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Primary 6 PSLE Mathematics Semestral Assessment 2 (End of Year) Paper 5

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Questions

TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

TuitionGoWhere Exam Practice (AI)

Subject: Mathematics
Level: Primary 6
Paper: SA2 Practice Paper (Version 5 of 5)
Topic Focus: Whole Numbers
Duration: 1 hour 15 minutes
Total Marks: 40

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates:

This paper consists of 20 questions.
Answer all questions.
Write your answers in the spaces provided.
For questions requiring working, show your working clearly. Marks may be awarded for method even if the final answer is incorrect.
Unless otherwise stated, give your answers in the simplest form or to the specified degree of accuracy.

Section A (10 marks)

Questions 1 to 10 carry 1 mark each.

1. Write the number six million, forty-five thousand, and eight in numerals.

Answer: __________________________

2. Round off 4,567,892 to the nearest hundred thousand.

Answer: __________________________

3. What is the value of the digit 7 in the number 2,704,159?

Answer: __________________________

4. Find the smallest 5-digit odd number that can be formed using the digits 0, 1, 3, 5, 9 without repetition.

Answer: __________________________

5. Express 360 as a product of its prime factors in index notation.

Answer: __________________________

6. Find the Greatest Common Divisor (GCD) of 24 and 36.

Answer: __________________________

7. Find the Least Common Multiple (LCM) of 8 and 12.

Answer: __________________________

8. Evaluate: $15 + 25 \times 4 - 10$ .

Answer: __________________________

9. A number is divisible by both 4 and 9. What is the smallest possible value of this number?

Answer: __________________________

10. If $A = 2^3 \times 3^2$ and $B = 2^2 \times 3^3$ , find the GCD of A and B.

Answer: __________________________

Section B (20 marks)

Questions 11 to 15 carry 2 marks each. Questions 16 to 18 carry 3 marks each. Question 19 carries 4 marks. Question 20 carries 5 marks.

11. Mr. Tan has $4,500**. He buys a laptop for **$ 1,299 and a printer for $345. How much money does he have left?

Answer: $__________________________

12. There are 1,200 students in a school. $\frac{3}{5}$ of them are boys. $\frac{1}{4}$ of the boys wear spectacles. How many boys wear spectacles?

Answer: __________________________

13. Find the sum of all the common factors of 18 and 24.

Answer: __________________________

14. A factory produces 2,400 bottles of juice every day. The bottles are packed into cartons of 12. How many cartons are produced in 5 days?

Answer: __________________________

15. The product of two numbers is 360. Their GCD is 6. What is their LCM?

Answer: __________________________

16. Three bells ring at intervals of 6 minutes, 8 minutes, and 12 minutes respectively. If they ring together at 9:00 a.m., at what time will they next ring together?

Answer: __________________________

17. A number is between 4,000 and 5,000. It is divisible by 5 and 9. The sum of its digits is 18. What is the number?

Answer: __________________________

18. Mrs. Lim wants to cut a ribbon of length 180 cm and another ribbon of length 240 cm into pieces of equal length without any remainder. What is the greatest possible length of each piece? How many pieces will she have in total?

Answer: Length: __________________________ cm
Total pieces: __________________________

19. Study the pattern below.

Figure 1: 1 dot
Figure 2: 4 dots
Figure 3: 9 dots
Figure 4: 16 dots

(a) How many dots are there in Figure 10?
(b) Which figure has 225 dots?

Answer:
(a) __________________________
(b) Figure __________________________

20. A cinema has 800 seats. On Saturday, $\frac{3}{4}$ of the seats were occupied. Of the occupied seats, $\frac{2}{5}$ were taken by adults. The rest were taken by children. Each adult ticket costs $15** and each child ticket costs **$ 10.

(a) How many children watched the movie?
(b) What was the total amount of money collected from the tickets sold?

Answer:
(a) __________________________
(b) $__________________________

End of Paper

Answers

Answer Key and Marking Scheme - Primary 6 Mathematics (Whole Numbers)

Paper: SA2 Practice Paper (Version 5)
Total Marks: 40

Section A (1 mark each)

1. 6,045,008
Teaching Note: Break down the number by place value.

Millions: 6
Hundred Thousands: 0
Ten Thousands: 4
Thousands: 5
Hundreds: 0
Tens: 0
Ones: 8
Combine: 6,045,008.
Common Mistake: Writing 6,450,008 or missing the zeros in the hundreds/tens places.

2. 4,600,000
Teaching Note: Look at the digit in the ten-thousands place (6). Since $6 \ge 5$ , round up the hundred-thousands digit (5 becomes 6). Replace all digits to the right with zeros.

3. 700,000
Teaching Note: The digit 7 is in the hundred-thousands place. Its value is $7 \times 100,000 = 700,000$ .

