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

Class 10 Maths - First Mid Term Test 2025 | Solved Paper

Shining Star Matric Hr Sec School, Kalpakkam

Class: 10 | FIRST MID TERM TEST - 2025

MATHEMATICS

Time Allowed: 1.30 Hours | Max. Marks: 50

Question Paper

PART-A 7 x 1 = 7
I. Choose the correct Answer.
  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
  2. The range of the relation \( R = \{ (x, x^2) \mid x \text{ is a prime number less than 13} \} \) is
    • a) \( \{2, 3, 5, 7\} \)
    • b) \( \{2, 3, 5, 7, 11\} \)
    • c) \( \{4, 9, 25, 49, 121\} \)
    • d) \( \{1, 4, 9, 25, 49, 121\} \)
  3. If \( f : A \to B \) is a bijective function and if \( n(B) = 7 \), then \( n(A) \) is equal to
    • a) 7
    • b) 49
    • c) 1
    • d) 14
  4. \( f(x) = (x + 1)^3 - (x - 1)^3 \) represents a function which is
    • a) linear
    • b) cubic
    • c) reciprocal
    • d) quadratic
  5. If the HCF of 65 and 117 is expressible in the form of \( 65m - 117 \), then the value of m is
    • a) 4
    • b) 2
    • c) 1
    • d) 3
  6. Given \( F_1 = 1, F_2 = 3 \) and \( F_n = F_{n-1} + F_{n-2} \) then \( F_5 \) is
    • a) 3
    • b) 5
    • c) 8
    • d) 11
  7. In an A.P. the first term is 1 and the common difference is 4. How many terms of the A.P. must be taken for their sum to be equal to 120?
    • a) 6
    • b) 7
    • c) 8
    • d) 9
PART-II 5 x 2 = 10
II. Answer any five questions only. [Q.No. 14 is compulsory].
  1. Let \( A = \{1, 2, 3\} \) and \( B = \{x \mid x \text{ is a prime number less than 10}\} \). Find \( A \times B \) and \( B \times A \).
  2. A Relation R is given by the set \( \{ (x, y) \mid y = x+3, x \in \{0, 1, 2, 3, 4, 5\}\} \). Determine its domain and range.
  3. 'a' and 'b' are two positive integers such that \( a^b \times b^a = 800 \). Find 'a' and 'b'.
  4. Solve \( 5x \equiv 4 \pmod{6} \).
  5. Find the number of terms in the AP. 3, 6, 9, 12, ..., 111.
  6. If \( f(x) = 2x - 1, g(x) = \frac{x+1}{2} \), show that \( fog = gof = x \).
  7. Represent the function \( f = \{ (1, 2), (2, 2), (3, 2), (4, 3), (5, 4)\} \) through
    i) an arrow diagram   ii) a table form   iii) a graph
PART-III 5 x 5 = 25
III. Answer any five of the following questions and Q.No. 21 is compulsory.
  1. Let \( A = \{x \in N \mid 1 < x < 4 \} \), \( B = \{ x \in W \mid 0 \le x < 2 \} \) and \( C = \{x \in N \mid x < 3 \} \). Then verify that,
    (i) \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)
    (ii) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)
  2. If the function \( f \) is defined by \( f(x) = \begin{cases} x+2 & \text{if } x > 1 \\ 2 & \text{if } -1 \le x \le 1 \\ x-1 & \text{if } -3 < x < -1 \end{cases} \), find the value of:
    (i) \( f(3) \)   (ii) \( f(0) \)   (iii) \( f(-1.5) \)   (iv) \( f(2) + f(-2) \)
  3. Find the HCF of 396, 504, 636.
  4. If \( p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} = 113400 \) where \( p_1, p_2, p_3, p_4 \) are primes in ascending order and \( x_1, x_2, x_3, x_4 \) are integers, find the value of \( p_1, p_2, p_3, p_4 \) and \( x_1, x_2, x_3, x_4 \).
  5. The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
  6. Find the sum of all natural numbers between 300 and 600 which are divisible by 7.
  7. If \( f(x) = x^2 \), \( g(x) = 2x \) and \( h(x) = x + 4 \). Prove that \( (fog) \circ h = f \circ (g \circ h) \).
PART-IV 1 x 8 = 8
IV. Answer any one of the following questions.
  1. Construct a triangle similar to a given triangle PQR with its sides equal to \( \frac{2}{3} \) of the corresponding sides of the triangle PQR (Scale factor \( \frac{2}{3} < 1 \)).
  2. Construct a triangle similar to a given triangle PQR with its sides equal to \( \frac{7}{4} \) of the corresponding sides of the triangle PQR (Scale factor \( \frac{7}{4} > 1 \)).

