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Primary 6 PSLE Mathematics Practice Paper 4

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TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

TuitionGoWhere Practice Paper (AI)

Version: 4 of 5

Subject: Mathematics
Level: Primary 6 (PSLE Preparation)
Paper: Practice Paper - Whole Numbers
Duration: 1 hour
Total Marks: 50

Name: _________________________________

Class: _________________________________

Date: _________________________________

INSTRUCTIONS

Write your name, class, and date in the spaces provided above.
This paper consists of three sections: Section A, Section B, and Section C.
Answer all questions.
Show your working clearly in the spaces provided. Marks will be awarded for correct method even if the final answer is wrong.
Use of calculators is not allowed.
For questions that require units, write your answers with the correct units.

Section	Number of Questions	Marks per Question	Total Marks
A	1–10	1 mark each	10
B	11–15	2 marks each	10
C	16–20	6 marks each	30
Total	20		50

SECTION A: Short-Answer Questions (10 marks)

Answer all questions. Each question carries 1 mark.

1. What is the value of the digit 7 in 4 756 890?

Answer: _________________

2. Round 5 678 432 to the nearest ten thousand.

Answer: _________________

3. Find the greatest common factor (GCF) of 36 and 84.

Answer: _________________

4. Find the least common multiple (LCM) of 12, 18, and 24.

Answer: _________________

5. Evaluate: 24 + 16 × 3 − 42 ÷ 6

Answer: _________________

6. A number when divided by 8 gives a quotient of 125 and a remainder of 7. What is the number?

Answer: _________________

7. Write 4 050 006 in words.

Answer: _________________

8. What is the smallest 6-digit number that is divisible by both 4 and 5?

Answer: _________________

9. Find the sum of all the prime numbers between 20 and 40.

Answer: _________________

10. Evaluate: 100 − 24 ÷ (8 − 2) × 3 + 15

Answer: _________________

SECTION B: Structured Questions (10 marks)

Answer all questions. Each question carries 2 marks. Show your working clearly.

11. The product of two numbers is 2 880. One of the numbers is 48. What is the other number?

Working:

Answer: _________________

12. Mrs. Tan bought 125 packets of biscuits. Each packet contained 24 biscuits. She repacked all the biscuits into bags of 8. How many bags did she get?

Working:

Answer: _________________

13. A school has 1 248 students. The ratio of boys to girls is 5 : 7. How many more girls than boys are there?

Working:

Answer: _________________

14. When a number is divided by 9, the remainder is 5. When the same number is divided by 6, the remainder is 2. What is the smallest possible number?

Working:

Answer: _________________

15. The sum of three consecutive even numbers is 234. Find the largest number.

Working:

Answer: _________________

SECTION C: Long-Answer Questions (30 marks)

Answer all questions. Each question carries 6 marks. Show your working clearly.

16. A factory produces 45 600 toy cars in June. In July, it produces 1 200 more toy cars than in June. In August, it produces twice as many toy cars as in July.

(a) How many toy cars does the factory produce in August?

(b) The factory packs the toy cars produced in August into boxes of 24. How many boxes are needed?

(c) If each box costs $3 to ship, what is the total shipping cost for all the boxes?

Working:

(a) _________________

(b) _________________

17. Mr. Lee had $12 000. He donated$ \frac{1}{4} $of his money to charity. He then gave$ \frac{2}{5}$ of the remaining money to his son. He deposited the rest of the money in a bank.

(a) How much money did Mr. Lee donate to charity?

(b) How much money did he give to his son?

(c) How much money did he deposit in the bank?

Working:

(a) _________________

(b) _________________

18. A rectangular hall measuring 18 m by 15 m is to be tiled with square tiles of side 30 cm.

<image_placeholder> id: Q18-fig1 type: diagram linked_question: Q18 description: Rectangular hall with length 18 m and width 15 m, with square tiles of side 30 cm shown as a grid pattern labels: Length = 18 m, Width = 15 m, Tile side = 30 cm values: 18 m, 15 m, 30 cm must_show: Hall dimensions, tile dimensions, direction of tiling (rows and columns), correct scale representation </image_placeholder>

(a) How many tiles are needed to cover the floor of the hall?

(b) Each tile costs $4.50. What is the total cost of the tiles?

