First Mid Term Test - 2023 | Solved
Virudhunagar District | Standard 9 - MATHEMATICS
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The set P contains integers $x$ such that $-1 < x < 1$.
The only integer that satisfies this condition is 0.
Therefore, the set is $P = \{0\}$.
A set containing exactly one element is called a singleton set.
Answer: a) Singleton set
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The set difference $B - A$ consists of elements that are in B but not in A.
If $B - A = B$, it means that none of the elements of A are present in B. This implies that the sets A and B are disjoint.
For disjoint sets, the intersection is the empty set ($\emptyset$).
Therefore, $A \cap B = \emptyset$.
Answer: d) $\emptyset$
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Given $n(A) = 10$ and $n(B) = 15$.
Minimum value of $n(A \cap B)$: The minimum intersection occurs when the sets have no elements in common (disjoint sets). In this case, $n(A \cap B) = 0$.
Maximum value of $n(A \cap B)$: The maximum intersection occurs when one set is a subset of the other. Since $n(A) < n(B)$, the maximum overlap happens if $A \subset B$. In this case, all elements of A are also in B, so $A \cap B = A$. Thus, the maximum value is $n(A) = 10$.
So, the minimum is 0 and the maximum is 10.
Answer: d) 0, 10
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This question relates to the properties of set operations.
The expression $P - (Q \cap R)$ represents the set of elements that are in P but not in the intersection of Q and R.
This is a standard set identity, known as De Morgan's law for set difference: $A - (B \cap C) = (A - B) \cup (A - C)$.
Applying this to our sets, we get $P - (Q \cap R) = (P - Q) \cup (P - R)$.
Answer: d) $(P - Q) \cup (P - R)$
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A rational number has a terminating decimal expansion if its denominator, in simplest form, has only prime factors of 2 and/or 5.
- a) $\frac{5}{64}$: The denominator is $64 = 2^6$. The only prime factor is 2. So, this will terminate.
- b) $\frac{8}{9}$: The denominator is $9 = 3^2$. It has a prime factor of 3. So, this will not terminate.
- c) $\frac{14}{15}$: The denominator is $15 = 3 \times 5$. It has a prime factor of 3. So, this will not terminate.
- d) $\frac{1}{12}$: The denominator is $12 = 2^2 \times 3$. It has a prime factor of 3. So, this will not terminate.
Answer: a) $\frac{5}{64}$
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We are looking for an irrational number between 2 and 2.5. Let's square these bounds: $2^2 = 4$ and $2.5^2 = 6.25$. So, we need an irrational number whose square is between 4 and 6.25.
- a) $\sqrt{11}$: $11$ is not between 4 and 6.25. ($\sqrt{11} \approx 3.32$)
- b) $\sqrt{5}$: $5$ is between 4 and 6.25. So, $\sqrt{5}$ is between 2 and 2.5. ($\sqrt{5} \approx 2.236$)
- c) $\sqrt{2.5}$: $2.5$ is not between 4 and 6.25. ($\sqrt{2.5} \approx 1.58$)
- d) $\sqrt{8}$: $8$ is not between 4 and 6.25. ($\sqrt{8} \approx 2.828$)
Answer: b) $\sqrt{5}$
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A number that is not rational is irrational. An irrational number cannot be expressed as a fraction $\frac{p}{q}$ and has a non-terminating, non-repeating decimal expansion.
- a) $\sqrt{\frac{8}{18}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$. This is a rational number.
- b) $\frac{7}{3}$. This is in the form $\frac{p}{q}$, so it is a rational number.
- c) $\sqrt{0.01} = \sqrt{\frac{1}{100}} = \frac{1}{10}$. This is a rational number.
- d) $\sqrt{13}$. Since 13 is a prime number, it is not a perfect square. The square root of any non-perfect square integer is irrational.
Answer: d) $\sqrt{13}$
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We want to multiply $\frac{1}{3}$ by a rational number $x$ to get a result that terminates after one decimal place. A number that terminates after one decimal place can be written in the form $\frac{k}{10}$ for some integer $k$.
So, $\frac{1}{3} \times x = \frac{k}{10}$.
This means $x = \frac{3k}{10}$. We need to find the smallest positive rational number $x$ from the given options.
Let's test the options:- a) $\frac{1}{3} \times \frac{1}{10} = \frac{1}{30} = 0.0\overline{3}$. Does not terminate.
- b) $\frac{1}{3} \times \frac{3}{10} = \frac{1}{10} = 0.1$. Terminates after one decimal place. This works.
- c) $\frac{1}{3} \times 3 = 1 = 1.0$. Terminates after one decimal place. This also works.
- d) $\frac{1}{3} \times 30 = 10 = 10.0$. Terminates. This also works.
