Road to PSLE

Tuition Classes for Academic Year 2010

noreply@blogger.com (Excel Eduservice) — Mon, 26 Oct 2009 14:26:00 +0000

Please note that we will start our new tuition classes for the Academic Year 2010 on 14 Nov 2009.

Our contact details can be found in this link.

Simple Machines - The Movable Pulley

noreply@blogger.com (Excel Eduservice) — Sun, 11 Oct 2009 01:10:00 +0000

This article was first published on 3 Sep 2008.

One of the functions of simple machines is to lighten our workload. A small effort can move a greater load. What is sacrificed is that the distance moved by the effort is greater than the distance moved by the load.

As an example, we will take a look at the movable pulley. So how does a movable pulley help to lighten our work?

Remember that in order to be able to move a load that is greater than the effort, the distance moved by the effort has to be greater than the distance moved by the load.

With reference to the diagram above:-

Pulley A - The distance from the pulley to the ceiling is 2m. This means that the cable attached to the pulley is also 2m.

Pulley B – The distance moved by the effort is 2 m. However, the load moves only 1 m from the original position of the pulley. This is confirmed by …. 2-m cable divided by 2 = 1m, which means that the distance from the pulley to the ceiling is 1m. Hence, the distance from the pulley now, is 1 m from where it was originally.

Since the effort moved twice the distance moved by the load, only ½ the effort is needed to raise the load. Hence if the load weighs 10 kg, an effort of slightly over 5 kg only is needed to move the load.

Simple Machines – The Inclined Plane

noreply@blogger.com (Excel Eduservice) — Sun, 11 Oct 2009 01:00:00 +0000

This article was originally published in Oct 2008.

As with all machines, if we want the force exerted by the effort to be less than the force exerted by the load, the distance travelled by the effort has to be more than the distance travelled by the load.

Below are 4 illustrations how the inclined plane works.

The more gradual the slope, the greater the distance travelled by the effort compared to the distance travelled by the load. This means that the more gradual the slope, the less effort is needed to move the load.

Examples of applications of the ramp in real life – wheelchair ramps for the disabled, gradual and winding slopes of a road found in mountainous terrain (e.g. roads to Gentling Highlands or Cameron Highlands).

It can be noted that the ‘sharper’ the wedge, the less the effort is needed because for the same distance travelled by the effort, the distance moved by the load is now less.

Examples of applications of the wedge in real life – axe head, blades of knives, metal wedges for prying open flanges of pipes in heavy industries.

Fig 1 - For every 1 turn the screw makes, the screw is driven down by the distance of ‘1 pitch’. Hence, the more gradual the slope of the thread, the smaller the distance it will be between 2 threads, as shown in Fig 2.

Fig 2 - the distance travelled by the load is less than in Fig 1, although in both cases, the effort moves by the same one turn. Since the load moves less in Fig 2, less effort is also needed.

In conclusion, for screws, the more the number of threads there is, OR the smaller the pitch, OR the more gradual the slope of the threads, the less the effort is needed to move the load.

Again, like the screw, the smaller the pitch, the less the ends of the ‘V’ move towards each other, which also means the less the distance the load moves.

Like the screw, for the screw jack, the more the number of threads there is, OR the smaller the pitch, OR the more gradual the slope of the threads, the less effort is needed to move the load.

Take note that for the screw jack, while the mechanism of the inclined plane is used to lift the vehicle off the ground, the handle of the screw jack works on the principle of the wheel and axle. In this case, the handle is the wheel, while the screw of the jack (the part with the threads) is the axle.

Important note to parents and students – The purpose of the above 4 illustrations is to help students understand the mechanics of the inclined plane. Most schools’ practice papers and science textbooks do not use the terms ‘pitch’ or ‘threads’. Those terms are technical terms used in the heavy and light industries.

In trying to make the illustration as simple as possible, I have found that I cannot avoid using those technical terms. The use of the above terms is to assist the student (or parent) to understand the mechanics of screws and screw jacks.

Take note that markers who will be marking your Science PSLE Paper, in all likelihood have been teachers all their lives and may not have worked in industries before, and would probably not be familiar with the terms ‘pitch’ and ‘threads’.

