🧮 10Th Maths First Mid Term Question Paper 2025 English medium | Tirunelveli

Standard 10 Maths - Common First Mid Term Test 2025 (Tirunelveli District) - Solved

Tirunelveli District

Common First Mid Term Test - 2025

Standard 10 - MATHEMATICS

Time: 1.30 Hrs. | Marks: 50

Part - A: Questions & Solutions

I. Choose the correct answer: 5 × 1 = 5

1) If A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then \(n[(A \cup C) \times B]\) is

a) 8 b) 20 c) 12 d) 16
Answer: c) 12 Step-by-step solution:
  1. Given sets: \(A = \{a, b, p\}\), \(B = \{2, 3\}\), \(C = \{p, q, r, s\}\).
  2. First, find the union of A and C: \(A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}\).
  3. Find the number of elements in \(A \cup C\): \(n(A \cup C) = 6\).
  4. Find the number of elements in B: \(n(B) = 2\).
  5. The number of elements in the Cartesian product \((A \cup C) \times B\) is given by \(n(A \cup C) \times n(B)\).
  6. Calculate the value: \(n[(A \cup C) \times B] = 6 \times 2 = 12\).

2) Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function \(f: A \to B\) given by \(f = \{(1, 4), (2, 8), (3, 9), (4, 10)\}\) is

a) Many-one function b) Identity function c) One-to-one function d) Into function
Answer: c) One-to-one function Step-by-step solution:
  1. The function is given by the set of ordered pairs: \(f = \{(1, 4), (2, 8), (3, 9), (4, 10)\}\).
  2. Check for One-to-one (Injective): A function is one-to-one if every distinct element in the domain A has a distinct image in the codomain B.
    • f(1) = 4
    • f(2) = 8
    • f(3) = 9
    • f(4) = 10
    Since all images (4, 8, 9, 10) are unique for each element in the domain, the function is one-to-one.
  3. Check for Into function: A function is 'into' if its range is a proper subset of its codomain. The range of f is {4, 8, 9, 10}. The codomain B is {4, 8, 9, 10}. Since the range is equal to the codomain, the function is 'onto', not 'into'.
  4. Many-one function: This is incorrect as we've established it is one-to-one.
  5. Identity function: This requires f(x) = x, which is not true here (e.g., f(1) ≠ 1).
  6. Therefore, the correct classification among the choices is a one-to-one function.

3) \(f(x) = (x+1)^3 - (x-1)^3\) represents a function which is

a) linear b) cubic c) reciprocal d) quadratic
Answer: d) quadratic Step-by-step solution:
  1. We need to simplify the expression for \(f(x)\).
  2. Recall the algebraic identities:
    • \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
    • \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\)
  3. Apply these to \(f(x)\) with a=x and b=1:
    • \((x+1)^3 = x^3 + 3(x^2)(1) + 3(x)(1^2) + 1^3 = x^3 + 3x^2 + 3x + 1\)
    • \((x-1)^3 = x^3 - 3(x^2)(1) + 3(x)(1^2) - 1^3 = x^3 - 3x^2 + 3x - 1\)
  4. Now, subtract the second expression from the first:

    \(f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)\)

    \(f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1\)

  5. Combine like terms:

    \(f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1)\)

    \(f(x) = 0 + 6x^2 + 0 + 2\)

    \(f(x) = 6x^2 + 2\)

  6. The resulting function is a polynomial of degree 2, which is a quadratic function.

4) The sum of the exponents of the prime factors in the prime factorization of 1729 is

a) 1 b) 2 c) 3 d) 4
Answer: c) 3 Step-by-step solution:
  1. We need to find the prime factorization of 1729.
  2. Start dividing by the smallest prime numbers. 1729 is not divisible by 2, 3, or 5.
  3. Try dividing by 7: \(1729 \div 7 = 247\).
  4. Now, find the prime factors of 247. It's not divisible by 7 or 11.
  5. Try dividing by 13: \(247 \div 13 = 19\).
  6. 19 is a prime number.
  7. So, the prime factorization of 1729 is \(7 \times 13 \times 19\).
  8. Writing this with exponents: \(1729 = 7^1 \times 13^1 \times 19^1\).
  9. The exponents of the prime factors are 1, 1, and 1.
  10. The sum of the exponents is \(1 + 1 + 1 = 3\).

