c. $ \frac{1}{5} $\
\
3 of 5 $ \{1,2,3,4,5\} $ and $ a = $\
and $ a $.\
a. $ \{1,3,5\} $\
b. $ \{2,4\} $\
c. $ u $\
d. $ \emptyset $\
16. which shows associative property of multiplication?\
a. $ 3 \times 4 = 4 \times 3 $\
b. $ (2 \times 3) \times 4 = 2 \times (3 \times 4) $\
c. $ 2(3 + 4) = 6 + 8 $\
21. identify the property: $ 7 + 0 = 7 $.\
a. commutative\
b. associative\
c. identity\
d. inverse\
22. if $ x = \{2,4,6,8\} $ and $ y = \{4,8,12\} $:\
a. $ x \subset y $\
b. $ y \subset x $\
c. disjoint\
d. $ x \cap y = \{4,8\} $\
23. using distributive property, $ 9 \times 11 = $\
a. 91\
b. 99\
c. 110\
d. 20\
24. proper subset of $ \{1,2,3,4\} $ is:\
a. $ \{1,2,3,4\} $\
b. $ \{1,2\} $\
c. $ \{5,6\} $\
d. $ \{1,2,3,4,5\} $\
25. if $ a = \{2,4,6\} $, then $ a \cup a = $\
a. $ a $\
b. $ a $\
c. $ u $\
d. $ \emptyset $\
26. $ (3 + 5) + 7 = 3 + (5 + 7) $ illustrates:\
a. commutative\
b. associative\
c. distributive\
d. identity\
27. 60 study maths, 50 english, 30 both. neither?\
a. 10\
b. 20\
c. 30\
d. 40\
28. $ p \cap q $ where $ p, q $ are factors of 12 and 18:\
a. $ \{1,2,3,6\} $\
b. $ \{12,18\} $\
c. $ \{4,9\} $\
d. $ \{1,2,3,4,6,9\} $\
29. integers are not closed under division because:\
a. always whole\
b. may give fractions\
c. division is commutative\
d. integers are negative\
30. 85 football, 70 basketball, 40 both. at least one?\
a. 115\
b. 125\
c. 155\
d. 195

Question

c. $ \frac{1}{5} $\
\
3 of 5 $ \{1,2,3,4,5\} $ and $ a = $\
and $ a $.\
  a. $ \{1,3,5\} $\
  b. $ \{2,4\} $\
  c. $ u $\
  d. $ \emptyset $\
16. which shows associative property of multiplication?\
  a. $ 3 \times 4 = 4 \times 3 $\
  b. $ (2 \times 3) \times 4 = 2 \times (3 \times 4) $\
  c. $ 2(3 + 4) = 6 + 8 $\
21. identify the property: $ 7 + 0 = 7 $.\
  a. commutative\
  b. associative\
  c. identity\
  d. inverse\
22. if $ x = \{2,4,6,8\} $ and $ y = \{4,8,12\} $:\
  a. $ x \subset y $\
  b. $ y \subset x $\
  c. disjoint\
  d. $ x \cap y = \{4,8\} $\
23. using distributive property, $ 9 \times 11 = $\
  a. 91\
  b. 99\
  c. 110\
  d. 20\
24. proper subset of $ \{1,2,3,4\} $ is:\
  a. $ \{1,2,3,4\} $\
  b. $ \{1,2\} $\
  c. $ \{5,6\} $\
  d. $ \{1,2,3,4,5\} $\
25. if $ a = \{2,4,6\} $, then $ a \cup a = $\
  a. $ a $\
  b. $ a $\
  c. $ u $\
  d. $ \emptyset $\
26. $ (3 + 5) + 7 = 3 + (5 + 7) $ illustrates:\
  a. commutative\
  b. associative\
  c. distributive\
  d. identity\
27. 60 study maths, 50 english, 30 both. neither?\
  a. 10\
  b. 20\
  c. 30\
  d. 40\
28. $ p \cap q $ where $ p, q $ are factors of 12 and 18:\
  a. $ \{1,2,3,6\} $\
  b. $ \{12,18\} $\
  c. $ \{4,9\} $\
  d. $ \{1,2,3,4,6,9\} $\
29. integers are not closed under division because:\
  a. always whole\
  b. may give fractions\
  c. division is commutative\
  d. integers are negative\
30. 85 football, 70 basketball, 40 both. at least one?\
  a. 115\
  b. 125\
  c. 155\
  d. 195

Accepted Answer

### Question 16:
# Explanation:
## Step1: Recall associative property of multiplication
The associative property of multiplication states that for any numbers $a$, $b$, and $c$, $(a\times b)\times c=a\times(b\times c)$.

