ___ 30
1 (× 15)
________ 2 (× 15) = 15 ___ 30
Descending order (from greatest to smallest):
4 __ 5 2 __ 3 1 __ 2 , ,
greatest smallest
26 $3 20-cent coins $3 = $0.20 + $0.20 + $0.20 + $0.20 + $0.20 + $0.20
+ $0.20 + $0.20 +$ 0.20 + $0.20 + $0.20 + $0.20 + $0.20 + $0.20 + $0.20
There are 15 20-cent coins in $3.
27 Lengthofthefield 1258 m Breadthofthefield 946 m Perimeterofthefield 1258 m + 946 m +1258 m + 946 m = 4408 m
28
60 minutes = 1 hour 60 minutes = 1 hour + 60 minutes = 1 hour
180 minutes = 3 hours Minutes left 205 minutes – 180 minutes = 25 minutes 205 minutes = 3 hours 25 minutes
29 Number of words Joshua can type in 1 minute 350 words Number of words Joshua can type in 5 minutes 350 words × 5 = 1750 words
30 Length of rope 276 m Length of rope left 80 m Length of 7 pieces of rope cut 276 m – 80 m = 196 m Length of 1 piece of rope cut 196 m ÷ 7 = 28 m
31 Jane’s age 9 years old Her sister’s age 7 years older than Jane 7 + 9 = 16 years old Jane’s age in 3 years’ time 9 + 3 = 12 years old Her sister’s age in 3 years’ time 16 + 3 = 19 years old Total age in 3 years' time 12 + 19 = 31 years
32 Numberofprawnsthefishermancaught 3368 Number of prawns sold on Monday 1238 Number of prawns sold on Tuesday 1015 Total number of prawns sold on Monday and Tuesday 1238 + 1015 = 2253 Number of prawns not sold 3368 – 2253 = 1115
33 Area of the square = 64 cm² Area of the square = length × length The length of a square is the same. 64 cm² = 8 cm × 8 cm Therefore, the length of the square is 8 cm.
34 Number of stamps Branson collected 2778 (Note: ‘He collected 1995 fewer stamps than Alvin.’ ‘He’
refers to Branson, hence Branson collected 1995 fewer stamps than Alvin. This also means that Alvin collected 1995 more stamps than Branson.)
Number of stamps Alvin collected 2778 + 1995 = 4773 Total number of stamps they collected 2778 + 4773 = 7551
35 Number of ice cream sticks Thomas used to build a house model 234
Number of ice cream sticks Thomas used to build 4 house models 234 × 4 = 936
36 Cost of each bar of chocolate $1.30 Cost of 2 bars of chocolate $1.30 + $1.30 = $2.60 Cost of each packet of apple juice $1.80 Cost of 3 packets of apple juice $1.80 + $1.80 + $1.80 = $5.40 Total amount Mrs Lim paid $2.60 + $5.40 = $8
37 Distance from Jason’s house to the library 1 km 205 m = 1205 m Distance from the library to the market 1 km 2 m = 1002 m Total distance Jason cycled from his house to the market 1205 m + 1002 m = 2207 m
38 Number of cakes sold from Monday to Wednesday 15 + 9 + 12 = 36 Number of cakes sold on Saturday 28 more than 36 28 + 36 = 64
39 Number of cakes sold from Monday to Sunday 15 + 9 + 12 + 18 + 26 + 64 + 39 = 183
40 Number of cakes sold in 1 week 183 Number of cakes sold in 3 weeks 183 × 3 = 549
Section C41 Amount of rice Maggie bought 1007 g Amount of rice Roland bought thrice (3 times) as
much as Maggie
Maggie
Roland?
1007 g
1 unit
1 unit 1 unit 1 unit
Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S3Mock Examination 1 / Mock Examination 2
Amount of rice Roland bought 1007 g × 3 = 3021 g Amount of rice they bought altogether 3021 g + 1007 g = 4028 g = 4 kg 28 g They bought 4 kg 28 g of rice altogether.
42 Number of seashells Ronnie had 357 Number of seashells Harvey had 159 more than Ronnie Number of seashells Davison had twice (2 times) as many as Harvey
Davison
1 unit
159 159
Harvey
357
159 ?
Ronnie
1 unit
Number of seashells Harvey had 159 + 357 = 516 Number of seashells Davison had 516 × 2 = 1032 Total number of seashells they had 1032 + 516 + 357 = 1905 They had 1905 seashells altogether.
43 Cost of 6 tables and 4 chairs $600 Cost of 3 tables and 4 chairs $360
6 tables and 4 chairs $600– 3 tables and 4 chairs $360
3 tables and 0 chair $240 Cost of 3 tables $240 Cost of 1 table $240 ÷ 3 = $80 He paid $80 for 1 table.
44 3 notebooks 1 group 9 notebooks 3 notebooks + 3 notebooks + 3 notebooks
1 group + 1 group + 1 group = 3 groups
Cost of 3 notebooks $4 (1 group) Cost of 9 notebooks $4 × 3 groups = $12 Amount of change Clarice received $50 – $12 = $38 She received $38 in change.
45 Perimeter of the square 48 m Length of the square 48 m ÷ 4 = 12 m Length of the rectangle twice (2 times) the length of the square 2 × 12 m = 24 m Perimeter of the rectangle length + breadth + length + breadth 24 m + 15 m + 24 m + 15 m = 78 m The perimeter of the rectangle is 78 m.
MOCK EXAMINATION 2
Section A1 (3) 3 thousands, 8 hundreds, 20 tens and 6 ones 3 thousands = 3000 8 hundreds = 800 20 tens = 200 6 ones = 6 3000 + 800 + 200 + 6 = 4006
2 (2) Difference between 4958 and 2069 Difference minus 4958 – 2069 = 2889
3 (4) 7016 g = 7000 g + 16 g = 7 kg 16 g
4 (3) In 7693, ones place 3 tens place 9 hundreds place 6 thousands place 7 The digit 6 is in the hundreds place.
5 (1) 659 cm = 600 cm + 59 cm = 6 m 59 cm
6 (4) $3.80 = 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ +
20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢ + 20¢
$3.80 19 twenty-cent coins
7 (2)
× = 64 = 8 × 8 (The same shapes have the same value.)