5) If 6 times of 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P is

a) 0 b) 6 c) 7 d) 13
Answer: a) 0 Step-by-step solution:
  1. Let the first term of the Arithmetic Progression (A.P) be 'a' and the common difference be 'd'.
  2. The nth term of an A.P is given by the formula \(t_n = a + (n-1)d\).
  3. The 6th term is \(t_6 = a + (6-1)d = a + 5d\).
  4. The 7th term is \(t_7 = a + (7-1)d = a + 6d\).
  5. According to the question, 6 times the 6th term equals 7 times the 7th term:

    \(6 \times t_6 = 7 \times t_7\)

    \(6(a + 5d) = 7(a + 6d)\)

  6. Expand the equation:

    \(6a + 30d = 7a + 42d\)

  7. Rearrange the terms to one side:

    \(0 = (7a - 6a) + (42d - 30d)\)

    \(0 = a + 12d\)

  8. We need to find the 13th term, \(t_{13}\).

    \(t_{13} = a + (13-1)d = a + 12d\)

  9. From step 7, we know that \(a + 12d = 0\).
  10. Therefore, the 13th term of the A.P is 0.

Part - B: Questions & Solutions

II. Answer any six questions only (Question 12 is compulsory) 6 × 2 = 12

6) If \(A \times B = \{(3, 2), (3, 4), (5, 2), (5, 4)\}\) then find A and B.

Solution:
  1. The set A is the set of all first elements in the ordered pairs of \(A \times B\).

    First elements are 3 and 5.

    Therefore, \(A = \{3, 5\}\).

  2. The set B is the set of all second elements in the ordered pairs of \(A \times B\).

    Second elements are 2 and 4.

    Therefore, \(B = \{2, 4\}\).

7) Let A = {1, 2, 3, 4, ..., 45} and R be the relation defined as "is a square of" on A. Write R as a subset of A×A. Also, find the domain and range of R.

Solution:

The relation R is defined as "is a square of" on set A. This means for an ordered pair (x, y) in R, x is the square of y, i.e., \(x = y^2\), where both x and y must belong to set A.

  1. We find pairs (x, y) that satisfy \(x = y^2\) with \(x, y \in \{1, 2, ..., 45\}\).
    • If y = 1, then x = 1² = 1. The pair is (1, 1).
    • If y = 2, then x = 2² = 4. The pair is (4, 2).
    • If y = 3, then x = 3² = 9. The pair is (9, 3).
    • If y = 4, then x = 4² = 16. The pair is (16, 4).
    • If y = 5, then x = 5² = 25. The pair is (25, 5).
    • If y = 6, then x = 6² = 36. The pair is (36, 6).
    • If y = 7, then x = 7² = 49. Since 49 is not in A, we stop here.
  2. The relation R as a set of ordered pairs:

    \(R = \{(1, 1), (4, 2), (9, 3), (16, 4), (25, 5), (36, 6)\}

  3. Domain of R: The set of all first elements of the pairs in R.

    Domain = {1, 4, 9, 16, 25, 36}

  4. Range of R: The set of all second elements of the pairs in R.

    Range = {1, 2, 3, 4, 5, 6}

8) Define: (i) One-one function (ii) Identity function

Solution:
  1. (i) One-one function (or Injective function):
    A function \(f: A \to B\) is called a one-to-one function if distinct elements of the domain A have distinct images in the codomain B.
    In other words, for all \(x_1, x_2 \in A\), if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\).
  2. (ii) Identity function:
    A function \(f: A \to A\) is called the identity function if every element of set A is mapped to itself.
    That is, \(f(x) = x\) for all \(x \in A\). The domain and codomain must be the same set.

