WASSCE Further Maths: Core Algebra and Logic Practice Questions & Answers

The intersection of sets $A$A and $B$B, denoted by $A \cap B$A∩B, is the set containing all elements that are common to both $A$A and $B$B. Since 3 and 4 appear in both sets, $A \cap B = \{3, 4\}$A∩B={3,4}.

The total number of subsets for a set with $n$n elements is given by $2^n$2n. The proper subsets include all subsets except the set itself. Therefore, the formula for the number of proper subsets is $2^n - 1$2n−1. For $n = 5$n=5, this is $2^5 - 1 = 32 - 1 = 31$25−1=32−1=31.

Let $U$U be the universal set, $P$P be Physics, and $C$C be Chemistry.
$n(U) = 40$n(U)=40, $n(P) = 25$n(P)=25, $n(C) = 18$n(C)=18, and $n(P \cup C)' = 8$n(P∪C)′=8.
The number of students offering at least one subject is $n(P \cup C) = n(U) - n(P \cup C)' = 40 - 8 = 32$n(P∪C)=n(U)−n(P∪C)′=40−8=32.
Using the set formula: $n(P \cup C) = n(P) + n(C) - n(P \cap C)$n(P∪C)=n(P)+n(C)−n(P∩C)
$32 = 25 + 18 - n(P \cap C) \implies 32 = 43 - n(P \cap C) \implies n(P \cap C) = 11$32=25+18−n(P∩C)⟹32=43−n(P∩C)⟹n(P∩C)=11.

The set builder notation indicates that $x$x is an integer ($\mathbb{Z}$Z). The inequality $-2 \le x < 3$−2≤x<3 means $x$x can take the value of $-2$−2 but must be strictly less than $3$3. The integers satisfying this condition are $-2, -1, 0, 1$−2,−1,0,1, and $2$2. Therefore, $A = \{-2, -1, 0, 1, 2\}$A={−2,−1,0,1,2}.

According to the distributive law of sets over intersection, taking the union of set $A$A with the intersection of sets $B$B and $C$C is equivalent to distributing the union operation across the intersection: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$A∪(B∩C)=(A∪B)∩(A∪C).

The complement of a set $X$X contains all elements in the universal set $\mu$μ that are not present in $X$X. Excluding the even numbers $\{2, 4, 6, 8\}${2,4,6,8} from $\mu$μ leaves the odd numbers $\{1, 3, 5, 7, 9\}${1,3,5,7,9}.

Since all 50 students play at least one sport, $n(F \cup B) = 50$n(F∪B)=50.
Using $n(F \cup B) = n(F) + n(B) - n(F \cap B)$n(F∪B)=n(F)+n(B)−n(F∩B):
$50 = 30 + 25 - n(F \cap B) \implies n(F \cap B) = 55 - 50 = 5$50=30+25−n(F∩B)⟹n(F∩B)=55−50=5.
The number of students who play only Basketball is $n(B) - n(F \cap B) = 25 - 5 = 20$n(B)−n(F∩B)=25−5=20.

First, find the union: $X \cup Y = \{a, b, c, d\}$X∪Y={a,b,c,d}.
Next, find the intersection: $X \cap Y = \{b, c\}$X∩Y={b,c}.
The symmetric difference represents elements in either $X$X or $Y$Y, but not both: $\{a, b, c, d\} - \{b, c\} = \{a, d\}${a,b,c,d}−{b,c}={a,d}.

To simplify $\sqrt{72}$72​, identify the largest perfect square factor of $72$72. Since $72 = 36 \times 2$72=36×2, we rewrite as $\sqrt{36 \times 2}$36×2​. This simplifies to $\sqrt{36} \times \sqrt{2} = 6\sqrt{2}$36​×2​=62​.

Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$(a+b)2=a2+2ab+b2, set $a = \sqrt{3}$a=3​ and $b = \sqrt{2}$b=2​.
The expansion gives: $(\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$(3​)2+2(3​)(2​)+(2​)2=3+26​+2=5+26​.