9th Standard Maths | கணிதம் - First Mid Term Exam 2025 - Question Papers, Answer Keys Official Original QP (EM) | Tirupattur

Class 9 - First Mid Term Test 2024 - Mathematics | Solved Paper

Class 9 Mathematics

FIRST MID TERM TEST - 2024

Time Allowed: 1.30 Hours [Max. Marks: 50]

Question Paper

PART - I

7 x 1 = 7

Choose the correct Answer.

  1. For any three sets A, B and C, \( (A - B) \cap (B - C) \) is equal to
    (a) A only (b) B only (c) C only (d) \(\phi\)
  2. If \( U = \{x:x \in \mathbb{N} \text{ and } x < 10\} \), \( A = \{1,2,3,5,8\} \) and \( B = \{2,5,6,7,9\} \), then \( n((A \cup B)') \) is
    (a) 1 (b) 2 (c) 4 (d) 8
  3. If \( A \cup B = A \cap B \), then
    (a) \(A \neq B\) (b) \(A = B\) (c) \(A \subset B\) (d) \(B \subset A\)
  4. If \( A = \{x,y,z\} \) then the number of non-empty subsets of A is
    (a) 8 (b) 5 (c) 6 (d) 7
  5. \( \sqrt{27} + \sqrt{12} = \) _____
    (a) \(\sqrt{39}\) (b) \(5\sqrt{6}\) (c) \(5\sqrt{3}\) (d) \(3\sqrt{5}\)
  6. Which one of the following is an irrational number?
    (a) \(\sqrt{25}\) (b) \(\frac{\sqrt{9}}{4}\) (c) \(\frac{7}{11}\) (d) \(\pi\)
  7. Which one of the following has a terminating decimal expansion?
    (a) \(\frac{5}{64}\) (b) \(\frac{8}{9}\) (c) \(\frac{14}{15}\) (d) \(\frac{1}{12}\)

PART - II

5 x 2 = 10

Answer any 5 Questions. Q.No. 14 is compulsory.

  1. Write the set of letters of the following words in Roster form.
    (i) ASSESSMENT   (ii) PRINCIPAL
  2. Write down the power set of the following set \(A = \{a,b\}\).
  3. If \(n(A) = 0\), find \(n(P(A))\).
  4. Represent \(A \Delta B\) through a Venn diagram.
  5. Find any three rational Numbers between \(-\frac{7}{11}\) and \(\frac{2}{11}\).
  6. Simplify the following: \(5\sqrt{3} + 18\sqrt{3} - 2\sqrt{3}\).
  7. (Compulsory) Represent the number 2000.57 in scientific notation.

PART - III

5 x 5 = 25

Answer any 5 questions. (Q.No.21 is compulsory)

  1. Verify \(A - (B \cup C) = (A - B) \cap (A - C)\) using Venn diagrams.
  2. From the Venn diagram, verify that \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\).
    Venn Diagram for Question 16
  3. In a party of 60 people, 35 had Vanilla ice cream, 30 had Chocolate ice cream. All the people had at least one ice cream. Then how many of them had,
    (i) both Vanilla and Chocolate ice cream.
    (ii) only Vanilla ice cream.
    (iii) only Chocolate ice cream.
  4. Find the value of a and b if \( \frac{\sqrt{7}-2}{\sqrt{7}+2} = a\sqrt{7} + b \).
  5. Arrange surds in descending order: \( \sqrt{5}, \sqrt[4]{4}, \sqrt[3]{3} \).
  6. Represent 4.863 on the number line.
  7. (Compulsory) If \(A = \{0,2,4,6,8\}\), \(B = \{x:x \text{ is a prime number and } x<11\}\) and \(C = \{x:x \in \mathbb{N} \text{ and } 5

PART - IV

1 x 8 = 8

Answer the following:

  1. (a) Construct the \(\Delta LMN\) such that LM = 7.5 cm, MN = 5 cm, and LN = 8 cm. Locate its centroid.
    (OR)
    (b) Construct the Centroid of \(\Delta PQR\) whose sides are PQ = 8cm, QR = 6cm, RP = 7cm.