= 8
8 ÷ = 4
= 8 ÷ 4
= 2
8 (2) Biggest odd number 6609
9 (4)
503 – 1 ? 8 = 335 503 – 335 = 168 ? = 6
10 (3)
Total number of rectangles 12 Number of shaded rectangles 5 Fraction of rectangles to be shaded to make 2 __ 3
2 (× 4)
_______ 3 (× 4) = 8 ___ 12
Fraction of shaded rectangles now 5 ___ 12
Fraction of more rectangles to be shaded
8 ___ 12 – 5 ___ 12 = 3 ___ 12
Number of more rectangles to be shaded 3
11 (4)
smaller number
?
bigger number
725225
1 unit
1 unit
2 units 725 – 255 = 470 1 unit 470 ÷ 2 = 235 bigger number 235 + 255 = 490
Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S4Mock Examination 2
12 (2)
E
F
K
J
EF // JK
13 (2)
12 cm
10 cm5 cm 7 cmA
B
10 cm
7 cm
12 cm Perimeterofthefigure 12 cm + 7 cm + 10 cm + 12 cm + 7 cm + 10 cm = 58 cm
14 (3) Number of stickers Sally gave away 1158 Number of stickers Tommy gave away 139 fewer than Sally
Tommy
1158
Sally
?139
? Number of stickers Tommy gave away 1158 – 139 = 1019 Total number of stickers Tommy and Sally gave away 1158 + 1019 = 2177
15 (1) Time Charlize completed the chores 6.15 pm Time taken for Charlize to complete 1 h 45 min Time Charlize started doing the chores
4.30 pm 5.15 pm 6.15 pm
start 45 min1 h completed
30 min 15 min
5 pm
16 (3) Cost of a jacket and 2 similar shirts $255 Cost of a jacket $105 Cost of 2 similar shirts $255 – $105 = $150 Cost of a shirt $150 ÷ 2 = $75
17 (4) Number of slices of pizza Andy cut 8 Number of slices Eve ate 4 Number of slices Carl ate 2 Total number of slices Eve and Carl ate 4 + 2 = 6 Number of slices left 8 – 6 = 2
Fraction of the pizza left 2 (÷ 2)
______ 8 (÷ 2) = 1 __ 4
18 (3) Total number of spectators 1500 Total number of groups 10 Number of spectators in each group 1500 ÷ 10 = 150
19 (2) Cost of a book $54.90 Cost of a pen $28.20 less than the book
pen
$54.90
book?
?
$28.20
Cost of the pen $54.90 – $28.20 = $26.70 Total cost of the book and pen $54.90 + $26.70 = $81.60 Amount of money Sue gave the cashier $100 Amount of change Sue received from the cashier $100.00 – $81.60 = $18.40
20 (4) Fraction of cake Ron ate
3 (× 2) _______ 5 (× 2) = 6 ___ 10
Fraction of cake Adam ate 3 ___ 10
Total fraction of cake Ron and Adam ate 6 ___ 10 + 3 ___ 10 = 9 ___ 10
Fraction of cake left 1 – 9 ___ 10 = 10 ___ 10 – 9 ___ 10 = 1 ___ 10
Section B21 5048 – Five thousand and forty-eight
22 209× 81672
7
23 10 tens more than 70 hundreds is 10 tens = 100 70 hundreds = 7000 7000 + 100 = 7100
24 Ascending order (from smallest to greatest): 5990 , 5999 , 6129 , 6162 smallest greatest
25 10, 10, 20, 60, , 1200, 7200
× 1 × 2 × 3 × 4 × 5 × 6
60 × 4 = 240
26 In order to compare the fractions, make all the denominators the same number.
5 (× 7)
_______ 8 (× 7) = 35 ___ 56 4 (× 8)
_______ 7 (× 8) = 32 ___ 56
Greater fraction 5 __ 8
27 ? ÷ 3 = 38 remainder 2 38 × 3 = 114 ? = 114 + 2 = 116
28
4 cm
5 cm
13 cm
12 cm 12 cm
13 cm
Perimeterofthefigure 12 cm + 13 cm + 12 cm + 13 cm = 50 cm
Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S5Mock Examination 2
29 The unit for the weighing scale is in kg. Each unit is 200 g. Mass of the basket of fruit 2800 g Mass of the empty basket 950 g Mass of the fruit 2800 g – 950 g = 1850 g
30 Mass of C 1 kg + 2 kg = 3 kg
Mass of C + 2 kg
= 3 kg + 2 kg = 5 kg
31 Cost of an exercise book $1.20 Cost of 2 exercise books $1.20 + $1.20 = $2.40 Cost of a pen $0.80 Cost of 2 pens $0.80 + $0.80 = $1.60 Cost of a stapler $2.30 Cost of 3 staplers $2.30 + $2.30 + $2.30 = $6.90 Total cost of all the items $2.40 + $1.60 + $6.90 = $10.90 Amount of money May gave the cashier $12 Amount of change May received from the cashier $12.00 – $10.90 = $1.10
32
CA
B
33 Total number of buns Melvin made 4000 Number of buns sold on Saturday 1087 Number of buns sold on Sunday 965 more than on Saturday
Saturday
Sunday
?
965
1087
?
Number of buns sold on Sunday 1087 + 965 = 2052 Total number of buns sold on Saturday and Sunday 1087 + 2052 = 3139 Number of buns left 4000 – 3139 = 861
34 Amount of money Alan saved $15.50 $15.50 = 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ +
50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢
Numberoffifty-centcoinshehad 31
35 Distance between the 2nd bush and 4th bush 250 m
250 m
2nd 3rd 4th
Distance between the 2nd bush and 3rd bush 250 m ÷ 2 = 125 m
125 m 125 m 125 m 125 m
5th 9th
Distance between the 5th bush and 9th bush 125 m × 4 = 500 m
36 Cost of a table $128.50 Cost of a chair $58.90 less than the table
chair
$128.50
table?
?