We need the smallest rational number among the choices that work. Comparing the working options: $\frac{3}{10}$, $3$, and $30$. The smallest is $\frac{3}{10}$.
Answer: b) $\frac{3}{10}$
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A set with $n$ elements has $2^n$ subsets.
Given $A = \{a, b\}$, $n=2$. So there are $2^2 = 4$ subsets.
The subsets are:
- The empty set: $\emptyset$
- Subsets with one element: $\{a\}$, $\{b\}$
- The set itself: $\{a, b\}$
The subsets of A are $\emptyset, \{a\}, \{b\}, \{a, b\}$.
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We need to find perfect cubes $x$ that are strictly greater than 27 and strictly less than 216.
We know that $3^3 = 27$ and $6^3 = 216$.
The perfect cubes between these are:
- $4^3 = 64$
- $5^3 = 125$
Both 64 and 125 satisfy the condition $27 < x < 216$.
In roster form, B = {64, 125}.
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Given $A = \{-3, -2, 1, 4\}$ and $B = \{0, 1, 2, 4\}$.
(i) A - B: This set contains elements that are in A but not in B.
$A - B = \{-3, -2, 1, 4\} - \{0, 1, 2, 4\}$
The elements 1 and 4 are common. Removing them from A leaves us with $\{-3, -2\}$.
(i) A - B = {-3, -2}
(ii) B - A: This set contains elements that are in B but not in A.
$B - A = \{0, 1, 2, 4\} - \{-3, -2, 1, 4\}$
The elements 1 and 4 are common. Removing them from B leaves us with $\{0, 2\}$.
(ii) B - A = {0, 2}
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Given sets:
$K = \{a, b, d, e, f\}$
$L = \{b, c, d, g\}$
$M = \{a, b, c, d, h\}$
(i) Find $K \cap (L \cup M)$
First, find $L \cup M$:
$L \cup M = \{b, c, d, g\} \cup \{a, b, c, d, h\} = \{a, b, c, d, g, h\}$
Now, find the intersection of K with this new set:
$K \cap (L \cup M) = \{a, b, d, e, f\} \cap \{a, b, c, d, g, h\}$
(i) $K \cap (L \cup M) = \{a, b, d\}$
(ii) Find $K \cup (L \cap M)$
First, find $L \cap M$:
$L \cap M = \{b, c, d, g\} \cap \{a, b, c, d, h\} = \{b, c, d\}$
Now, find the union of K with this new set:
$K \cup (L \cap M) = \{a, b, d, e, f\} \cup \{b, c, d\}$
(ii) $K \cup (L \cap M) = \{a, b, c, d, e, f\}$
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We need to find three rational numbers between $-\frac{7}{11}$ and $\frac{2}{11}$.
Since the denominators are the same, we can simply choose any three integers between the numerators -7 and 2.
The integers between -7 and 2 are: -6, -5, -4, -3, -2, -1, 0, 1.
We can choose any three. For example:
- $-\frac{6}{11}$
- $-\frac{5}{11}$
- $-\frac{4}{11}$
Three rational numbers are $-\frac{6}{11}, -\frac{5}{11}, \text{ and } 0$.
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The expression $81^{5/4}$ can be rewritten as $(81^{1/4})^5$.
First, we find the 4th root of 81, which is $81^{1/4}$.
We know that $3^4 = 3 \times 3 \times 3 \times 3 = 81$. So, $\sqrt[4]{81} = 3$.
Now, we raise this result to the power of 5:
$(3)^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243$.
The value of $81^{5/4}$ is 243.
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To find the decimal form, we perform the division $4 \div 11$.
0.3636...
_______
11 | 4.0000
-0
---
4 0
-3 3
----
70
-66
---
40
-33
---
7
The pattern '36' repeats. So, $\frac{4}{11} = 0.3636... = 0.\overline{36}$.
Therefore, $-\frac{4}{11} = -0.\overline{36}$.
The decimal form is $-0.\overline{36}$.
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Let $x = 0.999...$ which is $0.\overline{9}$.
Step 1: Write the equation.
$$ x = 0.999... \quad (1) $$Step 2: Multiply by 10 to shift the decimal point.
$$ 10x = 9.999... \quad (2) $$Step 3: Subtract equation (1) from equation (2).
$$ 10x - x = (9.999...) - (0.999...) $$ $$ 9x = 9 $$Step 4: Solve for x.
$$ x = \frac{9}{9} $$ $$ x = 1 $$Since we started with $x = 0.\overline{9}$, we have shown that $0.\overline{9} = 1$.
Hence, verified.
(ii) If $n[P(A)] = 256$, find $n(A)$.