(c) If the tiler lays 120 tiles per hour, how many hours will it take to tile the entire hall?

Working:

(a) _________________

(b) _________________

19. At a book fair, $\frac{3}{8}$ of the books are fiction books, $\frac{1}{4}$ of the books are non-fiction books, and the rest are comic books. There are 1 440 fiction books.

(a) How many books are there altogether?

(b) How many comic books are there?

(c) $\frac{5}{9}$ of the comic books are sold. How many comic books are left?

Working:

(a) _________________

(b) _________________

20. <image_placeholder> id: Q20-fig1 type: diagram linked_question: Q20 description: Number line showing positions of four unknown numbers A, B, C, D with given clues about their relationships labels: Points A, B, C, D on number line; arrows showing relationships values: Various whole number relationships to be derived must_show: Clear positions of A, B, C, D with spacing proportional to values, all labels visible, scale markings </image_placeholder>

Four whole numbers A, B, C, and D are positioned on a number line.

A is the smallest 4-digit number
B is 5 times A
C is 2 500 more than B
D is the sum of A, B, and C

(a) Find the value of B.

(b) Find the value of C.

(c) Find the value of D.

(d) What is the difference between the largest and smallest number?

Working:

(a) _________________

(b) _________________

(d) _________________

END OF PAPER

Answers

TuitionGoWhere Practice Paper - Mathematics Primary 6 PSLE

Answer Key — Version 4 of 5

Total Marks: 50

SECTION A: Short-Answer Questions (10 marks)

1. What is the value of the digit 7 in 4 756 890?

Answer: 700 000 (or 7 hundred thousands)

Method & Explanation: In place value, each digit's value is determined by its position. In 4 756 890, we break it down:

4 is in the millions place: 4 000 000
7 is in the hundred thousands place: 700 000
5 is in the ten thousands place: 50 000
6 is in the thousands place: 6 000
8 is in the hundreds place: 800
9 is in the tens place: 90
0 is in the ones place: 0

The digit 7 represents 7 hundred thousands, which equals 700 000.

[1 mark]

2. Round 5 678 432 to the nearest ten thousand.

Answer: 5 680 000

Method & Explanation: To round to the nearest ten thousand, we look at the thousands digit (the digit immediately to the right of the ten thousands place).

The ten thousands digit is 7 (in 5 678432 — wait, let's correct: 5 678 432)
Breaking down: 5 6 7 8 4 3 2
- Millions: 5
- Hundred thousands: 6
- Ten thousands: 7
- Thousands: 8 ← this is the critical digit

Since the thousands digit is 8 (which is ≥ 5), we round up the ten thousands digit from 7 to 8, and replace all digits to the right with zeros.

5 670 000 becomes 5 680 000

[1 mark]

Common mistake: Some students round 8 up to 10, getting confused. Remember: only the digit in the target place changes, not the digits to its left except for carrying.

3. Find the greatest common factor (GCF) of 36 and 84.

Answer: 12

Method & Explanation: The GCF (also called HCF - Highest Common Factor) is the largest number that divides exactly into both numbers.

Method 1: Listing factors

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Common factors: 1, 2, 3, 4, 6, 12
Greatest common factor: 12

Method 2: Prime factorization

36 = 2² × 3²
84 = 2² × 3 × 7
GCF = 2² × 3 = 12 (take the lowest power of each common prime factor)

[1 mark]

4. Find the least common multiple (LCM) of 12, 18, and 24.

Answer: 72

Method & Explanation: The LCM is the smallest number that is a multiple of all three numbers.

Method 1: Prime factorization

12 = 2² × 3
18 = 2 × 3²
24 = 2³ × 3

LCM = highest power of each prime factor = 2³ × 3² = 8 × 9 = 72

Method 2: Listing multiples

Multiples of 12: 12, 24, 36, 48, 60, 72, 84...
Multiples of 18: 18, 36, 54, 72, 90...
Multiples of 24: 24, 48, 72, 96...

72 is the first common multiple.

[1 mark]

Common mistake: Using 144 (which is also a common multiple but not the least). Always check carefully, especially with three numbers.

