Tirunelveli District
Common First Mid Term Test - 2025
Standard 9 - MATHEMATICS
Part - A
I. Choose the correct answer:
5 × 1 = 5
1) If \(B \subset A\), then \(n(A \cap B)\) is
Answer: b) \(n(B)\)
Explanation:
If B is a subset of A (\(B \subset A\)), it means all elements of set B are also present in set A. The intersection of A and B (\(A \cap B\)) is the set of elements common to both A and B. Since all elements of B are in A, the common elements are exactly the set B itself.
So, \(A \cap B = B\).
Therefore, the number of elements in \(A \cap B\) is equal to the number of elements in B, which is written as \(n(A \cap B) = n(B)\).
2) Let \(A = \{\emptyset\}\) and \(B = P(A)\), then \(A \cap B\) is
Answer: b) \(\{\emptyset\}\)
Explanation:
Given \(A = \{\emptyset\}\). This is a set containing one element, which is the empty set.
The power set of A, \(P(A)\), is the set of all subsets of A. The subsets of A are the empty set \(\emptyset\) and the set A itself, \(\{\emptyset\}\).
So, \(B = P(A) = \{\emptyset, \{\emptyset\}\}\).
Now we find the intersection of A and B:
\(A \cap B = \{\emptyset\} \cap \{\emptyset, \{\emptyset\}\}\).
The common element between A and B is \(\emptyset\).
Therefore, \(A \cap B = \{\emptyset\}\).
3) For any three sets A, B and C, \((A-B) \cap (B-C)\) is equal to
Answer: d) \(\emptyset\)
Explanation:
The set \(A-B\) contains elements that are in A but not in B.
The set \(B-C\) contains elements that are in B but not in C.
By definition, any element in \(A-B\) cannot be in B. Conversely, every element in \(B-C\) must be in B.
Since the two sets have no elements in common, their intersection is the empty set (\(\emptyset\)).
4) Which one of the following is an irrational number?
Answer: d) \(\pi\)
Explanation:
- a) \(\sqrt{25} = 5\), which is a rational number.
- b) \(\sqrt{\frac{9}{4}} = \frac{3}{2}\), which is a rational number.
- c) \(\frac{7}{11}\) is a fraction of two integers, so it is a rational number.
- d) \(\pi\) is a non-terminating, non-repeating decimal (3.14159...), which is the definition of an irrational number.
5) \(0.\overline{34} + 0.3\overline{4}\) =
(Note: Based on visual analysis, the question is interpreted as \(0.\overline{34} + 0.3\overline{4}\). There might be a typo in the question or options as this sum doesn't perfectly match any option. The closest intended question might be \(0.\overline{34} + 0.\overline{34}\) which equals \(0.\overline{68}\) (Option b). We solve the question as visually represented.)
Solution:
Let's solve based on the most likely interpretation of the bars in the question image, which appears to be \(0.\overline{34} + 0.3\overline{4}\).
First, convert the repeating decimals to fractions.
Let \(x = 0.\overline{34} = 0.343434...\)
$$100x = 34.3434...$$
$$100x - x = 34 \implies 99x = 34 \implies x = \frac{34}{99}$$
Let \(y = 0.3\overline{4} = 0.3444...\)
$$10y = 3.444...$$
$$100y = 34.444...$$
$$100y - 10y = 31 \implies 90y = 31 \implies y = \frac{31}{90}$$
Now, add the two fractions:
$$ x + y = \frac{34}{99} + \frac{31}{90} $$
The least common multiple (LCM) of 99 and 90 is 990.
$$ = \frac{34 \times 10}{990} + \frac{31 \times 11}{990} $$
$$ = \frac{340}{990} + \frac{341}{990} = \frac{681}{990} $$
To convert this back to a decimal, we divide 681 by 990:
$$ \frac{681}{990} = 0.6878787... = 0.6\overline{87} $$
This matches option (d) if we interpret it as a bar over 87.
Part - B
II. Answer ANY FIVE questions only: (Question number 12 is compulsory)
6 × 2 = 12
6) Define Equivalent sets with an example.
Answer:
Equivalent Sets: Two sets A and B are said to be equivalent sets if they contain the same number of elements. The order of elements or the elements themselves do not have to be the same. This is denoted as \(n(A) = n(B)\).
Example: Let \(A = \{1, 2, 3\}\) and \(B = \{a, b, c\}\).
Here, the number of elements in set A is \(n(A) = 3\), and the number of elements in set B is \(n(B) = 3\).
Since \(n(A) = n(B)\), sets A and B are equivalent.
Example: Let \(A = \{1, 2, 3\}\) and \(B = \{a, b, c\}\).
Here, the number of elements in set A is \(n(A) = 3\), and the number of elements in set B is \(n(B) = 3\).