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(i) Power set of B = {1, 2, 3}
The power set, $P(B)$, is the set of all subsets of B. Since $n(B)=3$, there will be $2^3 = 8$ subsets.
- Subset with 0 elements: $\emptyset$
- Subsets with 1 element: $\{1\}, \{2\}, \{3\}$
- Subsets with 2 elements: $\{1, 2\}, \{1, 3\}, \{2, 3\}$
- Subset with 3 elements: $\{1, 2, 3\}$
$P(B) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$
(ii) If $n[P(A)] = 256$, find $n(A)$
The number of elements in the power set of A is given by the formula $n[P(A)] = 2^{n(A)}$.
We are given $n[P(A)] = 256$.
So, $2^{n(A)} = 256$.
We need to find the power of 2 that equals 256. We can do this by prime factorization or by knowing powers of 2:
$2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64, 2^7=128, 2^8=256$.
So, $2^{n(A)} = 2^8$.
By comparing the exponents, we get $n(A) = 8$.
$n(A) = 8$
(i) A' (ii) B' (iii) A' $\cup$ B' (iv) A' $\cap$ B' (v) (A $\cup$ B)'
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Given: $U = \{a, b, c, d, e, f, g, h\}$, $A = \{b, d, f, h\}$, $B = \{a, d, e, h\}$.
(i) A' (Complement of A)
A' = U - A = $\{a, b, c, d, e, f, g, h\} - \{b, d, f, h\} = \{a, c, e, g\}$.
A' = {a, c, e, g}
(ii) B' (Complement of B)
B' = U - B = $\{a, b, c, d, e, f, g, h\} - \{a, d, e, h\} = \{b, c, f, g\}$.
B' = {b, c, f, g}
(iii) A' $\cup$ B'
A' $\cup$ B' = $\{a, c, e, g\} \cup \{b, c, f, g\} = \{a, b, c, e, f, g\}$.
A' $\cup$ B' = {a, b, c, e, f, g}
(iv) A' $\cap$ B'
A' $\cap$ B' = $\{a, c, e, g\} \cap \{b, c, f, g\} = \{c, g\}$.
A' $\cap$ B' = {c, g}
(v) (A $\cup$ B)'
First, find A $\cup$ B:
A $\cup$ B = $\{b, d, f, h\} \cup \{a, d, e, h\} = \{a, b, d, e, f, h\}$.
Now, find the complement:
(A $\cup$ B)' = U - (A $\cup$ B) = $\{a, b, c, d, e, f, g, h\} - \{a, b, d, e, f, h\} = \{c, g\}$.
(A $\cup$ B)' = {c, g}
Note: This also verifies De Morgan's Law, as $(A \cup B)' = A' \cap B'$.
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First, let's list the elements of each set:
- P = {x : x $\in$ W, 0 < x < 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9}
- Q = {x : x=2n+1, n $\in$ W, n < 5}. (Here n = 0, 1, 2, 3, 4)
- n=0: x = 2(0)+1 = 1
- n=1: x = 2(1)+1 = 3
- n=2: x = 2(2)+1 = 5
- n=3: x = 2(3)+1 = 7
- n=4: x = 2(4)+1 = 9
- R = {2, 3, 5, 7, 11, 13}
LHS: $P - (Q \cap R)$
1. Find $Q \cap R$:
$Q \cap R = \{1, 3, 5, 7, 9\} \cap \{2, 3, 5, 7, 11, 13\} = \{3, 5, 7\}$.
2. Find $P - (Q \cap R)$:
$P - (Q \cap R) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{3, 5, 7\} = \{1, 2, 4, 6, 8, 9\}$.
RHS: $(P - Q) \cup (P - R)$
1. Find $P - Q$:
$P - Q = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 3, 5, 7, 9\} = \{2, 4, 6, 8\}$.
2. Find $P - R$:
$P - R = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 3, 5, 7, 11, 13\} = \{1, 4, 6, 8, 9\}$.
3. Find $(P - Q) \cup (P - R)$:
$(P - Q) \cup (P - R) = \{2, 4, 6, 8\} \cup \{1, 4, 6, 8, 9\} = \{1, 2, 4, 6, 8, 9\}$.
Conclusion:
Since LHS = {1, 2, 4, 6, 8, 9} and RHS = {1, 2, 4, 6, 8, 9}, the identity is verified.
LHS = RHS. Hence, verified.
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Let C, F, and H be the sets of students who play Cricket, Football, and Hockey, respectively.
