Hi all,
our tuition classes for Year 2013 at Block 487 Admiralty Link have started. Please call to enquire. By the way, I have not posted the year end results of my students yet because ... I just got back from my 3 week reservice in the SAF. Hope to hear from you soon :)
Rgds,
Kenny Wong
Tuesday, December 4, 2012
Monday, August 20, 2012
P6 Maths - Fractions
Wendy cleaned 1/6 of a room in 2h. Lenna cleaned 1/3 of the same room in 1/2h. If they worked together, how long do they need to clean the room?
For this question, we need to find the 'speed' that they cleaned the room in the same unit of time. Let's use 1 hour here.
Wendy -> 2h -> 1/6 room
-------> 1h -> 1/6 divided by 2 = 1/6 x 1/2 = 1/12 room
Lenna -> 1/2h -> 1/3 room
-------> 1 h -> 1/3 x 2 = 2/3 room
Together -> 1h -> 1/12 + 2/3 = 3/4 room
Room is taken as 1 whole.
Time taken to clean room = 1 divided by 3/4 = 4/3h = 1&1/3h
For this question, we need to find the 'speed' that they cleaned the room in the same unit of time. Let's use 1 hour here.
Wendy -> 2h -> 1/6 room
-------> 1h -> 1/6 divided by 2 = 1/6 x 1/2 = 1/12 room
Lenna -> 1/2h -> 1/3 room
-------> 1 h -> 1/3 x 2 = 2/3 room
Together -> 1h -> 1/12 + 2/3 = 3/4 room
Room is taken as 1 whole.
Time taken to clean room = 1 divided by 3/4 = 4/3h = 1&1/3h
Monday, July 16, 2012
P6 Maths - Volume
Water drain out from Tap A into 2 containers at 800 cm3 per minute such that the height of water in the 2 containers is the same. The base area of the 2 containers are 500 cm3 and 300 cm3. What will be the height of the water level in the 2 containers after 20 minute?
This question requires a bit of algebra. Also, the student should be able to get the answer by " Guess & Check " method.
Total volume of water from tap A in 20 minutes = 800 x 20 = 16000
Volume of water in 2 containers --> 500 x height + 300 x height = 16000
--> 800 x height = 16000
height = 16000 / 800 = 20cm
This question requires a bit of algebra. Also, the student should be able to get the answer by " Guess & Check " method.
Total volume of water from tap A in 20 minutes = 800 x 20 = 16000
Volume of water in 2 containers --> 500 x height + 300 x height = 16000
--> 800 x height = 16000
height = 16000 / 800 = 20cm
Wednesday, June 27, 2012
P6 Maths - Whole Numbers, Fractions
Mr Lee spent $5190 on some watches and clocks. The amount spent on watches was $2310 more than the amount spent on clocks. He bought 4/5 times as many clocks as watches.
Each clock cost $13 less than each watch. What was the total number of watches and clocks bought by Mr Lee?
Amount spent on clocks = (5190 - 2310)/ 2 = 1440
Amount spent on watches = 1440 + 2310 = 3750
Most students should be able to do the above steps.
The important thing is to find the cost of 1 unit of both watches and clocks for comparison.
4/5 as many clocks as watches means clocks equal to 4 units while watches equal to 5 units.
Cost of 1 unit of clocks = 1440 / 4 = 360
Cost of 1 unit of watches = 3750 / 5 = 750
Difference between 1 unit of clock and watches = 750-360 = 390
Number of watches or clocks in 1 unit = 390 / 13 = 30
Hence total ( 9 units of clocks and watches ) = 9 x 30 = 270
Each clock cost $13 less than each watch. What was the total number of watches and clocks bought by Mr Lee?
Amount spent on clocks = (5190 - 2310)/ 2 = 1440
Amount spent on watches = 1440 + 2310 = 3750
Most students should be able to do the above steps.
The important thing is to find the cost of 1 unit of both watches and clocks for comparison.
4/5 as many clocks as watches means clocks equal to 4 units while watches equal to 5 units.
