Tag Archives: problem solving

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 4 July Problem Sums 3

P4-July-PSP-Ex 3

Ans:

1–  $9153

$2178 + $6975 = $9153

 

2–  5  5/12 kg

The total mass of the nails–>  9   11/12 kg —   11/12 kg = 9 kg

Mass of the nails when they are half-filled –>  9 /2 = 4   1/2 kg

4   1/2 kg +  11/12 kg = 5   5/12 kg

 

3–  3456 canned drinks

128 × 27 = 3456 canned drinks

 

4–  56.58 kg

Qn 4

6u –>  169.74 kg

1u –>  28.29 kg

28.29 kg × 2 = 56.58 kg

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 4 July Problem Sums 2

P4-July-PSP-Ex 2

Ans:

1–  $144

1/2 u –> $36

1u –>  $36 × 2 = $72

$72 × 2 = $144

 

2–  $487.80

3–  6  4/5 kg

4–  2637 more seashells
Qn 4

7u –>  6328 — 623 = 5705

1u –> 815

Jan­ice –>  815 × 7 = 4075 seashells

Car­o­line –>  815 + 623 =1438 seashells

4075 — 1438 =  2637 more seashells

Con­tinue read­ing

  

Primary 3 July Review 1

P3-Jul-Review 1

Ans:

1–  3

2–  3

3–  2

4–  4

5–  4

6–  2

Grandmother’s age now —  9 × 9 = 81  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

  

Primary 4 July Problem Sums 1

P4-July-PSP-Ex 1

Ans:

1–  63 sweets

2–  $90

Qn 2

2u –>  $75 — $45 = $30
1u –> $15

$15 + $75 = $90

 

3–  96 cook­ies
Qn 3

5u –>  80 cookies

1u –>  16 cookies

6u –>  16 cook­ies × 6 = 96 cook­ies

 

4–  $68
Qn 4

3u –>  $48 + $12 = $60

1u –> $20

$48 + $28 = $68

Con­tinue read­ing

  

Primary 4 June Problem Sums 1

P4-Jun-PSP-Ex1

Ans:

1–  1120 stick­ers
Qn1

9u –>  1260 stickers

1u –>  140 stickers

8u –>  140 × 8 = 1120 stick­ers

 

2–  $28 125

Total num­ber of TV sets sold –>  15 × 5 = 75

$375 × 75 = $28 125 

 

3–  a ->  4 ostriches  ,   b ->  76 more legs

Qn3

 

84 — 8 = 76 more legs   (b)

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%

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