Solutions

PART - I: Solutions

  1. 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, these two sets have no elements in common (one contains elements not in B, the other contains elements that are in B). Therefore, their intersection is the empty set, \(\phi\).
    (d) \(\phi\)
  2. Explanation: Given: \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\)
    \(A = \{1, 2, 3, 5, 8\}\)
    \(B = \{2, 5, 6, 7, 9\}\)
    First, find \(A \cup B\): \(A \cup B = \{1, 2, 3, 5, 8\} \cup \{2, 5, 6, 7, 9\} = \{1, 2, 3, 5, 6, 7, 8, 9\}\).
    Next, find the complement \((A \cup B)'\): \((A \cup B)' = U - (A \cup B) = \{1,..,9\} - \{1,2,3,5,6,7,8,9\} = \{4\}\).
    Finally, find the number of elements: \(n((A \cup B)') = 1\).
    (a) 1
  3. Explanation: The condition \( A \cup B = A \cap B \) means that the union of the sets is equal to their intersection. This is only possible if the sets are identical. If there were any element in A but not in B, it would be in the union but not the intersection. Similarly, for any element in B but not in A. Thus, A and B must contain exactly the same elements.
    (b) \(A = B\)
  4. Explanation: For a set A with \(n(A)\) elements, the total number of subsets is \(2^{n(A)}\).
    Here, \(A = \{x, y, z\}\), so \(n(A) = 3\).
    Total number of subsets = \(2^3 = 8\).
    These 8 subsets include the empty set (\(\phi\)). The question asks for the number of non-empty subsets.
    Number of non-empty subsets = Total subsets - 1 (the empty set) = \(8 - 1 = 7\).
    (d) 7
  5. Explanation: We simplify each term first.
    \( \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \)
    \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \)
    Now, we add them: \( 3\sqrt{3} + 2\sqrt{3} = (3+2)\sqrt{3} = 5\sqrt{3} \).
    (c) \(5\sqrt{3}\)
  6. Explanation:
    (a) \(\sqrt{25} = 5\), which is a rational number.
    (b) \(\frac{\sqrt{9}}{4} = \frac{3}{4}\), which is a rational number.
    (c) \(\frac{7}{11}\) is a fraction of integers, which is a rational number.
    (d) \(\pi\) (Pi) is a famous non-terminating, non-repeating decimal, which is an irrational number.
    (d) \(\pi\)
  7. Explanation: A rational number \(\frac{p}{q}\) has a terminating decimal expansion if the prime factorization of the denominator \(q\) consists only of powers of 2 and 5 (i.e., \(q = 2^m \times 5^n\)).
    (a) \(\frac{5}{64}\): Denominator is \(64 = 2^6\). This will terminate.
    (b) \(\frac{8}{9}\): Denominator is \(9 = 3^2\). This will not terminate.
    (c) \(\frac{14}{15}\): Denominator is \(15 = 3 \times 5\). This will not terminate.
    (d) \(\frac{1}{12}\): Denominator is \(12 = 2^2 \times 3\). This will not terminate.
    (a) \(\frac{5}{64}\)

PART - II: Solutions

  1. Explanation: In Roster form, we list the unique elements (letters) of the word.

    (i) ASSESSMENT → The unique letters are A, S, E, M, N, T.
    Set = {A, S, E, M, N, T}

    (ii) PRINCIPAL → The unique letters are P, R, I, N, C, A, L.
    Set = {P, R, I, N, C, A, L}

  2. Explanation: The power set \(P(A)\) is the set of all possible subsets of A. For \(A = \{a,b\}\), the subsets are:
    • The empty set: \(\phi\)
    • Subsets with one element: \(\{a\}\), \(\{b\}\)
    • Subset with two elements: \(\{a,b\}\)
    Power set, \(P(A) = \{\phi, \{a\}, \{b\}, \{a,b\}\}\)
  3. Explanation: If \(n(A) = 0\), it means A is the empty set, \(A = \phi\). The number of elements in the power set, \(n(P(A))\), is given by the formula \(2^{n(A)}\).
    \(n(P(A)) = 2^0 = 1\).
    The only subset of an empty set is the empty set itself. So, \(P(\phi) = \{\phi\}\).
    \(n(P(A)) = 1\)
  4. Explanation: The symmetric difference \(A \Delta B\) is the set of elements which are in either of the sets, but not in their intersection. It is equivalent to \((A \cup B) - (A \cap B)\) or \((A - B) \cup (B - A)\). In a Venn diagram, this corresponds to shading the regions of A and B that do not overlap.
    Venn Diagram for \(A \Delta B\):
    U A B
  5. Explanation: We need to find three rational numbers between \(-\frac{7}{11}\) and \(\frac{2}{11}\). Since they have a common denominator, we can simply choose any three integers between -7 and 2 for the numerator.
    Possible integers: -6, -5, -4, -3, -2, -1, 0, 1.
    Choosing three, we get:
    \(-\frac{6}{11}, 0, \frac{1}{11}\)
  6. Explanation: All terms are like surds (contain \(\sqrt{3}\)). We can combine their coefficients. $$ 5\sqrt{3} + 18\sqrt{3} - 2\sqrt{3} = (5 + 18 - 2)\sqrt{3} $$ $$ = (23 - 2)\sqrt{3} $$ $$ = 21\sqrt{3} $$
    \(21\sqrt{3}\)
  7. Explanation: Scientific notation expresses a number as a product \( a \times 10^n \), where \(1 \le |a| < 10\).
    For the number 2000.57, we need to move the decimal point to the left until there is only one non-zero digit before it.
    We move the decimal point 3 places to the left: 2.00057.
    Since we moved it 3 places to the left, the power of 10 is +3.
    \(2.00057 \times 10^3\)

