This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.

## Friday, November 26, 2010

### Rosyth boy tops PSLE

http://www.straitstimes.com/BreakingNews/Singapore/Story/STIStory_607535.html

ALEX Tan Kian Hye of Rosyth School is the top Primary School Leaving Examination (PSLE) pupil this year with a score of 282.

Fu Wan Ying from Tao Nan School had the second highest score of 279.

The top Malay pupil is Aquilah Dariah Mohd Zulkarnain, of Coral Primary School who scored 278, while the top Indian pupil is Muhammad Hameem, of Henry Park Primary School with a score of 274, and the top Eurasian pupil is Lendermann Monika Jiz-xin, of CHIJ Our Lady Queen of Peace, with a 269 score.

## Thursday, November 25, 2010

### Top PSLE 2010 Score - 282 (unofficial)

News from Kiasu Parents Forum.

Top Score - 282
School - Rosyth

Above is unofficial.

## Sunday, October 24, 2010

### Year 2011 Classes

We are open for registration for the Year 2011

Classes and Subjects (Small group tuition)

Primary 1 and 2 - English, Maths
Primary 3 to 6 - English, Maths, Science

We will be starting classes early or mid November.

Taught by a husband and wife team.

Home based tuition, located at Tampines, along Tampines Avenue 5.

For enquiries, you may call Mrs Song at

6260 4258 or 9424 7940

Alternatively, you email us at - FreeMathSample@gmail.com

## Sunday, August 22, 2010

### RGS Primary 2009 PSLE Math Prelim Paper 2 Q18

At first, 25% of Kumar's money was the same as 33 and 1/3 % of Lily's money. Lily's father gave her \$80 later, while Kumar spent \$325. In the end, Lily had 2 and 1/2 times as much money as Kumar.
a) How much money did Kumar have at first?
b) How much money did Lily have in the end?

Solution (Kumar) 4 units --> 2 parts + 325 (x5)**
(Lily) 3 units + 80 --> 5 parts (x2)**

**Kumar (x5) and Lily (x2) to make both of them 10 equal parts each.

(K) 20 units --> 10 parts + 1625
(L) 6 units + 160 --> 10 parts

(K) 20 units --> 10 parts + 1625
(L) 6 units --> 10 parts - 160

(K) - (L)
20 units - 6 units --> 10 parts - 10 parts + 1625 - (-160)
14 units --> 1625 + 160
14 units --> 1785
1 unit --> 1785 divided by 14 = 127.5

(a)
Kumar at first
4 units --> 4 x 127.5 = 510

(b)
Lily in the end
--> 3 units + 80
(3 x 127.5) + 80
= 382.5 + 80
= 462.5

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q17

At 9.30 am, Train A which was 200 m long, pulled out of Nanas Station and travelled towards Dadas Station at a uniform speed of 80 km/h. Half an hour later, Train B which was 150 m long, left Dadas Station and travelled towards Nanas Station at a uniform speed of 90 km/h.

a) How far has Train A travelled when Train B left Dadas Station?
b) The two trains met each other in a tunnel. Both trains took 15 minutes to completely travel through the tunnel. Calculate the length of the tunnel.

Solution

a)
Distance = speed x time
= 80 km/h x (1/2) hour
= 40 km

b)
(Train A) distance
= 80 km x (1/4) hr
= 20 km

(Train B) distance
= 90 km x (1/4) hr
= 22.5 km

Length of Tunnel
--> 20km + 22.5km - 0.2km (Train A's length) - 0.15km (Train B's length)
= 42.15 km

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q16

There were some sweets in Boxes X, Y and Z. Box X contained 20% of the total number of sweets in Boxes X, Y and Z. The ratio of the number of sweets in Box Y to the total number of sweets in Boxes X and Z is 2:1. If there are 24 more sweets in Box Y than Box Z, find the total number of sweets in Boxes X, Y and Z.

Solution

Box X --> 20% of total sweets
= 20/100
= 1/5 --> (3/15) of total sweets

Box Y --> 2 units
Box X + Z --> 1 unit
(Total 3 units)

Box Y --> (2 units)/(3 units)
= 2/3 --> (10/15) of total sweets Z --> 2 units
Y - Z --> 24 sweets
10 units - 2 units --> 24 sweets
8 units --> 24 sweets
1 unit --> 24 sweets divided by 8 = 3 sweets
15 units --> 15 x 3 sweets = 45 sweets

Printer Friendly Version

### RGS Primary 2009 PSLE Math Prelim Paper 2 Q15

The equilateral triangles below are formed using 2 cm sticks. a) How many sticks are needed to form Pattern 5?
b) In which pattern will each side of the triangle measure 30 cm?
c) Calculate the number of shaded triangles in Pattern 100.

