Number & Algebra and Sets — 10 Questions with Detailed Explanations
Below are 10 subjective questions (Level 100 Colleges of Education, on Number & Algebra and Sets. Each question is followed by a careful, step-by-step explanation
Question 1 — Simplify the expression
Q: Simplify \(3(2x - 5) - 4(x + 1)\).
Solution & Explanation
Start by applying the distributive law (multiply each bracket by the factor outside it).
\(3(2x - 5) = 3 \cdot 2x - 3 \cdot 5 = 6x - 15.\)
\( -4(x + 1) = -4\cdot x -4\cdot 1 = -4x - 4.\)
Now combine the two results:
\(6x - 15 - 4x - 4.\)
Group like terms (the \(x\)-terms and the constants): \((6x - 4x) + (-15 - 4) = 2x + (-19).\)
Answer: \(\,2x - 19.\)
Question 2 — Solve the equation
Q: Solve \(5x - 7 = 3x + 9\).
Solution & Explanation
We want to isolate \(x\). First collect the \(x\)-terms on one side and constants on the other.
Subtract \(3x\) from both sides: \(5x - 3x - 7 = 9.\) That gives \(2x - 7 = 9.\)
Add \(7\) to both sides: \(2x = 9 + 7 = 16.\)
Divide both sides by \(2\): \(x = \dfrac{16}{2} = 8.\)
Answer: \(x = 8.\)
Question 3 — Factorize completely
Q: Factorize \(x^2 - 9.\)
Solution & Explanation
Recognize the expression as a difference of squares: \(x^2 - 9 = x^2 - 3^2.\) The identity \(a^2 - b^2 = (a - b)(a + b)\) applies.
So \(x^2 - 9 = (x - 3)(x + 3).\)
Answer: \((x - 3)(x + 3).\)
Question 4 — Expand and simplify
Q: Expand \((2x + 3)(x - 4)\).
Solution & Explanation
Use the distributive property (FOIL for two binomials): multiply each term in the first bracket by each term in the second.
\( (2x + 3)(x - 4) = 2x\cdot x + 2x\cdot(-4) + 3\cdot x + 3\cdot(-4).\)
Compute each product: \(2x\cdot x = 2x^2,\; 2x\cdot(-4) = -8x,\; 3\cdot x = 3x,\; 3\cdot(-4) = -12.\)
Combine like terms: \(2x^2 + (-8x + 3x) - 12 = 2x^2 - 5x - 12.\)
Answer: \(2x^2 - 5x - 12.\)
Question 5 — Make \(x\) the subject
Q: Make \(x\) the subject from \(y = 4x - 5\).
Solution & Explanation
We want \(x\) in terms of \(y\). Add \(5\) to both sides: \(y + 5 = 4x.\) Then divide by \(4\).
\(x = \dfrac{y + 5}{4}.\)
Answer: \(x = \dfrac{y + 5}{4}.\)
Question 6 — Word problem (linear equation)
Q: A number is multiplied by 6 and then reduced by 5 to give 31. Find the number.
Solution & Explanation
Let the number be \(x\). The sentence "multiplied by 6 and then reduced by 5" gives the equation:
\(6x - 5 = 31.\)
Solve it: add \(5\) to both sides \(6x = 31 + 5 = 36.\) Divide by \(6\): \(x = \dfrac{36}{6} = 6.\)
Answer: The number is \(6.\)
Question 7 — Intersection of sets
Q: Given \(A = \{2,4,6,8\},\; B = \{4,8,10\}\). Find \(A \cap B\).
Solution & Explanation
The intersection \(A \cap B\) contains elements that are in both \(A\) and \(B\).
Compare the lists: elements in \(A\) are \(2,4,6,8\). Elements in \(B\) are \(4,8,10\). The numbers that appear in both are \(4\) and \(8\).
Answer: \(A \cap B = \{4,8\}.\)
Question 8 — Complement of a set
Q: If \(U = \{1,2,3,4,5,6,7,8\}\) is the universal set and \(A = \{2,3,5,7\}\), find the complement \(A'\).
Solution & Explanation
The complement \(A'\) (or \(U \setminus A\)) contains the elements of the universal set \(U\) that are not in \(A\).
List the elements of \(U\): \(1,2,3,4,5,6,7,8\). Remove the elements that appear in \(A\) (which are \(2,3,5,7\)).
The remaining elements are \(1,4,6,8\).
Answer: \(A' = \{1,4,6,8\}.\)
Question 9 — Union of sets
Q: For the sets \(P = \{a,b,c,d\},\; Q=\{b,d,e\}\). Find \(P \cup Q\).
Solution & Explanation
The union \(P \cup Q\) contains all elements that are in \(P\) or \(Q\) (without repetition).
Combine the elements: from \(P\) we have \(a,b,c,d\). From \(Q\) add \(b,d,e\). \(b\) and \(d\) are already present; add the new element \(e\) and ensure each element appears once.
Answer: \(P \cup Q = \{a,b,c,d,e\}.\)
Question 10 — Venn application (at least one)
Q: In a class of 40 students, 25 offer Mathematics (M), 18 offer Science (S), and 10 offer both. Find the number who offer at least one of the two subjects.
Solution & Explanation
Use the principle for two sets:
\(n(M \cup S) = n(M) + n(S) - n(M \cap S).\)
Substitute the values: \(n(M \cup S) = 25 + 18 - 10\).
Compute step by step: \(25 + 18 = 43.\) Then \(43 - 10 = 33.\)
So \(33\) students offer at least one of the two subjects (Mathematics or Science). If asked, the number who offer neither would be total minus this: \(40 - 33 = 7.\)
Answer: \(33\) students offer at least one (and \(7\) offer neither).
Extras & Notes
- These solutions show standard algebraic manipulations (distributive law, combining like terms, solving linear equations), factorization (difference of squares), and basic set operations (union, intersection, complement, and Venn counting).
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