$58.9
Cost of a chair $128.50 – $58.90 = $69.60 Total cost of a table and a chair $128.50 + $69.60 = $198.10
37 Amount of water a pot can hold 5850mℓ Amount of water Jane poured into the pot 1250mℓ Amount of water Alan poured into the pot 2555mℓ Total amount of water Jane and Alan poured into the pot 1250mℓ+2555mℓ=3805mℓ Amount of water to be added to reach its brim 5850mℓ–3805mℓ=2045mℓ=2 ℓ 45 mℓ
38 Number of seashells Adam collected 45 Number of seashells Eden collected 80 Number of seashells Leo collected 60 Number of seashells Colin collected 70 Number of seashells Ben collected 15 Therefore, Eden collected the most number of seashells.
39 Number of seashells Ben collected 15 4 times as many seashells as Ben 15 × 4 = 60 Number of seashells Leo collected 60 Therefore, Leo collected 4 times as many seashells as
Ben.
40 Total number of seashells the boys collected 45 + 80 + 60 + 70 + 15 = 270
Section C41 Total number of stamps Jenny, Clarice and Zoe
collected 870 Number of stamps Jenny collected 3 times as many as Clarice Number of stamps Zoe collected 120 more than Clarice (Note: ‘Clarice collected 120 fewer stamps than Zoe’
means Zoe collected 120 more stamps than Clarice.)
Jenny
?
1 unit 1 unit 1 unit
Clarice 1 unit
1 unitZoe 120
870
5 units 870 – 120 = 750 1 unit 750 ÷ 5 = 150 Number of stamps Jenny collected 150 × 3 = 450 Jenny collected 450 stamps.
Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S6Mock Examination 2 / Mock Examination 3
42 Before:
Dora
Eve
$85
$85
give
1 unit
1 unit $85
1 unit After:
Dora
Eve
1 unit
1 unit $85 $85
1 unit
1 unit $85 + $85 = $170 Before:
Dora 1 unit
$170
$85
? AmountofmoneyDorahadatfirst $170 + $85 = $255 Dora had $255 atfirst.
43 Number of hens and horses in a farm 24 Number of legs altogether 64Number of hens
Number of legs
Number of horses
Number of legs
Total number of animals
Total number of legs Check
14 14 × 2 = 28 10 10 × 4 = 40 24 28 + 40 = 68
15 15 × 2 = 30 9 9 × 4 = 36 24 30 + 36 = 66
16 16 × 2 = 32 8 8 × 4 = 32 24 32 + 32 = 64
There are 16 hens and 8 horses.
44 Distance from the library to the school 2025 m Distance from the school to the library 2025 m Total distance Ryan has cycled 2025 m + 2025 m = 4050 m = 4 km 50 m He has cycled 4 km 50 m.
45 Number of pins Milton had 178 Number of pins Kelly had 28 fewer than Milton Number of pins Timothy had twice (2 times) as many as Kelly Number of pins Sammy had 2 times as many as Timothy
Sammy
Timothy
Kelly
Milton 178
1 u
1 u 1 u
1 u 1 u 1 u 1 u
28
1 unit
1 unit 1 unit
? Number of pins Kelly had 178 – 28 = 150 1 u 150 Number of pins Sammy had 4 u 150 × 4 = 600 Sammy had 600 pins.
MOCK EXAMINATION 3
Section A1 (2)
438× 62628
42
2 (4) 5525, 5375, , 5075, 4925
– 150 – 150 – 150 – 150
5375 – 150 = 5225
3 (1)
4 (× 4)
_______ 9 (× 4) = 16 ___ 36
The missing numerator is 16.
4 (3) 2 h 50 min = min 1 h = 60 min 2 h = 60 min × 2 = 120 min 120 min + 50 min = 170 min
5 (3) 624 ÷ 9 = 69 R (Note: Use long division method to answer the question.)
remainder
699 624 – 54
84– 81
3
1
5
The remainder is 3.
6 (4) 155 tens more than 20 hundreds is . 155 tens = 1550 20 hundreds = 2000 2000 + 1550 = 3550
7 (1) In order to compare the fractions, make all the
denominators the same number.
3 (× 12)
________ 5 (× 12) = 36 ___ 60
1 (× 20)
________ 3 (× 20) = 20 ___ 60 3 (× 15)
________ 4 (× 15) = 45 ___ 60
2 (× 30)
________ 2 (× 30) = 60 ___ 60 2 (× 20)
________ 3 (× 20) = 40 ___ 60
Therefore, 1 __ 3 is smaller than 3 __ 5 .
8 (2)
3 4
12
12 3
36
36 2
?
3 × 4 = 12 12 × 3 = 36 36 × 2 = 72
9 (4) The unit for the beaker is in litres. Eachunitis200mℓ. Volume of the water in the beaker 1400mℓ=1 ℓ 400 mℓ
Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S7Mock Examination 3
10 (4) $6.50 = 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ +
50¢ + 50¢ + 50¢ + 50¢ + 50¢ + 50¢ $6.50 = 13fifty-centcoins11 (3) Thefigureismadeupof4similarrectangles.
4 cm
4 cm 4 cm
12 cm 12 cm
12 cm
4 cm × 3 = 12 cm
Perimeterofthefigure 12 cm + 12 cm + 4 cm + 12 cm + 4 cm + 12 cm = 56 cm12 (2) Time shown 12.45 pm (lunch time is in pm) Time taken for Mr Tan to have his lunch half an hour TimeMrTanwillfinishhislunch
12.45 pm 1 pm 1.15 pm
half an hour = 30 min
15 min 15 min
13 (4) Height of John 112 cm Height of Dave 22 cm taller than John
John 112 cm
Dave
?
?22 cm
Height of Dave 112 cm + 22 cm = 134 cm Total height of John and Dave 112 cm + 134 cm = 246 cm14 (1) Distance Alan has run 1 km 6 m = 1006 m Distance Ronnie has run 1580 m Distance Ronnie has run further than Alan 1580 m – 1006 m = 574 m15 (3) Number of pies Melvin made 308 Number of boxes 7 Number of pies in each box 308 ÷ 7 = 44
16 (4) Total number of students 38 Number of balloons for each student 3 Total number of balloons needed 38 × 3 = 114 Number of balloons Miss Tan had 110 Number of more balloons Miss Tan needed 114 – 110 = 417 (1) Number of adults at the carnival 1206 Number of children at the carnival 964
children
1206adults
964?