Solution b)
Pattern 1 --> 1 x 2cm = 2 cm
Pattern 2 --> 2 x 2cm = 4 cm
Pattern 3 --> 3 x 2cm = 6 cm
:
:
:
Pattern 15 --> 15 x 2cm = 30 cm From above
1 + 2 + 3 ..... + 100

1 + 99 = 100
2 + 98 = 100
3 + 97 = 100
:
:
49 + 51 = 100
(only 50 is not paired)

49 groups of 100 --> 4900
Add the unpaired 50 --> 4900 + 50 = 4950

### RGS Primary 2009 PSLE Math Prelim Paper 2 Q14

John shifted the decimal point of a number twice to the left to obtain a new number. The difference between the new number and the original number was 136.62.
a) How many times the new number is the original number?
b) What is the sum of the 2 numbers?

Solution

a)

b)
Original number --> 1 unit
New number --> 100 units

100 units - 1 unit --> 136.62
99 units --> 136.62
1 unit --> 136.62 divided by 99 = 1.38

Sum of the two numbers
--> 100 units + 1 unit = 101 units

101 units --> 101 x 1.38 = 139.38

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q13

The figure below is made up of a big circle, square and a small circle. The area of the square is 400 square cm. Find the area of the shaded region. (Correct your answer to 2 decimal places) Solution Area of square --> 400 square cm
Length of square --> 20 cm (square root of 400 sq cm)

Area of 1/4 square
--> 400 square cm divided by 4
= 100 square cm

Area of one right-angle triangle -->(1/2)(b)(h)
100 square cm = (1/2)(r)(r)
(r)(r) = (100 square cm) x 2 = 200 square cm

Area of circle = (3.14)(r)(r)
=(3.14)(200 square cm)
= 628 square cm

Shaded area -->(628 - 400) square cm
= 228 square cm
= 228.00 square cm (correct to 2 decimal places)

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q12

Tap A, Tap B, Tap C and an empty rectangle tank are shown below. Lily turned on Tap A with water flowing at a rate of 5 litres per minute. After 2 minutes, she placed a rock of volume 1250 cubic cm in the tank and turned on Tap B and Tap C as well. Tap C drains the tank at the rate of 2 litres per minute. After 5 more minutes, Lily turned off all the taps and noted that the height of the water level was 30 cm. Find the rate of the flow of water from Tap B.

Solution

Tap A (water filled into tank)
--> 5 litres x 7 (min) = 35 litres

Tap B (water drained from tank)
-->2 litres x 5 (min) = 10 litres

Volume of water in tank in the end
--> (50 x 30 x 30) cubic cm - 1250* cubic cm
= (45 000 - 1250) cubic cm
= 43 750 cubic cm
* volume of stone

43 750 cubic cm is the total volume of water filled through Tap A and Tap B less the volume of water drained from Tap C.
Tap A + Tap B - Tap C --> 43 750 ml
Tap B --> 43 750 ml - Tap A + Tap C
= 43 750 ml - 35 000 ml + 10 000 ml
= 18 750 ml (after 5 min)

1 min --> 18 750 ml divided by 5
= 3750 ml/min
= 3.75 litres/min

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q11

Mrs Kee baked some cookies and packed all the cookies in 12 small boxes and 5 big boxes. There were equal number of cookies in each small box and equal number of cookies in each big box. Each big box contained 14 more cookies than each small box. 18/29 of the cookies baked were packed in small boxes. How many cookies were there in each small box?

Solution

12 small boxes --> 18/29 of cookies
1 small box --> 18/29 divided by 12

5 big boxes --> 11/29 of cookies (29/29 - 18/29)
1 big box --> 11/29 divided by 5

There were 14 more cookies in each big box than in each small box
11/145 - 3/58 --> 14 cookies
1/290 --> 14 divided by 7 = 2

(Small box) 3/58 --> 15/290
15/290 --> 15 x 2 = 30

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q10

Some secondary one boys were asked to name their favourite sport. Their choices were represented on the pie chart below. There was an equal number of boys who liked athletics and swimming. 80 boys chose football as their favourite sport.
a) What fraction of the boys liked swimming?
b) Find the total number of secondary one pupils who took part in the survey.

Solution

100% --> 360 degrees
1% --> 360 degrees divided by 100 = 3.6 degrees
(Rugby) 5% --> 5 x 3.6 degrees = 18 degrees

a)
Athletics --> (90 - 18) degrees = 72 degrees
Athletics --> same number as Swimming
Swimming --> 72 degrees

Fraction of boys who liked swimming
--> 72/360 = 1/5

b)
Football --> (100 - 25* - 20** - 15)% = 40%
40% --> 80
10 % --> 80 divided by 4 = 20
100% --> 10 x 20 = 200

*25% --> Athletics + Rugby (90 degress is 25% of 360 degrees)
**20% --> Swimming is 1/5 (20%)

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q9

Tricia has some pink, red and yellow ribbons. 1/3 of them are pink ribbons. Four fewer than 1/3 of the remainder are red ribbons and the remaining 24 are yellow ribbons. How many pink ribbons does Tricia have?

