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.
Friday, November 26, 2010
Rosyth boy tops PSLE
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
Solution
(K) 20 units --> 10 parts + 1625
1 unit --> 1785 divided by 14 = 127.5
(a)
(b)
Lily in the end
--> 3 units + 80
(3 x 127.5) + 80
= 382.5 + 80
= 462.5
Answer: $462.50
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RGS Primary 2009 PSLE Math Prelim Paper 2 Q17
(Train B) distance
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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
Answer: 45 sweets
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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
Answer: Pattern 15
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
Answer: 4950 shaded triangles
RGS Primary 2009 PSLE Math Prelim Paper 2 Q14
99 units --> 136.62
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RGS Primary 2009 PSLE Math Prelim Paper 2 Q13
Solution
Length of square --> 20 cm (square root of 400 sq cm)
= 228.00 square cm (correct to 2 decimal places)
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RGS Primary 2009 PSLE Math Prelim Paper 2 Q12
Solution
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
= 43 750 ml - 35 000 ml + 10 000 ml
Answer: 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
= 3/58 of cookies
5 big boxes --> 11/29 of cookies (29/29 - 18/29)
1 big box --> 11/29 divided by 5
= 11/145 of cookies
There were 14 more cookies in each big box than in each small box
11/145 - 3/58 --> 14 cookies
7/290 --> 14 cookies
1/290 --> 14 divided by 7 = 2
(Small box) 3/58 --> 15/290
15/290 --> 15 x 2 = 30
Answer: 30 cookies
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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
Answer: 1/5 of the boys
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%)
Answer: 200 boys
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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
Answer: 15 pink ribbons
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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
Answer: 33 degrees
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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
Answer: 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
Answer: 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
Answer: 5
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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
Answer: $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
Answer: 14
b)
(6) + 2(6)/3
= 6 + 4
= 10
Answer: 10
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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.
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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?
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.
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
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?
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.
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
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.
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
=====
Please post solutions here.
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
Answer: 75 rubies
(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 %
Answer: 14 and two-sevenths %
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
Answer: 295 children
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
Answer: 66 cards
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
Answer: 30 cash prizes
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%
Answer: 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
Answer: $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
Answer: 105 square 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
Answer: 80 stickers
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
Answer: $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
Answer: 240 carnations
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
Answer: $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
Answer: 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.
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
Answer: 37.25 degrees
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
Answer: 14 580 visitors
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
Answer: $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?
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
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
Answer: 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
Answer: 184 $8-tickets
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
Answer: 64 red marbles
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
Answer: 120 blue marbles
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?
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
Answer: 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
Answer: 480 girls
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
Answer: 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
Answer: 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
Answer: $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
Answer: 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)
Answer: 30 marbles
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
Answer: 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)
Answer: 37.5 degrees
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
Answer: 178 durians
(b)
(Pears) 1 unit - 40
--> 89 - 40 = 49
Answer: 49 pears
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
Answer: $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
Answer: (18x - 3) years
(b)
18(4) - 3
= 72 - 3
= 69
Answer: 69 years
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
Answer: 1.16 kg
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
Answer: $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?
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
Wednesday, February 10, 2010
Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q18
Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q17
From the table above