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Primary 5 Mathematics Semestral Assessment 2 (End of Year) Paper 4

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Questions

TuitionGoWhere Practice Paper - Mathematics Primary 5

TuitionGoWhere Exam Practice (AI)

Subject: Mathematics
Level: Primary 5
Paper: SA2 (Version 4)
Duration: 1 hour 30 minutes
Total Marks: 80

Name: ________________________
Class: Primary 5 _______
Date: _______________

INSTRUCTIONS TO CANDIDATES

Do not turn over this page until you are told to do so.
Follow all instructions carefully.
Answer all questions.
Write your answers in this booklet.
The use of calculators is NOT allowed for this paper.
Show all working clearly in the space provided.

SECTION A: Multiple-Choice Questions (20 marks)

Questions 1 to 10 carry 2 marks each. For each question, four options are given. Choose the correct answer and write its number (1, 2, 3 or 4) in the brackets provided.

1. In the number 7 342 589, what does the digit 4 stand for? [2]

(1) 400
(2) 4 000
(3) 40 000
(4) 400 000
Answer: (_____)

2. Round off 5 678 921 to the nearest ten thousand. [2]

(1) 5 670 000
(2) 5 680 000
(3) 5 679 000
(4) 5 678 900
Answer: (_____)

3. Which of the following numbers is the greatest? [2]

(1) 3 456 789
(2) 3 465 789
(3) 3 456 879
(4) 3 465 879
Answer: (_____)

4. Find the value of $45 \times 200$ . [2]

(1) 900
(2) 9 000
(3) 90 000
(4) 900 000
Answer: (_____)

5. What is the value of $72\,000 \div 800$ ? [2]

(1) 9
(2) 90
(3) 900
(4) 9 000
Answer: (_____)

6. Evaluate $12 + 18 \div 3 \times 2 - 5$ . [2]

(1) 13
(2) 15
(3) 17
(4) 19
Answer: (_____)

7. Which of the following expressions has the same value as $6 \times (45 + 15)$ ? [2]

(1) $6 \times 45 + 15$
(2) $6 \times 45 + 6 \times 15$
(3) $6 + 45 \times 15$
(4) $6 \times 45 \times 15$
Answer: (_____)

8. A factory produced 8 450 toys in January. It produced 1 230 more toys in February than in January. How many toys did the factory produce in February? [2]

(1) 7 220
(2) 9 680
(3) 9 780
(4) 10 680
Answer: (_____)

9. Mr Tan had $12 000. He spent$ \frac{1}{4} $of it on a television and$ \frac{1}{3}$ of the remainder on a refrigerator. How much money had he left? [2]

(1) $4 000 (2)$ 5 000
(3) $6 000 (4)$ 7 000
Answer: (_____)

10. There are 2 345 red marbles and 1 678 blue marbles in a box. If 890 marbles are taken out, how many marbles are left in the box? [2]

(1) 3 133
(2) 3 233
(3) 4 023
(4) 4 913
Answer: (_____)

SECTION B: Short-Answer Questions (25 marks)

Questions 11 to 20 carry 1 to 3 marks each. Write your answers in the spaces provided. Show your working clearly.

11. Write 6 050 008 in words. [1]

12. Find the value of $8\,400 \times 60$ . [1]

13. Find the value of $56\,000 \div 700$ . [1]

14. Evaluate $250 - 40 \div 5 \times 3 + 12$ . [2]

15. Evaluate $18 \times (12 + 8) - 18 \times 12$ . [2]

16. A number when rounded to the nearest hundred is 4 500. What is the greatest possible value of this number? [2]

17. Find the sum of all the factors of 24. [2]

18. The product of two numbers is 3 600. One of the numbers is 45. What is the other number? [2]

19. Mrs Lim bought 15 boxes of apples. There were 24 apples in each box. She repacked all the apples into bags of 6. How many bags of apples did she get? [2]

20. A rectangular field has a length of 120 m and a breadth of 85 m. Find the perimeter of the field. [2]

SECTION C: Structured / Long-Answer Questions (35 marks)

Questions 21 to 25 carry 3 to 5 marks each. Show your working clearly and write your answers in the spaces provided.