9) Find the value of K, such that \(fog = gof\). \(f(x) = 3x+2\), \(g(x) = 6x-k\).

Solution:
  1. Given functions: \(f(x) = 3x+2\) and \(g(x) = 6x-k\).
  2. First, find the composition \(fog(x)\):

    \(fog(x) = f(g(x)) = f(6x-k)\)

    \(= 3(6x-k) + 2\)

    \(= 18x - 3k + 2\)

  3. Next, find the composition \(gof(x)\):

    \(gof(x) = g(f(x)) = g(3x+2)\)

    \(= 6(3x+2) - k\)

    \(= 18x + 12 - k\)

  4. We are given that \(fog = gof\), so we set their expressions equal:

    \(18x - 3k + 2 = 18x + 12 - k\)

  5. Subtract \(18x\) from both sides:

    \(-3k + 2 = 12 - k\)

  6. Solve for k:

    \(2 - 12 = -k + 3k\)

    \(-10 = 2k\)

    \(k = \frac{-10}{2} = -5\)

  7. The value of k is -5.

10) Solve: \(5x \equiv 4 \pmod{6}\)

Solution:
  1. The congruence \(5x \equiv 4 \pmod{6}\) means that \(5x - 4\) is a multiple of 6.
  2. We can write this as an equation: \(5x - 4 = 6n\) for some integer n.
  3. Rearranging, we get \(5x = 6n + 4\).
  4. We can test integer values for x to find a solution:
    • If x = 0, \(5(0) = 0\). \(0 \not\equiv 4 \pmod{6}\).
    • If x = 1, \(5(1) = 5\). \(5 \not\equiv 4 \pmod{6}\).
    • If x = 2, \(5(2) = 10\). \(10 = 1 \times 6 + 4\). So, \(10 \equiv 4 \pmod{6}\). This is a solution.
  5. Therefore, the smallest positive integer solution is \(x = 2\). The general solution is given by \(x \equiv 2 \pmod{6}\), which means x can be ..., -4, 2, 8, 14, ...
  6. The solution is \(x=2\).

11) Which term of the A.P. 16, 11, 6, 1, ... is -54?

Solution:
  1. The given Arithmetic Progression (A.P) is 16, 11, 6, 1, ...
  2. The first term is \(a = 16\).
  3. The common difference is \(d = 11 - 16 = -5\).
  4. We want to find the term number 'n' for which the term value \(t_n\) is -54.
  5. Using the formula for the nth term of an A.P: \(t_n = a + (n-1)d\).
  6. Substitute the known values:

    \(-54 = 16 + (n-1)(-5)\)

  7. Solve for n:

    \(-54 - 16 = -5(n-1)\)

    \(-70 = -5(n-1)\)

    \(\frac{-70}{-5} = n-1\)

    \(14 = n-1\)

    \(n = 14 + 1 = 15\)

  8. Therefore, the 15th term of the A.P. is -54.

12) (Compulsory) Find the sum of first 28 terms of an A.P. whose nth term is 4n-3.

Solution:
  1. The nth term of the A.P is given by \(t_n = 4n - 3\).
  2. To find the sum of the first 28 terms, we can use the formula \(S_n = \frac{n}{2}(a + l)\), where 'a' is the first term and 'l' is the last term.
  3. Find the first term (a) by setting n=1:

    \(a = t_1 = 4(1) - 3 = 1\)

  4. Find the 28th term (l) by setting n=28:

    \(l = t_{28} = 4(28) - 3 = 112 - 3 = 109\)

  5. Now calculate the sum of the first 28 terms (\(S_{28}\)) with n=28:

    \(S_{28} = \frac{28}{2}(1 + 109)\)

    \(S_{28} = 14(110)\)

    \(S_{28} = 1540\)

  6. The sum of the first 28 terms is 1540.

Part - C: Questions & Solutions

III. Answer any five questions only (Question 18 is compulsory) 5 × 5 = 25

13) Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that \(A \times (B - C) = (A \times B) - (A \times C)\).