Cost of 1 unit of clocks = 1440 / 4 = 360
Cost of 1 unit of watches = 3750 / 5 = 750
Difference between 1 unit of clock and watches = 750-360 = 390
Number of watches or clocks in 1 unit = 390 / 13 = 30
Hence total ( 9 units of clocks and watches ) = 9 x 30 = 270
Tuesday, May 22, 2012
P6 Maths - Percentage
Cathy has $3600 more money than Bob. 60% of Bob’s money is equal to 15% of Cathy’s money.
How much money does Cathy has?
For some questions, to see the picture more clearly, its better to change the percentage to fraction.
60/100 of Bob’s money = 15/100 of Cathy
Then we make the numerator the same,
60/100 of Bob’s money = 60/400 of Cathy
Taking the denominator, we now know that Bob has 100 units while Cathy has 400 units.
Difference = 400 – 100 = 300 units
300 units --> $3600
100 units --> 3600 / 3 = $1200
400 units --> 1200 * 4 = $4800 (Cathy)
How much money does Cathy has?
For some questions, to see the picture more clearly, its better to change the percentage to fraction.
60/100 of Bob’s money = 15/100 of Cathy
Then we make the numerator the same,
60/100 of Bob’s money = 60/400 of Cathy
Taking the denominator, we now know that Bob has 100 units while Cathy has 400 units.
Difference = 400 – 100 = 300 units
300 units --> $3600
100 units --> 3600 / 3 = $1200
400 units --> 1200 * 4 = $4800 (Cathy)
Monday, April 23, 2012
P6 Maths - Fractions ( "Before and After" Method )
Chloe had 120 more stickers than Gladys. After Chloe lost 1/5 and Gladys lost 3/4, Chloe had 184 more than Gladys. How many did Gladys have at first?
To use this method, we have to identify a certain number of units for both Chloe and Gladys. Since 20 units can both be divided into 5 parts and 4 parts respectively, it is used here.
Chloe --> Before
= 20 units + 120
- 4 units – 24 ( lost 1/5)
--> After
= 16 units + 96
Gladys --> Before
= 20 units
-15 units ( lost ¾ )
--> After
= 5 units
Using the “After” units, Chloe will have (11 units + 96) more stickers than Gladys.
11 u --> 184 – 96 = 88
1 u --> 88 / 11 = 8
20 u --> 20 * 8 = 160.
Wednesday, April 18, 2012
P5 Maths - Ratio
The number of stickers Raymond had to Cadence was 5 : 3. After Cadence gave away 39 stickers, the ratio became 6 : 1. How many stickers did both of them had altogether in the end?
For this qn, there is a "Before and After" ratio given. One of the subject in the ratio will remain unchanged.
Before, Raymond : Cadence
= 5 : 3 (multiply by 6)
= 30 : 18
After, Raymond : Cadence
= 6 : 1 ( multiply by 5)
= 30 : 5
Raymond did not have any changes made to his stickers. Hence the before and after units are made the same.
Cadence,
18u – 5u = 13u
13u --> 39 stickers ( Cadence gave away)
1 u --> 3 stickers
In the end,
30u + 5u = 35u
35u --> 35 * 3 = 105 stickers
For this qn, there is a "Before and After" ratio given. One of the subject in the ratio will remain unchanged.
Before, Raymond : Cadence
= 5 : 3 (multiply by 6)
= 30 : 18
After, Raymond : Cadence
= 6 : 1 ( multiply by 5)
= 30 : 5
Raymond did not have any changes made to his stickers. Hence the before and after units are made the same.
Cadence,
18u – 5u = 13u
13u --> 39 stickers ( Cadence gave away)
1 u --> 3 stickers
In the end,
30u + 5u = 35u
35u --> 35 * 3 = 105 stickers
Tuesday, April 17, 2012
P6 Maths - Speed
Mr Tan sends his son to school every morning at 6.15am. If he drives at an average speed of 60km/h, his son will be 10 min late. If he speeds at 120km/h, his son will be 10 min early. What speed should he drive at to arrive on exactly on time?
When the distance covered is the same, the ratio of time taken is opposite to the ratio of the speed.