Difference in number of adults and children 1206 – 964 = 242
18 (3) Amount of money Alice and Tom saved $960 Amount of money Alice saved twice (2 times) as much as Tom
Alice 1 unit 1 unit
Tom 1 unit$960
? 3 units $960 1 unit $960 ÷ 3 = $320 Amount of money Tom saved $320
19 (3) Cost of a table $24.50 Cost of a chair $19.60 Total cost of a table and a chair $24.50 + $19.60 = $44.10 Amount of money Mr Lee paid to the cashier $50 Amount of change Mr Lee received from the cashier $50.00 – $44.10 = $5.90
20 (2) Fraction of pizza Ray ate
2 (× 3) _______ 5 (× 3) = 6 ___ 15
Fraction of pizza Christine ate 1 (× 5)
_______ 3 (× 5) = 5 ___ 15
Total fraction of pizza Ray and Christine ate
6 ___ 15 + 5 ___ 15 = 11 ___ 15 Fraction of pizza left 1 – 11 ___ 15 = 15 ___ 15 – 11 ___ 15 = 4 ___ 15
Section B21 In 9732: 9 – thousands place 7 – hundreds place 3 – tens place 2 – ones place The digit 3 is in the tens place.
22 2150 less than tens is 1350. – 2150 = 1350 2150 + 1350 = 3500 = 350 tens
23 Biggest 4-digit even number 9850
24 7 ___ 12 + 1 __ 4 = 7 ___ 12 + 1 (× 3)
_______ 4 (× 3)
= 7 ___ 12 + 3 ___ 12
= 10 (÷ 2)
________ 12 (÷ 2)
= 5 __ 6
25 Sum of $39.05 and $28.35 $39.05 + $28.35 = $67.40
26 A quarter 1 __ 4
2 wholes
Number of quarters in 1 whole 4
Number of quarters in 2 wholes
2 × 4 = 827 8209 g = 8000 g + 209 g = 8 kg 209 g
Solutions to Primary 3 Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL RIGHTS RESERVED.
S8Mock Examination 3
28 12 + 12 + 6 + 6 + 6 = × 6 (6 + 6) + (6 + 6) + 6 + 6 + 6 = × 6 = number of groups of 6 6 + 6 + 6 + 6 + 6 + 6 + 6 = 7 sixes = 7 groups of 6 Therefore, = 7
29 Numberoffivesin100 100 ÷ 5 = 2030 Cost of each notebook $2.80 Cost of 3 notebooks $2.80 + $2.80 + $2.80 = $8.40 Amount of money Mr Chan gave the cashier $10 Amount of change Mr Chan received from the cashier $10.00 – $8.40 = $1.6031 Number of stamps Beatrice has 195 Number of stamps Alice has 5 times as many as
Beatrice
Alice
Beatrice 1 unit
195
1 unit 1 unit 1 unit 1 unit 1 unit?
? 1 unit 195 5 units 195 × 5 = 975 Number of stamps Alice has 975 Total number of stamps Beatrice and Alice have 195 + 975 = 1170
32 194
254 60
269
383 ?
375
496 121
254 – 194 = 60 383 – 269 = 114 496 – 375 = 12133 Amount of rose syrup Lily poured into the jug 907mℓ Amount of water Lily poured into the jug 1ℓ90mℓ =1090mℓ Total volume of mixture in the jug 1090mℓ+907mℓ =1997mℓ = 1 ℓ 997 mℓ34 Cost of 1 pencil $1.20 Cost of 3 pencils $1.20 + $1.20 + $1.20 = $3.60 (Note: Bought 3 pencils, free 1 pencil.) Therefore, amount paid for 3 pencils will get him 4 pencils. 4 pencils (1 group) $3.60 Number of groups for 20 pencils 20 ÷ 4 = 5 1 group $3.60 5 groups $3.60 + $3.60 + $3.60 + $3.60 + $3.60 = $1835 Letters with at least one pair of parallel lines H, M, E36 Letters with at least one pair of perpendicular lines T, H, E37 Number of containers Milton had 29 Number of erasers in each container 9 Total number of erasers 29 × 9 = 261 Number of erasers repacked in each packet 6 Number of packets packed 261 ÷ 6 = 43 R 3 Therefore, number of erasers not packed 338 ‘At least 59 stamps’ means the number must be 59 and
above. Sally (66 stamps) Bernard (94 stamps) Andy (59 stamps) Colin (86 stamps) Students who have collected at least 59 stamps Sally, Bernard, Andy, Colin
39 Number of stamps Sally collected 66 Number of stamps Bernard collected 94 Number of stamps Pearlie collected 37 Number of stamps Andy collected 59 Number of stamps Eileen collected 48 Number of stamps Colin collected 86 Total number of stamps they collected 66 + 94 + 37 + 59 + 48 + 86 = 390
40 + = 230
+ = 180
Therefore, = 230 – 180= 50
+ = 180
50 + 50 + = 180
= 180 – 50 – 50 = 80
= 80 ÷ 2 = 40
Section C41 Number of stamps Pete collected 863 Number of stamps Alan collected 229 fewer than Pete Number of stamps Ryan collected 195 more than Alan (Note: ‘Alan collected 229 fewer stamps than Pete and
195 fewer stamps than Ryan’ tells us that Alan collected 195 fewer stamps than Ryan. Hence, Ryan collected 195 more stamps than Alan.)
Pete
Alan ?
863
?
Ryan
229
195
?
Number of stamps Alan collected 863 – 229 = 634 Number of stamps Ryan collected 634 + 195 = 829 Total number of stamps they collected altogether 863 + 634 + 829 = 2326 The three boys collected 2326 stamps in all.
42 Amount of money Sam and Alan had $1650 Amount of money Alan had twice (2 times) as much as Sam
Alan 1 unit 1 unit
Sam 1 unit
$1650
3 units $1650 1 unit $1650 ÷ 3 = $550
Alan 1 unit
Sam 1 unit
$550
?
Amount of money Alan should give Sam so that they have equal amount of money $550 ÷ 2 = $275
Alan should give Sam $275.