In other words, it is highly advised that you do not use the terms ‘pitch’ and ‘threads’, unless you are able to draw and label the diagrams as accurately as the above.

Parents up in arms again over PSLE Mathematics paper

noreply@blogger.com (Excel Eduservice) — Sat, 10 Oct 2009 06:48:00 +0000

From Channel News Asia

As requested by some students, here is one possible workout for the solution.

Question

Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim's sweets to chocolates became 1:7 and the ratio of Ken's sweets to chocolates became 1:4. How many sweets did Ken buy?

Solution

Chocolates ----- C
Sweets ----- S

Jim bought 2 units of chocolates ----- C, C
Ken bought 2 units of sweets ---- S, S

Jim gave 1 unit of chocolates to Ken and Ken gave 1 unit of sweets to Jim.
Jim ----- S, C
Ken ----- S, C

Jim ate 12 sweets and Ken ate 18 chocolates.
Jim ----- S – 12; C
Ken ----- S; C – 18

Jim’s sweet to chocolate ratio 1:7. Therefore for number of sweets to be same as number of chocolates, multiply sweets by 7.
C ----- 7 x (S – 12)
C ----- 7S – 84

Ken’s sweet to chocolate ratio 1:4. Therefore, for number of sweets to be same as number of chocolates, multiply sweets by 4.
4 x S ---- C – 18
4S ---- C – 18

From Jim, (C ----- 7S – 84), replace Ken's C....
(Ken) 4S ----- C - 18
4S ----- (7S – 84) – 18
4S ----- 7S – 102
3S ----- 102
1S ----- 34

Number of sweets Jim had at first
2 units of sweets ----- 2 x 34 = 68

Answer: 68 sweets.

Rosyth Sch 2007 PSLE Math Prelim Q48

noreply@blogger.com (Excel Eduservice) — Mon, 28 Sep 2009 12:27:00 +0000

A bus was travelling at a constant speed from Town A to Town B. It passed a car travelling at a constant speed of 90 km/h in the opposite direction. 1 and 1/2 hours later, the bus reached Town B but the car was still 25 km away from Town A. If the bus took 4 hours to complete the whole journey, what is the distance between the two towns?

Solution

Answer: 256 km

Rosyth Sch 2007 PSLE Math Prelim Q47

noreply@blogger.com (Excel Eduservice) — Mon, 28 Sep 2009 12:24:00 +0000

The perimeter of a rectangle to that of a square is in the ratio of 11:6. If the square has an area of 144 square m and the length and breadth of the rectangle are in the ratio of 6:5, find the length and breadth of the rectangle.

Solution

Perimeter
(6 + 6 + 5 + 5) units ----- 88 m
22 units ----- 88 m
1 unit ----- 88m divided by 22 = 4 m

(Length) 6 units ----- 6 x 4 m = 24 m
(Breadth) 5 units ----- 5 x 4 m = 20 m

Answer: Length = 24 m; Breadth = 20 m

Rosyth Sch 2007 PSLE Math Prelim Q46

noreply@blogger.com (Excel Eduservice) — Mon, 28 Sep 2009 12:21:00 +0000

The total cost of 28 textbooks and workbooks is $784.
3/4 of the books are textbooks and the remaining books are workbooks. A workbook cost half as much as a textbook. Find the difference in the price of a textbook and a workbook.

Solution

Workbook ----- 1 unit
Textbook ------ 2 units

3/4 of the books are textbooks ----- 3/4 x 28 = 21

Workbooks ----- 28 – 21 = 7

21 textbooks + 7 workbooks ---- $784

(21 x 2 units) + 7 units ----- $784

42 units + 7 units ----- $784

49 units ----- $784

1 unit ------ $784 divided 49 = $16

Answer: $16

Rosyth Sch 2007 PSLE Math Prelim Q45

noreply@blogger.com (Excel Eduservice) — Fri, 25 Sep 2009 03:52:00 +0000

The figure shows a cube with 3 painted parts A, B and C. These painted parts are of the same area and they are touching the midpoints of the sides of the cube. The total area of the painted parts is 54 square cm. Find the volume of the cube.

