Tag Archives: Primary 6

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

  

Primary 6 SA1 Revision 6

P6-SA1-Ex 6

Ans:

1–  21 trucks

Assume all 25 vehi­cles are toy cars

25 × 4 = 100 wheels

142 — 100 = 42

42 ÷ 2 = 21 trucks 

 

2–  West

3–  1550 more pupils

Qn 3

5600 + 120 = 5720

5720 ÷ 4 = 1430

1430 — 120 = 1310 (School C)

1430 × 2 = 2860 (School A)

2860 — 1310 = 1550 more pupils

Con­tinue read­ing

  

Primary 6 SA1 Revision 5

P6-SA1-Ex 5

Ans:

1–  5.05

2–  45 786 ,   45 876 ,   54 678 ,   54 768

3–  1/5

4–  18.22

Con­tinue read­ing

  

Primary 6 SA1 Revision 4

P6-SA1-Ex 4

Ans:

1–  4

2–  1

3–  3

4–  2

Total of the 2 num­bers –>  30 × 2 = 60

Smaller num­ber ->  1u
Big­ger num­ber -> 2u

3u –>  60
1u –>  60 ÷ 3 = 20

Con­tinue read­ing

  

Primary 6 SA1 Revision 3

P6-SA1-Ex 3

Ans:

1–  18 kg

18 kg + 27 kg + 12 kg + 9 kg + 24 kg = 90 kg

90 kg + 36 kg = 126 kg

126 kg ÷ 7 = 18 kg

 

2–  56 cm²

8 cm² × 7 = 56 cm²

 

3–  350 km

11.25 a.m. to 2.45 p.m. ————>  3 h 20 min

3  20/60  × 105 =  350 km

Con­tinue read­ing