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S9Mock Examination 3 / Mock Examination 4
43 Number of packets of apples Mrs Soo bought 4 Number of packets of oranges Mrs Soo bought 6 Cost of a packet of apples $1.80 more than a packet of oranges
$27.20
1 unit
1 unit
1 unit
1 unit
1 unit1 unit
$1.80
$1.80
$1.80Packets of apples
1 unit
1 unit
1 unit
1 unit $1.80
Packets of oranges
10 units $27.20 – $1.80 – $1.80 – $1.80 – $1.80 = $20 1 unit $20 ÷ 10 = $2 Cost of a packet of apples $2 + $1.80 = $3.80 The cost of a packet of apples was $3.80.
44 Sue and Andy caught an equal number of crabs each. Number of crabs Sue sold 120 Number of crabs Andy sold 38 Number of crabs Andy left thrice (3 times) as many as Sue
Sue
Andy
sold 120Sue 1 unit
Andy 1 unit 1 unit 1 unitsold 38
2 units 120 – 38 = 82 1 unit 82 ÷ 2 = 41 Number of crabs Andy left 41 × 3 = 123 Andy had 123 crabs left.
45 Area of the rectangular piece of paper length × breadth = 18 cm × 8 cm = 144 cm² Area of the square length × length = 6 cm × 6 cm
= 36 cm² Area of the circle 28 cm² Area of the square and circle 36 cm² + 28 cm² = 64 cm² Area of the remaining piece of paper 144 cm² – 64 cm² = 80 cm² The area of the remaining piece of paper is 80 cm².
MOCK EXAMINATION 4
Section A1 (2)
138× 5690
41
2 (2) In 9328: 9 – thousands 3 – hundreds 2 – tens 8 – ones The digit 2 is in the tens place.
3 (4) Sum of 6 thousands, 30 hundreds and 40 tens 6 thousands = 6000 30 hundreds = 3000 40 tens = 400 6000 + 3000 + 400 = 9400
4 (3) 1ℓ=1000mℓ 3ℓ10mℓ=3010 mℓ
5 (3) 100¢ = $1.00 5385¢ = $53.85
6 (1) In order to compare the fractions, make all the
denominators the same number.
1 (× 20)
________ 3 (× 20) = 20 ___ 60 3 (× 15)
________ 4 (× 15) = 45 ___ 60
3 (× 12)
________ 5 (× 12) = 36 ___ 60 5 (× 10)
________ 6 (× 10) = 50 ___ 60
Therefore, 1 __ 3 is the smallest fraction.
7 (2) 4fifty-centcoins 50¢ + 50¢ + 50¢ + 50¢ = $2 8 twenty-cent coins 20¢ + 20¢ + 20¢ + 20¢ + 20¢ +
20¢ + 20¢ + 20¢ = $1.60 Total value $2 + $1.60 = $3.60
8 (2) Total number of triangles 9 Number of triangles shaded 4 Fraction of triangles to be shaded
2 (× 3) _______ 3 (× 3) = 6 __ 9
Fraction of triangles shaded 4 __ 9 Fraction of more triangles to be shaded 6 __ 9 – 4 __ 9 = 2 __ 9 Number of more triangles to be shaded 2
9 (1) Perimeterofthefigure 5 m + 3 m + 2 m + 5 m + 4 m + 3 m = 22 m = 2200 cm
10 (4) The unit for the weighing scale is in g. Each unit is 200 g. Mass of 1 mango 1200 g Mass of 3 mangoes 1200 g × 3 = 3600 g
11 (3) Amount of money Alice spent $82.50 Amount of money Felicia spent $49.35 less than Alice
Alice
Felicia
?
$49.35
$82.50
Amount of money Felicia spent $82.50 – $49.35 = $33.15
12 (3) Number of pins Lily has 1956 Number of pins Cindy has 380 fewer than Lily (Note: ‘She has 380 more pins than Cindy’. ‘She’
refers to Lily, hence Lily has 380 more pins than Cindy. Therefore, Cindy has 380 fewer pins than Lily.)
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S10Mock Examination 4
Lily
Cindy
?
1956
380?
Number of pins Cindy has 1956 – 380 = 1576 Total number of pins they have 1956 + 1576 = 353213 (3) Number of stamps Joey collected 96 Number of albums she had 8 Number of stamps in each album 96 ÷ 8 = 12 Total number of stamps she gave away 96 – 60 = 36 Number of albums she gave away 36 ÷ 12 = 314 (1) Time Mrs Kim started baking a cake 1.50 pm Time taken for Mrs Kim to bake a cake 1 h 25 min TimeMrsKimfinishedbakingthecake
2.50 pm 3.15 pm
1 h
1.50 pm
25 min
10 min 15 min
3 pm
15 (4) Amount of money Mr Toh paid for a table $38.50 Amount of money Mr Toh paid for a chair $15.90 Amount of money Mr Toh left $58.60 TotalamountofmoneyMrTohhadatfirst $38.50 + $15.90 + $58.60 = $11316 (2) Mass of Milton’s crate 80 kg Mass of Benny’s crate twice (2 times) the mass of Milton’s crate
Benny 1 unit 1 unit
Milton 1 unit
80 kg
?
?
1 unit 80 kg 2 units 80 kg × 2 = 160 kg Mass of Benny’s crate 160 kg Total mass of two crates 80 kg + 160 kg = 240 kg
17 (3) Amount of water in Bottle A 350mℓmorethaninBottleB Amount of water poured from Bottle A to Bottle B 400mℓ
Bottle A
Bottle B
350mℓ
Bottle A
Bottle B
400mℓpouredtoBottleB
50mℓ 350mℓ
Bottle A
Bottle B 50mℓ
50mℓ
?
350mℓ
Amount of water in Bottle B more than in Bottle A in the end 50mℓ+50mℓ+350mℓ=450 mℓ
18 (1)
578
480 340
98 238
A160
269418
578 – 160 = 418 578 – A = 269
578 – 98 = 480 578 – 238 = 340
578 – A = 269 A 578 – 269 = 30919 (2) Number of children who like blue 115 Number of children who like orange 90 Number of children who like blue more than those who
like orange 115 – 90 = 2520 (3) Number of children who like yellow 60 Number of children who like blue 115 Number of children who like orange 90 Number of children who like green 45 Total number of children who took part in the survey 60 + 115 + 90 + 45 = 310
Section B21 Eight thousand and nine – 8009
22 2 __ 5 + ? ___ 15 = 13 ___ 15
2 (× 3)
_______ 5 (× 3) + ? ___ 15 = 13 ___ 15
6 ___ 15 + ? ___ 15 = 13 ___ 15
? ___ 15 = 13 ___ 15 – 6 ___ 15 = 7 ___ 15
The missing number in the box is 7.