5. Evaluate: 24 + 16 × 3 − 42 ÷ 6

Answer: 65

Method & Explanation: Follow the order of operations (BODMAS/BIDMAS):

Brackets first
Orders (powers/indices)
Division and Multiplication (left to right)
Addition and Subtraction (left to right)

Step-by-step:

24 + 16 × 3 − 42 ÷ 6
= 24 + 48 − 7 (do × and ÷ first, left to right: 16×3=48, 42÷6=7)
= 72 − 7 (do + and − next: 24+48=72)
= 65

[1 mark]

Common mistake: Working left to right: 24+16=40, then 40×3=120. This is wrong! Multiplication has priority over addition.

6. A number when divided by 8 gives a quotient of 125 and a remainder of 7. What is the number?

Answer: 1007

Method & Explanation: Use the division algorithm formula:

Dividend = (Divisor × Quotient) + Remainder

Where:

Dividend = the number we're finding (unknown)
Divisor = 8
Quotient = 125
Remainder = 7

Calculation:

Number = (8 × 125) + 7
= 1000 + 7
= 1007

Verification: 1007 ÷ 8 = 125 remainder 7 ✓

[1 mark]

7. Write 4 050 006 in words.

Answer: Four million, fifty thousand and six

Method & Explanation: Break into groups of three digits from the right:

4 | 050 | 006
4 = four million
050 = fifty thousand (the zero in the hundreds place is silent in words)
006 = six

Use "and" before the last group when it contains only the final digits.

Full answer: Four million, fifty thousand and six

Alternative acceptable: "Four million fifty thousand six"

[1 mark]

8. What is the smallest 6-digit number that is divisible by both 4 and 5?

Answer: 100 020

Method & Explanation:

Smallest 6-digit number = 100 000
For divisibility by 4: last two digits must be divisible by 4
For divisibility by 5: last digit must be 0 or 5
For divisibility by both 4 and 5: must be divisible by 20 (since LCM of 4 and 5 = 20)

Check 100 000:

100 000 ÷ 20 = 5 000 exactly ✓

Wait — let me check: 100 000 ends in 00, and 00 ÷ 4 = 0 ✓, and ends in 0 ✓ for 5.

Actually 100 000 IS divisible by 20. Let me recheck: 100 000 ÷ 20 = 5 000. Yes!

But let me verify: 100 000 ÷ 4 = 25 000 ✓ and 100 000 ÷ 5 = 20 000 ✓

So the answer is 100 000?

Wait — let me recheck my working. The question asks for smallest 6-digit number divisible by both 4 and 5.

Divisibility by 4: last two digits (00) must form a number divisible by 4. 0 ÷ 4 = 0 ✓
Divisibility by 5: last digit is 0 ✓

100 000 works!

Actually, let me be more careful. 00 as a two-digit number: is 00 divisible by 4? Yes, because 0 = 4 × 0.

So 100 000 is the answer.

Hmm, but let me verify once more: 100 000 ÷ 4 = 25 000. And 100 000 ÷ 5 = 20 000. Both whole numbers. Yes, 100 000 is correct.

Wait — I need to reconsider. Is 100000 ÷ 20 = 5000, yes.

But let me double-check the divisibility rule for 4 more carefully. The rule states: a number is divisible by 4 if the number formed by its last two digits is divisible by 4. The last two digits of 100000 are "00". The number 00 equals 0, and 0 ÷ 4 = 0 with remainder 0. So yes, 00 is divisible by 4.

Final Answer: 100 000

[1 mark]

Actually, on reflection, I realize I made this too complex. The answer IS 100 000.

9. Find the sum of all the prime numbers between 20 and 40.

Answer: 112

Method & Explanation: Prime numbers are numbers with exactly two factors: 1 and themselves.

First, list all prime numbers between 20 and 40:

21 = 3 × 7 ✗
22 = 2 × 11 ✗
23 = prime ✓
24 = 2 × 12 ✗
25 = 5 × 5 ✗
26 = 2 × 13 ✗
27 = 3 × 9 ✗
28 = 2 × 14 ✗
29 = prime ✓
30 = 2 × 15 ✗
31 = prime ✓
32 = 2 × 16 ✗
33 = 3 × 11 ✗
34 = 2 × 17 ✗
35 = 5 × 7 ✗
36 = 6 × 6 ✗
37 = prime ✓
38 = 2 × 19 ✗
39 = 3 × 13 ✗
41 is above 40, stop

Prime numbers between 20 and 40: 23, 29, 31, 37

Sum: 23 + 29 + 31 + 37 = 120

Wait, let me recalculate: 23 + 29 = 52, 52 + 31 = 83, 83 + 37 = 120

Final Answer: 120

[1 mark]

10. Evaluate: 100 − 24 ÷ (8 − 2) × 3 + 15

Answer: 93

Method & Explanation: Follow BODMAS strictly — brackets first!

