Question 1

Let set \( M = \{s, k, y\} \).

• a) Cardinality of set \( M \): \( n(M) = 3 \)
• b) Subsets of set \( M \):
  – Proper subsets: \( \emptyset, \{s\}, \{k\}, \{y\}, \{s, k\}, \{s, y\}, \{k, y\} \)
  – Improper subset: \( \{s, k, y\} \)
• c) Overlapping set example: Let \( N = \{s, a, m\} \). Since \( M \cap N = \{s\} \neq \emptyset \), it is overlapping.

Question 2

Given (where \( U \) is the counting number less than 10):

• \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \)
• \( X = \{1, 2, 4, 5\} \)
• \( Y = \{5\} \)

• a) Listing method:
  \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \)
  \( X = \{1, 2, 4, 5\} \)
  \( Y = \{5\} \)
• b) Is Y a subset of X? Give a reason:
  Yes, \( Y \) is a subset of \( X \) because all the elements of set \( Y \) is also present in set \( X \).
• c) Represent sets U, X, and Y in a Venn diagram.
• d) Suggest an element that can be removed from set X so that X and Y become disjoint set:
  (Hint: element can be kept from universal set only)
  Answer: Remove element \( 5 \) from set \( X \) so that \( X \cap Y = \emptyset \) (disjoint sets).

Question 3

Given set \( A = \{1, 3, 5\} \) and \( B = \{y \mid y = 2x – 1, x \in A\} = \{1, 5, 9\} \).

• a) Improper subset of A: \( \{1, 3, 5\} \)
• b) Three subsets of B with 2 elements: \( \{1, 5\}, \{1, 9\}, \{5, 9\} \)
• c) Represent set A and B in a Venn diagram.

Question 4

• a) Express 43 into binary number:
  \( 43_{10} = 101011_2 \)
• b) Express 25 into quinary number system:
  \( 25_{10} = 100_5 \)
• c) Value of \( A \) in \((1A111)_2 = (23)_{10}\):
  \((1 \times 2^4) + (A \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) = 23\) \(= 16 + 8A + 4 + 2 + 1 = 23\) \(= 23 + 8A = 23\) \(\implies 8A = 23 – 23\) \(\implies 8A = 0 \implies \mathbf{A = 0}\)
• d) Express \( 1001_2 \) into decimal:
  \((1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)\) \(= 8 + 0 + 0 + 1 = \mathbf{9_{10}}\)
• e) Express \( 1100101_2 \) into decimal:
  \((1 \times 2^6) + (1 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)\) \(= 64 + 32 + 0 + 0 + 4 + 0 + 1 = \mathbf{101_{10}}\)
• f) Express \( 42_5 \) into decimal:
  \((4 \times 5^1) + (2 \times 5^0)\) \(= 20 + 2 = \mathbf{22_{10}}\)

Question 5: Representation on Number Line

a) Represent \(\sqrt{2}\) on the number line:
Construct a right-angled triangle with base 1 unit and perpendicular 1 unit. Hypotenuse = \(\sqrt{1^2+1^2} = \sqrt{2}\). Mark it with a compass.

b) Represent \(\sqrt{3}\) on the number line:
Using the \(\sqrt{2}\) position as base, construct a perpendicular of 1 unit to get \(\sqrt{3}\).

Question 6: Recurring Decimals

• a) Convert \( 0.\overline{24} \) into a fraction:
  Let \( x = 0.\overline{24} \) \( x = 0.242424\dots \) — (i) Multiplying both sides by 100: \( 100x = 24.242424\dots \) — (ii) Subtracting (i) from (ii): \( 100x – x = 24.2424\dots – 0.2424\dots \) \( 99x = 24 \) \( x = \frac{24}{99} \) \( \mathbf{x = \frac{8}{33}} \)
• b) Convert \( 4.\overline{78} \) into a fraction:
  Let \( x = 4.\overline{78} \) \( x = 4.787878\dots \) — (i) Multiplying both sides by 100: \( 100x = 478.787878\dots \) — (ii) Subtracting (i) from (ii): \( 100x – x = 478.7878\dots – 4.7878\dots \) \( 99x = 474 \) \( x = \frac{474}{99} \) \( \mathbf{x = \frac{158}{33}} \)
• c) Identify whether 3400 is rational or irrational:
  \( 3400 = \frac{3400}{1} \) Since it can be expressed in the form \( \frac{p}{q} \), it is a rational number.