23 Descending order (from greatest to smallest): 9660 , 9606 , 6999 , 6996 greatest smallest
24 10 × 8 = 10 × 6 + 10 × ?
80 60 ? 80 – 60 = 20 10 × ? = 20 20 ÷ 10 = 2
25 1 h 60 min 3 h 60 min × 3 = 180 min 180 min + 47 min = 227 minutes
26 37 ÷ ? = 7 remainder 2 37 – 2 = 35 35 ÷ 7 = 5
27
B 3 6 9
+ 1 4 7 A
5 8 4 5
11
15 – 9 = 6 5 – 1 = 4 A = 6, B = 4
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S11Mock Examination 4
28 Number of cars 5 people can sit in 1 Number of cars 128 people can sit in 128 ÷ 5 = 25 R 3 128 people can sit in 25 cars with 3 people remaining. Therefore, another car is needed for the remaining
3 people. Total number of cars for all the people 25 + 1 = 26
29 Breadth of a rectangle 9 cm Length of a rectangle 3 times as long as its breadth
length 1 unit 1 unit 1 unit
breadth 1 unit
9 cm 1 unit 9 cm 3 units 9 cm × 3 = 27 cm Length of a rectangle 27 cm Area of the rectangle length × breadth = 27 cm × 9 cm = 243 cm²
30 Total number of blue and red marbles 51 Total number of bottles 3 Number of blue marbles in 1 bottle 8 Number of blue marbles in 3 bottles 8 × 3 = 24 Number of red marbles 51 – 24 = 27 Number of red marbles in 1 bottle 27 ÷ 3 = 9
31 Amount of water a jar can hold 1350mℓ Amount of water a glass can hold 50mℓ Numberofglassesofwatertofillupajar 1350mℓ÷50mℓ=27 Numberofglassesofwatertofillup2jars 27 × 2 = 54
32 Fractionofflourusedtobakeacake 2 __ 5 Fractionofflourusedtobakesomebread 3 ___ 10 Fractionofflourusedtobakecakeandbread
2 (× 2)
_______ 5 (× 2) + 3 ___ 10
= 4 ___ 10 + 3 ___ 10
= 7 ___ 10
Fractionofflourleft 1 – 7 ___ 10 = 10 ___ 10 – 7 ___ 10 = 3 ___ 10
33 Amount of money Ryan gave the cashier $50 Amount of change Ryan received from the cashier $15.50 Amount of money Ryan spent on 1 book and 2 calculators $50.00 – $15.50 = $34.50 Cost of a book $15.50 Cost of 2 calculators $34.50 – $15.50 = $19 Cost of 1 calculator $19 ÷ 2 = $9.50
34 Amount of money Dave and Chloe saved $1708 Amount of money Dave saved $795
Dave
Chloe
?
?
$795
$1708
Amount of money Chloe saved $1708 – $795 = $913 Amount of money Chloe saved more than Dave $913 – $795 = $118
35 Mass of a dog 24 kg Mass of a puppy six times as light as the dog (Note: ‘It is six times as heavy as a puppy’. ‘It’ refers
to the dog, hence the dog is six times as heavy as the puppy. Therefore, the puppy is six times as light as the dog.)
dog 1 unit
puppy 1 unit
?
1 unit 1 unit 1 unit 1 unit 1 unit
24 kg
6 units 24 kg 1 unit 24 kg ÷ 6 = 4 kg Mass of a puppy 4 kg Total mass of the dog and the puppy 24 kg + 4 kg
= 28 kg
36 Since the 2 shapes are squares, the lengths and the breadths have the same measurement.
20 cm
20 cm 20 cm
20 cm Perimeterofthefigure 20 cm + 20 cm + 20 cm + 20 cm = 80 cm Length of the wire used 80 cm
37 Length of the rope 6 m = 600 cm Length of the rope cut away 600 cm – 28 cm = 572 cm Number of equal pieces being cut from it 4 Length of each piece of rope 572 cm ÷ 4 = 143 cm
38 Cost of an eraser $0.80 Cost of 3 erasers $0.80 + $0.80 + $0.80 = $2.40 Cost of a pen twice (2 times) the cost of 3 erasers $2.40 + $2.40 = $4.80
39 Find the common multiple of 7 and 8. The number must be between 50 and 80.
7 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77 8 8, 16, 24, 32, 40, 48, 56, 64, 72, 80 He bought 56 sweets.
40 Cost of a pack of 5 apples $3 Amount of money Clarice had $20 Number of packs of apples she could buy $20 ÷ $3 = 6 R 2 Maximum pack of apples she could buy 6 1 pack 5 apples 6 packs 6 × 5 apples = 30 apples
Section C41 Number of chocolate and vanilla cookies Cindy baked
658 Number of cookies Cindy gave Kenny 63 Number of cookies left 658 – 63 = 595 Number of chocolate cookies left four times the vanilla cookies
chocolate cookies 1 unit
vanilla cookies 1 unit
1 unit 1 unit 1 unit
?
595
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S12Mock Examination 4 / Mock Examination 5
5 units 595 1 unit 595 ÷ 5 = 119 Number of chocolate cookies left 119 × 4 = 476 476 chocolate cookies were left.
42 Time taken for Fanny to do household chores on Monday 1 h 45 min
Time taken for Fanny to do household chores on Tuesday 2 h 50 min
2 h 50 min– 1 h 45 min
1 h 05 min
4 1
She took 1 h 5 min longer to do the household chores on Tuesday than on Monday.
43 Cost of 1 shirt and 1 pair of trousers $50 Cost of 1 shirt and 2 pairs of trousers $90
1 shirt and 2 pairs of trousers $90– 1 shirt and 1 pair of trousers $50
0 shirt and 1 pair of trousers $40 Cost of 1 shirt + cost of 1 pair of trousers $50 Cost of 1 shirt + $40 $50 Cost of 1 shirt $50 – $40 = $10 The cost of 1 shirt is $10.