Solution 4 units --> 24 - 4 = 20
1 unit --> 20 divided by 4 = 5
3 units --> 3 x 5 = 15

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q8

In the figure below, O is the centre of the circle where OCD is an equilateral triangle. Given that Angle OAB = 20 degrees and Angle AOD = 127 degrees, find Angle BOC. Solution

Angle DOC --> 60 degrees (Triangle OCD is equilateral)
Angle AOB --> (180 - 20 - 20) degrees
= 140 degrees (Triangle OAB is isosceles)

Angle BOC --> (360 - 140 - 127 - 60) degrees
= 33 degrees

Printer Friendly Version

## Thursday, August 19, 2010

### RGS Primary 2009 PSLE Math Prelim Paper 2 Q7

Last year the total age of Mr Tan and his wife was p years old. His wife is 1 year younger than him. What is his wife's age 2 years from now? Express your answer in terms of p.

Solution

Last year 2 units + 1 --> p
2 units --> p - 1
1 unit --> (p -1)/2

Two years from now
--> (p - 1)/2 + 3
= (p - 1)/2 + 6/2
= (p -5)/2

Answer: [(p - 5)/2] years old

### RGS Primary 2009 PSLE Math Prelim Paper 2 Q6

Alice, Beth and Claire had 600 stamps altogether. After Beth had given 30 stamps to Alice, Beth had twice as many stamps as Claire and Alice had to 20 stamps more than Claire. How many stamps did Claire have?

Solution 4 units --> 600 - 20 = 580
1 unit --> 580 divided by 4 = 145

### RGS Primary 2009 PSLE Math Prelim Paper 2 Q4

Find the number that is exactly between 4/5 and 1 and 1/3.

Solution

4/5 + 1 and 1/3
= 12/15 + 1 and 5/15
= 2 and 2/15

2 and 2/15 divided
= 1 and 1/15

## Tuesday, August 17, 2010

### RGS Primary 2009 PSLE Math Prelim Paper 2 Q3

When a number is divided by 8, the quotient is 121 with no remainder. What is the remainder when the same number is divided by 9?

Solution

? divided by 8 --> 121
? --> 121 x 8
? --> 968

968 divided by 9
= 107 R5

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q2

A dress cost \$139.95. During a sale, a 20% discount was given. Calculate the sale price of the dress.

Solution

Discount --> 20%
Cost after discount --> 80%

Sale price after discount
--> (80/100) x \$139.95 = \$111.96

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### RGS Primary 2009 PSLE Math Prelim Paper 2 Q1

Find the value of each of the following expressions when y = 6.
a) 3y - 4
b) y + 2y/3

Solution

a)

3(6) - 4
= 18 - 4
= 14

b)
(6) + 2(6)/3
= 6 + 4
= 10

Printer Friendly Version

## 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.

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

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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

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

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

### ACS Primary 2009 PSLE Math Prelim Paper 2 Q16

a) the area of A
b) the perimeter of B  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

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

### 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

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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

## 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)

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

Printer Friendly Version

### 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

Printer Friendly Version

### 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

Printer Friendly Version

### 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 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)

### 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

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

Printer Friendly Version

### 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

### 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

### 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)

### 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

### 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

### 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

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## Sunday, July 11, 2010

### Request for help from a reader - 10 Jul 2010

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.

## Tuesday, June 22, 2010

### Request for help - CHIJ St Nicholas 2009 Prelim Paper 2 Q17

Hello everyone, I received an email from a reader seeking some help. Below is the email reproduced unedited. Please feel free to help.

======================

Hi,
I am a regular reader of your blog "road-to-psle". I have 1 question from CHIJ St Nicholas 2009 Prelim Paper 2 Q17 that I cannot solve. I have asked many people including tutors and teachers but they also cannot solve. Can you please help me? The question:

At Carpark P. the number of lorries to that of vans was in the ralio 3: 7.
At Carpark Q, the number of lorries to that of vans was in the ratio 8: 9.
When 40% more lorries from an industrial park entered Carpark P and
20% of the vans at Carpark Q moved to Carpark P. there were 76 fewer
Iorries at Carpark P than at Carpark Q. How many vehicles were there
altogether at the two carparks finally? Ans:564

Thanks
John

=====

http://www.atfreeforum.com/excel/viewtopic.php?p=5&mforum=excel#5

## Saturday, May 01, 2010

### Time for a Breather

While it is good to be consistently studying, there must also be short break periods to relax. Here are two video clips for you to enjoy.

Johann Pachelbel's Canon in D (rock violin - duet)

(rock violin - solo)

## Wednesday, April 14, 2010

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q18

Sparkles Jewellery Shop sold diamonds, rubies and emeralds. 3/5 of the gemstones were diamonds. There were 168 fewer rubies than diamonds. The ratio of the number of rubies to the number of emeralds was 7:3. After some rubies were sold, 30% of the remaining gemstones in the shop were rubies and emeralds.
a) How many rubies were sold?
b) Find the percentage decrease in the number of gemstones. Leave your answer as a fraction.

