Tag Archives: Paper 2

Primary 6 Prelims Revision 9

P6-Prelims-Ex 9

Ans:

1– $45

Julian –> 1u

Fred­er­ick –> 2u

Mar­cus –> 6u

5u –> $25

1u –> $5

$5 × 9u = $45

 

2–  16 more tricycles

Assume all are tri­cy­cles –> 56 × 3 = 168 wheels

168 — 148 = 20 wheels extra

Each bicy­cle has 1 lesser wheel than each tricycle.

There­fore, 20 ÷ 1 = 20 bicycles

56 — 20 00= 36 tricycles

36 — 20 = 16 more tri­cy­cles than bicycles

 

3–  $44

4–  22.5 km/h faster

Mar­cus’ speed –>  270 /4 = 67.5 km/h

Mar­cus’ new speed –>  270/3 = 90 km/h

90 km/h — 67.5 km/h = 22.5 km/h faster

 

5–  396 boys

Girls –>  1u × 5 = 5u

Boys –>  (1u + 16) × 3 = 3u + 48

8u + 48 –>  976 pupils

8u –> 928

1u –> 116

116 × 3 + 48 = 396 boys

 

6–  $2945

7 hand­bags + 5 watches = 3 hand­bags + 11 watches

7 hand­bags — 3 hand­bags = 11 watches — 5 watches

4 hand­bags = 6 watches

6 watches –>  $285 × 4 = $1140

1 watch –>  $1140 ÷ 6 = $190

$285 × 7 + $190 × 5 = $2945

 

7–  300 stick­ers

Maxwell –> 100%

Jeanet­ter –> 120%

20/100 × 120% = 24%

120%  - 24% = 96% left

96% –>  288 stickers

1% — 3 stickers

3 × 100 = 300 stick­ers

 

8–  153 cm

17 + 13 = 30 cm — 3 pieces

1 piece –>  10 cm

14 × 10 cm + 13 cm = 153 cm

Con­tinue read­ing

Primary 6 Prelims Revision 3

P6-Prelims-Ex 3

Ans:

1–  3 : 12 : 17

2–  4.95 cm

Length of the tank –>  180 cm³ ÷ 12 edges = 15 cm

33/100 × 15% = 4.95 cm

 

3–  100 balls

Total units of balls –>  5u + 4u = 9u

Units of balls in each bag in the end –>  9u/2 = 4.5u

5u — 4.5u = 0.5u —>  10 balls

1u –>  10 × 2 = 20 balls

20 × 5u = 100 balls

 

4–  540 km

5–  $24

1 blouse –> 1u

1 pair of jeans –>  1u + $12

3 pairs of jeans + 4 blouses –>  (1u + $12) × 3 + 4u = 7u + $36

7u + $36 -> $204

7u –>  $204 — $36 = $168

1u –>  $24

 

6–  891 beads

Red : Blue + Green
4   : 5

Blue : Green
3    :    8     –> 11u

Adjust the units for the blue and green beads such that they are the same in both cases.

Com­mon mul­ti­ple of 5 and 11 is 55u

There­fore,

Red  :  Blue + Green
44   : 55

Blue  :  Green
15   : 40

Red  :  Blue  :  Green
44 :    15   : 40

44u — 40u = 4u –>  36 beads

1u –>  9 beads

44u + 55u = 99u —->  9 beads × 99 = 891 beads

Con­tinue read­ing

Primary 6 SA1 Revision 12

P6-SA1-Ex 12

Ans:

1–  175

2–  $43.50

70% of the price –>  $101.50

30% –>  $101.50/70 × 30 = $43.50

 

3–  (3+ 6) kg

Alan’s mass ->  (+ 3) kg

Nicholas’ mass ->  (m + 3) — 2 = (+ 1) kg

Brett’s mass ->  (+ 1) + 1 = (+ 2) kg

(m + 3) + (+ 1) + (+ 2) = (3+ 6) kg

 

4–  195 km

1  1/2 h× 80 km/h = 120 km

45/60 h × 100 km/h = 75 kg

120 km + 75 km = 195 km

 

5–  $6441.60

Earn­ings in Jan­u­ary –>  122/100 × $6000 = $7320

Earn­ings in Feb­ru­ary –>  88/100 × $7320 = $6441.60

 

6–  30 peo­ple

Frac­tion of men  –>  30/100 = 3/10 = 9/30

Frac­tion of men if 10 women alighted –>  45/100 = 9/20

Women –>  30u — 9u = 21u

20u — 9u = 11u

10u –>  10 women

30 u –>  30 people

Con­tinue read­ing

Primary 2 Whole Numbers Level 3 Ex 1

P2-Whole Numbers-L3-Ex 1

Ans:

1–  241 mar­bles

2–  538 stamps

Num­ber of stamps Irene has –>  325 — 112 = 213

325 + 213 = 538 stamps

 

3–  202 mar­bles

4–  40 pupils

5–  30 apples

Num­ber of apples in each bas­ket –>  2 + 3 = 5

5 × 6 = 30 apples

Con­tinue read­ing

Primary 5 October Problem Sums 5

P5-PSP-Oct-Ex5

Ans:

1–  $6588

2–  2400 cm³

Amount of water at first –>  4/5 × 30 cm × 20 cm × 30 cm = 14 400 cm³

Amount of water in the end –>  2/3 × 30 cm × 20 cm × 30 cm = 12 000 cm³

14 400 cm³ — 12 000 cm³ = 2400 cm³

 

