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PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
1. Mandy has 23 one-dollar coins.
Nicole has 148 one-dollar coins.
Every day, Nicole gives Mandy 12 one-dollar coins each time while Mandy gives
Nicole 6 one-dollar coins.
After how many days would Mandy have 25% more money than Nicole?
Solution:
Mandy: Nicole Total
At first 23 148 171
-6 -12
+12 +6
Change (+6) : (-6) for every day...
In the end 5 units : 4 units 9 units
25%
; Mandy has 1 unit more than Nicole
9 units 171
1 unit 19
4 units 76 (Nicole)
148 – 76 = 72 one-dollar coins given to Mandy
Since the number of Mandy’s one-dollar coins increased by 6 every day,
72 ÷ 6 = 12 days
Ans: After 12 days, Mandy will have 25% more money than Nicole.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
2. The ratio of the number of $2 notes to the number of $5 notes to the number of
$10 notes Roy had was 1 : 6 : 4.
After spending
of the $2 notes and $210 using only $5 notes, the value of the
$10 notes Roy had became
of the total amount of money he had left.
Find the total amount of money Roy had at first.
Solution:
Method 1:
Number of $5 notes spent = $210 ÷ $5 = 42
$2 : $5 : $10
Ratio of the number of notes 1 : 6 : 4 x3
(At first) 3 units : 18 units : 12 units
Change -1 unit -42
___________________________________________________________
Ratio of the number of notes 2 units : : 12 units
(In the end)
Ratio of the value of money 4 units : : 120 units
(In the end)
Since the value of $10 notes in the end was
of the total amount of money left,
; the total amount of money left 144 units
144 – 120 – 4 = 20 units (value of $5 notes)
Number of $5 notes 20 ÷ 5 = 4 units
18 – 4 = 14 units
14 units 42
1 unit 3
3 units 9 ($2 notes) ; Value $18
18 units 54 ($5 notes) ; Value $270
12 units 36 ($10 notes) ; Value $360
Total amount of money at first $18 + $270 + $360 = $648
Ans: Roy had $648 at first.
2 $2 notes = $4 12 $10 notes = $120 Ratio of value of $2 : $10 = 4 : 120
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
Solution:
Method 2:
$2 : $5 : $10
Ratio of the number of notes 1 : 6 : 4 Grouping
$2 : $30 : $40
Ratio of the value of money 1 unit : 15 units : 20 units
x3
(At first) 3 units : 45 units : 60 units
Change -1 unit -$210
___________________________________________________________
Ratio of the value of money
(In the end) 2 units : 10 units : 60 units
Since the value of $10 notes in the end was
of the total amount of money left,
; the total amount of money left 72 units
72 – 60 – 2 = 10 units (value of $5 notes)
35 units $210
1 unit $6
Total number of units (value of money) at first 3 + 45 + 60 = 108 units
108 units $648
Ans: Roy had $648 at first.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
3. In the final year examination, Janet sat for a total of 5 papers based on 5 different
subjects.
Her best subject was Mathematics, followed by Mother Tongue, Science, Social
Studies and English.
Janet obtained the sum of the scores of 2 subjects.
The following are all the possible sums:
157, 163, 166, 170, 171, 173, 178, 179, 184, 187
What score did Janet obtain for her Science paper?
Solution:
Given 5 different subjects as A, B, C, D and E, the 10 possible sums include
A+B, A+C, A+D, A+E,
B+C, B+D, B+E,
C+D, C+E, and
D+E.
Since each subject is counted 4 times among the 10 possible sums which means
4A + 4B + 4C + 4D + 4E = total of 10 sums
Total of 10 sums 157+163+166+170+171+173+178+179+184+187 = 1728
total score of all 5 papers 1728 ÷ 4 = 432
Combined score of English and Social Studies papers 157
Combined score of Mother Tongue and Mathematics papers 187
Score of Janet’s Science paper 432 – 157 – 187 = 88
Ans: Janet obtained a score of 88 for her Science paper.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
4. Boys and girls attended a birthday party.
If 36 boys left the party,
of the children remaining would be girls.