Question 7: Scientific Notation & Simplification

• a) Express 3400 into scientific notation:
  \( 3400 = 3.4 \times 1000 \) \( \mathbf{= 3.4 \times 10^3} \)
• b) Simplify:
  \((4.3 \times 10^8) \times (2.0 \times 10^6)\) \(= (4.3 \times 2.0) \times (10^8 \times 10^6)\) \(= 8.6 \times 10^{8+6}\) \(= \mathbf{8.6 \times 10^{14}}\)
• c) Simplify:
  \(\frac{1.20 \times 10^{-8}}{0.3 \times 10^{-3}}\) \(= \left(\frac{1.20}{0.3}\right) \times \left(\frac{10^{-8}}{10^{-3}}\right)\) \(= 4.0 \times 10^{-8 – (-3)}\) \(= \mathbf{4.0 \times 10^{-5}}\)
• d) Simplify \((1.2 \times 10^5) + (5.35 \times 10^6)\):
  Method 1 (Using Common Power): \(= (1.2 \times 10^5) + (5.35 \times 10^1 \times 10^5)\) \(= (1.2 \times 10^5) + (53.5 \times 10^5)\) \(= (1.2 + 53.5) \times 10^5\) \(= 54.7 \times 10^5 = \mathbf{5.47 \times 10^6}\)
  Method 2 (Using Standard Form & Reconverting): \(= 120000 + 5350000 = 5470000\) \(= \mathbf{5.47 \times 10^6}\)
• e) Simplify \((5.35 \times 10^6) – (1.2 \times 10^5)\):
  Method 1 (Using Common Power): \(= (5.35 \times 10^1 \times 10^5) – (1.2 \times 10^5)\) \(= (53.5 \times 10^5) – (1.2 \times 10^5)\) \(= (53.5 – 1.2) \times 10^5\) \(= 52.3 \times 10^5 = \mathbf{5.23 \times 10^6}\)
  Method 2 (Using Standard Form & Reconverting): \(= 5350000 – 120000 = 5230000\) \(= \mathbf{5.23 \times 10^6}\)

• a) \((2 – \sqrt{3})(2 + \sqrt{3})\):
  \(= (2)^2 – (\sqrt{3})^2\) \(= 4 – 3 = \mathbf{1}\)
• b) \((\sqrt{5} + 2)(\sqrt{5} – 2)\):
  \(= (\sqrt{5})^2 – (2)^2\) \(= 5 – 4 = \mathbf{1}\)
• c) \(\sqrt{8} \times \sqrt{72}\):
  \(= \sqrt{8 \times 72}\) \(= \sqrt{576} = \mathbf{24}\)
• d) \(\sqrt[3]{125} + \sqrt{64}\):
  \(= \sqrt[3]{5^3} + \sqrt{8^2}\) \(= 5 + 8 = \mathbf{13}\)
• e) \(\sqrt{400} – 2\sqrt[3]{343}\):
  \(= \sqrt{20^2} – 2\sqrt[3]{7^3}\) \(= 20 – 2(7)\) \(= 20 – 14 = \mathbf{6}\)
• f) \(\sqrt[3]{8000} – 2\sqrt{25}\):
  \(= \sqrt[3]{20^3} – 2\sqrt{5^2}\) \(= 20 – 2(5)\) \(= 20 – 10 = \mathbf{10}\)