PART - III: Solutions

  1. Verification of \(A - (B \cup C) = (A - B) \cap (A - C)\) using Venn diagrams.

    LHS: \(A - (B \cup C)\)

    1. Draw three intersecting circles for A, B, and C.
    2. Shade the region for \(B \cup C\). This includes all of B and all of C.
    3. The final region for \(A - (B \cup C)\) is the part of A that is NOT in the shaded \(B \cup C\) region. This is the part of A that is completely outside of B and C.

    RHS: \((A - B) \cap (A - C)\)

    1. Step 1: Find \(A - B\). Shade the region of A that does not overlap with B.
    2. Step 2: Find \(A - C\). Shade the region of A that does not overlap with C.
    3. Step 3: Find the intersection. The final region is where the shadings from Step 1 and Step 2 overlap. This intersection is exactly the part of A that is outside both B and C.

    Since the final shaded region is the same for both LHS and RHS, the identity is verified.

  2. Verification from Venn Diagram.
    1. Find the sets and their cardinalities:
      • Set A = {5, 15, 10, 20} \(\implies n(A) = 4\)
      • Set B = {10, 20, 30, 40} \(\implies n(B) = 4\)
      • Intersection \(A \cap B\) = {10, 20} \(\implies n(A \cap B) = 2\)
      • Union \(A \cup B\) = {5, 15, 10, 20, 30, 40} \(\implies n(A \cup B) = 6\)
    2. Verify the formula \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\):
      • LHS: \(n(A \cup B) = 6\)
      • RHS: \(n(A) + n(B) - n(A \cap B) = 4 + 4 - 2 = 6\)
    Since LHS = RHS (6 = 6), the formula is verified.
  3. Let V be the set of people who had Vanilla and C be the set of people who had Chocolate. Given:
    • Total people, \(n(V \cup C) = 60\) (since everyone had at least one)
    • \(n(V) = 35\)
    • \(n(C) = 30\)

    (i) How many had both Vanilla and Chocolate ice cream?

    We use the formula: \(n(V \cup C) = n(V) + n(C) - n(V \cap C)\) $$ 60 = 35 + 30 - n(V \cap C) $$ $$ 60 = 65 - n(V \cap C) $$ $$ n(V \cap C) = 65 - 60 = 5 $$
    5 people had both.

    (ii) How many had only Vanilla ice cream?

    This is \(n(V - C) = n(V) - n(V \cap C)\) $$ n(V - C) = 35 - 5 = 30 $$
    30 people had only Vanilla.

    (iii) How many had only Chocolate ice cream?

    This is \(n(C - V) = n(C) - n(V \cap C)\) $$ n(C - V) = 30 - 5 = 25 $$
    25 people had only Chocolate.
  4. To find a and b, we first rationalize the denominator of the given expression. $$ \frac{\sqrt{7}-2}{\sqrt{7}+2} $$ Multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{7}-2 \): $$ = \frac{(\sqrt{7}-2)}{(\sqrt{7}+2)} \times \frac{(\sqrt{7}-2)}{(\sqrt{7}-2)} $$ $$ = \frac{(\sqrt{7}-2)^2}{(\sqrt{7})^2 - (2)^2} $$ Numerator: \( (\sqrt{7}-2)^2 = (\sqrt{7})^2 - 2(\sqrt{7})(2) + 2^2 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7} \)
    Denominator: \( (\sqrt{7})^2 - (2)^2 = 7 - 4 = 3 \) $$ = \frac{11 - 4\sqrt{7}}{3} = \frac{11}{3} - \frac{4}{3}\sqrt{7} $$ Now, we compare this with \( a\sqrt{7} + b \): $$ -\frac{4}{3}\sqrt{7} + \frac{11}{3} = a\sqrt{7} + b $$ By comparing the coefficients, we get:
    \(a = -\frac{4}{3}\) and \(b = \frac{11}{3}\)
  5. To arrange surds in descending order, we must make their orders (indices) equal. The surds are \( \sqrt{5} = 5^{1/2} \), \( \sqrt[4]{4} = 4^{1/4} \), and \( \sqrt[3]{3} = 3^{1/3} \). The orders are 2, 4, and 3. The LCM of (2, 4, 3) is 12.