Solution Rubies --> 7/10 of 2/5 = 14/50
Diamonds - Rubies --> 168 (more diamonds than rubies)
3/5 - 14/50 = 168
16/50 --> 168
1/50 --> 168 divided by 16 = 10.5

Rubies + Emeralds (2/5 of whole)
20/50 --> 20 x 10.5 = 210
2/5 of whole --> 210
1/5 of whole --> 210 divided by 2 = 105
(Diamonds) 3/5 of whole --> 3 x 105 = 315

After selling, Rubies and Emeralds make up 30%, while Diamonds make up 70%.
Diamonds 70% --> 315
10% --> 315 divided by 7 = 45
Rubies and Emeralds 30% --> 3 x 45 = 135

(a)
(Rubies and Emeralds before selling) - (Rubies and Emeralds after selling)
--> 210 - 135 = 75

(b)

Percentage decrease in number of gemstones

1/50 of gemstones --> 10.5
50/50 of gemstones -> 50 x 10.5 = 525

75 rubies sold
--> (75/525) x 100% = 14 and 2/7 %

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q17

The number of boys to the number of girls in Happy Kindergarten was 3:5. The number of boys to the number of girls in Merry Kindergarten was 4:5. Both kindergartens had the same number of boys. When 10 girls from Happy Kindergarten joined Merry Kinderten, the ratio of the number of boys to the number of girls in Happy Kindergarten became 2:3. Find the total enrolment of the 2 kindergartens.

Solution

Happy
B : G
3 : 5
(multiply both by 4)*
12 : 20
(Total units --> 12 + 20 = 32)

Merry
B : G
4 : 5
(multiply both by 3)*
12 : 15
(Total units --> 12 + 15 = 27)

* Happy (x4) and Merry (x3) to make the number of units for boys in both kindergartens the same, because the number of boys are the same for both kindergartens.

After 10 girls left Happy to join Merry,
Ratio in Happy
B : G
2 : 3
(multiply by 6)**
12 : 18

** (x6) to make units for boys 12 again.

Number of units for girls in Happy before transfer --> 20 units
Number of units for girls in Happy after transfer --> 18

20 units - 18 units --> 10
2 units --> 10
1 unit --> 10 divided by 2 = 5

Total for enrolment for both kindergartens
Happy --> 32 units
Merry --> 27 units
32 units + 27 units --> 59 units
59 units --> 59 x 5 = 295

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q16

Ali and Minhua were playing a card game using 96 cards. In the first game, Minghua lost 1/5 of his cards to Ali. In the second game, Ali lost 1/3 of his cards to Minghua. After the second game, both boys had the same number of cards. How many cards did Ali have at first?

Solution Before 1st game
Minghua --> 5 units
Ali --> 11 units
Total 16 units --> 96
1 unit --> 96 divided by 16 = 6
11 units --> 66

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q15

There were some prizes ot be won at a Shop and Win Contest. 35% were cash prizes and there rest were household items. Some cash prizes were given out and the percentage of cash prizes decreased to 22%. If there were 54 more household items than cash prizes at first, find the number of cash prizes being removed.

Solution

Before
Cash --> 35%
Household Items consists of
--> 35% + 30%(54 items) = 65%

30% --> 54
1% --> 54 divided by 30 = 1.8
(Cash before) 35% --> 35 x 1.8 = 63
(Household Items) 65% --> 65 x 1.8 = 117

After
Cash became --> 22%
Household Items --> (remaining) 78%

78% --> 117
1% --> 117 divided by 78 = 1.5
(Cash after) 22% --> 22 x 1.5 = 33

Cash before - Cash after --> Cash prizes removed
63 - 33 = 30

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q14

There were ducks, goats and hens at a farm. 15% of the animals were ducks. There were 180 fewer ducks than hens. The remaining 121 animals were goats. What percentage of the animals at the farm were goats? Levae your answer correct to 2 decimal places.

Solution * 180 more hens than ducks, therefore hens --> 15% + 180.
** Ducks (15%) and hens (15% + 180), therefore goats give the remaining 70% less 180.

70% - 180 --> 121
70% --> 121 + 180 = 301
10% --> 301 divided by 7 = 43
100% 10 x 43 = 430

Percentage of animals were goats
--> (121/430) x 100% = 28.139% ~ 28.14%

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### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q13

Abbie, Ellen and Faheem sold umbrellas to raise funds for charity. Each umberlla was priced at \$22.50. Abbie sold 1/6 of the umbrellas while Ellen and Faheem sold the remaining umbrellas in the ratio of 2:7 respectively. If Faheem sold 364 umbrellas more than Abbie, what was the total amount of money collected by them?

Solution

Abbie --> 1/6
Remaining --> 5/6

Since ratio for Ellen : Faheem is 2 : 7,
Faheem sold 7/9 of remainder

Faheem (7/9 of the remaning 5/6)
--> 7/9 x 5/6 = 35/54

(Faheem - Abbie) --> 35/54 - 1/6 = 13/27
13/27 --> 364 (Faheem sold 364 more than Abbie)
1/27 --> 364 divided by 13 = 28
27/27 --> 27 x 28 = 756

756 x \$22.50 = \$17 010

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q12

The figure below is made up of 2 squares and a rectangle. The ratio of the area of A to the area of B to the area of C is 9:4:3. The ratio of the unshaded part of B to the unshaded part of C is 4:1. If the shaded part is 56 square cm, find the area of A that is not covered by B. Solution (C) 3 parts --> 56 sq cm + 1 unit
(multiply by 4)
(C) 12 parts --> 224 sq cm + 4 units
(B) 4 parts --> 56 sq cm + 4 units

12 parts - 4 parts --> 224 sq cm - 56 sq cm
8 parts --> 168 sq cm
1 part --> 168 sq cm divided by 8 = 21 sq cm

(A - B) 5 parts --> 5 x 21 sq cm = 105 sq cm

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q11

Jai, Maira and Lynn have 760 stickers altogether. Jai has 8/9 as many stickers as Maira and 6/11 as many stickers as Lynn. If Jai and Lynn want to have an equal number of stickers, how many stickers must Lynn give to Jai?