4. 10,359
Teaching Note:

Smallest 5-digit number: Start with the smallest non-zero digit (1) in the ten-thousands place.
Next smallest digits in ascending order: 0, 3, 5.
Must be odd: The last digit must be odd. The remaining digit is 9, which is odd.
Arrangement: 1, 0, 3, 5, 9.
Check: 10,359 is odd. If we put 9 earlier, e.g., 10,395, it is larger. If we put 5 at the end, 10,395 > 10,359. Wait, let's re-evaluate.
Digits: 0, 1, 3, 5, 9.
Ten-thousands: 1 (smallest non-zero).
Remaining: 0, 3, 5, 9.
To be smallest, arrange remaining in ascending order: 0, 3, 5, 9 -> 10,359. Is it odd? Yes (ends in 9).
Alternative: If we tried to end with 5 (next smallest odd), we would have 1, 0, 3, 9, 5 -> 10,395. This is larger than 10,359.
So, 10,359 is correct.

5. $2^3 \times 3^2 \times 5$
Teaching Note: Use a factor tree or repeated division.
$360 = 36 \times 10 = (6 \times 6) \times (2 \times 5) = (2 \times 3) \times (2 \times 3) \times 2 \times 5 = 2^3 \times 3^2 \times 5$ .

6. 12
Teaching Note:
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
Common factors: 1, 2, 3, 4, 6, 12.
Greatest is 12.

7. 24
Teaching Note:
Multiples of 8: 8, 16, 24, 32...
Multiples of 12: 12, 24, 36...
Least Common Multiple is 24.

8. 105
Teaching Note: Follow Order of Operations (BODMAS/PEMDAS).

Multiplication: $25 \times 4 = 100$ .
Addition/Subtraction (left to right): $15 + 100 - 10$ .
$115 - 10 = 105$ .

9. 36
Teaching Note: A number divisible by 4 and 9 must be divisible by their LCM. Since 4 and 9 are coprime, $LCM(4, 9) = 4 \times 9 = 36$ . The smallest positive multiple is 36.

10. $2^2 \times 3^2$ (or 36)
Teaching Note: GCD takes the lowest power of common prime factors.
$A = 2^3 \times 3^2$
$B = 2^2 \times 3^3$
Common bases: 2 and 3.
Lowest power of 2: $2^2$ .
Lowest power of 3: $3^2$ .
$GCD = 2^2 \times 3^2 = 4 \times 9 = 36$ .

Section B

**11. $2,856** (2 marks) *Working:* Total spent =$ 1,299 + 345 = 1,644 $. Remaining =$ 4,500 - 1,644 $.$ 4,500 - 1,600 = 2,900 $.$ 2,900 - 44 = 2,856 $. *Answer:*$ 2,856.

12. 180 boys (2 marks)
Working:
Number of boys = $\frac{3}{5} \times 1,200 = 3 \times 240 = 720$ .
Boys with spectacles = $\frac{1}{4} \times 720 = 180$ .
Answer: 180.

13. 15 (2 marks)
Working:
Factors of 18: 1, 2, 3, 6, 9, 18.
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
Common factors: 1, 2, 3, 6.
Sum = $1 + 2 + 3 + 6 = 12$ .
Correction in thought process: Wait, did I miss any?
18: 1, 2, 3, 6, 9, 18.
24: 1, 2, 3, 4, 6, 8, 12, 24.
Common: 1, 2, 3, 6.
Sum: $1+2+3+6 = 12$ .
Answer: 12.
(Self-Correction: The initial draft answer said 15, but calculation shows 12. 1+2+3+6=12. 9 is not a factor of 24. 4 is not a factor of 18. Correct sum is 12.)

14. 1,000 cartons (2 marks)
Working:
Bottles per day = 2,400.
Cartons per day = $2,400 \div 12 = 200$ .
In 5 days = $200 \times 5 = 1,000$ .
Alternative: Total bottles = $2,400 \times 5 = 12,000$ .
Total cartons = $12,000 \div 12 = 1,000$ .
Answer: 1,000.

15. 60 (2 marks)
Working:
Formula: $Product \ of \ two \ numbers = GCD \times LCM$ .
$360 = 6 \times LCM$ .
$LCM = 360 \div 6 = 60$ .
Answer: 60.

16. 9:24 a.m. (3 marks)
Working:
Find LCM of 6, 8, and 12.
Multiples of 12: 12, 24, 36...
24 is divisible by 6 ( $6 \times 4$ ) and 8 ( $8 \times 3$ ).
LCM = 24 minutes.
They ring together every 24 minutes.
Next time = 9:00 a.m. + 24 mins = 9:24 a.m.
Answer: 9:24 a.m.