3 painted parts on 3 faces of the cube ----- 54 square cm

1 painted part on 1 face of the cube
54 square cm divided by 3 = 18 square cm

Area of 1 triangle is 18 square cm divided by 2
= 9 square cm

For area of triangle to be 9 square cm, the base has to be 6 cm and height has to be 3 cm as worked out below
(1/2 x 6 cm x 3 cm = 9 square cm)

1 side of the cube is therefore 6 cm as seen from above diagram.

Volume of cube
(6 x 6 x 6) cubic cm = 216 cubic cm

Answer: 216 cubic cm

Rosyth Sch 2007 PSLE Math Prelim Q44

noreply@blogger.com (Excel Eduservice) — Fri, 25 Sep 2009 03:42:00 +0000

The figure is not drawn to scale. What fraction of the figure is the total area of the shaded regions A, B, C, D, E, F, G and H?

Solution

The base of the whole figure is (4 + 3 + 6 + 3 + 2) cm = 18 cm

1st row from the top
Fig A has a base of 6 cm.

2nd row
The sum of the bases of Figs B and C is 6 cm.

3rd row
The sum of the bases of Figs D and E is 6 cm.

4th row
The sum of the bases of Figs F and G is 6 cm.

5th row
Fig H has a base of 6 cm.

The average length of the bases of all shaded figures is 6 cm.

The length of the base of the whole figure is 18 cm.

Therefore, the fraction of whole figure that is shaded is
6 cm divided by 18 cm = 1/3

Answer: 1/3

Note that the above method can be used because the vertical columns also correspond to the 6 cm shaded out of the 18 cm total vertical length of the figure.

Rosyth Sch 2007 PSLE Math Prelim Q43

noreply@blogger.com (Excel Eduservice) — Fri, 25 Sep 2009 03:31:00 +0000

The area of the figure ABCD is 48.5 square cm . Find the length of BE.
(Give your answer to the nearest tenth)

Solution

Area of ACD
1/2 x 18 cm x 10.4 cm = 93.6 square cm

Area of ABC
93.6 square cm - 48.5 square cm = 45.1 square cm

Area of Triangle ABC ----- 1/2 x base x height
45.1 square cm = 1/2 x 18 cm x height

height = (45.1 sq cm x 2 ) divided by18 cm
= 5.01 cm ~ 5.0 cm (to the nearest tenth)

Answer: 5.0 cm

Rosyth Sch 2007 PSLE Math Prelim Q42

noreply@blogger.com (Excel Eduservice) — Fri, 25 Sep 2009 02:43:00 +0000

Find the area of the figure.

Answer: 196 square cm

Rosyth Sch 2007 PSLE Math Prelim Q41

noreply@blogger.com (Excel Eduservice) — Fri, 25 Sep 2009 02:35:00 +0000

The table below shows the rates at which a construction worker is being paid daily.

How many hours must he work in a day to earn $140?

Solution

First 8 hours of work ----- $10 x 8 = $80

To earn $140, he needs another $60.

Overtime rate is ----- $10 per hour x 1.5 = $15 per hour.

For overtime pay to be equal to $60, he must work
$60 divided by $15 per hour = 4 hours

Total time he needs to work
8h + 4h = 12h

Answer: 12 hours

Rosyth Sch 2007 PSLE Math Prelim Q40

noreply@blogger.com (Excel Eduservice) — Thu, 24 Sep 2009 04:18:00 +0000

Machine A can do a printing job in 2 hours. Machine B can do the same job in 3 hours. If both machines are used at the same time, how long would it take to complete half the printing job?

Solution

Since Machine A takes 2 hours and Machine B takes 3 hours, Machine A will be able to get more work done than Machine B, if both are used at the same time, in the ratio of 3:2 as shown below.

(Machine A)
Whole job ----- 2 hours
3/5 of job ----- 3/5 x 2 hours = 6/5 hours

Time taken if both machines are used at the same time is 6/5 hours because while Machine A completes 3/5 of the job, Machine B completes the remaining 2/5 of the job.

Time taken for whole job done if both machines are used at same time is therefore 6/5 hours

Time taken for half the job done if both machines are used at the same time is
1/2 x 6/5 hours = 3/5 hour.

Answer: 3/5 hour

Rosyth Sch 2007 PSLE Math Prelim Q39

noreply@blogger.com (Excel Eduservice) — Thu, 24 Sep 2009 04:07:00 +0000

Find the sum of the six marked angles in the diagram.