Since \(n(A) = n(B)\), sets A and B are equivalent.
7) If \(n(A) = 0\), find \(n[P(A)]\).
Answer:
Given that \(n(A) = 0\). This means that set A is an empty set, \(A = \emptyset\).
The number of elements in the power set of A, denoted as \(n[P(A)]\), is given by the formula:
$$ n[P(A)] = 2^{n(A)} $$
Substituting the given value \(n(A) = 0\):
$$ n[P(A)] = 2^0 $$
$$ n[P(A)] = 1 $$
The power set \(P(A)\) contains only one element, which is the empty set itself: \(P(A) = \{\emptyset\}\).
8) Let \(A = \{x:x \text{ is an even natural number } 1
Answer:
First, let's list the elements of each set.
Set A consists of even natural numbers greater than 1 and less than or equal to 12.
$$ A = \{2, 4, 6, 8, 10, 12\} $$
Set B consists of multiples of 3 that are natural numbers less than or equal to 12.
$$ B = \{3, 6, 9, 12\} $$
Now, we find the intersection of A and B (\(A \cap B\)), which contains elements common to both sets.
$$ A \cap B = \{2, 4, 6, 8, 10, 12\} \cap \{3, 6, 9, 12\} $$
The common elements are 6 and 12.
$$ A \cap B = \{6, 12\} $$
9) Find any three rational numbers between \(-\frac{7}{11}\) and \(\frac{2}{11}\).
Answer:
We need to find three rational numbers between \(-\frac{7}{11}\) and \(\frac{2}{11}\). Since the denominators are the same, we can look at the integers between the numerators -7 and 2.
Integers between -7 and 2 are -6, -5, -4, -3, -2, -1, 0, 1.
We can choose any three corresponding rational numbers.
Three rational numbers are:
$$ -\frac{6}{11}, \quad -\frac{5}{11}, \quad \text{and} \quad \frac{1}{11} $$
10) Verify that \(1 = 0.\overline{9}\).
Answer:
To verify that \(1 = 0.\overline{9}\), we convert the repeating decimal \(0.\overline{9}\) into a fraction.
Let \(x = 0.\overline{9}\), which means \(x = 0.999...\)
Multiply by 10 to shift the decimal point:
$$ 10x = 9.999... $$
Now, subtract the original equation from this new equation:
$$ 10x - x = (9.999...) - (0.999...) $$
$$ 9x = 9 $$
Divide by 9:
$$ x = \frac{9}{9} = 1 $$
Since we started with \(x = 0.\overline{9}\) and found that \(x = 1\), we have verified that \(1 = 0.\overline{9}\).
11) Find the value of \((243)^{\frac{2}{5}}\).
Answer:
We need to evaluate \((243)^{\frac{2}{5}}\). This can be written as \(((243)^{\frac{1}{5}})^2\).
First, find the 5th root of 243. We look for a number which, when multiplied by itself 5 times, gives 243.
$$ 3 \times 3 \times 3 \times 3 \times 3 = 3^5 = 243 $$
So, \((243)^{\frac{1}{5}} = 3\).
Now, we square this result:
$$ ((243)^{\frac{1}{5}})^2 = (3)^2 = 9 $$
Therefore, \((243)^{\frac{2}{5}} = 9\).
12) Represent \(A \Delta B\) through Venn diagram.
Answer:
The symmetric difference of two sets A and B, denoted by \(A \Delta B\), is the set of elements which are in either of the sets, but not in their intersection. It is defined as \(A \Delta B = (A - B) \cup (B - A)\).
The Venn diagram below represents \(A \Delta B\). The shaded region shows the elements that are in A only or in B only.
Shaded region represents \(A \Delta B\)
Part - C
III. Answer ANY FIVE questions only: (Question number 18 is compulsory)
5 × 5 = 25
13) Let \(U = \{0, 1, 2, 3, 4, 5, 6, 7\}\), \(A = \{1, 3, 5, 7\}\) and \(B = \{0, 2, 3, 5, 7\}\). Find the following sets: (i) A' (ii) B' (iii) A' \(\cup\) B' (iv) A' \(\cap\) B' (v) (A \(\cup\) B)'.