Solution (48 + 54 + 88) units --> 760 stickers
190 units --> 760
1 unit --> 760 divided by 190 = 4

Lynn must give Jai 20 units for Lynn and Jai to have equal number of stickers.
20 units --> 20 x 4 = 80

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q10

In September, Rafael received a monthly salary of \$2698 and she saved 20% of it. In December, her salary was reduced by 5% and she continued to save 20% of her salary. How much less was her savings in December compared to that in September? Leave your answer correct to the nearest dollar.

Solution

Sep --> 20% x \$2698 = \$539.60

5% x 539.6% = \$26.98 ~ \$27

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q9

At a shop, the ratio of the number of stalks of roses to the number of stalks of carnation was 4:5. Mrs Tan sold 176 stalks of carnations and was left with thrice as many stalks of roses as carnations. How many stalks of carnations did she have at first?

Solution * 'Before' x3 and 'After' x4 to make the Roses have an equal number of units of 12 each because no roses were sold.

Carnations before --> 15 units
Carnations after --> 4 units
15 units - 4 units --> 176 (carnations sold)
11 units --> 176
1 unit --> 176 divided by 11 = 16
15 units --> 15 x 16 = 240

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q8

Kumar spent \$1729 on a set of encyclopedia. H spent 1/4 of the remainder on a camera and still had 2/5 of his money left. Find the total amount of money he had at first.

Solution 3/5 of money --> \$1729 + 1 unit
(multiply above by 3)
9/5 of money --> \$5187 + 3 units
2/5 of money --> 3 units

9/5 - 2/5 --> \$5187 + 3 units - 3 units
7/5 --> \$5187
1/5 --> \$5187 divided by 7 = \$741
5/5 --> 5 x \$751 = \$3705

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### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q7

The figure below is made up of a square and a rectangle. The perimeter of the figure is 178 cm and the area of the square, PQRS, is 784 square cm. What is the length of ST? Solution

Square root of 784 square cm = 28 cm (length of 1 side of square)
2 lengths of rectangle --> (178 - 28 - 28 - 28 - 28) cm = 66 cm
1 length --> 66 cm divided by 2 = 33 cm
ST --> 33 cm + 28 cm = 61 cm

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q6

ABE is an isosceles triangle and BCDE is a rhombus. AED is a straight linie. Given that Angle ECG = 16 degrees, find Angle GCD. Solution

Ange BEA --> (180 - 33) divided by 2 = 73.5
Angle BED --> 180 - 73.5 = 106.5
Angle CED --> 106.5 divided by 2 = 53.25
Angle ECD = Angle CED = 53.5
Angle GCD = 53.25 - 16 = 37.25

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q5

The number of visitors to the zoo has been increasing by 8% each year since the year 2003. If the number of visitors in 2003 was 12 500, how many visitors went to the zoo in 2005?

Solution

Yr 2003 --> 12 500
Yr 2004 --> 108% x 12 500 = 13 500
Yr 2005 --> 108% x 13 500 = 14 580

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q4

A department store was offering a promotion for customers who bought T-shirts. Any customer who bought 3-T shirts would get a 50% discount on the 3rd T-shirt. Pamela bought 3 T-shirts and paid an average cost of \$11.90 for each T-shirt. Find the orginal price of one T-shirt.

Solution

Original Price of 1 T-shirt --> 2 units
Price of 3rd T-shirt --> 1 unit

2 T-shirts at original price + 1 discounted price T-shirt --> \$11.90 (avg)
(2 x 2 units) + 1 unit --> \$11.90 x 3 = \$37.50 (total)
4 units + 1 unit --> \$37.50
5 units --> \$37.50
1 unit --> \$37.50 divided by 5 = \$7.14
2 units --> 2 x \$7.14 = \$14.28

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### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q3

Mr Ong had 5 kg or rice. After he sold 5/12 of it, he repacked the remaining rice into smaller packets of 2/3 kg each. How much rice had he left?

Solution

7/12 left --> 7/12 x 5 kg = 35/12 kg (35 twelfths kg)

Repacked this 35 twelfths kg into packets of 8 twelfths kg (2/3 kg --> 8/12 kg)
35 twelfths divided by 8 packets --> 4 wholes and 3 twelfths
3 twelfths (left unpacked) --> 3/12 --> 1/4 or 0.25

Answer: 1/4 kg or 0.25 kg left

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### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q2

CHIJ St Nicholas Girls' Sch 2009 SA1 Paper 2 Q2

Evelyn left her house for a concert at 17 55. She took 1 h 5 min to reach the theatre but was 10 minutes late. What time did the concert begin?