If 36 girls left the party, 48% of the children remaining would be boys.
How many children were there at the birthday party?
Solution:
In both cases, the total number of children remaining does not change.
Case 1: If 18 boys left the party
Boys : Girls Total
? ?
Change -36
_______________________________________________
Case 1 3 : 7 10 units
x10
30 units : 70 units 100 units
Case 2: If 18 girls left the party
Boys : Girls Total
? ?
Change -36
_______________________________________________
Case 2 48 units : 52 units 100 units
Comparing the number of boys in both cases, 48 – 30 = 18 units
18 units 36
1 unit 2
Total number of children 70 units + 48 units = 118 units
118 units 236
Ans: There were 236 children at the birthday party.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
5. Mr Chew had 3 large wooden cubes painted grey on all faces.
However, he decided to cut each large wooden cube into 8 identical smaller
wooden cubes and paint the unpainted faces of the smaller cubes grey.
He realized that he had painted a total area of 145 800 cm2 on the unpainted faces
of the smaller wooden cubes.
What was the total volume, in cubic metres, of the 3 large wooden cubes Mr Chew
had at first?
Solution:
Since each smaller wooden cube has 3 unpainted faces, and each large wooden cube
can be cut into 8 smaller wooden cubes,
Total number of square faces unpainted 3 × 3 × 8 = 72
Area of 1 square face 145 800 ÷ 72 = 2025 cm2
Length of the edge of a small wooden cube = 45 cm
Length of the edge of a large wooden cube 45 × 2 = 90 cm = 0.9 m
Total volume of 3 large wooden cubes 0.9 m × 0.9 m × 0.9 m × 3 = 2.187 m3
Ans: The total volume of the 3 large wooden cubes was 2.187 m3.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
6. Mrs Smith and Mrs Tate went shopping with a total sum of $4580.
They went to a boutique and each bought the same dress.
Mrs Smith, who had membership, was entitled to a 20% discount on the dress.
After the purchase, she had 30% of her money left.
Mrs Tate, who was not a member, was entitled to a 5% discount. After the
purchase, she had 40% of her money left.
(a) Who had more money at first, and how much more?
(b) What was the price of the dress they bought?
Solution:
Mrs Smith: 30% of money left ; 70% or
spent
Paid 80% of dress’ price
of her money
÷4
20% of dress’ price
÷ 4 =
of her money
×5
100% of dress’ price
× 5 =
of her money
---------------------------------------------------------------------------------------------
Mrs Tate: 40% of money left ; 60% or
=
spent
Equivalent
Paid 95% of dress’ price
of her money (Dress’ price)
÷19
5% of dress’ price
÷ 19 =
of her money
×20
100% of dress’ price
× 20 =
of her money
of Mrs Smith’s money
of Mrs Tate’s money
Mrs Smith Mrs Tate
=
=
96 + 133 = 229 units Difference 133 – 96 = 37 units
229 units $4580
1 unit $20
37 units $740
Ans: (a) Mrs Tate has $740 more than Mrs Smith at first.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
Solution: (continued)
Dress’ price was
of Mrs Smith’s money at first
96 units $1920 (Mrs Smith’s money at first)
Dress’ price
× $1920 = $1680
Ans: (b) The price of the dress was $1680.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
7. Ivan and Jeth shared a bag of marbles. Ivan received 65% of the marbles.
Later, Ivan bought another 40 marbles while Jeth lost 2 of his marbles.
In the end, 30% of Ivan’s marbles was equal to 80% of Jeth’s marbles.
What was the total number of marbles both of them had in the end?
Solution:
65% =
; 30% =
; 80% =
In the end,
of Ivan’s marbles was equal to
of Jeth’s marbles.
