Tag Archives: difficult

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 5 Average Level 3 Ex 4

P5-Average-L3-Ex4

Ans:

1–  432 more stamps

Total num­ber of stamps Jamie and Patrick had ->  360 × 2 = 720

Patrick -> 1u

Jamie -> 4u

5u –>  720 stamps

1u –>  144 stamps

4u — 1u = 3u

3 × 144 = 432 more stamps

 

2–  71.5 kg

Total mass of 3 boys if they weigh as much as Derek –>  75 × 3 + 12 + 9 = 246 kg

Derek’s mass  –>  246 kg ÷ 3 = 82 kg

Joe’s actual mass ->    82 kg — 12 kg = 70 kg

Ron’s actual mass ->  82 kg — 9 kg = 73 kg

Aver­age mass of Joe and Ron –>  (70 + 73) ÷ 2 =  71.5 kg

3–  5.8 m

Total mass of the pole, plank and rod –>  4.8 m × 3 = 14.4 m

Rod –> 1u

Pole –> 2u

Plank –>  2u + 0.4 m

5u +  0.4 m ->  14.4 m

5u –>  14.4 m — 0.4 m = 14 m

Rod  –>  14 m ÷ 5 = 2.8 m

Pole –>  2.8 m × 2 = 5.6 m

Plank –>  2.8 m × 2 + 0.4 = 6 m

Aver­age length of pole and plank –>  (5.6 m + 6 m ) ÷ 2 = 5.8 m

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 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

  

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