Tag Archives: Primary 6

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

  

Primary 6 June Problem Sums 6

P6-PSP-Jun-Ex 6

Ans:

1–  60 stamps
Qn 1

4u –>  26 + 14 = 40

1u –>  40 ÷ 4 = 10 stamps

6u –> 10 × 6 = 60 stamps

 

2–  6.05 m

Total length of rope Edmond and Mandy had ->  24.6 m + 32.5 m = 57.1 m

57.1 m — 4.2 m = 52.9 m

Length of rope Mandy had in the end –>  52.9 m ÷ 2 = 26.45 m

32.5 m — 26.45 m = 6.05 m

 

3–  13 big water­mel­ons and 5 small watermelons

Assume all 18 water­mel­ons were small –>  18 × $3 = $54

$67 — $54 = $13 (excess)

Since each big water­melon cost $1 more–>  $13 ÷ $1 = 13 big watermelons

18 — 13 = 5 small watermelons

 

4–  See below
Qn 4

Con­tinue read­ing

  

Primary 6 SA1 Revision 9

P6-SA1-Ex 9

Ans:

1–  4 cm

Length of square –>  12 cm

12 ÷ 2 = 6 cm ( length of square B)

Are of square B –>  144 ÷ 6 = 24 cm²

24 ÷ 6 = 4 cm

 

2–  3.43 p.m.

135 ÷ 75 = 1  4/5 h

1  4/5 h + 1/4 h = 2  1/20 h

2  1/20 h = 2 h 3 min

2 h 3 min after 1.40 p.m. –>  3.43 p.m.

 

3–  343 cm³ 

Area of 1 sur­face of the cube –>  294 ÷ 6 = 49 cm²

Length of cube –>  7 cm

Vol­ume –>  7 × 7 × 7 = 343 cm³

 

4–  11%

$440 — $395 = $45

45/395 × 100 ≈ 11%

Con­tinue read­ing

  

Primary 6 SA1 Revision 8

P6-SA1-Ex 8

Ans:

1–  2/5 , 2/3 , 1/2 , 3/4

2–  3.08 kg

3–  264 min

4h –>  4 × 60 = 240 min

2/5 × 60 = 24 min

240 + 24 = 264 min

 

4–  $128

3/4 of Irene’s money is equal to 4/5 of Adam’s.
[make the numer­a­tor common)

-> 12/16 of Irene’s =  12/15 of Adam’s

Irene   :   Adam
16     : 15

15u –> $120

1u –> $8

16u –>  $8 × 16 = $128 

Con­tinue read­ing

  

Primary 6 SA1 Revision 7

P6-SA1-Ex 7

Ans:

1–  2

2–  2

3–  4

4–  2

10 + 5 + 6/1000 = 15.006

15.056 — 15.006 = 0.05

0.05 = 5/100 = 1/20

Con­tinue read­ing