=
Ivan’s marbles : 40 units ;
=
Jeth’s marbles : 15 units
Ivan : Jeth (In the end) 40 : 15 = 8 : 3
Ivan : Jeth
At first 13 units : 7 units
Change +40 -2 OR 30% of Ivan = 80% of Jeth
______________________________
Ivan Jeth
In the end 8 : 3 80% 30%
= 8 : 3
3 × (13 units + 40) 8 × (7 units – 2)
39 units + 120 56 units – 16
17 units 120 + 16 = 136
1 unit 8
At first, the total number of marbles (20 units) 160
Total number of marbles in the end 160 + 40 – 2 = 198
Ans: Both of them had 198 marbles in the end.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
8. The diagram below, not drawn to scale, shows the floor plan of a walkway in a
park. The walkway is 1.12 m wide and consists of circular, identical cemented
tiles of negligible thickness. Two identical grass patches, made up of artificial
grass mats which cost $16.50 per square metre, are placed alongside the
walkway.
Given that the grass patches cost a total of $6468,
(a) how many circular cemented tiles are there altogether; and
(b) what is the total area, in square centimetres, of the walkway covered by all the
circular cemented tiles?
(Take π =
)
Grass Patch
... ... ...
... ... ...
... ... ...
Walkway
1.12 m
Grass Patch
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
Solution:
Diameter of a circular tile 1.12 m ÷ 2 = 0.56 m
Cost of a grass patch $6468 ÷ 2 = $3234
Area of 1 grass patch $3234 ÷ $16.50 = 196 m2
Breadth of a grass patch (5 circular tiles) 0.56 m × 5 = 2.8 m
Length of a grass patch 196 m2 ÷ 2.8 m = 70 m
Number of circular tiles forming up the length of a grass patch 70 m ÷ 0.56 m = 125
Number of circular tiles forming up the breadth of a grass patch 5
Total number of circular cemented tiles
(125 + 125 + 125 + 5 + 5) × 2 + 6 + 4 + 4 + 4 + 6 = 794
Ans: (a) There are 794 circular cemented tiles altogether.
Grass Patch
... ... ...
... ... ...
... ... ...
Grass Patch
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
Solution: (continued)
Radius of a circular tile 28 cm
Area of a circular tile
× 28 cm × 28 cm = 2464 cm2
Total area of walkway covered by all the circular cemented tiles
2464 cm2 × 794 = 1 956 416 cm2
Ans: (b) The total area of the walkway covered by all the circular cemented tiles is
1 956 416 cm2.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
9. Brandon and Carell can complete a painting assignment together in 20 minutes.
If both of them started the painting assignment for 5 minutes, followed by Brandon
who continued painting for 21 minutes, Carell would take another 7
minutes to
complete the entire painting assignment.
How many minutes would it take each of them to complete the painting
assignment alone?
Solution:
20 mins 1 painting
÷4 ÷4
5 mins
painting
painting 21 mins 7
mins
After the first 5 mins, if C had continued painting with B for 7
mins, his part will
have been completed.
1 min
painting (B and C together)
7
mins =
mins
×
=
painting
Remaining part of painting to be completed by B 1 –
–
=
Brandon :
painting 21 – 7
= 13
mins
painting 13
÷ 3 =
×
=
mins
painting
× 8 = 36 mins
Carell : 1 min
–
=
–
=
=
painting
painting 45 mins
Ans: Brandon : 36 mins ; Carell : 45 mins
B + C B
C
C
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
10. At 6.30 p.m., Jack and Jill were at the park and they started to run towards the
seafood centre along the same route. Jill was running at a speed of 180 m/min.
After 12
minutes, Jack arrived at a bicycle kiosk to get a bicycle. He cycled back
at a speed 200 m/min faster than his running speed to meet Jill.
At 6.45 p.m., both of them met at a point mid-way between the park and the
seafood centre. Jack picked Jill up and continued cycling to the seafood centre at
a speed 3 times as fast as Jill’s running speed.
(a) At what time did Jack and Jill arrive at the seafood centre?
(b) Find the ratio of Jack’s running speed to Jill’s running speed.
Solution:
6.30 p.m. 15 min 6.45 p.m.
Park Meeting Bicycle Seafood
6.30 p.m. Point Kiosk Centre
6.45 p.m.