3–  38 mar­bles

Qn 3

2u –>  96 — 60 = 36

Barry –>  36 ÷ 2 = 18 marbles

Joe –>  60 — 18 = 42 marbles

Tom –>  18 × 3 = 54 marbles

Total –> 54 + 42 + 18= 114 marbles

114 ÷ 3 = 38 mar­bles

 

4–  Cup Y —  160 ml    ,   Cup Z —  60 ml

Y   :   Z
8   : 3

8u + 3 = 11u

11u ÷ 2 = 5.5u

8u — 5.5u = 2.5u

2.5u –>  50 ml

1u –>  20 ml

Cup Y –>  20 ml × 8u = 160 ml

Cup Z –>  20 ml × 3 = 60 ml

 

5–  $651.80

Qn 5

2u –>  $1505 — $201.40 = $1303.60

$1303.60 ÷ 2 = $651.80

 

6–  $15.50

Qn 6

 

2u–  $108 — $71 = $37

1u —  $18.50

Megan –> $18.50 × 3 = $55.50

$71 — $55.50 = $15.50

Con­tinue read­ing

Primary 3 Fractions Level 1 Ex 1

P3-Fractions-L1-Ex1

Ans:

1–  4

2–  2

3–  4

4/16 = 1/4

 

4–  1

5–  3

6–  2

7–  1

Total num­ber of daisies –>  2 + 3 + 4 = 9

3/9 = 1/3 

Con­tinue read­ing

Primary 6 July Problem Sums 6

P6-July-PSP-Ex 6

Ans:

1–  $800

Frac­tion of money spent on food and shoes –>  1/4 + 1/5 = 9/20

20/20 — 9/20 = 11/20

11u –>  $340 + $100 = $440

1u –> $40

20u –>  $40 ×  20 = $800

 

2–  14 cook­ies

Muffins   :   Cookies

At first  ——>           [   5             :         6   ]  × 2

In the end   ——>    [   2             :        1  ]  × 5

[Make the units of muffins at first and in the end the same]

Muffins   :   Cookies

At first  ———>             10          : 12

In the end  ——->           10          : 5

12u –>  24 cookies

1u –>  2 cookies

5u –>  10 cook­ies left

24 — 10 = 14 cook­ies 

 

3–  Joseph spent more by $2000

3/5 of Terry’s money is equal to 1/3 of Joseph’s money.

[Make the numer­a­tor common]

3/5 of Terry’s is equal to  3/9 of Joseph’s.

Terry   :   Joseph
5         : 9

14u –> $7000

1u –> $500

Amount Terry spent –>  $500 × 5 = $2500

Amount Joseph spent –>  $500 × 9 = $4500

$4500 — $2500 = $2000

Con­tinue read­ing

Primary 6 July Problem Sums 5

P6-July-PSP-Ex 5

Ans:

1–  95

[Use trial and error from the largest 2-digit mul­ti­ple of 9]

Great­est pos­si­ble 2-digit mul­ti­ple of 9 –> 99

99 + 5 = 104 (wrong: answer is not 2-digit)

90 + 5 = 95 (correct)

 

2–  100 grey squares and + 99 white tiles

Num­ber of grey tiles –> Pat­tern Num­ber × 2

50 × 2 = 100 grey tiles

Num­ber of white tiles –>  Num­ber of grey tiles — 1

100 — 1 = 99 white tiles

 

3–  108 teach­ers

Pupils  –>                      Boys  :  Girls
4     : 5

5u –> 2400

1u –> 280

9u –>  280 × 9 = 4320 children

Teach­ers  :  Pupils
1          : 40

40u –> 4320

1u –>  108 teach­ers

 

4–  14 years old

Nicholas’ age now –>  35 — 8 = 27 years old

Timothy’s age now –>  1/3 × 27 = 9 years old

9 + 5 = 14 years old

Con­tinue read­ing

Primary 6 Circles Level 3 Ex 1

P6-Circles-L3-Ex 1

Ans:

1–  1152 m

Perime­ter of each semi­cir­cle–>  1/2 × 22/7 × 28 = 44 m

2 semi­cir­cles –>  44 × 2 = 88 m

Total perime­ter of track –>  100 + 100 + 88 = 288 m

4 rounds –>  288 m × 4 = 1152 m

 

2–  Area ->  87.5 cm² , Perime­ter –>  39 cm
Area of square –>  7 cm × 7 cm = 49 cm²

Area of quad­rant –>  1/4 ×22/7 × 7 × 7 = 38.5 cm²

Area of fig­ure –>  49 cm² + 38.5 cm² = 87.5 cm²

Perime­ter of quad­rant –>  1/4 × 22/7 × 14 cm = 11 cm

Perime­ter of fig­ure –>  11 cm + 4 × 7 cm = 39 cm

 

3–  34 cir­cles
Qn 3

 

 

 

 

 

Length ->  69 cm ÷ 4 ≈ 17

Breadth -> 9 cm÷ 4 ≈ 2

17 × 2 = 34 cir­cles 

Con­tinue read­ing

Primary 5 July Problem Sums 1

P5-July-PSP-Ex 1

Ans:

1–  $2425
Qn 1

1u –>  $1600 + $750 = $2350

$2350 + $750 = $2425

 

2–  29/45

Amount Diane saves -> $1800 — $640 = $1160

1160/1800 = 29/45

 

3–  4  1/2 kg

Mass of apples when the bas­ket is com­pleted filled –>  8 kg — 1 kg = 7 kg

Mass of apples when bas­ket is 1/2 filled –>  7 kg ÷ 2 = 3  1/2 kg

3  1/2 + 1  = 4  1/2 kg

 

Con­tinue read­ing