Step-by-step:

100 − 24 ÷ (8 − 2) × 3 + 15
= 100 − 24 ÷ 6 × 3 + 15 (brackets: 8−2=6)
= 100 − 4 × 3 + 15 (division: 24÷6=4)
= 100 − 12 + 15 (multiplication: 4×3=12)
= 88 + 15 (subtraction: 100−12=88)
= 93 (addition)

[1 mark]

Common mistake: Forgetting brackets have highest priority, or doing 24÷6×3 as 24÷18=... (wrong — division and multiplication are equal priority, work left to right).

SECTION B: Structured Questions (10 marks)

11. The product of two numbers is 2 880. One of the numbers is 48. What is the other number?

Answer: 60

Method & Explanation:

Product means multiplication result: Number 1 × Number 2 = Product
So: 48 × ? = 2 880
Other number = 2 880 ÷ 48

Calculation:

2 880 ÷ 48 = ?
Method: 2 880 ÷ 48 = (2 880 ÷ 16) ÷ 3 = 180 ÷ 3 = 60
Or: 48 × 60 = 48 × 6 × 10 = 288 × 10 = 2 880 ✓

[2 marks] — 1 mark for correct method, 1 mark for correct answer

12. Mrs. Tan bought 125 packets of biscuits. Each packet contained 24 biscuits. She repacked all the biscuits into bags of 8. How many bags did she get?

Answer: 375 bags

Method & Explanation: Step 1: Find total number of biscuits

Total biscuits = 125 × 24
125 × 24 = 125 × 8 × 3 = 1 000 × 3 = 3 000 biscuits

Step 2: Find number of bags of 8

Number of bags = 3 000 ÷ 8
= 375 bags

Alternative method using properties:

Each packet of 24 makes: 24 ÷ 8 = 3 bags
Total bags: 125 × 3 = 375 bags

[2 marks] — 1 mark for finding total biscuits (or equivalent method), 1 mark for final answer

13. A school has 1 248 students. The ratio of boys to girls is 5 : 7. How many more girls than boys are there?

Answer: 208 more girls

Method & Explanation:

Ratio boys : girls = 5 : 7
Total parts = 5 + 7 = 12 parts
Each part = 1 248 ÷ 12 = 104 students

Number of boys = 5 × 104 = 520 Number of girls = 7 × 104 = 728

Difference = 728 − 520 = 208 more girls

Alternative: Difference in parts = 7 − 5 = 2 parts

2 × 104 = 208

[2 marks] — 1 mark for finding value of one part or individual amounts, 1 mark for correct final answer

14. When a number is divided by 9, the remainder is 5. When the same number is divided by 6, the remainder is 2. What is the smallest possible number?

Answer: 32

Method & Explanation:

Condition 1: Number ÷ 9 gives remainder 5

Possible numbers: 5, 14, 23, 32, 41, 50, 59, 68, ...

Condition 2: Number ÷ 6 gives remainder 2

Possible numbers: 2, 8, 14, 20, 26, 32, 38, 44, 50, ...

Find the smallest common number in both lists:

14: 14 ÷ 9 = 1 R 5 ✓, but 14 ÷ 6 = 2 R 2 ✓ — wait, let me check: 14 ÷ 6 = 2 remainder 2. Yes!

Actually 14 appears in both! Let me verify:

14 ÷ 9 = 1 remainder 5 ✓
14 ÷ 6 = 2 remainder 2 ✓

So the answer is 14?

Wait, let me recheck my lists. For condition 1 (÷9, R5): start from 5, add 9s:

5, 14, 23, 32, 41, 50, 59, 68, 77, 86...

For condition 2 (÷6, R2): start from 2, add 6s:

2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68...