44 Amount of money Daryl, Timothy and Jacob shared $308 Amount of money Daryl received twice as much as Timothy Amount of money Jacob received half as much as Timothy
Daryl
Daryl
Timothy
Jacob
$308
?
1 unit 1 unit
Timothy
Jacob
1 unit $308
?
1 __ 2 unit
7 units $308 1 unit $308 ÷ 7 = $44 Amount of money Jacob received $44 Jacob received $44.
45 Figure 1 3 Figure 4 7
Figure 2 4 Figure 5 9
Figure 3 6 Figure 6 10
There are 10 in Figure 6.
MOCK EXAMINATION 5
Section A1 (3) In 2957: 2 2000 9 900 5 50 7 7 The value of digit 9 is 900.
2 (4) 6347 6000 (thousands place) 4672 600 (hundreds place) 9476 6 (ones place) 2763 60 (tens place)
3 (3) Sum of 4095 and 3904 4095 + 3904 = 7999
4 (4) 145 tens less than is 29 hundreds. 145 tens = 1450 29 hundreds = 2900 – 1450 = 2900 2900 + 1450 = 4350
5 (2) 784 = 4 groups of ? 784 ÷ 4 = 196
6 (2)
6 (÷2)
_______ 26 (÷2) = 3 ___ 13
7 (1) $1.00 = 100 cents (100 cents is equal to 100 1¢ coins.) Therefore, $1.00 = 20 5¢ coins
8 (2) 25, 50, 150, , 3000 × 2 × 3 × 4 × 5 150 × 4 = 600
9 (1) AB ⊥ AH HG ⊥ AH DC ⊥ AH FE ⊥ AH AB ⊥ BC HG ⊥ BC DC ⊥ BC FE ⊥ BC AB ⊥ DE HG ⊥ DE DC ⊥ DE FE ⊥ DE AB ⊥ GF HG ⊥ GF DC ⊥ GF FE ⊥ GF Number of pairs of perpendicular lines 16
10 (1)
? 5 8× 6
3 3 4 8
34
So, 33 – 3 = 30. 30 ÷ 6 = 5
11 (2)
Total number of parts 8Number of unshaded parts 5
Fraction of unshaded parts 5 __ 8
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S13Mock Examination 5
12 (3) Number of seashells Roland gave each of his friends
8 Number of friends Roland had 11 Number of seashells Roland gave all his friends 11 × 8 = 88 Number of seashells Roland left 12 TotalnumberofseashellsRolandhadatfirst 88 + 12 = 100
13 (4) Number of sheets in a pack 128 Number of packs 8 Total number of sheets in 8 packs 128 × 8 = 1024
14 (2) Cost of a camera $557.85 Cost of a computer $988.95 Total cost of the camera and computer $557.85 + $988.95 = $1546.80 Amount of money Miss Liu gave the cashier $1000 × 2 = $2000 Amount of change Miss Liu received from the cashier $2000.00 – $1546.80 = $453.20
15 (2) Number of packets of lollipops 6 Number of lollipops in each packet 12 Total number of lollipops 12 × 6 = 72 Number of small bags 8 Number of lollipops in each small bag 72 ÷ 8 = 9
16 (1) Number of pins Jordan gave Alice 1957 Number of pins Alice gave Sammy 964 Number of pins Alice left 2860 (Note: We have to work backwards to get the initial
number of pins Alice had.) Number of pins before Alice gave Sammy 2860 + 964 = 3824 Number of pins before Jordan gave Alice 3824 – 1957 = 1867 Therefore,numberofpinsAlicehadatfirst 1867
17 (3) Time taken for Mandy to clean the windows 85 minutes Time taken for Mandy to clean the toilet 30 minutes Total time taken for Mandy to do the chores 85 min + 30 min = 115 min = 1 h 55 min Time Mandy completed the chores 2.55 pm Time Mandy started doing the chores
1.55 pm 2.55 pm
55 min 1 h
1 pm
18 (2) Totalmassofapacketofflourand2packetsofrice 1 kg 220 g = 1220 g Mass of a packet of rice 550 g Mass of 2 packets of rice 550 g + 550 g = 1100 g Massofapacketofflour1220 g – 1100 g = 120 g
19 (4) Amount of orange juice Daniel sold on Saturday 6ℓ78mℓ=6078mℓ Amount of orange juice Daniel sold on Friday 750mℓlessthanonSaturday (Note:‘sold750mℓmoreorangejuiceonSaturdaythan
onFriday’tellsusthathesold750mℓlessorangejuiceon Friday than on Saturday.)
Saturday
Friday
?
750mℓ
6078mℓ
Amount of orange juice Daniel sold on Friday 6078mℓ–750mℓ=5328 mℓ20 (3)
8 cm
32 cm
28 cm
32 cm
28 cm
Total length of the wire 32 cm + 28 cm + 32 cm + 28 cm + 8 cm + 8 cm
+ 8 cm + 8 cm = 152 cm
Section B21 8906 – Eight thousand nine hundred and six22 Smallest 4-digit odd number – 100123 Difference (minus) between $50 and $39.40 $50.00 – $39.40 = $10.6024 Find a common multiple of 3, 4, 2 (numbers from the
denominators). Common multiple of 3, 4, 2 is 12. Convert all denominators to the same number.
2 (× 4)
_______ 3 (× 4) = 8 ___ 12 3 (× 3)
_______ 4 (× 3) = 9 ___ 12 1 (× 6)
_______ 2 (× 6) = 6 ___ 12
Ascending order (from smallest to greatest):
1 __ 2 2 __ 3 3 __ 4 , ,
smallest greatest25 Even numbers between 20 and 30 22, 24, 26 and 28 (Note: 20 and 30 are even numbers but they cannot
be included in the sum of the even numbers as the question is asking ‘between 20 and 30’ only.)
Sum of all numbers between 20 and 30 22 + 24 + 26 + 28 = 10026 (Note: Use long division method to answer the
question.)
2083 624 6 24 24
0 remainder The remainder is 0.
