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Thursday, July 22, 2010

Free Maths Worked Solutions - Nanyang P5 CA2 2009

Notice - Those who downloaded the P5 Maths CA2 Paper 2 Nanyang 2009, between 1000h to 1200h on 22 Jul 2010, please discard the copy as there are some errors in it.

Here is the amended copy - Download here.

For more samples, visit our - Free Maths Worked Solutions Page.

If you notice any errors in the free copy, you may highlight them here - Feedback.

Tuesday, July 20, 2010

ACS Primary 2009 PSLE Math Prelim Paper 2 Q18

There were some black and white marbles in a bag. If 20 black marbles are removed from the bag, the total number of marbles will be 7 times the number of black marbles left. If 50 white marbles are removed from the bag, the total number of marbles left will be 5 times the number of black marbles left. How many marbles are there in the bag?

Solution

20 black marbles (removed)
Black --> 1 unit + (20)
White --> 6 units

50 white marbles (removed)
Black --> 1 part
White --> 4 parts + (50)

(Black) 1 unit + 20 --> 1 part ...... (x4)*
(White) 6 units --> 4 parts + 50

(Black) 4 units + 80 --> 4 parts
(White) 6 units --> 4 parts + 50

*(x4) to make an equal number of 4 parts for both Black and White Marbles


(White Marbles) - (Black Marbles)
6 units - 4 units - 80 --> 4 parts + 50 - 4 parts
2 units - 80 --> 50
2 units --> 50 + 80
2 units --> 130
1 unit --> 130 divided by 2 = 65

Total number of marbles
Black --> 1 unit + 20
White --> 6 units
--> 7 units + 20
= (7 x 65) + 20
= 475

Answer: 475 marbles


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ACS Primary 2009 PSLE Math Prelim Paper 2 Q17

65% of the animals on a farm were cows and the rest were goats. When 240 more cows and goats were added to the farm, the percentage of cows increased by 20% and the number of goats doubled. How many goats were there on the farm at first?

Solution

At first
Cows --> 65%
Goats --> 35%

Cows 20% more
--> (20/100) x 35% = 13% more

Goats doubled
--> 35% more

Total added
13% (cows) + 35% (goats) --> 240
48% --> 240
1% --> 240 divided by 48 = 5

Goats at first
35% --> 35 x 5 = 175

Answer: 175 goats


ACS Primary 2009 PSLE Math Prelim Paper 2 Q16

The figure below is made up of semicircles and quadrants. Find
a) the area of A
b) the perimeter of B
Leave your answer correct to 1 decimal place.



Solution


a)
Area of unshaded semicircle (on the left of the figure above)
--> (1/2)(3.14)(3)(3)sq cm = 14.13 sq cm

Area of unshaded square (at the bottom left of the figure)
--> 6cm x 6cm = 36 sq cm
(The unshaded portion outside the shaded area B on the bottom right square, can be shifted left to form a square at the bottom left of the figure.)

Total Area of A
--> (14.13 + 36) sq cm
= 50.13 sq cm
~ 50.1 sq cm

Answer: 50.1 sq cm


b)

Perimeter of B
--> There are 3 quadrants of radius 6cm, 1 semicircle of radius 3 cm and one straight line measuring 6 cm
--> (3/4)(2)(3.14)(6)cm + (1/2)(2)(3.14)(3)cm + 6 cm
= (28.26 + 9.42 + 6) cm
= 43.68 cm
~ 43.7 cm

Answer: 43.7 cm


ACS Primary 2009 PSLE Math PrelimPaper 2 Q15

There were 192 apples and pears in a box. John removed 2/5 of the apples from the box and he added 24 pears into the box. As a result, there was an equal number of apples and pears left in the box. How many more apples than pears were there in the box at first?