Solutions

PART-A Solutions
1. Correct Answer: c) 12 Step-by-step solution:
  • Given sets: \( A = \{a, b, p\} \), \( B = \{2, 3\} \), \( C = \{p, q, r, s\} \).
  • First, find the union of A and C: \( A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\} \).
  • Find the number of elements in \( A \cup C \): \( n(A \cup C) = 6 \).
  • Find the number of elements in B: \( n(B) = 2 \).
  • The number of elements in the Cartesian product \( (A \cup C) \times B \) is \( n(A \cup C) \times n(B) \).
  • \( n[(A \cup C) \times B] = 6 \times 2 = 12 \).
2. Correct Answer: c) {4, 9, 25, 49, 121} Step-by-step solution:
  • The relation is \( R = \{ (x, x^2) \mid x \text{ is a prime number less than 13} \} \).
  • The prime numbers less than 13 are 2, 3, 5, 7, 11.
  • The domain (the set of x-values) is \( \{2, 3, 5, 7, 11\} \).
  • The range is the set of the second elements of the ordered pairs, which are \( x^2 \).
  • Range = \( \{2^2, 3^2, 5^2, 7^2, 11^2\} = \{4, 9, 25, 49, 121\} \).
3. Correct Answer: a) 7 Step-by-step solution:
  • A bijective function is a function that is both one-to-one (injective) and onto (surjective).
  • For a function between two finite sets to be bijective, the number of elements in the domain must be equal to the number of elements in the codomain.
  • Given \( f : A \to B \) is bijective and \( n(B) = 7 \).
  • Therefore, \( n(A) = n(B) = 7 \).
4. Correct Answer: d) quadratic Step-by-step solution:
  • Given function: \( f(x) = (x + 1)^3 - (x - 1)^3 \).
  • We use the algebraic identities: \( (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \) and \( (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \).
  • \( f(x) = (x^3 + 3(x^2)(1) + 3(x)(1^2) + 1^3) - (x^3 - 3(x^2)(1) + 3(x)(1^2) - 1^3) \)
  • \( f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) \)
  • \( f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 \)
  • Combine like terms: \( f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1) \)
  • \( f(x) = 6x^2 + 2 \).
  • This is a polynomial of degree 2, which is a quadratic function.
5. Correct Answer: b) 2 Step-by-step solution:
  • First, find the HCF of 65 and 117 using Euclid's division algorithm.
  • \( 117 = 1 \times 65 + 52 \)
  • \( 65 = 1 \times 52 + 13 \)
  • \( 52 = 4 \times 13 + 0 \)
  • The HCF is the last non-zero remainder, which is 13.
  • We are given that the HCF is expressible as \( 65m - 117 \).
  • So, \( 65m - 117 = 13 \).
  • \( 65m = 13 + 117 \implies 65m = 130 \).
  • \( m = \frac{130}{65} = 2 \).
6. Correct Answer: d) 11 Step-by-step solution:
  • Given: \( F_1 = 1 \), \( F_2 = 3 \), and the recurrence relation \( F_n = F_{n-1} + F_{n-2} \).
  • \( F_3 = F_{3-1} + F_{3-2} = F_2 + F_1 = 3 + 1 = 4 \).
  • \( F_4 = F_{4-1} + F_{4-2} = F_3 + F_2 = 4 + 3 = 7 \).
  • \( F_5 = F_{5-1} + F_{5-2} = F_4 + F_3 = 7 + 4 = 11 \).
7. Correct Answer: c) 8 Step-by-step solution:
  • Given A.P. has first term \( a = 1 \), common difference \( d = 4 \), and sum \( S_n = 120 \).
  • The formula for the sum of an A.P. is \( S_n = \frac{n}{2} [2a + (n-1)d] \).
  • \( 120 = \frac{n}{2} [2(1) + (n-1)4] \).
  • \( 240 = n [2 + 4n - 4] \).
  • \( 240 = n [4n - 2] \).
  • \( 240 = 4n^2 - 2n \).
  • Rearrange into a quadratic equation: \( 4n^2 - 2n - 240 = 0 \).
  • Divide by 2: \( 2n^2 - n - 120 = 0 \).
  • Solving the quadratic equation (by factorization or formula): \( 2n^2 - 16n + 15n - 120 = 0 \).