12
min
Jack
Jill
2
min
12
min 2
min (when Jack returned)
15 min
Distance between park and meeting point (mid-way) 180 m/min × 15 min = 2700 m
Distance between meeting point and seafood centre 2700 m
Cycling speed 180 × 3 = 540 m/min
Time taken to cycle from meeting point to seafood centre
= 5 min
6.45 p.m. 5 min 6.50 p.m.
Ans: (a) Jack and Jill arrived at the seafood centre at 6.50 p.m.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
Solution: (continued)
Let Jack’s running speed be 1 unit (distance) per min ;
and his cycling speed would be 1 unit + 200 m (distance) per min
For 12
min, Jack ran 1 unit × 12
min = 12
units
At bicycle kiosk, Jack cycled back (1 unit + 200 m/min) × 2
min = 2
units + 500 m
Distance between park and meeting point (mid-way) 2700 m
12
units – (2
units + 500 m) 2700 m
10 units 2700 m + 500 m = 3200 m
1 unit 320 m
Jack’s running speed 320 m/min
Jill’s running speed 180 m/min
320 : 180 = 16 : 9
Ans: (b) The ratio of Jack’s running speed to Jill’s running speed was 16 : 9.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
11. A school conducts a fund-raising event each year.
Compared to last year, the number of boys who participated remains the same this
year and there are 25% more girls who participated this year.
The total amount of funds raised last year was $78 800, whereas the total amount
of funds raised this year is $101 400.
Given that the average amount of funds raised by each boy and that of each girl
this year increased by 10% and 20% respectively as compared to last year, find
(a) the amount of funds raised by the girls this year, and
(b) the increase in the number of girls who participated, if the average amount of
funds raised by each girl last year was $115.
Solution:
Let the amount of funds raised by girls last year be 100 units
This year, assume average funds remains unchanged, due to 25% more girls
amount of funds raised by girls should be 100 units ×
= 125 units
As average funds raised by each girl increased by 20%,
amount of funds raised by girls this year 125 units ×
= 150 units
Last Year : This Year
Total $78 800 $101 400
– Funds raised by girls -100 units : -150 units
_____________________________________________
Funds raised by boys 10 : 11
11 × ($78 800 – 100 units) 10 × ($101 400 – 150 units)
$866 800 – 1100 units $1 014 000 – 1500 units
400 units $147 200
1 unit $368
150 units $55 200
Ans: (a) The girls raised a total of $55 200 this year.
Funds raised by boys
increased by 10% (this year
versus last year) without any
change in number of boys
110% =
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
Solution: (continued)
100 units $36 800 (raised by girls last year)
Number of girls who participated last year $36 800 ÷ $115 = 320
Increase in the number of girls (last year to this year)
× 320 = 80
Ans: (b) The number of girls who participated in fund-raising increased by 80.
PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)
www.matharena.com.sg email: admin@matharena.com.sg Blk 488B Tampines Street 45 #B1-147, Tampines, Singapore 521488
Tel: 6783 3218 / 6781 9325 / 8121 6628
12. A watermelon, with 93% of its weight being water, was left under the Sun at noon.
At 1 p.m., some of its water evaporated such that only 92% of the watermelon’s
weight is due to water.
Given that the amount of water evaporated every hour is the same, what will the
percentage of water by weight of the watermelon be at 4 p.m.?
Solution:
92% =
=
; 93% =
Watermelon : Water
At noon (12 p.m.) 7 units : 93 units
×8 ×8
56 units 744 units
-? 100 units of water
______________________________________ evaporated
At 1 p.m. 8 units : 92 units
×7 ×7
56 units 644 units
At 1 p.m. 644 units of water
Since each hour, amount of water evaporated is 100 units,
at 4 p.m. (3 hours later) 644 – (3 × 100) = 344 units
Weight of watermelon 56 units + 344 units = 400 units
Percentage of water
× 100% = 86%
Ans: The percentage of water at 4 p.m. will be 86%.
Weight of the fruit
itself without water
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