Solution:
  1. First, define the sets A, B, and C based on the given descriptions.
    • A = {Natural numbers less than 8} = \(\{1, 2, 3, 4, 5, 6, 7\}\)
    • B = {Prime numbers less than 8} = \(\{2, 3, 5, 7\}\)
    • C = {Even prime number} = \(\{2\}\)
  2. Calculate the Left Hand Side (LHS): \(A \times (B - C)\)
    • First find \(B - C\): \(B - C = \{2, 3, 5, 7\} - \{2\} = \{3, 5, 7\}\).
    • Now find \(A \times (B - C)\):
      \(A \times (B - C) = \{1, 2, 3, 4, 5, 6, 7\} \times \{3, 5, 7\}\)

      \(= \{(1,3), (1,5), (1,7), (2,3), (2,5), (2,7), (3,3), (3,5), (3,7), (4,3), (4,5), (4,7), (5,3), (5,5), (5,7), (6,3), (6,5), (6,7), (7,3), (7,5), (7,7)\}\)

  3. Calculate the Right Hand Side (RHS): \((A \times B) - (A \times C)\)
    • First find \(A \times B\):
      \(A \times B = \{(1,2), (1,3), (1,5), (1,7), (2,2), (2,3), (2,5), (2,7), ... , (7,7)\}\).
    • Next find \(A \times C\):
      \(A \times C = \{1, 2, 3, 4, 5, 6, 7\} \times \{2\}\)

      \(= \{(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)\}\)

    • Now find \((A \times B) - (A \times C)\). This means we remove all elements of \(A \times C\) from \(A \times B\). In other words, we remove all pairs from \(A \times B\) where the second element is 2.

      \(= \{(1,3), (1,5), (1,7), (2,3), (2,5), (2,7), (3,3), (3,5), (3,7), (4,3), (4,5), (4,7), (5,3), (5,5), (5,7), (6,3), (6,5), (6,7), (7,3), (7,5), (7,7)\}\)

  4. Verification:

    Comparing the results from step 2 and step 3, we see that the sets are identical.

    Thus, LHS = RHS. Hence, it is verified that \(A \times (B - C) = (A \times B) - (A \times C)\).

14) Let A = {1, 2, 3, 4} and B = {2, 5, 8, 11, 14} be two sets. Let \(f: A \to B\) be a function given by \(f(x) = 3x-1\). Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) in a graphical form.

Solution:

First, let's find the images for each element in the domain A using \(f(x) = 3x - 1\).

  • \(f(1) = 3(1) - 1 = 2\)
  • \(f(2) = 3(2) - 1 = 5\)
  • \(f(3) = 3(3) - 1 = 8\)
  • \(f(4) = 3(4) - 1 = 11\)

(i) By Arrow Diagram:

An arrow diagram shows the mapping from elements of set A to elements of set B.

A 1 2 3 4 B 2 5 8 11 14

(ii) In a Table Form:

x f(x) = 3x - 1
1 2
2 5
3 8
4 11

(iii) As a Set of Ordered Pairs:

The function f can be represented as the set of all its input-output pairs.

\(f = \{(1, 2), (2, 5), (3, 8), (4, 11)\}\)

(iv) In a Graphical Form:

We plot the ordered pairs as points on a Cartesian plane. Since the domain is a discrete set of points, we only plot the points and do not connect them with a line.

Plot the points:
(1, 2)
(2, 5)
(3, 8)
(4, 11)
on a graph with X and Y axes.

15) Find the HCF of 396, 504, 636.

Solution:

We will use Euclid's Division Algorithm to find the HCF.

  1. Step 1: Find the HCF of the first two numbers, 504 and 396.

    \(504 = 1 \times 396 + 108\)

    \(396 = 3 \times 108 + 72\)

    \(108 = 1 \times 72 + 36\)

    \(72 = 2 \times 36 + 0\)

    The last non-zero remainder is 36. So, HCF(504, 396) = 36.