First common number: 14

Let me verify:

14 = 9 × 1 + 5 = 9 + 5 = 14 ✓
14 = 6 × 2 + 2 = 12 + 2 = 14 ✓

Answer: 14

[2 marks] — 1 mark for systematic listing or correct method, 1 mark for correct answer

15. The sum of three consecutive even numbers is 234. Find the largest number.

Answer: 80

Method & Explanation:

Method 1: Using algebra Let the middle even number be n

Three consecutive even numbers: (n − 2), n, (n + 2)
Sum: (n − 2) + n + (n + 2) = 3n = 234
n = 234 ÷ 3 = 78
Largest number = n + 2 = 80

Method 2: Using average

Average of three numbers = 234 ÷ 3 = 78
Since they're consecutive even numbers, they are: 78 − 2, 78, 78 + 2
That is: 76, 78, 80

Verification: 76 + 78 + 80 = 154 + 80 = 234 ✓

[2 marks] — 1 mark for correct method (finding middle value or setting up equation), 1 mark for correct final answer

SECTION C: Long-Answer Questions (30 marks)

16. Factory toy car production (6 marks)

(a) How many toy cars does the factory produce in August?

Answer: 93 600 toy cars

(b) How many boxes are needed?

Answer: 3 900 boxes

(c) Total shipping cost?

Answer: $11 700

Method & Explanation:

(a) August production

June: 45 600
July: 45 600 + 1 200 = 46 800
August: 2 × 46 800 = 93 600 toy cars

(b) Number of boxes

93 600 ÷ 24 = 3 900 boxes

Calculation: 93 600 ÷ 24 = 93 600 ÷ 12 ÷ 2 = 7 800 ÷ 2 = 3 900

(c) Shipping cost

3 900 × $3 = **$ 11 700**

[6 marks] — 2 marks for each part

Part (a): 1 mark for correct July figure, 1 mark for correct August figure
Part (b): 1 mark for correct method, 1 mark for correct answer
Part (c): 1 mark for correct multiplication, 1 mark for correct answer with units

17. Mr. Lee's money (6 marks)

(a) Charity donation

Answer: $3 000

(b) Given to son

Answer: $3 600

(c) Bank deposit

Answer: $5 400

Method & Explanation:

Starting amount: $12 000

(a) Charity donation

$\frac{1}{4}$ × 12 000 = 12 000 ÷ 4 = $3 000

(b) Remaining after charity

12 000 − 3 000 = $9 000

Given to son (fraction of remaining, not original!):

$\frac{2}{5}$ × 9 000 = (9 000 ÷ 5) × 2 = 1 800 × 2 = $3 600

(c) Bank deposit

Remaining = 9 000 − 3 600 = $5 400

Or: fraction deposited = 1 − $\frac{2}{5}$ = $\frac{3}{5}$ of remainder

$\frac{3}{5}$ × 9 000 = $5 400

Verification: 3 000 + 3 600 + 5 400 = 12 000 ✓

[6 marks] — 2 marks for each part

Part (a): 2 marks for correct answer with working
Part (b): 1 mark for finding remainder, 1 mark for correct answer
Part (c): 1 mark for correct method, 1 mark for correct answer

Common mistake: Using original $12 000 for part (b):$ \frac{2}{5}$ × 12 000 = 4 800. This is wrong! The fraction applies to the remaining money after charity.

18. Rectangular hall tiling (6 marks)

(a) Number of tiles needed

Answer: 3 000 tiles

(b) Total cost

Answer: $13 500

(c) Hours to tile

Answer: 25 hours

Method & Explanation:

Important: Convert to same units first!

18 m = 1 800 cm
15 m = 1 500 cm
Tile side = 30 cm

(a) Number of tiles

Method 1: By area

Hall area = 1 800 × 1 500 = 2 700 000 cm²
Tile area = 30 × 30 = 900 cm²
Number of tiles = 2 700 000 ÷ 900 = 3 000 tiles

Method 2: By counting (often simpler!)

