Tag Archives: problem sums

Primary 6 Prelims Revision 6

P6-Prelims-Ex 6

Ans:

1–  $1225

18% = 9/50

90% = 9/10

There­fore,

Bran­don   :   Charles
50        :     10
5          : 1

6u –> $1470

1u –> $245

$245 × 5u = $1225

 

2–  9/10 h

Time taken to fill 1/9 of the tank –>  2/5 ÷ 4 = 1/10 h

1/10 h × 9 = 9/10 h

 

3–  8 pen­tagons

3 × 5 = 15 match­sticks for the 3 extra pentagons

60 — 15 = 45 match­sticks (The num­ber of squares and pen­tagons are now the same)

Num­ber of match­sticks needed to make 1 square and 1 pen­ta­gon -> 4 + 5
= 9

45/9 = 5

5 + 3 = 8 pen­tagons

 

4–  10 hours

55 min = 11/12 h

11/12 h + 3/4 h = 1   2/3 h

5  1/2 h ÷ 11/12 h = 6 sets

6 × 1   2/3 h = 10 hours

 

5–  $40

85% — 55% = 30%

30% –> $12

100% –>  12/30 × 100 = $40

Con­tinue read­ing

  

Primary 6 Prelims Revision 3

P6-Prelims-Ex 3

Ans:

1–  3 : 12 : 17

2–  4.95 cm

Length of the tank –>  180 cm³ ÷ 12 edges = 15 cm

33/100 × 15% = 4.95 cm

 

3–  100 balls

Total units of balls –>  5u + 4u = 9u

Units of balls in each bag in the end –>  9u/2 = 4.5u

5u — 4.5u = 0.5u —>  10 balls

1u –>  10 × 2 = 20 balls

20 × 5u = 100 balls

 

4–  540 km

5–  $24

1 blouse –> 1u

1 pair of jeans –>  1u + $12

3 pairs of jeans + 4 blouses –>  (1u + $12) × 3 + 4u = 7u + $36

7u + $36 -> $204

7u –>  $204 — $36 = $168

1u –>  $24

 

6–  891 beads

Red : Blue + Green
4   : 5

Blue : Green
3    :    8     –> 11u

Adjust the units for the blue and green beads such that they are the same in both cases.

Com­mon mul­ti­ple of 5 and 11 is 55u

There­fore,

Red  :  Blue + Green
44   : 55

Blue  :  Green
15   : 40

Red  :  Blue  :  Green
44 :    15   : 40

44u — 40u = 4u –>  36 beads

1u –>  9 beads

44u + 55u = 99u —->  9 beads × 99 = 891 beads

Con­tinue read­ing

  

Primary 2 Whole Numbers Level 3 Ex 2

P2-Whole Numbers-L3-Ex 2

Ans:

1–  8 mark­ers

 

2–  8 stamps

3–  569 stick­ers

4–  776 stamps

5–  40 pupils

Qn 5

Num­ber of girls –> 30 ÷ 3 = 10

30 + 10 = 40 pupils

 

6–  9 beads

Total num­ber of beads they had –>  76 + 45 = 121

130 — 121 = 9 beads

 

7–  720 stick­ers

Num­ber of stick­ers Tanya has –>  341 + 38 = 379

379 + 341 = 720 stick­ers

 

8–  537 stamps

  

Primary 6 SA1 Revision 12

P6-SA1-Ex 12

Ans:

1–  175

2–  $43.50

70% of the price –>  $101.50

30% –>  $101.50/70 × 30 = $43.50

 

3–  (3+ 6) kg

Alan’s mass ->  (+ 3) kg

Nicholas’ mass ->  (m + 3) — 2 = (+ 1) kg

Brett’s mass ->  (+ 1) + 1 = (+ 2) kg

(m + 3) + (+ 1) + (+ 2) = (3+ 6) kg

 

4–  195 km

1  1/2 h× 80 km/h = 120 km

45/60 h × 100 km/h = 75 kg

120 km + 75 km = 195 km

 

5–  $6441.60

Earn­ings in Jan­u­ary –>  122/100 × $6000 = $7320

Earn­ings in Feb­ru­ary –>  88/100 × $7320 = $6441.60

 

6–  30 peo­ple

Frac­tion of men  –>  30/100 = 3/10 = 9/30

Frac­tion of men if 10 women alighted –>  45/100 = 9/20

Women –>  30u — 9u = 21u

20u — 9u = 11u

10u –>  10 women

30 u –>  30 people

Con­tinue read­ing

  

Primary 2 Whole Numbers Level 3 Ex 1

P2-Whole Numbers-L3-Ex 1

Ans:

1–  241 mar­bles

2–  538 stamps

Num­ber of stamps Irene has –>  325 — 112 = 213

325 + 213 = 538 stamps

 

3–  202 mar­bles

4–  40 pupils

5–  30 apples

Num­ber of apples in each bas­ket –>  2 + 3 = 5

5 × 6 = 30 apples

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 4 March Problem Sums 5

P4-PSP-Mar-Ex 5

Ans:

1–  $96

2–  28 sweets

Qn 2

If 1 add 12 more sweets to Jacque­line, she will have 1 unit.

4u –>  100 + 12 = 112

112 ÷ 4 = 28 sweets

 

3–  a ->  $8250   b ->  $2475

Amount of money Melanie spent in Jan­u­ary –>  $825 × 2 = $1650

$1650 × 5 = $8250          (a)

$825 + $1650 = $2475      (b)

 

4–  5/8 m

5– 1175 stamps

Roger –> 1u

Joshua –> 4u

4u — 1u = 3u

3u –>  705 stamps

1u –>  235 stamps

235 × 5u = 1175 stamps

Con­tinue read­ing

  

Primary 3 Whole Numbers Level 3 Ex 4

P3-Whole Numbers-L3-Ex 4

Ans:

1–  4200 beads

Qn 1

Num­ber of beads Jen­nifer has –>  1866 + 468 = 2334

2334 + 1866 = 4200 beads

 

2–  195 steps

3–  67 cook­ies

4–  a –>  1277 men   ,   b ->  1085 more men

Num­ber of men who par­tic­i­pated –>  1277 — 89 = 1188

Num­ber of women who par­tic­i­pated –>  123 — 20 = 103

1188 — 103 =1085 more men 

 

5–  5460 stick­ers 

Num­ber of months from Feb­ru­ary to Octo­ber –> 9

9 × 20 = 180 more than January

5280 + 180 = 5460 stick­ers

 

6–  $712

Total amount of money Melody and Lisa have –>  $2835 + $4259 = $7094

Amount of money each of them has in the end –>  $7094 ÷ 2 = $3547

$4259 — $3547 = $712

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 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