Answer:
Given:
\(U = \{0, 1, 2, 3, 4, 5, 6, 7\}\)
\(A = \{1, 3, 5, 7\}\)
\(B = \{0, 2, 3, 5, 7\}\)
(i) A' (Complement of A): Elements in U but not in A.
$$ A' = U - A = \{0, 1, 2, 3, 4, 5, 6, 7\} - \{1, 3, 5, 7\} = \{0, 2, 4, 6\} $$
(ii) B' (Complement of B): Elements in U but not in B.
$$ B' = U - B = \{0, 1, 2, 3, 4, 5, 6, 7\} - \{0, 2, 3, 5, 7\} = \{1, 4, 6\} $$
(iii) A' \(\cup\) B': Union of A' and B'.
$$ A' \cup B' = \{0, 2, 4, 6\} \cup \{1, 4, 6\} = \{0, 1, 2, 4, 6\} $$
(iv) A' \(\cap\) B': Intersection of A' and B'.
$$ A' \cap B' = \{0, 2, 4, 6\} \cap \{1, 4, 6\} = \{4, 6\} $$
(v) (A \(\cup\) B)': Complement of (A \(\cup\) B).
First, find A \(\cup\) B:
$$ A \cup B = \{1, 3, 5, 7\} \cup \{0, 2, 3, 5, 7\} = \{0, 1, 2, 3, 5, 7\} $$
Now, find the complement:
$$ (A \cup B)' = U - (A \cup B) = \{0, 1, 2, 3, 4, 5, 6, 7\} - \{0, 1, 2, 3, 5, 7\} = \{4, 6\} $$
(Note: This also verifies De Morgan's Law: \((A \cup B)' = A' \cap B'\)).
14) If \(A = \{p, q, r, s\}\), \(B = \{m, n, q, s, t\}\) and \(C = \{m, n, p, q, s\}\), then verify the associative property of union of sets.
Answer:
The associative property of union of sets is \( (A \cup B) \cup C = A \cup (B \cup C) \).
Given sets:
\(A = \{p, q, r, s\}\)
\(B = \{m, n, q, s, t\}\)
\(C = \{m, n, p, q, s\}\)
LHS: \( (A \cup B) \cup C \)
First, find \(A \cup B\):
$$ A \cup B = \{p, q, r, s\} \cup \{m, n, q, s, t\} = \{m, n, p, q, r, s, t\} $$
Now, find the union of this result with C:
$$ (A \cup B) \cup C = \{m, n, p, q, r, s, t\} \cup \{m, n, p, q, s\} = \{m, n, p, q, r, s, t\} $$
RHS: \( A \cup (B \cup C) \)
First, find \(B \cup C\):
$$ B \cup C = \{m, n, q, s, t\} \cup \{m, n, p, q, s\} = \{m, n, p, q, s, t\} $$
Now, find the union of A with this result:
$$ A \cup (B \cup C) = \{p, q, r, s\} \cup \{m, n, p, q, s, t\} = \{m, n, p, q, r, s, t\} $$
Since LHS = RHS, the associative property of union of sets is verified.
15) Let \(A = \{b, d, e, g, h\}\) and \(B = \{a, c, e, h\}\). Verify that \(n(A-B) = n(A) – n(A \cap B)\).
Answer:
Given sets:
\(A = \{b, d, e, g, h\}\)
\(B = \{a, c, e, h\}\)
We need to verify the property \(n(A-B) = n(A) – n(A \cap B)\).
LHS: \(n(A-B)\)
First, find the set \(A-B\) (elements in A but not in B):
$$ A-B = \{b, d, e, g, h\} - \{a, c, e, h\} = \{b, d, g\} $$
The number of elements in \(A-B\) is:
$$ n(A-B) = 3 $$
RHS: \(n(A) – n(A \cap B)\)
First, find \(n(A)\):
$$ n(A) = 5 $$
Next, find the set \(A \cap B\) (elements common to A and B):
$$ A \cap B = \{b, d, e, g, h\} \cap \{a, c, e, h\} = \{e, h\} $$
The number of elements in \(A \cap B\) is:
$$ n(A \cap B) = 2 $$
Now calculate the RHS:
$$ n(A) – n(A \cap B) = 5 - 2 = 3 $$
Since LHS = 3 and RHS = 3, we have LHS = RHS. The property is verified.
16) Represent \(\sqrt{9.3}\) on a number line.
Answer:
To represent \(\sqrt{9.3}\) on the number line, we use a geometric construction.
Steps of Construction:
- Draw a horizontal line and mark a point A on it. Let A represent the number 0.
- From A, measure a distance of 9.3 units and mark the point as B. So, \(AB = 9.3\).
- From B, measure a distance of 1 unit and mark the point as C. So, \(BC = 1\).
- Find the midpoint of the line segment AC. Let's call this midpoint O. The length of AC is \(9.3 + 1 = 10.3\), so the midpoint O is at \(10.3 / 2 = 5.15\) from A.
- With O as the center and OA (or OC) as the radius, draw a semicircle.
- At point B, draw a line perpendicular to AC that intersects the semicircle at a point D.
- The length of the line segment BD is equal to \(\sqrt{9.3}\).