Solution



5 units + 1 part --> 192
3 units --> 1 part + 24

5 units + 1 part --> 192
3 units - 1 part --> 24

(5 units + 3 units) + (1 part - 1part) --> 192 + 24
8 units --> 216
1 unit --> 216 divided by 8 = 27

At first
(Apples) 5 units --> 5 x 27 = 135
(Pears) 192 - 135 = 57

Apples more than pears
--> 135 - 57 = 78

Answer: 78 more apples


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Monday, July 19, 2010

ACS Primary 2009 PSLE Math Prelim Paper 2 Q14

At 9.30 am, Mr Yeo left Town A for Town B driving at a speed of 75 km/h throughout his journey. At 10.30 am, Mr Lee also left Town A for Town B driving at a certain speed. He kept to the same speed throughout his journey. At 1.30 pm, both of them passed a Shopping Mall that was 150 km way from Town B. How many minutes earlier did Mr Lee reach Town B than Mr Yeo?

Solution



(Town A to Shopping Mall)
Yeo --> distance = 75 km/h x 4h = 300 km
Lee ---> speed = 300 km divided by 3h = 100 kmh

(Shopping Mall to Town B)
Lee --> time = 150 km divided by 100 km/h = 1.5 h
Yeo --> time = 150 km divided by 75 km/h = 2 h

2 h - 1.5 h
= 0.5 h
= 30 min

Answer: 30 minutes

Sunday, July 18, 2010

ACS Primary 2009 PSLE Math Prelim Paper 2 Q13

In the figure below, O is the centre of the circle and AE is parallel to BC. DF = DE, Angle OAB = 58 degrees and Angle FED = 50 degrees.


a) Find Angle GBC
b) Find Angle DCB

Solution


a)
Angle ABO = 58 deg (isosceles triangle)
Angle BOG = (58 + 58) deg = 116 deg (exterior angles)
Angle OGB = [(180 - 116) divided by 2] = 32 deg
Angle GBC = 32 deg (alternate angles)

Answer: 32 degrees


b)
Angle FDE = (180 - 50 - 50) deg = 80 deg
Angle FDE = Angle GDC = 80 deg
Angle DCB = (180 - 80) deg = 100 deg

Answer: 100 degrees

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ACS Primary 2009 PSLE Math Prelim Paper 2 Q12

Water flows from Tap A at a rate of 250 ml per minute and from Tap B at a rate of 350 ml per minute. When both taps are turned on for 9 minutes, the water from both taps fill a container with a square base of side 25 cm. What is the height of the water level?



Solution

Tap A
1 min --> 250 ml
9 min --> 9 x 250 ml = 2250 ml

Tap B
1 min --> 350 ml
9 min --> 9 x 350 ml - 3150 ml

Total --> 2250 ml + 3150 ml = 5400 ml


Volume of cuboid = Base area x height
5400 cubic cm = 25 cm x 25 cm x height
5400 cubic cm = 625 square cm x height
height = 5400 cubic cm divided by 625 square cm
height = 8.64 cm

Answer: 8.64 cm


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ACS Primary 2009 PSLE Math Prelim Paper 2 Q11

Container A has 150 more marbles than Container B. If 30 marbles are being transferred from Container B to Container A, there will be thrice as many marbles in Container A as Container B. How many marbles are there in Container A in the beginning?

Solution


2 units --> 30 + 150 + 30 = 210
1 unit --> 210 divided by 2 = 105

Container A in the beginning
--> 3 units - 30
= (3 x 105) - 30
= 315 - 30
= 285

Answer: 285 marbles


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ACS Primary 2009 PSLE Math Prelim Paper 2 Q10

Given that the length of DB is 14 cm and the shaded region is 115 square cm, find the lenght of AC, correct to 2 decimal places.


Solution

Area of Triangle = Half base x height.
Area of Triangle ABC --> (1/2)(AC)(DX + 14)
Area of Triangle ACD --> (1/2)(AC)(DX)

Area of shaded area
--> Area of Triangle ABC - Area of Triangle ACD = 115

(1/2)(AC)(DX + 14) - (1/2)(AC)(DX) = 115

(1/2)(AC)(DX) + (1/2)(AC)(14) - (1/2)(AC)(DX) = 115

(1/2)(AC)(14) = 115
7(AC) =115
AC = 115 divided by 7
= 16.428
~ 16.43 (nearest to 2 decimal places)

Answer: 16.43 cm

ACS Primary 2009 PSLE Math Prelim Paper 2 Q9

The pie chart below shows the different types of muffins sold in a bakery. A total of 240 muffins were sold.


a) How many banana muffins were sold?
b) If a chocolate muffin cost $1.60, how much did the bakery collect from the sale of chocolate muffins?