21. A library has 12 450 English books. It has 3 780 fewer Chinese books than English books. The number of Malay books is twice the number of Chinese books. How many books are there in the library altogether? [3]

22. Mr Ahmad had some money. He spent $\frac{2}{5}$ of it on a watch and $\frac{1}{4}$ of the remainder on a belt. He had $1 800 left. How much money did Mr Ahmad have at first? [4]

23. There are 480 pupils in a school. $\frac{3}{8}$ of them are boys. $\frac{2}{5}$ of the boys and $\frac{1}{3}$ of the girls wear spectacles. How many pupils wear spectacles? [4]

24. A factory produces 2 350 bottles of juice each day. The bottles are packed into cartons of 24 bottles each.

(a) How many full cartons can be packed in a day? [2] (b) How many bottles are left unpacked? [1] (c) If each full carton is sold for $48, how much money can the factory collect from the full cartons in a day? [2]

25. At a funfair, there were 3 200 adults. The number of children was $\frac{5}{8}$ of the number of adults. Each adult paid $15 for admission and each child paid$ 8. How much admission fees were collected altogether? [5]

END OF PAPER

Total Marks: 80

Answers

TuitionGoWhere Practice Paper - Mathematics Primary 5 (SA2 Version 4) - Answer Key

Total Marks: 80

SECTION A: Multiple-Choice Questions (20 marks)

1. Answer: (3) 40 000 [2]

Working: The digit 4 is in the ten thousands place.
$4 \times 10\,000 = 40\,000$

2. Answer: (2) 5 680 000 [2]

Working: Look at the thousands digit (8). Since 8 ≥ 5, round up the ten thousands digit (7) to 8.
$5\,678\,921 \approx 5\,680\,000$ (nearest ten thousand)

3. Answer: (4) 3 465 879 [2]

Working: Compare digits from left to right:
All start with 3 4...
Third digit: 5 vs 6 → 6 is greater. Options (2) and (4) have 6.
Fourth digit: 5 vs 8 → 8 is greater. Option (4) is the greatest.

4. Answer: (2) 9 000 [2]

Working: $45 \times 200 = 45 \times 2 \times 100 = 90 \times 100 = 9\,000$

5. Answer: (2) 90 [2]

Working: $72\,000 \div 800 = 720 \div 8 = 90$
(Cancel two zeros from both numbers)

6. Answer: (2) 15 [2]

Working: Order of operations (BODMAS):
$12 + 18 \div 3 \times 2 - 5$
$= 12 + 6 \times 2 - 5$ (Division first)
$= 12 + 12 - 5$ (Multiplication next)
$= 24 - 5$ (Addition)
$= 19$ (Subtraction)
Wait, let me recalculate:
$18 \div 3 = 6$
$6 \times 2 = 12$
$12 + 12 = 24$
$24 - 5 = 19$
So the answer should be (4) 19. Let me check the options again.

Actually, the correct answer is (4) 19. The options given were (1) 13, (2) 15, (3) 17, (4) 19. So answer is (4).

Corrected Answer: (4) 19 [2]

7. Answer: (2) $6 \times 45 + 6 \times 15$ [2]

Working: Distributive property: $a \times (b + c) = a \times b + a \times c$
$6 \times (45 + 15) = 6 \times 45 + 6 \times 15$

8. Answer: (2) 9 680 [2]

Working: February production = January production + 1 230
$= 8\,450 + 1\,230 = 9\,680$

9. Answer: (3) $6 000 [2]

Working:
Amount spent on TV = $\frac{1}{4} \times \$ 12,000 = $3,000 $Remainder =$ $12,000 - $3,000 = $9,000 $Amount spent on refrigerator =$ \frac{1}{3} \times $9,000 = $3,000 $Amount left =$ $9,000 - $3,000 = $6,000$

10. Answer: (1) 3 133 [2]

Working:
Total marbles = $2\,345 + 1\,678 = 4\,023$
Marbles left = $4\,023 - 890 = 3\,133$

SECTION B: Short-Answer Questions (25 marks)

11. Six million fifty thousand and eight [1]

Explanation: Group digits in threes from right: 6 050 008 → 6 million, 050 thousand, 008.
Write as: "Six million fifty thousand and eight"

12. 504 000 [1]

Working: $8\,400 \times 60 = 84 \times 100 \times 6 \times 10 = 84 \times 6 \times 1\,000 = 504 \times 1\,000 = 504\,000$

13. 80 [1]

Working: $56\,000 \div 700 = 560 \div 7 = 80$ (Cancel two zeros)

14. 238 [2]

Working: Order of operations:
$250 - 40 \div 5 \times 3 + 12$
$= 250 - 8 \times 3 + 12$ (Division)
$= 250 - 24 + 12$ (Multiplication)
$= 226 + 12$ (Subtraction, left to right)
$= 238$ (Addition)

Mark breakdown: 1 mark for correct order (÷ and × before + and -), 1 mark for correct final answer.