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S14Mock Examination 5
27 tens = 350 35 tens = 350
28 Product (multiply) of 746 and 5
746× 53730
32
29 Total number of squares 12 (Note: Since there are 12 squares in all, the denominator
for 3 __ 4 has to be 12.)
3 (× 3)
_______ 4 (× 3) = 9 ___ 12
Therefore, number of shaded squares needed 9
30 1 – ? = 4 __ 5 – 1 __ 3
4 __ 5 – 1 __ 3 = 4 (× 3)
_______ 5 (× 3) – 1 (× 5)
_______ 3 (× 5)
= 12 ___ 15 – 5 ___ 15
= 7 ___ 15
1 – ? = 7 ___ 15
? = 1 – 7 ___ 15
= 15 ___ 15 – 7 ___ 15
= 8 ___ 15
31 Number of units from 0 to 1 12 (Note: Since there are 12 units in all, the denominator
for 2 __ 3 has to be 12.)
2 (× 4)
_______ 3 (× 4) = 8 ___ 12
?0 1 2 __ 3
1 ___ 12 2 ___ 12 3 ___ 12 4 ___ 12 5 ___ 12 6 ___ 12 7 ___ 12 8 ___ 12 9 ___ 12 10 ___ 12 11 ___ 12 12 ___ 12
Therefore, the missing fraction in the box 3 (÷ 3)
_______ 12 (÷3) = 1 __ 4
32 Fraction of cake Jane gave her mother 2 __ 3
Fraction of cake Jane ate 1 __ 6
Fraction of cake left 1 – 2 __ 3 – 1 __ 6
= 1 – 2 (× 4)
_______ 3 (× 4) – 1 (× 2)
_______ 6 (× 2)
= 12 ___ 12 – 8 ___ 12 – 2 ___ 12
= 2 (÷ 2)
________ 12 (÷ 2)
= 1 __ 6
33 The unit for the container is in litres. Eachunitis100mℓ. Volume of the water in the container shown 200mℓ Volume of the water poured into the container 180mℓ Final volume of the water in the container 200mℓ+180mℓ=380 mℓ
34 Number of pages of storybook Clarice reads per day 5 Total number of pages Clarice reads for 5 days 5 × 5 = 25 Total number of pages in a storybook 90 Number of unread pages 90 – 25 = 6535 Bag A 1 kg 6 g = 1006 g Bag B 1500 g Bag C 3 kg 90 g = 3090 g Bag D 2 kg 100 g = 2100 g Bag A is the lightest.36 Bag A 1006 g Bag C 3090 g Total mass of bags A and C 1006 g + 3090 g = 4096 g37 MassofflourMrsChanusedtobake3cakes 1950 g MassofflourMrsChanusedtobake1cake 1950 g ÷ 3 = 650 g MassofflourMrsChanusedtobake7cakes 650 g × 7 = 4550 g38 Number of stamps Farah had 290 Number of stamps Farah gave her brother 32 Number of stamps Farah left 290 – 32 = 258 (Note: ‘She shared the remaining stamps with Betty
and Alice’. ‘She’ refers to Farah. It means the remaining stamps are shared among Farah, Betty and Alice.)
Number of stamps Betty received 258 ÷ 3 = 8639 Cost of a pencil $0.80 Cost of 3 pencils $0.80 + $0.80 + $0.80 = $2.40 Cost of 4 erasers $6.00 – $2.40 = $3.60 Cost of 1 eraser $3.60 ÷ 4 = $0.9040 Number of vanilla cupcakes Carl sold 675 3 times as many as chocolate cupcakes
vanilla cupcakes 1 unit
chocolate cupcakes 1 unit
1 unit 1 unit
675
?
3 units 675 1 unit 675 ÷ 3 = 225 Number of chocolate cupcakes Carl sold 225 Total number of cupcakes Carl sold 675 + 225 = 900
Section C41 Number of stamps Felicia collected 984 Number of Singapore stamps 7 times as many as Malaysia stampsSingapore stamps 1 u
Malaysia stamps ?984
1 u
1 u 1 u 1 u 1 u 1 u 1 u
8 units 984 1 unit 984 ÷ 8 = 123 Number of Singapore stamps more than Malaysia stamps 123 × 6 = 738 She collected 738 more Singapore stamps than
Malaysia stamps.42 Number of books in the library 1000 Number of Malay books in the library 259 Number of Chinese books in the library 285 Number of Malay and Chinese books 259 + 285 = 544 Number of English and Tamil books 1000 – 544 = 456 Number of English books three times as many as Tamil books
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S15Mock Examination 5
English books 1 unit
Tamil books 1 unit
1 unit 1 unit
?
456
4 units 456 1 unit 456 ÷ 4 = 114 Number of English books 114 × 3 = 342 There were 342 English books.
43 (a) Length of each side of triangle 12 cm Length of 3 sides of triangle 12 cm × 3 = 36 cm Length of wire used for the triangle 36 cm twice the length of wire used for the rectangle
triangle 1 unit
rectangle 1 unit
1 unit
36 cm
2 units 36 cm 1 unit 36 cm ÷ 2 = 18 cm Length of wire used for the rectangle 18 cm Length of the rectangle 6 cm Total length of the rectangle 6 cm + 6 cm = 12 cm Total breadth of the rectangle 18 cm – 12 cm = 6 cm Breadth of the rectangle 6 cm ÷ 2 = 3 cm
12 cm 12 cm
12 cm
6 cm3 cm 3 cm
6 cm The breadth of the rectangle is 3 cm. (b) Length of wire used for the triangle 36 cm Length of wire used for the rectangle 18 cm Total length of the 2 wires 36 cm + 18 cm = 54 cm The total length of the 2 wires is 54 cm.
44 Number of slices of cake Jack cut 15
Fraction of cake Joshua ate 1 __ 5 = 1 (× 3)
_______ 5 (× 3) = 3 ___ 15
Fraction of cake May ate 2 ___ 15
leftJoshua
May 10 slices of the cake were left.
45 Cost of a table $60 thrice (3 times) as much as a chair Cost of a mat $15 less than a chair
table 1 unit
chair 1 unit
1 unit 1 unit
$60
mat$15
? 3 units $60 1 unit $60 ÷ 3 = $20 Cost of a chair $20 Cost of a mat $20 – $15 = $5 The cost of the mat is $5.