Tag Archives: Primary 6

Primary 6 SA1 Revision 10

P6-SA1-Ex 10

Ans:

1–  3

2–  3

3–  4

2u –> 18 cm

1u –>  9 cm

Length –>  9 cm × 3u = 27 cm

Perime­ter –>  (18 + 27) cm × 2 = 90 cm

 

4–  2

5–  2

6–  3

Frac­tion of remain­ing cake –> 5/6

5/6 ÷ 1/12 = 10 pupils

Con­tinue read­ing

  

Primary 6 April Problem Sums 5

P6-PSP-Apr-Ex 5

Ans:

1–  30 m

Spaces between 100 trees –> 99

Dis­tance between each tree –> 297 m ÷ 99 = 3 m

3 m × 10 spaces = 30 m

 

2–  28 pupils

Num­ber of pupils who passed one or both sub­jects –> 40 — 2 = 38

30 + 36 = 66

66 — 38 = 28 pupils

 

3–  a ->  (18s + 24) kg ,   b ->   16 kg

Mass of chicken –>  3s kg

Mass of puppy –>  (3s + 4) kg

Mass of dog –>  (3+ 3s + 4) kg × 2 = (12+ 8) kg

3s + 3s + 4 + 12s + 8 = (18s + 12) kg                   (a)

18 × 2 + 12 = 48 kg

48 kg ÷ 3 = 16 kg                           (b)

 

4–  $51.30

Total cost of 5 apples and 5 man­goes –>  $3.25 × 5 = $16.25

Cost of 3 apples –>  $17.60 — $16.25 = $1.35

Cost of 1 apple –> $1.35 ÷ 3 = $0.45

Cost of 1 mango –> $3.25 — $0.45 = $2.80

Cost of 2 man­goes and 7 apples –>  $2.80 × 2 + $0.45 × 7 = $8.70

$50 — $8.70 = $51.30

Con­tinue read­ing

  

Primary 6 March Review 7

P6-Mar-Review 7

Ans:

1–  3

2–  4

3–  4

3.30 p.m. to 4.30 p.m. –>  1st hour –> $3.50

4.30 p.m. to 5.45 p.m.–>  1 h 15 min ->  3 sets of half-hours
->  3 × $1 = $3

$3.50 + $3 = $6.50

 

4–  1

5–  2

6–  3

7–  2

75% — 50% = 25%

25% –>  105 ml

50% –>  105 ml × 2 = 210 ml

Con­tinue read­ing

  

Primary 6 Algebra Level 2 Ex 5

P6-Algebra-L2-Ex5

Ans:

1–  $(7y/2) /  $3.50y

Price of 1 durian –>  $(y/10)

$(y/10) × 35 = $(7y/2)/ $3.50y

 

2–  55

3–  (35 

4–  (4+ 10) years old

Brother’s age –>  3n + 9 — 2n = (n + 9) years old

Total age now –>  3n + 9 + n + 9 = 4n + 18 years old

4n + 18 – 4 – 4 = (4n + 10) years old

 

5–  $(m – 20)/3

6–  4.5z² cm²

Area of Tri­an­gle WXY –>  1/2 × 3z × 1.5= 2.25 z² cm²

2.25 z² × 2 = 4.5z² cm²

Con­tinue read­ing

  

Primary 6 Algebra Level 1 Ex 4

P6-Algebra-L1-Ex 4

Ans:

1–  1

2–  1

3–  4

12x + 19 + (4x × 6 ÷ 3) – 8
= 12x + 19 + 8x - 8 = 20x + 11

 

4–  2

5–  2

Amount left after giv­ing her daugh­ter —  $(800 — 40z)

Each son ->  $(800 — 40z) ÷ 3

Con­tinue read­ing

  

Primary 6 September Problem Sums 10

P6-PSP-Sep-Ex 10

Ans:

1–  See below

Qn 1

 

2–  9360 cm³

Height of water needed to fill the con­tainer –>  25 cm — 18.5 cm = 6.5 cm

45 cm × 32 cm × 6.5 cm = 9360 cm³

 

3–  68 but­tons

36 + 3 = 39 more buttons

8 — 5 = 3  buttons

∴ She needs 39 more but­tons because she is putting 3 more but­tons into each box.

39 ÷ 3 = 13 boxes

13 × 5 + 3 = 68 but­tons

 

4–  $580

107% –>  $465.45

100% –>  $435 (dis­counted price with­out GST)

75% –> $435

100% –>  435/75 × 100 = $580

 

5–  136 chil­dren

40% = 40/100 = 2/5

2/3 of the boys = 2/5 of the girls

Boys  :  Girls
3      : 5

2u –> 34

1u –> 17

17 × 8 = 136 chil­dren 

Con­tinue read­ing

  

Primary 6 August Review 6

P6-Aug-Review6

Ans:

1–  4

2–  1

Total of the 4 num­bers –>  4 × 22 = 88

88 — 24 — 26 — 28 = 10

 

3–  1

4–  4

15/60 × 300 km = 75 km

75 km = 75 000 m

 

5–  4

6–  4

5/5 — 2/5 = 4/5  (owned cats and dogs)

3/5 –>  48 + 72 = 120 children

1/5 –>  40 children

5/5 –>  40 × 5 = 200 chil­dren

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 Speed Level 2 Ex 1

P6-Speed-L2-Ex 1

Ans:

1–  52 km/h

11.25 a.m. to 2.40 p.m. –>  3 h 15 min = 3  1/4 h

169 km ÷ 3  1/4 h = 52 km/h

 

2–  2.4 m/s

Dis­tance between 1st floor and 25th floor –>  24 × 4 = 96 m

96 m ÷ 40 sec = 2.4 m/s

 

3–  60 km/h

Time taken 180 km ÷ 54 km/h = 3  1/3 h = 3 h 20 min

New time –>  3 h 20 min — 20 min = 3 h

180 km ÷ 3 h = 60 km/h

Con­tinue read­ing