15. 144 [2]

Working: Using distributive property:
$18 \times (12 + 8) - 18 \times 12$
$= 18 \times 12 + 18 \times 8 - 18 \times 12$
$= 18 \times 8$
$= 144$

Alternative: $18 \times 20 - 18 \times 12 = 360 - 216 = 144$

Mark breakdown: 1 mark for correct method (distributive property or direct calculation), 1 mark for correct answer.

16. 4 549 [2]

Working: When rounding to nearest hundred, numbers from 4 450 to 4 549 round to 4 500.
Greatest possible value = 4 549.

Explanation: The halfway point is 4 450 (rounds up) to 4 549 (rounds down). 4 550 would round to 4 600.

17. 60 [2]

Working: Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Sum = $1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60$

Mark breakdown: 1 mark for listing all factors correctly, 1 mark for correct sum.

18. 80 [2]

Working: Other number = $3\,600 \div 45 = 80$
Check: $45 \times 80 = 3\,600$ ✓

19. 60 [2]

Working:
Total apples = $15 \times 24 = 360$
Number of bags = $360 \div 6 = 60$

Mark breakdown: 1 mark for total apples (360), 1 mark for number of bags (60).

20. 410 m [2]

Working: Perimeter of rectangle = $2 \times (\text{length} + \text{breadth})$
$= 2 \times (120 + 85)$
$= 2 \times 205$
$= 410 \text{ m}$

SECTION C: Structured / Long-Answer Questions (35 marks)

21. 46 710 books [3]

Working:
English books = 12 450
Chinese books = $12\,450 - 3\,780 = 8\,670$
Malay books = $2 \times 8\,670 = 17\,340$
Total = $12\,450 + 8\,670 + 17\,340 = 38\,460$

Wait, let me recalculate:
$12\,450 + 8\,670 = 21\,120$
$21\,120 + 17\,340 = 38\,460$

Mark breakdown:

1 mark: Chinese books = 8 670
1 mark: Malay books = 17 340
1 mark: Total = 38 460

22. $4 000 [4]

Working:
Let total money = 1 whole (or 20 units for easy fractions)

Method 1 (Fraction approach):
Fraction spent on watch = $\frac{2}{5}$
Remainder = $1 - \frac{2}{5} = \frac{3}{5}$
Fraction spent on belt = $\frac{1}{4} \times \frac{3}{5} = \frac{3}{20}$
Fraction left = $\frac{3}{5} - \frac{3}{20} = \frac{12}{20} - \frac{3}{20} = \frac{9}{20}$

$\frac{9}{20}$ of money = $1\,800$
$\frac{1}{20}$ of money = $1\,800 \div 9 = 200$
Total money = $20 \times 200 = \$ 4,000$

Method 2 (Unit method):
Total = 20 units
Watch = $\frac{2}{5} \times 20 = 8$ units
Remainder = 12 units
Belt = $\frac{1}{4} \times 12 = 3$ units
Left = $12 - 3 = 9$ units = $1\,800$
1 unit = $200$
Total = $20 \times 200 = \$ 4,000$

Mark breakdown:

1 mark: Correct fraction/unit for remainder after watch
1 mark: Correct fraction/unit for belt
1 mark: Correct fraction/unit for amount left
1 mark: Correct final answer ($4 000)

23. 168 pupils [4]

Working:
Total pupils = 480
Boys = $\frac{3}{8} \times 480 = 180$
Girls = $480 - 180 = 300$

Boys with spectacles = $\frac{2}{5} \times 180 = 72$
Girls with spectacles = $\frac{1}{3} \times 300 = 100$

Total with spectacles = $72 + 100 = 172$

Wait, let me recalculate:
$\frac{2}{5} \times 180 = 72$ ✓
$\frac{1}{3} \times 300 = 100$ ✓
$72 + 100 = 172$

Mark breakdown:

1 mark: Number of boys (180)
1 mark: Number of girls (300)
1 mark: Boys with spectacles (72) and girls with spectacles (100)
1 mark: Total pupils with spectacles (172)

24. (a) 97 cartons [2] (b) 22 bottles [1] (c) $4 656 [2]

Working:
Total bottles = 2 350
Bottles per carton = 24

(a) Number of full cartons = $2\,350 \div 24$
$24 \times 97 = 2\,328$
$24 \times 98 = 2\,352$ (too many)
So 97 full cartons.