## Step2: Analyze each option
- Option A: $3\times4 = 4\times3$ shows the commutative property of multiplication (order of multiplication doesn't matter), not associative.
- Option B: $(2\times3)\times4=2\times(3\times4)$ follows the form $(a\times b)\times c = a\times(b\times c)$, so it shows the associative property of multiplication.
- Option C: $2(3 + 4)=6 + 8$ is an incorrect application (left side is $2\times7 = 14$, right side is $14$ but the form is not associative of multiplication, it's a wrong distributive - like attempt).

# Answer: B

### Question 21:
# Explanation:
## Step1: Recall number properties
- Commutative property: $a + b=b + a$ (for addition) or $a\times b=b\times a$ (for multiplication).
- Associative property: $(a + b)+c=a+(b + c)$ (for addition) or $(a\times b)\times c=a\times(b\times c)$ (for multiplication).
- Identity property of addition: $a+0 = a$ (0 is the additive identity).
- Inverse property: $a+(-a)=0$ (for addition) or $a\times\frac{1}{a}=1$ ($a
eq0$, for multiplication).

## Step2: Analyze the equation $7 + 0=7$
The equation $7+0 = 7$ is in the form $a + 0=a$, which is the identity property of addition.

# Answer: C

### Question 22:
# Explanation:
## Step1: Recall set operations and subset definitions
- A subset $A\subset B$ if every element of $A$ is an element of $B$.
- The intersection of two sets $X\cap Y$ is the set of elements common to both $X$ and $Y$.

## Step2: Analyze each option
- Option A: $X=\{2,4,6,8\}$, $Y = \{4,8,12\}$. The element $2\in X$ but $2
otin Y$, so $X$ is not a subset of $Y$.
- Option B: $Y=\{4,8,12\}$, $X=\{2,4,6,8\}$. The element $12\in Y$ but $12
otin X$, so $Y$ is not a subset of $X$.
- Option C: Disjoint sets have no common elements. But $X$ and $Y$ have $4$ and $8$ in common, so they are not disjoint.
- Option D: $X\cap Y=\{x:x\in X\text{ and }x\in Y\}$. The common elements of $X=\{2,4,6,8\}$ and $Y = \{4,8,12\}$ are $4$ and $8$, so $X\cap Y=\{4,8\}$.

# Answer: D

### Question 23:
# Explanation:
## Step1: Recall distributive property
The distributive property of multiplication over addition states that $a\times(b + c)=a\times b+a\times c$. We can write $11=10 + 1$.

## Step2: Apply distributive property
$9\times11=9\times(10 + 1)=9\times10+9\times1=90 + 9 = 99$.

# Answer: B

### Question 24:
# Explanation:
## Step1: Recall proper subset definition
A proper subset of a set $S$ is a subset that is not equal to $S$ itself, i.e., it contains some but not all elements of $S$. Let $S=\{1,2,3,4\}$.

## Step2: Analyze each option
- Option A: $\{1,2,3,4\}$ is equal to $S$, so it is not a proper subset.
- Option B: $\{1,2\}$ contains some elements of $S$ and is not equal to $S$, so it is a proper subset.
- Option C: $\{5,6\}$ has no elements in common with $S$, so it is not a subset of $S$, let alone a proper subset.
- Option D: $\{1,2,3,4,5\}$ contains elements not in $S$ (i.e., $5$), so it is not a subset of $S$.

# Answer: B

### Question 25:
# Explanation:
## Step1: Recall union of a set and its complement
For a universal set $U$, the complement of a set $A$ (denoted $A'$) contains all elements in $U$ that are not in $A$. The union of a set and its complement $A\cup A'$ contains all elements in $A$ or in $A'$, which is the entire universal set $U$.

## Step2: Apply the concept
Given $A = \{2,4,6\}$, $A'$ contains all elements in $U$ not in $A$. Then $A\cup A'$ will be all elements in $U$, so $A\cup A'=U$.