Solution

a) (1/4) x 240 = 60

Answer: 60 banana muffins



b)
Cheese --> (1/8) x 240 = 30
Blueberry --> (20/100) x 240 = 48
Banana --> 60
Chocolate Chip --> 68

Total for the above
--> 30 + 48 + 60 + 68 = 206

Chocolate --> 240 - 206 = 34

Sale collected from chocolate muffin
--> 34 x $1.60 = $54.40

Answer: $54.40

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ACS Primary 2009 PSLE Math Prelim Paper 2 Q8

The ratio of the number of pencils to the number of erasers in a box was 2:3. When 42 pencils were added and 15 erasers were removed, the ratio of the number of pencils to erasers became 3:4. How many erasers were there left in the box?

Solution


Before
--> Pencils 2 units, Erasers 3 units

After 42 pencils added and 15 erasers removed
--> Pencils 3 parts, Erasers 4 parts

3 parts --> 2 units + 42 (x3)*
4 parts --> 3 units - 15 (x2)*

*(x3) and (x2) to give an equal number of 6 units for both equations.

9 parts --> 6 units + 126
8 parts --> 6 units - 30

9 parts - 8 parts --> 126 - (-30)
1 part --> 156
4 parts --> 4 x 156 = 624

Answer: 624 erasers


ACS Primary 2009 PSLE Math Prelim Paper 2 Q7

A bar of chocolate is sold at $3.50 each or in packets of 4 at $12 per packet. Alice wants to buy exactly 38 bars of chocolate for a party. What is the least amount of money that Alice could have spent on the chocolates?

Solution

38 bars --> 36 bars + 2 bars
--> 9 packets of 4 bars + 2 individual bars
--> (9 x $12) + (2 x $3.50)
= $108 + $7
= $115

Answer: $115

ACS Primary 2009 PSLE Math Prelim Paper 2 Q5

If it takes 1 worker 4 days to paint a flat, how many days will it take 8 workers to paint 4 flats if they all work at the same rate?

Solution

1 worker --> 1 flat, 4 days

8 workers --> 8 flats, 4 days
(8 workers give an output 8 times more)

8 workers --> 4 flats, 2 days
(Same 8 workers need only half the time for 4 flats)

Answer: 2 days

ACS Primary 2009 PSLE Math Prelim Paper 2 Q4

Barry poured 230 ml of lemonade into a jug that already contained 1.6 litres of juice. He then poured out the drink equally into 3 cups. What is the volume of the drink in each cup?

Solution

0.23 + 1.6 = 1.83

1.83 divided by 3 = 0.61

Answer: 0.61 litre


ACS Primary 2009 PSLE Math Prelim Paper 2 Q3

The circumference of a circular disc is 154 cm What is the radius of the circular disc?



Solution


154 cm = (22/7) d

d = (154cm)(7/22) = 49 cm

d = 49 cm divided by 2 = 24.5 cm

Answer: 24.5 cm

ACS Primary 2009 PSLE Math Prelim Paper 2 Q2

Box X contains 6 times as many oranges as Box Y. Box Z contains 111 fewer oranges than Box X. If Box Y contains 1/3 the number of oranges in Box Z, find the total number of oranges in the 3 boxes.

Solution

X --> 6 units

Y --> 1 unit
Z --> 6 units - 111

(3x) Y --> Z (since Y contains 1/3 of Z, we need 3 of Ys to be equal to one of Z)
3 units --> 111
1 unit --> 111 divided by 3 = 37

(Total) 13 units - 111
--> (13 x 37) - 111
= 370

Answer: 370 oranges

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Sunday, July 11, 2010

Request for help from a reader - 10 Jul 2010

Posted by a reader which can be found in this link.

Aden and John started jogging along a circular track. Aden started at Point X while John started at Point Y where the line XY formed the diameter of the circle. Aden and John jogged toward each other along the circular track from their respective starting point and first met at Point W which was 80 m from Point X. After they met for the first time, they continued jogging along the track and finally met again for the second time at Point Z which was 60 m from Point Y. Find the distance of the circular track. (Answer : 360 m)

Any kind soul would like to help? Please post solution here. Registration is not required.