- To represent this on the number line, keep the compass point at B (which is at 9.3 on our line), set the radius to BD, and draw an arc that intersects the number line at a point E. Since our starting point A is 0, we should place the compass point at A, set the radius to BD, and draw an arc. A better way is to consider B as the origin for a moment. Let's re-state the final step for clarity: With B as the center and BD as the radius, draw an arc that intersects the main line at a point E. The point E on the line represents the number \(\sqrt{9.3}\) if B is considered the origin. Since A is our origin (0), the point we are looking for is at a distance of BD from B. So we transfer the length BD starting from A. Place compass point at A, radius equal to BD, and mark point P on the line. Point P represents \(\sqrt{9.3}\).
Geometric construction of \(\sqrt{9.3}\).
17) Find the 5th root of \(\frac{1024}{3125}\).
Answer:
We need to find the value of \( \left(\frac{1024}{3125}\right)^{\frac{1}{5}} \).
This is equal to \( \frac{1024^{\frac{1}{5}}}{3125^{\frac{1}{5}}} \).
First, find the 5th root of 1024. We need a number that, when multiplied by itself five times, equals 1024.
Let's try powers of 2: \(2^2=4, 2^4=16, 2^8=256\). We know \(2^{10} = 1024\).
So, \(1024^{\frac{1}{5}} = (2^{10})^{\frac{1}{5}} = 2^{10 \times \frac{1}{5}} = 2^2 = 4\).
Next, find the 5th root of 3125. This number ends in 5, so its root likely ends in 5.
Let's try powers of 5: \(5^1=5, 5^2=25, 5^3=125, 5^4=625, 5^5=3125\).
So, \(3125^{\frac{1}{5}} = (5^5)^{\frac{1}{5}} = 5^{5 \times \frac{1}{5}} = 5^1 = 5\).
Combining the results:
$$ \left(\frac{1024}{3125}\right)^{\frac{1}{5}} = \frac{4}{5} $$
18) Find the decimal expansion of \(\sqrt{3}\).
Answer:
To find the decimal expansion of \(\sqrt{3}\), we use the long division method. We will find it up to three decimal places.
1. 7 3 2
+----------------
1 | 3. 00 00 00
| 1
+----------------
27 | 2 00
| 1 89
+----------------
343 | 11 00
| 10 29
+----------------
3462 | 71 00
| 69 24
+----------------
1 76
Steps:
- Start with the number 3. The largest integer whose square is less than or equal to 3 is 1. So, the first digit is 1.
- Subtract \(1^2=1\) from 3, which leaves 2. Bring down a pair of zeros to make it 200.
- Double the current quotient (1), which gives 2. We need to find a digit 'x' such that 2x \(\times\) x is close to 200. \(27 \times 7 = 189\). So the next digit is 7.
- Subtract 189 from 200, which leaves 11. Bring down another pair of zeros to make it 1100.
- Double the current quotient (17), which gives 34. We need to find a digit 'y' such that 34y \(\times\) y is close to 1100. \(343 \times 3 = 1029\). So the next digit is 3.
- Subtract 1029 from 1100, which leaves 71. Bring down another pair of zeros to make it 7100.
- Double the current quotient (173), which gives 346. We need a digit 'z' such that 346z \(\times\) z is close to 7100. \(3462 \times 2 = 6924\). So the next digit is 2.
Part - D
IV. Answer the following:
1 × 8 = 8
19) a) In a party of 45 people, each one likes tea or coffee or both. 35 people like tea and 20 people like coffee. Find the number of people who
- like both tea and coffee
- do not like tea
- do not like coffee
Answer a):
Let T be the set of people who like tea, and C be the set of people who like coffee.
Given:
- Total number of people, \(n(T \cup C) = 45\) (since each one likes tea or coffee or both).
- Number of people who like tea, \(n(T) = 35\).
- Number of people who like coffee, \(n(C) = 20\).
(OR)
b) Verify \(A-(B \cup C) = (A-B) \cap (A-C)\) using Venn diagram.
Answer b):
We will verify the set identity \(A-(B \cup C) = (A-B) \cap (A-C)\) using Venn diagrams.
1. LHS: \(A-(B \cup C)\)
This represents elements that are in A but not in the union of B and C.
2. RHS: \((A-B) \cap (A-C)\)
This represents the intersection of two sets: (elements in A but not B) and (elements in A but not C).
Conclusion:
As we can see from the Venn diagrams, the final shaded region for \(A-(B \cup C)\) is identical to the final shaded region for \((A-B) \cap (A-C)\). Both represent the part of set A that is not in B and not in C.
Hence, the identity is verified.
\(B \cup C\)
Final: \(A-(B \cup C)\)
\(A-B\)
\(A-C\)
Final: \((A-B) \cap (A-C)\)