  2. Step 2: Find the HCF of the result from Step 1 (36) and the third number (636).

    \(636 = 17 \times 36 + 24\)

    \(36 = 1 \times 24 + 12\)

    \(24 = 2 \times 12 + 0\)

    The last non-zero remainder is 12. So, HCF(636, 36) = 12.

  3. The HCF of 396, 504, and 636 is 12.

16) The ratio of 6th and 8th term of an A.P is 7:9. Find the ratio of 9th term to 13th term.

Solution:
  1. Let 'a' be the first term and 'd' be the common difference of the A.P. The nth term is \(t_n = a + (n-1)d\).
  2. We are given that the ratio of the 6th term to the 8th term is 7:9.

    \(\frac{t_6}{t_8} = \frac{a + (6-1)d}{a + (8-1)d} = \frac{a+5d}{a+7d} = \frac{7}{9}\)

  3. Cross-multiply to find a relationship between 'a' and 'd':

    \(9(a+5d) = 7(a+7d)\)

    \(9a + 45d = 7a + 49d\)

    \(9a - 7a = 49d - 45d\)

    \(2a = 4d\)

    \(a = 2d\)

  4. Now, we need to find the ratio of the 9th term to the 13th term: \(\frac{t_9}{t_{13}}\).

    \(\frac{t_9}{t_{13}} = \frac{a + (9-1)d}{a + (13-1)d} = \frac{a+8d}{a+12d}\)

  5. Substitute \(a = 2d\) into this expression:

    \(\frac{t_9}{t_{13}} = \frac{(2d)+8d}{(2d)+12d} = \frac{10d}{14d}\)

  6. Simplify the fraction (cancel 'd'):

    \(\frac{10}{14} = \frac{5}{7}\)

  7. The ratio of the 9th term to the 13th term is 5:7.

17) Find the sum of all natural numbers between 300 and 600 which are divisible by 7.

Solution:
  1. We need to find the sum of an arithmetic progression. The numbers are between 300 and 600 and are divisible by 7.
  2. Find the first term (a): The first number greater than 300 that is divisible by 7.

    Divide 300 by 7: \(300 \div 7 = 42\) with a remainder of 6.

    The first term is \(300 + (7 - 6) = 301\).

    So, \(a = 301\).

  3. Find the last term (l): The last number less than 600 that is divisible by 7.

    Divide 600 by 7: \(600 \div 7 = 85\) with a remainder of 5.

    The last term is \(600 - 5 = 595\).

    So, \(l = 595\).

  4. The common difference \(d\) is 7. The A.P. is 301, 308, ..., 595.
  5. Find the number of terms (n):

    Using the formula \(l = a + (n-1)d\):

    \(595 = 301 + (n-1)7\)

    \(595 - 301 = 7(n-1)\)

    \(294 = 7(n-1)\)

    \(n-1 = \frac{294}{7} = 42\)

    \(n = 43\)

  6. Find the sum (S_n):

    Using the formula \(S_n = \frac{n}{2}(a+l)\):

    \(S_{43} = \frac{43}{2}(301 + 595)\)

    \(S_{43} = \frac{43}{2}(896)\)

    \(S_{43} = 43 \times 448\)

    \(S_{43} = 19264\)

  7. The sum of all natural numbers between 300 and 600 divisible by 7 is 19264.

18) (Compulsory) If \(f(x) = x^2\), \(g(x) = 3x\) and \(h(x) = x-2\), prove that \((fog)oh = fo(goh)\).

Solution:

We need to prove the associative property of function composition for the given functions.