Speed ratio = 60 : 120 = 1 : 2
Time ratio = 2 : 1
Difference = 2 - 1 = 1 unit
(Difference between 10 min early and 10 min late = 20 min)
1 unit --> 20 min
2 units --> 40 min
When he travels at 60 km/h, he will take 40 min, or 2/3 hr.
Distance to school = 60km * 2/3h
= 40 km
To be on time, the time taken is 30 min, (10 min less than 40min)
Hence Speed = 40km / 1/2h
= 80km/h.
When the distance covered is the same, the ratio of time taken is opposite to the ratio of the speed.
Speed ratio = 60 : 120 = 1 : 2
Time ratio = 2 : 1
Difference = 2 - 1 = 1 unit
(Difference between 10 min early and 10 min late = 20 min)
1 unit --> 20 min
2 units --> 40 min
When he travels at 60 km/h, he will take 40 min, or 2/3 hr.
Distance to school = 60km * 2/3h
= 40 km
To be on time, the time taken is 30 min, (10 min less than 40min)
Hence Speed = 40km / 1/2h
= 80km/h.
Wednesday, April 11, 2012
P6 Maths - Speed, Fractions
A motorist travelled 2/5 of his journey in 4/5 hr. He travelled the remaining 84 km in 1 1/5 hr.
Find the avergage speed for his whole journey.
To find average speed, we need 2 things, “ total distance” and “total time”.
1 – 2/5 = 3/5 -> remaining journey
3/5 of his journey = 84 km
1/5 of his journey = 84 / 3 = 28 km
5/5 of his journey = 28 * 5 = 140 km ( total distance )
Total time taken = 4/5 + 1 1/5 = 2 hrs
Hence
Average speed = Total distance / total time
= 140 / 2
= 70 km / h.
Find the avergage speed for his whole journey.
To find average speed, we need 2 things, “ total distance” and “total time”.
1 – 2/5 = 3/5 -> remaining journey
3/5 of his journey = 84 km
1/5 of his journey = 84 / 3 = 28 km
5/5 of his journey = 28 * 5 = 140 km ( total distance )
Total time taken = 4/5 + 1 1/5 = 2 hrs
Hence
Average speed = Total distance / total time
= 140 / 2
= 70 km / h.
Wednesday, April 4, 2012
P5 Maths - Numbers ( Going Backwards Method )
A, B & C has 204 erasers altogether. A gave some to B and B’s erasers doubled. B gave some to C and C’s erasers doubled. In the end, they have equal number of erasers. How many does A have at first?
For this qn, we have to use the “ Going Backwards” method. We have to start from the end, and move backwards step by step.
204 / 3 = 68 (A,B,C has the same at the end)
68 / 2 = 34 ( This is C before his erasers doubled)
68 + 34 = 102 ( This is B before he gave C some eraser)
102 / 2 = 51 (This is B before his erasers doubled)
68 + 51 = 119 ( This is A before he gave B some eraser)
Ans: 119
For this qn, we have to use the “ Going Backwards” method. We have to start from the end, and move backwards step by step.
204 / 3 = 68 (A,B,C has the same at the end)
68 / 2 = 34 ( This is C before his erasers doubled)
68 + 34 = 102 ( This is B before he gave C some eraser)
102 / 2 = 51 (This is B before his erasers doubled)
68 + 51 = 119 ( This is A before he gave B some eraser)
Ans: 119
Monday, April 2, 2012
P6 Maths - Percentage, Decimals
Conversion of percentage to decimal.
Convert ½% to decimal.
If your answer is 0.5, you are wrong. This question may look simple, but it always managed to get students tricked!
Whenever we convert percentage to a decimal, we have to divide by 100.
½% = ½ / 100
= ½ * 1/100
= 1/200
Hence the answer is 1/200 or 0.005.
Convert ½% to decimal.
If your answer is 0.5, you are wrong. This question may look simple, but it always managed to get students tricked!
Whenever we convert percentage to a decimal, we have to divide by 100.
½% = ½ / 100
= ½ * 1/100
= 1/200
Hence the answer is 1/200 or 0.005.