Solution

1 h 5 min after 1755 h --> 1900h
10 min before 1900h --> 1850 h
1850h --> 6.50 pm

### CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q1

The table below shows the difference between the digits in some 2-digit numbers. List all the 2-digit numbers, from 40 to 100, in which the digits have a difference of 3. Arrange them in ascending order.

Answer: 41, 47, 52, 58, 63, 69, 74, 86, 96

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## Sunday, February 21, 2010

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q18

Ace Drama Company sold some tickets for a children's performance. It sold the same number of \$8 and \$5 tickets in Week 1 and collected a total of \$1664. In Week 2, it sold 96 \$8 and \$5 tickets. If the company collected \$632 more from the sale of \$8 tickets than the \$5 tickets in the two weeks, how many \$8 tickets were sold altogether?

Solution

Week 1
\$8 tickets --> 8 units
\$5 tickets --> 5 units
Total 13 units --> \$1664
1 unit --> \$1664 divided by 13 = \$128

Value of tickets sold in Week 1
\$8 tickets --> 8 x \$128 = \$1024
\$5 tickets --> 5 x \$128 = \$640

Difference between \$8 tickets and \$5 tickets in Week 1
\$1024 - \$640 = \$384

Difference between \$8 tickets and \$5 tickets in Week 2
\$632 - \$384 = \$248

If 96 tickets in Week 2 were equally sold between \$5 tickets and \$8 tickets, there will be 48 tickets each.
\$5 tickets --> 48 x \$5 = \$240
\$8 tickets --> 48 x \$8 = \$384

Difference between \$8 tickets and \$5 tickets in Week 2 if equal number of \$5 tickets and \$8 sold, would be
\$384 - \$240 = \$144

But the difference is \$248 and not \$144. Instead, we have,
\$248 - \$144 = \$104 (more)

For every \$8 sold instead of \$5, there would be an increase of (\$8 + \$5) = \$13
\$104 (more) divided by \$13 = 8 (tickets more)

48 tickets + 8 tickets (more) = 56 \$8 tickets sold on 2nd day.
1st Week --> \$1024 divided by \$8 = 128 (tickets)

Total number of \$8 tickets
--> 128 + 56 = 184

## Friday, February 19, 2010

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q17

John had some red and blue marbles in a box. The sum of 1/4 of the red marbles and 2/5 of the blue marbles in the box is 64. The sum of 3/4 of the red marbles and 4/5 of the blue marble in the box is 144.
a) How many red marbles are there in the box?
b) How many blue marbles are there in the box?

Solution

Red (all marbles) --> R R R R
Blue (all marbles) --> B B B B B

1/4 or red marbles and 2/5 or blue marbles -- > 64
R + B B --> 64

3/4 of red marble and 4/5 of blue marbles --> 144
R R R + B B B B --> 144

R + B B --> 64 (multiply all by 2)
R R + B B B B --> 128

We now have
R R R + B B B B --> 144
R R + B B B B --> 128

R --> 144 - 128 = 16

a)
Number of Red Marbles
R R R R --> 4 x 16 = 64

b)
R + B B --> 64
16 + B B --> 64
B B --> 64 - 16 = 48
B --> 48 divided by 2 = 24

Number of Blue Marbles
B B B B B --> 5 x 24 = 120

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q16

The ratio of the number of boys to the number of girls in School A is 4:1. the ratio of the number of boys to the number of girls in School B is 2:3. School A had twice as many pupils as School B.

a) What is the ratio of boys in School A to the number of girls in School B?

b) After 30 girls left School A to join School B, the ratio of the number of boys to the number of girls in School B is now 5:8. How many girls are there in School B now?

Solution * School A is multiplied by 2 to give a total of 10 units. This is because School B has 5 units. School A has twice the number of pupils as School B.

(a)
Boys from School A --> 8 units
Girls from School B --> 3 units

Ratio of number of boys in School A to number of girls in School B
--> 8 : 3

b)
School B * Before is multiplied by 5 and After is multiplied by 2 to make the boys have a common unit of 10 for both Before and After, because there was no transfer of boys.

Girls increased by 1 unit after the transfer
1 unit--> 30

Number of girls in School B after transfer
16 units --> 16 x 30 = 480

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q15

Three men, A, B and C, worked together to paint a wall. If the painting was done by one man, the time taken to complete the wall for A, B and C would have been 6 hours, 8 hours and 12 hours respectively. A and B had painted for 3 hours after which A rested. B and C then continued with the painting. What would be the total number of hours taken to complete the wall? (Give your answer as a mixed number.)

