MATHWITHCYE SPM KSSM Mathematics

Chapter 2: Number Bases (Asas Nombor)

Focus: Place Values, Multi-Base Conversions, Base Arithmetic & Cryptographic Codes

Total Marks
/ 30 Marks
Suggested Time: 45 Mins

Section A • Bahagian A [12 Marks]

Question 1 [4 Marks]

(a) State the place value of the digit 3 in the number $2341_5$. [1 mark]
(b) Convert $342_8$ to a number in base 5, showing clear step-by-step working. [3 marks]

(a) Place value $= 5^2 = 25$

$5^2 \text{ or } 25$
P1

(b) Step 1: Convert $342_8$ to Base 10: $(3 \times 64) + (4 \times 8) + (2 \times 1) = 192 + 32 + 2 = 226_{10}$

$226_{10}$
K1

Step 2: Repeated division by 5: $226 \div 5 = 45 \text{ R } 1; 45 \div 5 = 9 \text{ R } 0; 9 \div 5 = 1 \text{ R } 4; 1 \div 5 = 0 \text{ R } 1$

$1401_5$
K1 N1
Question 2 [4 Marks]

Calculate the following arithmetic operations in the specified base:

(a) $11011_2 + 1010_2$ in base 2. [2 marks]
(b) $725_8 - 367_8$ in base 8. [2 marks]

(a) Column addition in base 2 with carry-overs

$100101_2$
K1 N1

(b) Column subtraction in base 8 with borrowing of 8

$336_8$
K1 N1
Question 3 [4 Marks]

Given that $p_7 - 143_5 = 10110_2$, find the value of $p$ in base 7.

Convert to Base 10: $143_5 = 25 + 20 + 3 = 48_{10}$; $10110_2 = 16 + 4 + 2 = 22_{10}$

$p_{10} = 22 + 48 = 70_{10}$
K2

Convert $70_{10}$ to Base 7: $70 \div 7 = 10 \text{ R } 0; 10 \div 7 = 1 \text{ R } 3; 1 \div 7 = 0 \text{ R } 1$

$p = 130_7$
K1 N1

Section B • Bahagian B (HOTS / KBAT) [18 Marks]

Question 4 (Computer Memory & Data Transfer) [9 Marks]

In digital computing, storage addresses are encoded in binary (base 2) and octal (base 8). A network router receives a packet with size encoded as $111010110_2\text{ bytes}$.

(a) Convert $111010110_2$ to octal (base 8) directly by grouping into triplets. [3 marks]
(b) Calculate the packet size in base 10 (decimal). [3 marks]
(c) If a buffer memory can hold $1200_5\text{ bytes}$, determine whether the packet will cause a buffer overflow. [3 marks]

(a) Grouping: $(111)_2 (010)_2 (110)_2$

$= 726_8$
K1 N2

(b) Expansion: $(7 \times 64) + (2 \times 8) + (6 \times 1) = 448 + 16 + 6 = 470_{10}$

Packet size $= 470\text{ bytes}$
K1 N2

(c) Buffer capacity in Base 10: $1200_5 = (1 \times 125) + (2 \times 25) = 125 + 50 = 175\text{ bytes}$.

Since Packet size (470 bytes) > Buffer capacity (175 bytes), YES, it will cause an overflow.
K1 N2
Question 5 (Vending Machine & Cryptography) [9 Marks]

A special coin counter in a banking system tallies coins in base 5. At the end of the day, Cashier $A$ deposited $3421_5$ coins and Cashier $B$ deposited $434_5$ coins.

(a) Find the total number of coins deposited in base 5. [3 marks]
(b) If each coin is worth RM0.50, calculate the total monetary value in RM in base 10. [4 marks]
(c) State the total number of coins in base 8. [2 marks]

(a) Addition in base 5: $3421_5 + 434_5$

$= 4410_5$
K1 N2

(b) Total coins in Base 10: $(4 \times 125) + (4 \times 25) + (1 \times 5) = 500 + 100 + 5 = 605\text{ coins}$.

$\text{Total Value} = 605 \times \text{RM } 0.50 = \text{RM } 302.50$
K2 N2

(c) Convert $605_{10}$ to Base 8: $605 \div 8 = 75 \text{ R } 5; 75 \div 8 = 9 \text{ R } 3; 9 \div 8 = 1 \text{ R } 1; 1 \div 8 = 0 \text{ R } 1$

$= 1135_8$
K1 N1
© MathWithCYE · SPM Form 4 KSSM Mathematics Revision Portal