The sum of the all the angles in the triangle in the centre is 180 degrees. The sum of all the 3 angles outside the centre triangle, in the figure above is therefore also 180 degrees (opposite angles).

The sum of all the six marked angles is hence,

Sum of marked angles in 1 triangle ----- 180 degrees
Sum of marked angles in 3 triangles ----- 180 degrees x 3 = 540 degrees
Sum of 6 marked angles ----- (540 – 180) degrees = 360 degrees

Answer: 360 degrees

Rosyth Sch 2007 PSLE Math Prelim Q38

noreply@blogger.com (Excel Eduservice) — Wed, 23 Sep 2009 06:52:00 +0000

In the figure not drawn to scale, the ratio of the area of the bigger square to the smaller square is 8:4. If 25% of the larger square is shaded, what percentage of the whole figure is not shaded?

Solution

Portion unshaded ----- 6 units + 2 units = 8 units
Total area ----- 6 units + 2 units + 2 units = 10 units

Percentage of figure that is not shaded
8/10 x 100% = 80%

Answer: 80%

Rosyth Sch 2007 PSLE Math Prelim Q37

noreply@blogger.com (Excel Eduservice) — Wed, 23 Sep 2009 06:50:00 +0000

A tank with a square base of side 5 m was 1/3 full of water. When another 100 cubic meters of water was poured into the tank, it became 1/2 full. Find the capacity of the tank.

Solution

1/2 capacity – 1/3 capacity = 1/6 capacity

1/6 capacity ----- 100 cubic metres

6/6 capacity ----- 6 x 100 cubic m = 600 cubic m

Answer: 600 cubic metres

Rosyth Sch 2007 PSLE Math Prelim Q36

noreply@blogger.com (Excel Eduservice) — Wed, 23 Sep 2009 06:48:00 +0000

The average weight of 5 boys is 8w kg. When 2 more children whose weights are 11w kg and 12 w kg respectively joined the group, what is the average weight of the boys now?

Solution

Average weight of 5 boys ----- 8w kg

Total weight of 5 boys ----- 8w kg x 5 = 40w kg

Total weight of 7 boys
(40w + 11w + 12w) kg = 63w kg

Average weight of 7 boys
63w kg divided by 7 = 9w kg

Answer: 9w kg

Easier Free Item Downloads

noreply@blogger.com (Excel Eduservice) — Tue, 22 Sep 2009 13:00:00 +0000

Excel Eduservice would like to thank everyone for supporting this website. We are pleased this site has helped many P6 students, parents and tutors the last 2 years.

We have added a new feature that makes downloading of free items easier. You can now download them without sending an email. Simply click the link below.

Free Maths Worked Solutions and PLSE Aggregate Score Calculator

There is a NEW free maths worked solution (Nanyang Pri P6 Prelim 2007) in the free section above.

At Excel Eduservice, we work to serve you better.

You're in good hands when you're with Excel!

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q48

noreply@blogger.com (Excel Eduservice) — Tue, 22 Sep 2009 12:07:00 +0000

Valley Department Store sold a dress for $585. This was 17% more than the price of a similar dress in Wiki Department Store. During a sale, both stores offered same percentage discount on the dress. Christine bought the dress in Wiki Department Store and found that she paid $68 less than the discounted price in Valley Department Store.

(a) Find the price of the dress in Wiki Department Store before the sale.
(b) What is the percentage discount given during the sale?

Solution

(a)
Valley ----- 100% + 17% = $585
117% ----- $585
1% ----- $585 divided by 117 = $5

(Wiki) 100% ----- 100 x $5 = $500

Answer: $500

(b)
Before discount

$85 is the difference in price between Valley and Wiki before the discount. After the discount, the difference was $68.

$85 - $68 = $17
Discount ----- ($17/$85) x 100% = 20%

Answer: 20%

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q47

noreply@blogger.com (Excel Eduservice) — Tue, 22 Sep 2009 08:07:00 +0000

Melissa spent 3/8 of her money on 3 pencils and 8 pens, and 4/5 of the remainder on 15 markers. Each pencil cost 2/3 as much as a pen. Each marker cost $0.20 more than a pencil. What is the cost of a pen?