Solution

First 3 hours
A painted for 3 h --> 3/6 = 1/2 (of the wall)*
B painted for 3 h --> 3/8 = 3/8 (of the wall)**
1/2 + 3/8 = 7/8 (of the wall was painted in first 3 hours)

* A takes 6h to paint the whole wall, therefore, in 3h, 3/6
** B takes 8h to paint the whole wall, therefore, in 3h, 3/8

A rests, B and C continue to paint remaining 1/8 of wall
Ratio of hours taken to paint whole wall
B : C
8 : 12
2 : 3

B takes less time than C, therefore B would have painted more of the wall
Ratio of amount of wall painted
B : C
3 : 2 --> Total units is 5

Time taken for B to paint whole wall
5/5 of wall --> 8h
Therefore, 3/5 of wall --> (3/5) x 8h = (24/5)h

Since only 1/8 of wall remains
1/8 of wall left --> (1/8) x (24/5)h = 3/5 h

Total time taken
--> 3h + 3/5 h = 3 and 3/5 hours

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q14

Mr Lim paid \$134.40 for some jackfruits and pomeloes. The cost of the pomelo was 0.8 that of a jackfruit. A pomelo cost \$5.60. If all the pomeloes cost \$22.40 more than the jackfruits, how many fruits did he buy?

Solution 2 units --> \$134.40 - \$22.40 = \$112
1 unit --> \$112 divided by 2 = \$56

Cost of Pomeloes
--> \$56 + \$22.40 = \$78.40

Number of Pomeloes bought
--> \$78.40 divided \$5.60 (per pomelo)
= 14 pomeloes

Cost of of 1 pomelo is 0.8 of 1 jackfruit
Pomelo --> 8 units
Jackfruit--> 10 units

(Pomelo) 8 units --> \$5.60
1 unit --> \$5.60 divided by 8 = \$0.70
(Jackfruit) 10 units --> 10 x \$0.70 = \$7

Number of Jackfruits bought
--> Total cost - cost of pomeloes
\$134.40 - \$78.40 = \$56
\$56 divided by \$7 (per jackfruit) = 8 (jackfruits)

Total number of fruits
14 (pomeloes) + 8 (jackfruits) = 22 fruits

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q13

The ratio of Jane's allowance to Olivia's allowance was 4:3. After Jane and Olivia were given \$15 and \$8 respectively, the ratio of Jane's allowance to Olivia's allowance became 3:2. How much allowance did Jane have at first?

Solution

Jane
4 units (at first) + \$15 (given) --> 3 parts

Olivia
3 units (at first) + \$8 --> 2 parts
2 parts --> 3 units + \$8
1 part --> 3/2 units + \$4
3 parts --> 3 x 3/2 units + 3 x \$4
= 9/2 units + \$12

(Jane) 3 parts --> 4 units + \$15
(Olivia) 3 parts --> 9/2 units + \$12

(Olivia) 9/2 units + \$12 --> (Jane) 4 units + \$15
9/2 units + \$12 --> 8/2 units + \$15
9/2 units - 8/2 units --> \$15 - \$12
1/2 unit --> \$3
1 unit --> \$6

Jane at first
4 units --> 4 x \$6 = \$24

## Wednesday, February 17, 2010

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q12

Rahim's age is 2/9 of his grandfather's. His grandfather will be 100 years old in 19 years' time. In how many years' time will Rahim's age be 1/4 of his grandfather?

Solution

Now
Rahim --> 2 units
Grandfather --> 9 units

19 years' time (Grandfather)
9 units + 19 --> 100
9 units --> 100 - 19 = 81
1 unit --> 81 divided by 9 = 9

Now
Rahim --> 2 x 9 years = 18 years
Grandfather --> 9 x 9 years = 81 years

Difference between Grandfather and Rahim
--> 81 years - 18 years = 63 years

Future (when Rahim 1/4 of Grandfather's age)
Rahim --> 1 unit
Grandfahter --> 4 units

But Rahim is 63 years younger
4 units - 1 unit --> 63 years
3 units --> 63 years
1 unit --> 63 years divided by 3 = 21 years

Rahim now --> 18 years
Rahim in future --> 21 years

21 years - 18 years = 3 years

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q11

There were some marbles in a box. Sofie took out 2/5 of the marbles and put in 6 more. Then John took out 1/6 of the remaining marbles and put in 5 more. there were 25 marbles left. How many marbles were in the box at first?

Solution 6 units - 1 unit + 6 - 1 + 5 --> 25 (left in box)
5 units + 10 --> 25
5 units --> 25 - 10 = 15
1 unit --> 15 divided by 5 = 3

Marbles at first
--> 10 units x 3
= 30

(At first, there were 10 units and not 5 units because the 5 units have been cut into halves, giving a total of 10 smaller units)

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q10

In the figure shown below, SUVX is a square. STU is an equilateral triangle and TXW is a straight line.
a) Find the value of Angle STX.
b) Find the value of Angle WVX. Solution a)
Line TZ passes through V, while line TY is passes through the centre of Line SU.
Angle STX is 1/4 of Angle STU.
Angle STU is 60 degrees (Triangle STU is equilateral)

Angle STX --> (1/4) x 60 degrees = 15 degrees

b)
Angle SXT = 15 degrees (Triangle STX is isosceles)
Angle TXV --> (90 - 15) degrees = 75 degrees
Angle WXV --> (180 - 75) degrees = 105 degrees

Angle WVX
--> (180 - 105) degrees divided by 2 = 37.5 degrees
(Triangle WVX is isosceles)

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q9

Nigel had a total of 227 durians and pears. He sold half of the durians and bought another 40 pears. As a result, he had an equal number of durians and pears.
a) How many durians were there at first?
b) How many pears were there at first?