Given data: $n(C) = 240$, $n(F) = 180$, $n(H) = 164$ $n(C \cap F) = 42$, $n(F \cap H) = 38$, $n(C \cap H) = 40$ $n(C \cap F \cap H) = 16$
(i) The number of students in the college
Since each student plays at least one game, the total number of students is $n(C \cup F \cup H)$. Using the principle of inclusion-exclusion:
$$ n(C \cup F \cup H) = n(C) + n(F) + n(H) - n(C \cap F) - n(F \cap H) - n(C \cap H) + n(C \cap F \cap H) $$ $$ = 240 + 180 + 164 - 42 - 38 - 40 + 16 $$ $$ = 584 - 120 + 16 $$ $$ = 464 + 16 = 480 $$(i) Total number of students = 480.
(ii) The number of students who play only one game
We can find this by calculating the number of students in each 'only' category.
- Only Cricket: $n(C \text{ only}) = n(C) - n(C \cap F) - n(C \cap H) + n(C \cap F \cap H) = 240 - 42 - 40 + 16 = 174$.
- Only Football: $n(F \text{ only}) = n(F) - n(C \cap F) - n(F \cap H) + n(C \cap F \cap H) = 180 - 42 - 38 + 16 = 116$.
- Only Hockey: $n(H \text{ only}) = n(H) - n(C \cap H) - n(F \cap H) + n(C \cap F \cap H) = 164 - 40 - 38 + 16 = 102$.
Total students playing only one game = $174 + 116 + 102 = 392$.
(ii) Number of students playing only one game = 392.
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(i) $0.\overline{24}$
Let $x = 0.242424... \quad (1)$
Since two digits are repeating, multiply by 100.
$100x = 24.242424... \quad (2)$
Subtract (1) from (2):
$100x - x = 24.2424... - 0.2424...$
$99x = 24$
$x = \frac{24}{99}$. Simplifying by dividing by 3, we get $x = \frac{8}{33}$.
(i) $0.\overline{24} = \frac{8}{33}$
(ii) $-5.1\overline{32}$
First, let's convert $5.1\overline{32}$ to a rational number. Let $y = 5.13232...$
Multiply by 10 to isolate the non-repeating part:
$10y = 51.3232... \quad (A)$
Since two digits are repeating, multiply (A) by 100:
$1000y = 5132.3232... \quad (B)$
Subtract (A) from (B):
$1000y - 10y = 5132.3232... - 51.3232...$
$990y = 5081$
$y = \frac{5081}{990}$
Since the original number was negative:
(ii) $-5.1\overline{32} = -\frac{5081}{990}$
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To represent $\sqrt{9.3}$ on a number line, we use a geometric construction.
Steps of Construction:
- Draw a horizontal line and mark a point A on it.
- Mark a point B on the line such that the length of the segment AB is 9.3 units.
- From B, mark a point C on the line such that the length of BC is 1 unit. Now, AC = 9.3 + 1 = 10.3 units.
- Find the midpoint of the segment AC. Let's call it O. The midpoint is at a distance of $\frac{10.3}{2} = 5.15$ from A.
- With O as the center and OA (or OC) as the radius, draw a semicircle.
- Draw a line perpendicular to AC passing through the point B. Let this perpendicular line intersect the semicircle at a point D.
- The length of the segment BD is exactly $\sqrt{9.3}$ units.
- To represent this on the number line, consider the initial line as the number line with B as the origin (0). With B as the center and BD as the radius, draw an arc that cuts the number line at a point E.
- The point E on the number line represents the value $\sqrt{9.3}$.
The construction described above correctly places the point representing $\sqrt{9.3}$ on the number line.
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The decimal expansion of $\sqrt{3}$ is found using the long division method for square roots. $\sqrt{3}$ is an irrational number, so its decimal expansion is non-terminating and non-repeating.
Calculation (first few steps):
1.732...
_________
1 | 3.00 00 00
| -1
|----------
27 | 2 00
| -1 89
|----------
343| 11 00
| -10 29
|----------
3462| 71 00
| -69 24
|----------
| 1 76
The process continues indefinitely.
The decimal expansion of $\sqrt{3}$ is approximately 1.73205...
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This is one of De Morgan's Laws. We will verify it by drawing Venn diagrams for the Left Hand Side (LHS) and the Right Hand Side (RHS) and showing they are identical.
LHS: $(A \cap B)'$
Step 2: Shade $(A \cap B)'$. This is everything *except* the region shaded in Step 1.
RHS: $A' \cup B'$
A' (shaded light blue):
B' (shaded light pink):
Step 4: Shade $A' \cup B'$. This is the union (combination) of all shaded areas from Step 3.
Conclusion:
The final shaded region for $(A \cap B)'$ in Step 2 is identical to the final shaded region for $A' \cup B'$ in Step 4. Both diagrams show the entire area outside of the intersection of A and B is shaded.
Thus, the law $(A \cap B)' = A' \cup B'$ is verified by Venn diagrams.