Along length: 1 800 ÷ 30 = 60 tiles
Along width: 1 500 ÷ 30 = 50 tiles
Total tiles: 60 × 50 = 3 000 tiles

(b) Total cost

3 000 × $4.50 = 3 000 ×$ 4 + 3 000 × $0.50 = 12 000 + 1 500 = **$ 13 500**

(c) Hours to tile

3 000 ÷ 120 = 3 000 ÷ 12 ÷ 10 = 250 ÷ 10 = 25 hours

[6 marks] — 2 marks for each part

Part (a): 1 mark for unit conversion or correct method, 1 mark for answer
Part (b): 1 mark for method, 1 mark for answer with dollar sign
Part (c): 1 mark for method, 1 mark for answer with units (hours)

Common mistakes:

Forgetting to convert metres to centimetres, getting 18 ÷ 30 = impossible (0.6 tiles)
Part (c): using 120 ÷ 3 000 instead of 3 000 ÷ 120

19. Book fair fractions (6 marks)

(a) Total books

Answer: 3 840 books

(b) Comic books

Answer: 1 440 comic books

(c) Comic books left

Answer: 640 comic books

Method & Explanation:

First, find the fraction of comic books:

Fiction: $\frac{3}{8}$
Non-fiction: $\frac{1}{4}$ = $\frac{2}{8}$
Total so far: $\frac{3}{8}$ + $\frac{2}{8}$ = $\frac{5}{8}$
Comic books: 1 − $\frac{5}{8}$ = $\frac{3}{8}$

(a) Total books

$\frac{3}{8}$ of total = 1 440 (fiction books)
Total = 1 440 ÷ $\frac{3}{8}$ = 1 440 × $\frac{8}{3}$ = 480 × 8 = 3 840 books

Verification: $\frac{3}{8}$ × 3 840 = 3 840 ÷ 8 × 3 = 480 × 3 = 1 440 ✓

(b) Comic books

$\frac{3}{8}$ × 3 840 = 1 440 comic books
Or: total − fiction − non-fiction = 3 840 − 1 440 − 960 = 1 440

(c) Comic books left after selling $\frac{5}{9}$

Sold: $\frac{5}{9}$ × 1 440 = 800
Left: 1 440 − 800 = 640
Or: $\frac{4}{9}$ × 1 440 = 160 × 4 = 640

[6 marks] — 2 marks for each part

Part (a): 1 mark for correct method (understanding fiction = 3/8 of total), 1 mark for correct calculation
Part (b): 1 mark for finding fraction or method, 1 mark for answer
Part (c): 1 mark for correct fraction ( $\frac{4}{9}$ left), 1 mark for correct answer

20. Number line problem (6 marks)

<image_placeholder> id: Q20-fig1 type: diagram linked_question: Q20 description: Number line showing positions of four unknown numbers A, B, C, D with given clues about their relationships labels: Points A, B, C, D on number line; arrows showing relationships values: Various whole number relationships to be derived must_show: Clear positions of A, B, C, D with spacing proportional to values, all labels visible, scale markings </image_placeholder>

(a) Value of B

Answer: 5 000

(b) Value of C

Answer: 7 500

(c) Value of D

Answer: 12 500

(d) Difference between largest and smallest

Answer: 7 500

Method & Explanation:

(a) Find A, then B

A = smallest 4-digit number = 1 000
B = 5 × A = 5 × 1 000 = 5 000

(b) Find C

C = B + 2 500 = 5 000 + 2 500 = 7 500

(c) Find D

D = A + B + C = 1 000 + 5 000 + 7 500 = 13 500

Wait, let me recheck: 1 000 + 5 000 + 7 500 = 6 000 + 7 500 = 13 500

Actually, let me reread: "D is the sum of A, B, and C"

D = 1 000 + 5 000 + 7 500 = 13 500

(d) Largest and smallest

Numbers: A = 1 000, B = 5 000, C = 7 500, D = 13 500
Largest = D = 13 500
Smallest = A = 1 000
Difference = 13 500 − 1 000 = 12 500

[6 marks] — marks allocated as follows:

Part (a): 1 mark for A identified as 1 000, 1 mark for B = 5 000
Part (b): 2 marks for C = 7 500
Part (c): 1 mark for correct addition, 1 mark for D = 13 500
Part (d): includes in part (c)'s follow-through, or 1 additional mark if separate marking used

Note: If working is correct throughout but arithmetic error in early part, method marks may be awarded for follow-through.

END OF ANSWER KEY

Total Marks: 50

Section	A	B	C	Total
Marks	10	10	30	50