17. 4,059 (3 marks)
Working:

Between 4,000 and 5,000: First digit is 4. Number is $4,abc$ .
Divisible by 5: Last digit ( $c$ ) is 0 or 5.
Divisible by 9: Sum of digits is divisible by 9.
Sum of digits is 18.
Case 1: Last digit is 0. Number is $4,ab0$ . Sum $4+a+b+0=18 \Rightarrow a+b=14$ .
Possible pairs for $(a,b)$ : (5,9), (6,8), (7,7), (8,6), (9,5).
Smallest number? We want the smallest number between 4000-5000. So we want smallest $a$ .
If $a=5, b=9 \Rightarrow 4,590$ .
If $a=6, b=8 \Rightarrow 4,680$ .
Wait, can $a$ be smaller? $a+b=14$ . Min $a$ is 5 (since max $b=9$ ). So 4,590 is the smallest ending in 0.
Case 2: Last digit is 5. Number is $4,ab5$ . Sum $4+a+b+5=18 \Rightarrow a+b=9$ .
We want the smallest number, so smallest $a$ .
Min $a=0 \Rightarrow b=9$ . Number: 4,095.
Check divisibility by 5: Ends in 5. Yes.
Check divisibility by 9: Sum 18. Yes.
Check range: 4,000 < 4,095 < 5,000. Yes.
Compare 4,095 and 4,590. 4,095 is smaller.
Is there a smaller one? $a=0, b=9$ is the minimum $a$ .
Let's check if 4,005 works? Sum $4+0+0+5=9$ (Not 18).
Let's check 4,015? Sum 10.
Let's check 4,095? Sum 18.
So 4,095 is a candidate.
Wait, did I miss any constraints? "Divisible by 5 and 9".
Is 4,095 divisible by 5? Yes.
Is 4,095 divisible by 9? $4+0+9+5=18$ . Yes.
Is it the smallest?
Let's check numbers starting with 4,0..
$4,005$ (Sum 9, No).
$4,015$ (Sum 10, No).
...
$4,095$ (Sum 18, Yes).
What about ending in 0?
$4,050$ (Sum 9, No).
$4,140$ (Sum 9, No).
$4,230$ (Sum 9, No).
$4,320$ (Sum 9, No).
$4,410$ (Sum 9, No).
$4,500$ (Sum 9, No).
$4,590$ (Sum 18, Yes).
Comparing 4,095 and 4,590. 4,095 is smaller.
Answer: 4,095.

18. Length: 60 cm; Total pieces: 7 (3 marks)
Working:
Greatest possible length = GCD of 180 and 240.
$180 = 60 \times 3$ .
$240 = 60 \times 4$ .
$GCD(180, 240) = 60$ .
Length of each piece = 60 cm.
Number of pieces from 180 cm ribbon = $180 \div 60 = 3$ .
Number of pieces from 240 cm ribbon = $240 \div 60 = 4$ .
Total pieces = $3 + 4 = 7$ .
Answer: 60 cm, 7 pieces.

19. (a) 100; (b) Figure 15 (4 marks)
Working:
Pattern: Figure $n$ has $n^2$ dots.
Fig 1: $1^2=1$ . Fig 2: $2^2=4$ . Fig 3: $3^2=9$ .
(a) Figure 10: $10^2 = 100$ dots.
(b) $n^2 = 225$ .
$n = \sqrt{225} = 15$ .
Answer: (a) 100, (b) 15.

**20. (a) 360 children; (b) $6,300** (5 marks) *Working:* Total seats = 800. Occupied seats =$ \frac{3}{4} \times 800 = 600 $. Adults =$ \frac{2}{5} $of occupied seats. Children = Rest of occupied seats =$ 1 - \frac{2}{5} = \frac{3}{5} $of occupied seats. (a) Number of children =$ \frac{3}{5} \times 600 = 3 \times 120 = 360 $. Number of adults =$ 600 - 360 = 240 $(or$ \frac{2}{5} \times 600 = 240$).

(b) Cost from adults = $240 \times 15$ .
$240 \times 10 = 2,400$ .
$240 \times 5 = 1,200$ .
Total adult cost = $3,600$ .

Cost from children = $360 \times 10 = 3,600$ .

Total money collected = $3,600 + 3,600 = 7,200$ .

Wait, let me re-calculate.
Adults: 240. Cost: $240 \times 15 = 3,600$ .
Children: 360. Cost: $360 \times 10 = 3,600$ .
Total: $3,600 + 3,600 = 7,200$ .

Let me double check the "Rest were taken by children" part.
"Of the occupied seats, 2/5 were taken by adults. The rest were taken by children."
Yes, 3/5 are children.
$3/5 \times 600 = 360$ . Correct.
$2/5 \times 600 = 240$ . Correct.
$240 \times 15 = 3,600$ .
$360 \times 10 = 3,600$ .
Sum = 7,200.

Answer:
(a) 360
(b) $7,200