15 Common Types of Primary 4 Maths Questions Every Singapore Student Must Know

Q1. Lowest Common Multiple (LCM)

A factory has a container of flour. The container of flour can be repacked into smaller identical 2−kg packets, 6−kg packets or 16−kg packets without any leftover.

(a) What is the smallest possible mass (in kilograms) of flour in the container?

(b) Assume that the mass of flour in the container is the value stated in (a). Besides using 2-kg, 6-kg, or 16-kg packets, state six other packet sizes (in whole kilograms) that can also be used to repack all the flour equally into smaller identical packets without any leftover.

Solution:

(a)
We are looking for a number that can be divided exactly by 2 or 6 or 16. Hence this number is a common multiple of 2, 6 and 16.

Listing the multiples of 2, 6 and 16, we have:

Multiples of 16: 16, 32, 48, …
Multiples of 6: …, 30, 36, 42, 48, …
Multiples of 2: …, 40, 42, 44, 46, 48, …

See that the smallest common multiple of 2, 6 and 16 is 48.

Hence the smallest possible mass of flour in the container is 48 kg. (ans)

(b)
Factors of 48 are 1, 3, 4, 8, 12 and 24.

The mass of each packet could be 1 kg, 3 kg, 4 kg, 8 kg, 12 kg or 24 kg (ans)

Note: 48 is not included since we are repacking into smaller packets.

Q2. Highest Common Factor (HCF)

A cardboard measures 80 cm by 64 cm. Identical squares are to be cut out from cardboard such that there will be no leftover cardboard.

(a) For the least number of squares to be cut out, what is the area of each square?

(b) How many such squares can be cut out from the cardboard?

Solution:

(a)
To minimise the number of squares cut out, each square should be as large as possible. Also the length of the square must divide each side of the cardboard exactly.

We need to identify the HCF of 80 and 64.

1 × 80                        1 × 64
2 × 40                       2 × 32
4 × 20                       4 × 16
5 × 16                        8 × 8
8 × 10

See that the largest number that divides 80 and 64 is 16.
When each side of the square is 16 cm, we will cut out the least possible number of squares from the cardboard.

Area of each 16–cm square = 16 cm × 16 cm = 256 cm2 (ans)

(b)
The number of 16–cm squares that can be cut along the length of the cardboard
= 80 ÷ 16 = 5

The number of 16–cm squares that can be cut along the breadth of the cardboard
= 64 ÷ 16 = 4.

Total squares cut from the cardboard = 5 × 4 = 20 (ans)

Q3. Stacking Model 1 (remove excess)

An adult ticket costs $6 more than a child ticket. 2 adults and 3 children paid $82. What is the cost of a child’s ticket?

Solution:

$6 × 2 = $12
$82 – $12 = $70
5 units = $70
1 unit = $70 ÷ 5 = $14 (ans)

Q4. Stacking Model 2 (completing unit)

Four boys received a sum of money. Mike received $10 more than Nate. Owen received twice as much money as Nate. Peter received $6 less than Owen. The four boys received $76 in total. How much money did Peter receive?

Solution:

Note: A common error for pupils in this question would be to subtract $10 and $6 from $76. Then equate the value to 6 units. This is wrong because Peter is actually short of $6 to form 2 complete units. Subtracting $6 from Peter would not enable him to form 2 units. The correct method would be to add $6 and subtract $10 from the total.

Solution:

$76 – $10 + $6 = $72
6 units = $72
1 unit = $72 ÷ 6 = $12

Peter has $6 less than 2 units.
2 units = $12 × 2 = $24
Amount received by Peter = $24 – $6 = $18 (ans)

Q5. Model Drawing – More than/Less than (with Changes)

There were 40 more boys than girls in a badminton club at the start of January. When 10 girls left the badminton club at the end of January, there were 3 times as many boys as girls remaining in the badminton club. How many children were in the badminton at the start of January?

Solution:

2 units = 40 + 10 = 50
1 unit = 50 ÷ 2 = 25

At the start of January, there were 4 units and 10.

4 units = 4 × 25 = 100
100 + 10 = 110 (ans)

Q6. Model Drawing – Equal Stage

A class library has a collection of fiction and non–fiction books. There were 4 times as many fiction books as non–fiction books. When 18 more non–fiction books were donated to the class library and 27 fiction books were loaned out, there were an equal number of fiction books and non–fiction books in the class library. How many books were in the class library at first?

Solution:

See that the difference of 27 and 18 represents a total of 3 units.

3 units = 18 + 27 = 45
1 unit = 45 ÷ 3 = 15
5 units = 5 × 15 = 75 (ans)

Q7. Model Drawing – Internal Transfer

Two tanks, X and Y, contain some amount of water. Tank X contains 8 more litres of water than tank Y. When 20 ℓ of water is poured from tank X into tank Y, tank Y will contain 3 times as much water as tank X. How much water do both tanks contain altogether?

Solution:

X has 8 ℓ more. After transferring 8 ℓ to Y, the remaining amount transferred to Y is
20 – 8 = 12 ℓ
2 units = 12 + 8 + 12 = 32 ℓ
1 unit = 32 ÷ 2 = 16 ℓ
No. of units in the end = 1 unit (from X) + 3 units (from Y) = 4 units
4 units = 4 × 16 = 64 ℓ (ans)

Q8. Model Drawing – Constant Difference

When Singapore celebrates its 60th year of independence in 2025, James is 42 years old. In which year did Singapore have four times as many years of independence as James’ age?

Solution:

The age gap between Singapore and James = 60 – 42 = 18 years

When Singapore was four times James’ age:

3 units = 18 years
1 unit = 18 ÷ 3 = 6 years
4 units = 4 × 6 = 24 years

When Singapore celebrated 24 years of independence, James was 6 years old.
This happened in 2025 – 60 + 24 = 1989 (ans)

Q9. Model Drawing – Repeated Identity

A fruit crate has 3 times as many mangoes as guavas, and 2 times as many guavas as apples. There are 135 fruits in the crate. How many mangoes are in the crate?

Solution:

The number of guavas is the repeated identity in both comparisons and so the quantity of guavas in both comparisons must be the same. First, draw 3 units for mangoes and 1 unit for guavas. Then cut each unit into 2 smaller units.

9 units = 135
1 unit = 135 ÷ 9 = 15
6 units = 6 × 15 = 90 (ans)

Q10. Grouping – Quantity and Value

Five times as many children as adults attended a carnival. Each adult received two drink coupons while each child received four drink coupons. A total of 7700 coupons were given out. How many children attended the carnival?

Solution:

See that for every adult present, there would be 5 children present. This is the smallest possible group that can be formed. In this context, quantity refers to the number of people, while value refers to the number of coupons.

In a group there will be 6 people (1 adult and 5 children) and 22 coupons given out.
Since 7700 coupons were given out, then the total number of groups must be
7700 ÷ 22 = 350 groups

There are 5 children in each group. So in 350 groups, there will be
350 × 5 = 1750 children (ans)

Q11. Excess and Shortage (unfixed containers)

A fruit seller has some apples. If he packs 4 apples into a bag, he will have 2 apples left. If he packs 7 apples into a bag, he will be short of 2 apples. What is the smallest possible number of apples the fruit seller has?

Solution:

Depending on the packing method, the final number of bags will be different.
By Listing Method, we have the following.

Ans: 54 apples

 Q12. Excess and Shortage (fixed containers)

Mrs Devi bought some pencils for her pupils. If she gives 7 pencils to each pupil, there will be 35 stickers left. If she gives 5 pencils to each pupil, there will be 105 pencils left. How many pupils are there in Mrs Devi’s class?

Solution:
The number of pupils in the class is fixed, regardless of the number of pencils each pupil receives.

After giving 5 pencils to each pupil, there is a leftover of 105.

Each pupil already has 5 pencils. To give 7 pencils to each pupil, each pupil takes 2 more from the leftover of 105. And there is a final remainder of 35.

When each pupil takes 2 pencils, 105 – 35 = 70 pencils are given out.

Total no. of pupils = 70 ÷ 2 = 35 (ans)

This question can also be solved using Listing Method from Q11 but it will take more time.

Ans: 35 pupils

Q13. Assumption

There were 125 crystals to be used for making 25 pieces of necklaces and bracelets. A necklace will use up 8 crystals while a bracelet will use up 3 crystals. How many bracelets were made?

Solution:
Assume all 25 pieces are necklaces. (Tip: always assume the item that is not required by the question)

Total no. of crystals to be used = 25 × 8 = 200

No. of crystals in excess = 200 – 125 = 75

To reduce the excess, we need to exchange some necklaces for bracelets.

When one necklace is removed, the total no. used crystals reduces by 8 and when replaced by one bracelet, the total no. of crystals used up will be increased by 3. So the overall reduction is 5.

Total no. of replacements = 75 ÷ 5 = 15
This means 15 necklaces are replaced by 15 bracelets.

Ans: 15 bracelets were made.

Q14. Fraction of a Quantity

This refers to part of a given amount, usually obtained by multiplying a fraction with a whole number.

For example: “What is 3/4 of 20?”
This is asking for a fraction of a quantity →  3/4 × 20 = 15

It is an operation (usually multiplication) — taking a portion of something.

Question:
A packet of rice weighs 10 kg. A family uses 1/4 of the packet of rice. How much rice is left? Express your answer in kilograms and grams.

Method 1 (Part of a whole:
Fraction of rice that is left = 1 – 1/4 = 3/4
3/4 × 10 kg = 7.5 kg                     =7 kg 500 g (ans)

Method 2 (Model Drawing):

4 units → 10 kg
1 unit → 10 kg ÷ 4 = 2.5 kg
3 units → 2.5 kg × 3 = 7.5 kg
= 7 kg 500 g (ans)

Q15. Fractional Quantity (Fraction of a Set)

This refers to a quantity (or measurement) that is itself a fraction, not a whole number.

For example: “The bottle contains 1 1/2 ℓ of juice.”

It means the value is not entirely a whole number — it includes a fractional part, which is 1/2 .

It is a description — stating that a value (or measurement) is a fraction or mixed number.

Question:

Mrs Lee bought 1 1/2 kg of rice. She used 4/5 kg of it. What was the mass of rice left? Express your answer in grams.

Solution:

1 1/2 kg – 4/5 kg = 1 5/10 – 8/10 = 15/10 – 8/10 = 7/10 kg

7/10 kg means 7/10 of one kilogram (= 1000 g)

To convert  7/10 kg to grams:
7/10 × 1000 g = 700 g (ans)

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