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Primary 4 Maths 2024 Test Papers

2024 Primary 4 Maths Test Papers

All eTest paper downloads are free and come with answers.

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P4 Maths EOY 2024 ACSJ Paper
P4 Maths EOY 2024 Ai Tong Paper
P4 Maths EOY 2024 Catholic High Paper
P4 Maths EOY 2024 Henry Park Paper
P4 Maths EOY 2024 Maha Bodhi Paper
P4 Maths EOY 2024 Nan Chiau Paper
P4 Maths EOY 2024 Nan Hua Paper
P4 Maths EOY 2024 Nanyang Paper
P4 Maths EOY 2024 Pei Chun Paper
P4 Maths EOY 2024 Raffles Paper
P4 Maths EOY 2024 Red Swastika Paper
P4 Maths EOY 2024 Rosyth Paper
P4 Maths EOY 2024 Rulang Paper
P4 Maths EOY 2024 St Hilda Paper
P4 Maths EOY 2024 Tao Nan Paper

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P4 Maths WA1 2024 Ai Tong Paper
P4 Maths WA1 2024 Catholic High Paper
P4 Maths WA1 2024 Henry Park Paper
P4 Maths WA1 2024 Maha Bodhi Paper
P4 Maths WA1 2024 Nan Hua Paper
P4 Maths WA1 2024 Nanyang Paper
P4 Maths WA1 2024 Raffles Paper
P4 Maths WA1 2024 Red Swastika Paper
P4 Maths WA1 2024 Rulang Paper
P4 Maths WA1 2024 St Hilda Paper
P4 Maths WA1 2024 Tao Nan Paper
P4 Maths WA2 2024 Ai Tong Paper
P4 Maths WA2 2024 Henry Park Paper
P4 Maths WA2 2024 Nan Hua Paper
P4 Maths WA2 2024 Nanyang Paper
P4 Maths WA2 2024 Pei Chun Paper
P4 Maths WA2 2024 Raffles Paper
P4 Maths WA2 2024 Rosyth Paper
P4 Maths WA2 2024 Rulang Paper
P4 Maths WA2 2024 St Hilda Paper
P4 Maths WA3 2024 Ai Tong Paper
P4 Maths WA3 2024 Catholic High Paper
P4 Maths WA3 2024 Nan Hua Paper
P4 Maths WA3 2024 Pei Chun Paper
P4 Maths WA3 2024 Rulang Paper
P4 Maths WA3 2024 St Hilda Paper
P4 Maths WA3 2024 Tao Nan Paper

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Sample Maths Question & Worked Solutions

Nanyang P4 EOY 2024 Maths Q43

There were 76 more belts than caps at first. After 158 belts and 30 caps were sold, there were 3 times as many caps as belts left.
How many belts were there in the end?

Let's define variables for the number of belts and caps at first:
Let b be the number of belts at first.
Let c be the number of caps at first.

Step 1: Establish Equations

From the problem statement:
There were 76 more belts than caps at first:
b=c+76

After selling 158 belts and 30 caps, there were 3 times as many caps as belts left:
Belts left: b−158
Caps left: c−30
Given that caps left are three times the belts left:
c−30=3(b−158)

Step 2: Solve for b and c
Substitute b=c+76 into the second equation:
c−30=3((c+76)−158)
c−30=3(c+76−158)
c−30=3(c−82)
c−30=3c−246
−30+246=3c−c
216=2c
c=108
Find b:
b=108+76=184
Step 3: Find the number of belts left
b−158=184−158=26
Final Answer: There were 26 belts left in the end.

Raffles P4 EOY 2024 Maths Q42

The total mass of an empty school bag and 5 similar books was 5.4 kg. The total mass of the same school bag and 2 similar books was 3 kg.
What was the mass of 1 book?

Let's define variables:
Let B be the mass of one book (kg).
Let S be the mass of the empty school bag (kg).

Step 1: Establish Equations
From the problem statement:
The total mass of the school bag and 5 books was 5.4 kg:
S+5B=5.4
The total mass of the school bag and 2 books was 3 kg:
S+2B=3

Step 2: Subtract the Two Equations
(S+5B)−(S+2B)=5.4−3
S−S+5B−2B=2.4
3B=2.4
B=2.4 ÷ 3 = 0.8
Final Answer: The mass of one book is 0.8 kg.

Raffles P4 EOY 2024 Maths Q43

Ravi had $92 more than Peter. After Ravi gave $100 to Peter, Peter had 5 times as much as Ravi.
How much money did Ravi have at first?

Let's define variables:
Let R be the amount of money Ravi had at first.
Let P be the amount of money Peter had at first.

Step 1: Establish Equations
From the problem statement:
Ravi had $92 more than Peter:
R=P+92
After Ravi gave $100 to Peter:
Ravi’s new amount: R−100
Peter’s new amount: P+100
Given that Peter now has 5 times as much as Ravi:
P+100=5(R−100)

Step 2: Solve for R and P
Substitute R=P+92 into the second equation:
P+100=5((P+92)−100)
P+100=5(P+92−100)
P+100=5(P−8)
P+100=5P−40
100+40=5P−P
140=4P
P=140 ÷ 4=35
P=35
Find R:
R=P+92=35+92=127
Final Answer: Ravi had $127 at first.




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