PSLE Math Problem Sums that Most P6 Pupils Stumble · PDF filePSLE Math – Problem Sums...

PSLE Math – Problem Sums that Most P6 Pupils Stumble Over (Part 7)

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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.



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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



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



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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.



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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.



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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.



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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.



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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


× 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


÷ 19 =

of her money

×20


× 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.



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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.



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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.



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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



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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



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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.



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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



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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.



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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.



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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% =



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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.



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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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