# Answer: C

### Question 26:
# Explanation:
## Step1: Recall associative property of addition
The associative property of addition states that for any numbers $a$, $b$, and $c$, $(a + b)+c=a+(b + c)$.

## Step2: Analyze the equation $(3 + 5)+7=3+(5 + 7)$
The equation $(3 + 5)+7=3+(5 + 7)$ is in the form $(a + b)+c=a+(b + c)$, so it illustrates the associative property of addition.

# Answer: B

### Question 27:
# Explanation:
## Step1: Recall principle of inclusion - exclusion for two sets
Let $M$ be the set of students studying Maths, $E$ be the set of students studying English. The number of students studying either Maths or English is $n(M\cup E)=n(M)+n(E)-n(M\cap E)$. The total number of students is not given, but we assume a universal set (let's say the total number of students is $N$). The number of students studying neither is $N - n(M\cup E)$. But we can assume the universal set is the set of all students considered. Wait, actually, if we assume the total number of students is, say, $T$ (but since we are not given, maybe there is a typo or we assume it's a class where we can calculate the number of students studying neither as $T-(n(M)+n(E)-n(M\cap E))$. But looking at the options, let's assume the total number of students is, for example, if we consider that the number of students studying Maths is $60$, English is $50$, both is $30$. Then $n(M\cup E)=60 + 50-30=80$. But the question is about "Neither?". Wait, maybe the original problem has a total number of students, but since it's not shown, maybe it's a mistake. But looking at the options (10,20,30,40), let's re - check. Wait, maybe the total number of students is $100$ (a common assumption). Then number of students studying neither is $100-(60 + 50 - 30)=100 - 80 = 20$? Wait, no, the options are 10,20,30,40. Wait, maybe the total number of students is $80$? No, $60+50 - 30=80$, so neither would be $0$ which is not an option. Wait, maybe the question is "How many study neither?" and the total number of students is, say, $100$. Then $n(\text{neither})=100-(60 + 50-30)=20$.

## Step2: Calculate using inclusion - exclusion
$n(M\cup E)=n(M)+n(E)-n(M\cap E)=60 + 50-30 = 80$. If we assume the total number of students is $100$ (a common case), then the number of students studying neither is $100 - 80=20$.

# Answer: B

### Question 28:
# Explanation:
## Step1: Find factors of 12 and 18
- Factors of $12$: $1,2,3,4,6,12$
- Factors of $18$: $1,2,3,6,9,18$

## Step2: Find intersection $P\cap Q$
The intersection of the set of factors of $12$ (let $P$ be factors of $12$) and factors of $18$ (let $Q$ be factors of $18$) is the set of elements common to both. So $P\cap Q=\{1,2,3,6\}$.

# Answer: A

### Question 29:
# Explanation:
## Step1: Recall closure property
A set is closed under an operation if the result of the operation on any two elements of the set is also an element of the set.

## Step2: Analyze integer division
Integers are not closed under division because when we divide two integers, we may get a fraction. For example, $2\div3=\frac{2}{3}$, which is not an integer.
- Option A: "Always whole" is wrong because division of integers does not always give a whole number.
- Option B: "May give fractions" is correct as shown in the example.
- Option C: Division is not commutative (e.g., $4\div2 = 2$, $2\div4 = 0.5$), but this is not the reason for non - closure.
- Option D: "Integers are negative" is irrelevant to the closure of integers under division.

# Answer: B

### Question 30:
# Explanation:
## Step1: Recall principle of inclusion - exclusion for two sets
Let $F$ be the set of football players, $B$ be the set of basketball players. The number of players who play at least one sport is $n(F\cup B)=n(F)+n(B)-n(F\cap B)$.

## Step2: Apply inclusion - exclusion
Given $n(F)=85$, $n(B)=70$, $n(F\cap B)=40$. Then $n(F\cup B)=85 + 70-40=115$.

# Answer: A

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 16:

Step1: Recall associative property of multiplication

Step2: Analyze each option

Step1: Recall number properties

Step2: Analyze the equation \(7 + 0=7\)

Step1: Recall set operations and subset definitions

Step2: Analyze each option

Answer:

Question 21: