P4 MATHS - Fractions

The number of girls in a school was 2/5 of the number of pupils in the school. There were 48 more boys than girls. After some girls
went for a camp, the number of boys in the school was 4 times the number of the remaining girls. How many girls went camping?

To start off,

the student must know that since 2/5 of the pupils are girls, 3/5 must be boys.

1 - 2/5 = 3/5

We change the fraction to 2 units for girls, and 3 units for boys.

1 unit ---> 48
3 units ---> 48 * 3 = 144 ( Boys )
2 units ---> 48 * 2 = 96 ( Girls)


After, some girls went for a camp, the student must know that the number of boys remained unchanged.

Since the boys now become 4 times the number of girls,

144 / 4 = 38 ( girls now)

96 - 38 = 58 ( girls that went camping )

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

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

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)

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.

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.

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.

Primary 6 Maths - Fractions

Alice, Ben and Cathy shared some money. Ben's share is 1/3 of the money. 1/2 of Alice's share is 2/3 of Cathy's share. If Ben's share is $56, how much must Alice give to Cathy so that both Alice and Cathy would have the same amount of money?

1. Look for the values that is given in qn. The value given here is $56. Hence we must try to match the value to the unit.

In this qn,

Ben’s share = 1/3 of money = $56.
Hence Alice and Cathy must make up 2/3 of money = $56 x 2 = $112.

2. Relate the fractions given in qn.

Look at this statement “1/2 of Alice's share is 2/3 of Cathy's share. “

This means 1 unit of A’s share = 2 units B’s. A has 2 units in total, we have to change to B’s equal, that means A has 2x2=4 units. Hence now, A to C = 4 : 3.
Alice and Cathy = 7 units = $112.

A = 112 / 7 x 4 = $168
C = 112 / 7 x 3 = $126

Alice must give ½ the difference btw them to make same amount for both.

Hence Alice must give [168-126] / 2 = $21.