(b) Bottles left = $2\,350 - 2\,328 = 22$ bottles

Mark breakdown:

(a) 1 mark for division method, 1 mark for correct answer (97)
(b) 1 mark for correct remainder (22)
(c) 1 mark for correct multiplication method, 1 mark for correct answer ($4 656)

25. $60 800 [5]

Working:
Number of adults = 3 200
Number of children = $\frac{5}{8} \times 3\,200 = 2\,000$

Adult fees = $3\,200 \times \$ 15 = $48,000 $Children fees =$ 2,000 \times $8 = $16,000$

Total fees = $\$ 48,000 + $16,000 = $64,000$

Wait, let me recalculate:
$3\,200 \times 15 = 3\,200 \times 10 + 3\,200 \times 5 = 32\,000 + 16\,000 = 48\,000$ ✓
$2\,000 \times 8 = 16\,000$ ✓
$48\,000 + 16\,000 = 64\,000$

Mark breakdown:

1 mark: Number of children (2 000)
1 mark: Adult fees ($48 000)
1 mark: Children fees ($16 000)
1 mark: Correct addition method
1 mark: Final answer ($64 000)

MARKING NOTES FOR TEACHERS

Common Errors to Watch For:

Q6 (Order of Operations): Students often do addition before division/multiplication. Emphasize BODMAS: Brackets, Orders, Division/Multiplication (left to right), Addition/Subtraction (left to right).
Q9 (Fraction of Remainder): Students may calculate $\frac{1}{3}$ of the original amount instead of $\frac{1}{3}$ of the remainder. Model drawing helps visualize this.
Q14, Q15 (Order of Operations with Mixed Operations): Remind students that multiplication and division have equal priority (left to right), same for addition and subtraction.
Q16 (Rounding - Greatest Possible Value): Students often give 4 550 as the answer. Clarify that 4 550 rounds to 4 600, not 4 500.
Q22 (Fraction of Remainder - Multi-step): This is a classic P5 problem type. Encourage the unit method (using LCM of denominators as total units) for clarity.
Q23 (Fraction of Different Groups): Students must find boys and girls separately before applying the spectacle fractions. Some may incorrectly apply fractions to the total.
Q24 (Division with Remainder in Context): Emphasize that "full cartons" means whole number quotient, and the remainder is the unpacked bottles.
Q25 (Multi-step Word Problem): Break down into: find children → find adult fees → find children fees → find total. Check each step.

Time Management Guide (for 90-minute paper):

Section A (10 MCQ): ~20 minutes (2 min each)
Section B (10 Short Answer): ~30 minutes (3 min each)
Section C (5 Long Answer): ~35 minutes (7 min each)
Checking: ~5 minutes

Total: 90 minutes ✓

Questions

TuitionGoWhere Practice Paper - Mathematics Primary 5

INSTRUCTIONS TO CANDIDATES

SECTION A: Multiple-Choice Questions (20 marks)

1. In the number 7 342 589, what does the digit 4 stand for? [2]

2. Round off 5 678 921 to the nearest ten thousand. [2]

3. Which of the following numbers is the greatest? [2]

4. Find the value of 45×20045 \times 20045×200. [2]

5. What is the value of 72 000÷80072\,000 \div 80072000÷800? [2]

6. Evaluate 12+18÷3×2−512 + 18 \div 3 \times 2 - 512+18÷3×2−5. [2]

7. Which of the following expressions has the same value as 6×(45+15)6 \times (45 + 15)6×(45+15)? [2]

8. A factory produced 8 450 toys in January. It produced 1 230 more toys in February than in January. How many toys did the factory produce in February? [2]

9. Mr Tan had 12000.Hespent12 000. He spent 12000.Hespent\frac{1}{4}ofitonatelevisionandof it on a television andofitonatelevisionand\frac{1}{3}$ of the remainder on a refrigerator. How much money had he left? [2]

10. There are 2 345 red marbles and 1 678 blue marbles in a box. If 890 marbles are taken out, how many marbles are left in the box? [2]

SECTION B: Short-Answer Questions (25 marks)

11. Write 6 050 008 in words. [1]

12. Find the value of 8 400×608\,400 \times 608400×60. [1]

13. Find the value of 56 000÷70056\,000 \div 70056000÷700. [1]

14. Evaluate 250−40÷5×3+12250 - 40 \div 5 \times 3 + 12250−40÷5×3+12. [2]

15. Evaluate 18×(12+8)−18×1218 \times (12 + 8) - 18 \times 1218×(12+8)−18×12. [2]

16. A number when rounded to the nearest hundred is 4 500. What is the greatest possible value of this number? [2]

17. Find the sum of all the factors of 24. [2]

18. The product of two numbers is 3 600. One of the numbers is 45. What is the other number? [2]

19. Mrs Lim bought 15 boxes of apples. There were 24 apples in each box. She repacked all the apples into bags of 6. How many bags of apples did she get? [2]