Solution

40 units – 30 units ----- $3
10 units ----- $3
1 unit ----- $3 divided by 10 = $0.30

(Pen) 3 units ----- $0.30 x 3 = $0.90

Answer: $0.90

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q46

noreply@blogger.com (Excel Eduservice) — Mon, 21 Sep 2009 05:39:00 +0000

At first, the ratio of the number of marbles received by John and Peter was 4:7. The ratio of the number of marbles received by Peter and Sam was 9:5. Then, John gave 1/12 of his marbles to Sam, and Peter gave 1/9 of his marbles to Sam. As a result, Sam had 135 marbles in the end.

(a) Find the ratio of the number of John’s marbles to the number of Sam’s marbles at first.

(b) Find the total number of marbles received by the 3 boys.

Solution

45 units ----- 135 (marbles)
1 unit ----- 135 divided by 45 = 3
134 units ----- 134 x 3 = 402

(a) Answer: 36 : 35
(b) Answer: 402 marbles

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q45

noreply@blogger.com (Excel Eduservice) — Thu, 17 Sep 2009 12:22:00 +0000

The figure below shows a rectangle PQRS. The lines are extended from point P, Q, R and S and they meet at point Y. The length of QR is 20 cm and the length of XY is 4 cm. Given that QS is a straight line, the area of Triangle PQY is 72 square cm and the area of Triangle SRY is 84 square cm, find the shaded area of Triangle QSY.

Solution

Area of Triangle PQY + Area of Triangle SRY
72 sq cm + 84 sq cm = 156 square cm

Area covered by Triangles PQY and SRY is also 1/2 of area of Rectangle PQRS. This also means that Area of Triangle PQS is also 156 sq cm.

Area of Triangle PSY = 1/2 x 20 cm x 4 cm = 40 sq cm

Area of Triangle PQY = 72 sq cm

Shaded area
Area of PQS – Area of PQY – Area of PSY
=156 sq cm - 72 sq cm - 40 sq cm= 44 square cm

Answer: 44 square cm

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q44

noreply@blogger.com (Excel Eduservice) — Thu, 17 Sep 2009 04:52:00 +0000

A group of pupils sat for two tests, test A and test B. The number of pupils who failed test A was 1/7 of those who passed test A. Given that there were 20 pupils who failed test A,

(a) How may pupils passed test A?
(b) The number of pupils who failed test B was 1/4 of those who failed test A. What fraction of the pupils passed test B?

Solution

(a)
1 unit ----- 20
7 units ----- 20 x 7 = 140

Answer: 140 pupils

(b)
Failed Test A ----- 20
Failed Test B ----- 1/4 x 20 = 5

(Total number of pupils)
8 units ----- 8 x 20 = 160

Fraction of pupils who failed Test B
5/160 = 1/32

Fraction of pupils who passed Test B
1 – 1/32 = 31/32

Answer: 31/32

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q43

noreply@blogger.com (Excel Eduservice) — Wed, 16 Sep 2009 09:22:00 +0000

There are some red and green beads in a container. If 75 more green beads are put into the container, the percentage of red beads will decrease from 30% to 20%. How many red beads are in the container?

Solution

12 units ----- 7 units + 75
5 units ----- 75
1 unit ----- 75 divided by 5 = 15

Red at first
3 units ----- 3 x 15 = 45

Answer: 45 red beads

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q41

noreply@blogger.com (Excel Eduservice) — Tue, 15 Sep 2009 04:01:00 +0000

Jane started driving at 9.30 am from Town A to Town B. At 11.30 am, Jane had covered only 2/5 of the distance. She had to cover another 144 km before she reached Town B.

(a) What was the distance between Town A and Town B?

(b) If Jane were to travel at an average speed of 72km/h after 11.30 am, at what time would she reach Town B?
(Express your answer using the 24-hour clock)

Solution

(a)
3 units ----- 144 km
1 unit ----- 144 km divided by 3 = 48 km
5 units ----- 5 x 48 km = 240 km

Answer: 240 km

(b)
Time = 144km divided by 72 km/h = 2 h
2 hours after 11.30 am is 1.30 pm or 1330 h.

Answer: 1330 h