Solution

(a) 3 units --> 267
1 unit --> 267 divided by 3 = 89

(Durians) 2 units --> 2 x 89 = 178

(b)
(Pears) 1 unit - 40
--> 89 - 40 = 49

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q8

Wilson and Yi Lin had \$71 altogether. Yi Lin and Patrick had \$105 altogether. Wilsom hand 3/5 of the money that Patrick had. How much money did Yi Lin have?

Solution Yi Lin + Patrick
? + 5 units --> \$105

Yi Lin + Wilson
? + 3 units --> \$71

5 units - 3 units --> \$105 - \$71
2 units --> \$34
1 unit --> \$34 divided by 2 = \$17

Yi Lin + Wilson
? + 3 units --> \$71
? + (3 x \$17) --> \$71
? + \$51 --> \$71
? --> \$71 - \$51 = \$20

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q6

(a) Arif is 2x years old. His father is 4 times as old as he. His mother is 3 years younger than his father. What is their total age in terms of x?

(b) If x is = 4, find their total age.

Solution

(a)
Arif --> 2x
Father --> 4 x 2x = 8x
Mother --> 8x - 3

Total --> 2x + 8x + 8x - 3 = 18x - 3

(b)
18(4) - 3
= 72 - 3
= 69

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q5

A container with Bottle A placed in it has a mass of 4.27 kg. An identical container with Bottle B placed in it has a mass of 6.58 kg. The mass of Bottle A is 1/3 that of Bottle B. What is the mass of the Bottle A? Give your answer correct to 2 decimal places.

Solution

Bottle A --> 1 unit
Bottle B --> 3 unit

Container + Bottle A (1 unit) --> 4.27 kg
Container + Bottle B (3 units) --> 6.58 kg

3 units - 1 unit --> 6.58 kg - 4.27 kg
2 units --> 2.31 kg
1 unit --> 2.31 kg divided by 2 units = 1.155 kg
1.155 kg --> 1.16 kg correct to 2 decimal places

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q4

Box A contains only one-dollar coins. Box B contains only fifty-cent coins and Box C contains only twenty-cent coins. Box A has 5 times as many coins as Box C. Box B contains 12 coins fewer than Box A. Box C contains half the number of coins in Box B. How much money is there in Box B?

Solution 2 units --> 5 units - 12
3 units --> 12
1 unit --> 12 divided by 3 = 4

(Box B)
5 units - 12
-->(5 x 4) - 12
= 20 - 12
= 8 (coins)

8 coins x \$0.50 = \$4

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q3

The ratio of the number of Chloe's stickers to the number of Faith's stickers is 3:5. The ratio of the number of Faith's stickers to Melissa's stickers is 6:7. If Melissa has 204 stickers more than Chloe, how many stickers do they have altogether?

Solution Melissa - Chloe --> 204
35 units - 18 units --> 204
17 units --> 204
1 unit --> 204 divided by 17 = 12

All stickers
Chloe + Faith + Melissa
18 + 30 + 35 = 83
83 units --> 83 x 12 = 996

### Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q1

LMNO is a square. PQN and PLQ are isosceles triangles. Angle QNM is 23 degrees. Find Angle NPQ. Solution

Angle PNO --> 23 deg (mirror image of Angle MNQ)
Angle OPN --> 90 deg - 23 deg = 67 deg
Angle LPQ --> 45 deg (Triangle LQP is isosceles and Angle LPQ is a rt angle)
Angle NPQ --> (180 - 45 - 67) deg = 68 deg

## Wednesday, February 10, 2010

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q18

Ai Tong School organized a 2-day camp. On the first day, the number of boys was 600 more than the girls. On the second day, the number of boys decreased by 10% but the number of girls increased by 10%. If there were 2540 children on the second day, how many children were there on the first day?

Solution 2nd Day
Boys --> 0.9 unit + 540
Girls --> 1.1 unit
Total --> 2540

0.9 unit + 540 + 1.1 unit --> 2540
2 units + 540 --> 2540
2 units --> 2540 - 540 = 2000
1 unit --> 2000 divided by 2 = 1000

1st Day
2 units + 600
(2 x 1000) + 600
= 2000 + 600
= 2600

### Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q17

75% of the children in the stadium were girls. After 52 girls and 4 boys left, the remaining children formed groups of 8. In each group, there were 3 boys. How many children were there in the stadium at first?

Solution Note - "Before" for girls is 3 parts, while "Before" for boys is 1 part. This is to represent 75% girls and 25% boys, before 52 girls and 4 boys left.

From the table above
4 units --> 40
1 unit --> 40 divided by 4 = 10

Total boys (after)
3 units --> 3 x 10 = 30

Total girls (after)
5 units --> 5 x 10 = 50

Total children (after)
30 + 50 = 80

Total children at first
80 + 52 + 4 = 136