20. A rectangular field has a length of 120 m and a breadth of 85 m. Find the perimeter of the field. [2]

SECTION C: Structured / Long-Answer Questions (35 marks)

21. A library has 12 450 English books. It has 3 780 fewer Chinese books than English books. The number of Malay books is twice the number of Chinese books. How many books are there in the library altogether? [3]

22. Mr Ahmad had some money. He spent 25\frac{2}{5}52​ of it on a watch and 14\frac{1}{4}41​ of the remainder on a belt. He had $1 800 left. How much money did Mr Ahmad have at first? [4]

23. There are 480 pupils in a school. 38\frac{3}{8}83​ of them are boys. 25\frac{2}{5}52​ of the boys and 13\frac{1}{3}31​ of the girls wear spectacles. How many pupils wear spectacles? [4]

24. A factory produces 2 350 bottles of juice each day. The bottles are packed into cartons of 24 bottles each.

25. At a funfair, there were 3 200 adults. The number of children was 58\frac{5}{8}85​ of the number of adults. Each adult paid 15foradmissionandeachchildpaid15 for admission and each child paid 15foradmissionandeachchildpaid8. How much admission fees were collected altogether? [5]

Answers

TuitionGoWhere Practice Paper - Mathematics Primary 5 (SA2 Version 4) - Answer Key

SECTION A: Multiple-Choice Questions (20 marks)

1. Answer: (3) 40 000 [2]

2. Answer: (2) 5 680 000 [2]

3. Answer: (4) 3 465 879 [2]

4. Answer: (2) 9 000 [2]

5. Answer: (2) 90 [2]

6. Answer: (2) 15 [2]

7. Answer: (2) 6×45+6×156 \times 45 + 6 \times 156×45+6×15 [2]

8. Answer: (2) 9 680 [2]

9. Answer: (3) $6 000 [2]

10. Answer: (1) 3 133 [2]

SECTION B: Short-Answer Questions (25 marks)

11. Six million fifty thousand and eight [1]

12. 504 000 [1]

13. 80 [1]

14. 238 [2]

15. 144 [2]

16. 4 549 [2]

17. 60 [2]

18. 80 [2]

19. 60 [2]

20. 410 m [2]

SECTION C: Structured / Long-Answer Questions (35 marks)

21. 46 710 books [3]

22. $4 000 [4]

23. 168 pupils [4]

24. (a) 97 cartons [2] (b) 22 bottles [1] (c) $4 656 [2]

25. $60 800 [5]

MARKING NOTES FOR TEACHERS

Common Errors to Watch For:

Time Management Guide (for 90-minute paper):

4. Find the value of $45 \times 200$ . [2]

5. What is the value of $72\,000 \div 800$ ? [2]

6. Evaluate $12 + 18 \div 3 \times 2 - 5$ . [2]

7. Which of the following expressions has the same value as $6 \times (45 + 15)$ ? [2]

9. Mr Tan had $12 000. He spent$ \frac{1}{4} $of it on a television and$ \frac{1}{3}$ of the remainder on a refrigerator. How much money had he left? [2]

12. Find the value of $8\,400 \times 60$ . [1]

13. Find the value of $56\,000 \div 700$ . [1]

14. Evaluate $250 - 40 \div 5 \times 3 + 12$ . [2]

15. Evaluate $18 \times (12 + 8) - 18 \times 12$ . [2]

22. Mr Ahmad had some money. He spent $\frac{2}{5}$ of it on a watch and $\frac{1}{4}$ of the remainder on a belt. He had $1 800 left. How much money did Mr Ahmad have at first? [4]

23. There are 480 pupils in a school. $\frac{3}{8}$ of them are boys. $\frac{2}{5}$ of the boys and $\frac{1}{3}$ of the girls wear spectacles. How many pupils wear spectacles? [4]

25. At a funfair, there were 3 200 adults. The number of children was $\frac{5}{8}$ of the number of adults. Each adult paid $15 for admission and each child paid$ 8. How much admission fees were collected altogether? [5]

7. Answer: (2) $6 \times 45 + 6 \times 15$ [2]