Tuesday, March 27, 2012
Tips to improve in PSLE Mathematics
The 3 points mentioned below are important to get a good score in mathematics. Students must first improve on these fundamentals first. Then they go on to individual topics practice. Maths is one subject that a 15 to 20 mark improvement can be made in a few months with close coaching.
1. Multiplication Tables
At primary 4, students have to learn addition, subtraction, multiplication and division of fractions. He will be at a disadvantage as he cannot find the common denominator or simplify the fraction to a simpler form easily compared to a fellow student who has his multiplication tables at his fingertips.
As in every maths exam, time plays a crucial factor, especially for paper 1 of PSLE maths. If the student has to do working for simple calculations, he will have less time for the more difficult questions and for checking his work.
2. Concept of Fractions.
”Fraction” questions play an important part in P5 and P6 maths. For students that do not have a strong foundation of this topic, they will not be able to do problems sums that are worth 4 or 5 marks each.
Some problem sums mix topics like percentage and fraction, or even algebra and fraction together, hence it is crucial to grasp the concept of Fractions and its operations early.
3. Modeling skills.
Modeling techniques are taught to students as early as primary 3. It helps the students to simplify the question by using visualization. Some problem sums require models to be shown as part of the working.
When it comes to primary 5 and 6, as the level of questions get higher, modeling skills will definitely be useful.
1. Multiplication Tables
At primary 4, students have to learn addition, subtraction, multiplication and division of fractions. He will be at a disadvantage as he cannot find the common denominator or simplify the fraction to a simpler form easily compared to a fellow student who has his multiplication tables at his fingertips.
As in every maths exam, time plays a crucial factor, especially for paper 1 of PSLE maths. If the student has to do working for simple calculations, he will have less time for the more difficult questions and for checking his work.
2. Concept of Fractions.
”Fraction” questions play an important part in P5 and P6 maths. For students that do not have a strong foundation of this topic, they will not be able to do problems sums that are worth 4 or 5 marks each.
Some problem sums mix topics like percentage and fraction, or even algebra and fraction together, hence it is crucial to grasp the concept of Fractions and its operations early.
3. Modeling skills.
Modeling techniques are taught to students as early as primary 3. It helps the students to simplify the question by using visualization. Some problem sums require models to be shown as part of the working.
When it comes to primary 5 and 6, as the level of questions get higher, modeling skills will definitely be useful.
Monday, March 26, 2012
Primary 6 Maths - Percentage
Sandra had 360 pears and apples at her store. After selling 30% of the apples and 40% of the pears, she had 230 fruits left.
How many apples did she sell?
How many apples did she have at first?
1. For percentage questions, its easier to view it like a units question.
30% of apples – 3 unit out of 10 is sold.
40% of pears – 4 parts out of 10 is sold.
70% of apples – 7 units is left.
60% of pears – 6 parts is left.
1)------7 units + 6 parts = 230
(multiply by 2 throughout) --> 14 u + 12 p = 460
2)------3units + 4 parts = 360-230 = 130
(multiply by 3 throughout) -->9 u + 12 p = 390
Take the first eqn minus the second, (get rid of one unknown)
14 u + 12 p – 9 u – 12 p = 460-390
5u = 70 ( we have successfully get rid of 1 unknown)
1u = 14
3u = 42--- apples sold
10 u = 140 ----apples at first
How many apples did she sell?
How many apples did she have at first?
1. For percentage questions, its easier to view it like a units question.
30% of apples – 3 unit out of 10 is sold.
40% of pears – 4 parts out of 10 is sold.
70% of apples – 7 units is left.
60% of pears – 6 parts is left.
1)------7 units + 6 parts = 230
(multiply by 2 throughout) --> 14 u + 12 p = 460
2)------3units + 4 parts = 360-230 = 130
(multiply by 3 throughout) -->9 u + 12 p = 390
Take the first eqn minus the second, (get rid of one unknown)
14 u + 12 p – 9 u – 12 p = 460-390
5u = 70 ( we have successfully get rid of 1 unknown)
1u = 14
3u = 42--- apples sold
10 u = 140 ----apples at first
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