  1. Calculate the Left Hand Side (LHS): \((fog)oh\)
    • First, find \(fog(x)\):
      \(fog(x) = f(g(x)) = f(3x) = (3x)^2 = 9x^2\).
    • Now, apply \(h(x)\) to this result: \((fog)oh(x) = (fog)(h(x))\).
      \((fog)(h(x)) = (fog)(x-2) = 9(x-2)^2\).
    • Expand the expression:
      \(9(x-2)^2 = 9(x^2 - 4x + 4) = 9x^2 - 36x + 36\).
  2. Calculate the Right Hand Side (RHS): \(fo(goh)\)
    • First, find \(goh(x)\):
      \(goh(x) = g(h(x)) = g(x-2) = 3(x-2) = 3x-6\).
    • Now, apply this result to \(f(x)\): \(fo(goh)(x) = f(goh(x))\).
      \(f(goh(x)) = f(3x-6) = (3x-6)^2\).
    • Expand the expression:
      \((3x-6)^2 = (3(x-2))^2 = 3^2(x-2)^2 = 9(x^2 - 4x + 4) = 9x^2 - 36x + 36\).
  3. Conclusion:

    From our calculations, LHS = \(9x^2 - 36x + 36\) and RHS = \(9x^2 - 36x + 36\).

    Since LHS = RHS, it is proved that \((fog)oh = fo(goh)\).

Part - D: Question & Solution

Answer the following: 1 × 8 = 8

19) a) Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac{7}{3}\) of the corresponding sides of the triangle PQR. (Scale factor \(\frac{7}{3} > 1\))

(OR)

b) Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac{3}{5}\) of the corresponding sides of the triangle PQR. (Scale factor \(\frac{3}{5} < 1\))

Solution:

These questions require geometric constructions. The steps are described below.

a) Construction for Scale Factor \(\frac{7}{3} > 1\)

Here we will construct a triangle P'QR' whose sides are \(\frac{7}{3}\) times the sides of triangle PQR. The new triangle will be larger than the original.

Steps of Construction:
  1. Draw any triangle PQR (the given triangle).
  2. Draw a ray QX starting from Q and making an acute angle with the side QR, on the side opposite to vertex P.
  3. Since the scale factor is \(\frac{7}{3}\), we need to mark 7 (the greater of 7 and 3) points on the ray QX. Use a compass to mark points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆, Q₇ on QX such that all segments are equal: \(QQ₁ = Q₁Q₂ = \dots = Q₆Q₇\).
  4. Join the point corresponding to the denominator (3), which is Q₃, to the vertex R. This forms the line segment Q₃R.
  5. Now, draw a line through the point corresponding to the numerator (7), which is Q₇, that is parallel to the line Q₃R. To do this, construct an angle at Q₇ equal to ∠QQ₃R. This new line will intersect the extended side QR at a point. Call this point R'.
  6. Finally, draw a line through R' that is parallel to the side PR. This line will intersect the extended side QP at a point. Call this point P'.
  7. The triangle P'QR' is the required similar triangle, with each side being \(\frac{7}{3}\) times the corresponding side of triangle PQR.
(OR)

b) Construction for Scale Factor \(\frac{3}{5} < 1\)

Here we will construct a triangle P'QR' whose sides are \(\frac{3}{5}\) times the sides of triangle PQR. The new triangle will be smaller than the original.

Steps of Construction:
  1. Draw any triangle PQR (the given triangle).
  2. Draw a ray QX starting from Q and making an acute angle with the side QR, on the side opposite to vertex P.
  3. Since the scale factor is \(\frac{3}{5}\), we need to mark 5 (the greater of 3 and 5) points on the ray QX. Use a compass to mark points Q₁, Q₂, Q₃, Q₄, Q₅ on QX such that all segments are equal: \(QQ₁ = Q₁Q₂ = \dots = Q₄Q₅\).
  4. Join the point corresponding to the denominator (5), which is Q₅, to the vertex R. This forms the line segment Q₅R.
  5. Now, draw a line through the point corresponding to the numerator (3), which is Q₃, that is parallel to the line Q₅R. This new line will intersect the side QR at a point. Call this point R'.
  6. Finally, draw a line through R' that is parallel to the side PR. This line will intersect the side QP at a point. Call this point P'.
  7. The triangle P'QR' is the required similar triangle, with each side being \(\frac{3}{5}\) times the corresponding side of triangle PQR.
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