    1. Convert \(5^{1/2}\) to the order 12: \( 5^{1/2} = 5^{6/12} = \sqrt[12]{5^6} = \sqrt[12]{15625} \)

    2. Convert \(4^{1/4}\) to the order 12: \( 4^{1/4} = 4^{3/12} = \sqrt[12]{4^3} = \sqrt[12]{64} \)

    3. Convert \(3^{1/3}\) to the order 12: \( 3^{1/3} = 3^{4/12} = \sqrt[12]{3^4} = \sqrt[12]{81} \)

    Now compare the numbers inside the 12th root: 15625, 64, and 81. In descending order: \( 15625 > 81 > 64 \). Therefore, \( \sqrt[12]{15625} > \sqrt[12]{81} > \sqrt[12]{64} \).
    Descending Order: \( \sqrt{5}, \sqrt[3]{3}, \sqrt[4]{4} \)
  6. Representing 4.863 on the number line using successive magnification.
    1. Step 1: Locate 4.8. The number 4.863 lies between 4 and 5. We divide the interval [4, 5] into 10 equal parts. The eighth mark represents 4.8. Our number lies between 4.8 and 4.9.
    2. Step 2: Locate 4.86. We magnify the interval [4.8, 4.9] and divide it into 10 equal parts. The sixth mark represents 4.86. Our number lies between 4.86 and 4.87.
    3. Step 3: Locate 4.863. We magnify the interval [4.86, 4.87] and divide it into 10 equal parts. The third mark in this interval is precisely 4.863.
  7. Verify the distributive law: \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) Given sets:
    • \(A = \{0, 2, 4, 6, 8\}\)
    • \(B = \{x:x \text{ is prime, } x<11\} = \{2, 3, 5, 7\}\)
    • \(C = \{x:x \in \mathbb{N}, 5

    LHS: \(A \cup (B \cap C)\)

    First, find \(B \cap C\): \(B \cap C = \{2, 3, 5, 7\} \cap \{6, 7, 8\} = \{7\}\).
    Now, find \(A \cup (B \cap C)\): \(A \cup \{7\} = \{0, 2, 4, 6, 8\} \cup \{7\} = \{0, 2, 4, 6, 7, 8\}\).
    LHS = {0, 2, 4, 6, 7, 8}

    RHS: \((A \cup B) \cap (A \cup C)\)

    First, find \(A \cup B\): \(A \cup B = \{0, 2, 4, 6, 8\} \cup \{2, 3, 5, 7\} = \{0, 2, 3, 4, 5, 6, 7, 8\}\).
    Next, find \(A \cup C\): \(A \cup C = \{0, 2, 4, 6, 8\} \cup \{6, 7, 8\} = \{0, 2, 4, 6, 7, 8\}\).
    Finally, find their intersection: \((A \cup B) \cap (A \cup C) = \{0, 2, 3, 4, 5, 6, 7, 8\} \cap \{0, 2, 4, 6, 7, 8\} = \{0, 2, 4, 6, 7, 8\}\).
    RHS = {0, 2, 4, 6, 7, 8}

    Since LHS = RHS, the property is verified.

PART - IV: Solution

  1. This question requires geometric construction. Below are the steps for both options.

    (a) Construct \(\Delta LMN\) and its centroid. (LM=7.5, MN=5, LN=8)

    1. Construct the Triangle:
      1. Draw a line segment LN of length 8 cm.
      2. With L as the center, use a compass to draw an arc with a radius of 7.5 cm.
      3. With N as the center, draw another arc with a radius of 5 cm.
      4. Let the point where the two arcs intersect be M. Join LM and MN. \(\Delta LMN\) is the required triangle.
    2. Locate the Centroid:
      1. The centroid is the point of intersection of the medians. A median connects a vertex to the midpoint of the opposite side.
      2. Find the midpoint of any two sides, for example, LN and MN. Let's call them P and Q respectively. (To find a midpoint, draw perpendicular bisectors).
      3. Draw the medians MP (from vertex M to midpoint P of LN) and LQ (from vertex L to midpoint Q of MN).
      4. The point where the medians MP and LQ intersect is the centroid, G.

    (b) (OR) Construct \(\Delta PQR\) and its centroid. (PQ=8, QR=6, RP=7)

    1. Construct the Triangle:
      1. Draw a line segment PQ of length 8 cm.
      2. With P as the center, use a compass to draw an arc with a radius of 7 cm.
      3. With Q as the center, draw another arc with a radius of 6 cm.
      4. Let the point where the two arcs intersect be R. Join PR and QR. \(\Delta PQR\) is the required triangle.
    2. Locate the Centroid:
      1. Find the midpoint of any two sides, for example, PQ and QR. Let's call them S and T respectively.
      2. Draw the medians RS (from vertex R to midpoint S of PQ) and PT (from vertex P to midpoint T of QR).
      3. The point where the medians RS and PT intersect is the centroid, G.