  • \( 2n(n - 8) + 15(n - 8) = 0 \).
  • \( (2n + 15)(n - 8) = 0 \).
  • The possible values for n are \( n = 8 \) or \( n = -15/2 \). Since the number of terms cannot be negative or a fraction, \( n = 8 \).
PART-II Solutions
8. Solution
  • Given \( A = \{1, 2, 3\} \).
  • The set B is the set of prime numbers less than 10. So, \( B = \{2, 3, 5, 7\} \).
  • To find \( A \times B \):
    \( A \times B = \{1, 2, 3\} \times \{2, 3, 5, 7\} \)
    \( = \{(1,2), (1,3), (1,5), (1,7), (2,2), (2,3), (2,5), (2,7), (3,2), (3,3), (3,5), (3,7)\} \)
  • To find \( B \times A \):
    \( B \times A = \{2, 3, 5, 7\} \times \{1, 2, 3\} \)
    \( = \{(2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (5,1), (5,2), (5,3), (7,1), (7,2), (7,3)\} \)
9. Solution
  • The relation R is given by \( R = \{ (x, y) \mid y = x+3, x \in \{0, 1, 2, 3, 4, 5\}\} \).
  • The Domain is the set of all possible x-values.
    Domain = \( \{0, 1, 2, 3, 4, 5\} \).
  • The Range is the set of all resulting y-values. We calculate y for each x:
    If \(x=0, y = 0+3=3\)
    If \(x=1, y = 1+3=4\)
    If \(x=2, y = 2+3=5\)
    If \(x=3, y = 3+3=6\)
    If \(x=4, y = 4+3=7\)
    If \(x=5, y = 5+3=8\)
    Range = \( \{3, 4, 5, 6, 7, 8\} \).
10. Solution
  • We are given \( a^b \times b^a = 800 \), where 'a' and 'b' are positive integers.
  • First, find the prime factorization of 800.
    \( 800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2 \).
  • Now we compare this with the given expression: \( a^b \times b^a = 2^5 \times 5^2 \).
  • By direct comparison, we can see two possibilities:
    Case 1: Let \( a = 2 \) and \( b = 5 \). This gives \( 2^5 \times 5^2 \), which matches.
    Case 2: Let \( a = 5 \) and \( b = 2 \). This gives \( 5^2 \times 2^5 \), which also matches.
  • Therefore, the values for 'a' and 'b' are 2 and 5.
11. Solution
  • We need to solve the congruence \( 5x \equiv 4 \pmod{6} \).
  • This means \( 5x - 4 \) is a multiple of 6.
    \( 5x - 4 = 6k \) for some integer \( k \).
  • Rearranging for x: \( 5x = 6k + 4 \).
  • We can test values of x starting from 1 to find a solution:
    If \(x=1\), \( 5(1) = 5 \). \( 5 \pmod{6} \equiv 5 \neq 4 \).
    If \(x=2\), \( 5(2) = 10 \). \( 10 \pmod{6} \equiv 4 \). This is a solution.
  • Thus, a solution is \( x=2 \).
    The general solution is of the form \( x = 2 + 6n \), where n is any integer. For example, 2, 8, 14, ... are all solutions.
    The simplest positive solution is x = 2.
12. Solution
  • The given Arithmetic Progression (AP) is 3, 6, 9, 12, ..., 111.
  • First term, \( a = 3 \).
  • Common difference, \( d = 6 - 3 = 3 \).
  • The last term, \( t_n = 111 \).
  • Using the formula for the nth term of an AP: \( t_n = a + (n-1)d \).
  • \( 111 = 3 + (n-1)3 \).
  • \( 111 - 3 = (n-1)3 \).
  • \( 108 = (n-1)3 \).
  • \( \frac{108}{3} = n-1 \implies 36 = n-1 \).
  • \( n = 36 + 1 = 37 \).
  • There are 37 terms in the AP.
13. Solution
  • Given functions: \( f(x) = 2x - 1 \) and \( g(x) = \frac{x+1}{2} \).
  • To find \( fog(x) \):
    \( fog(x) = f(g(x)) = f\left(\frac{x+1}{2}\right) \).
    \( = 2\left(\frac{x+1}{2}\right) - 1 \).
    \( = (x+1) - 1 = x \).
  • To find \( gof(x) \):
    \( gof(x) = g(f(x)) = g(2x - 1) \).
    \( = \frac{(2x-1)+1}{2} \).
    \( = \frac{2x}{2} = x \).
  • Since \( fog(x) = x \) and \( gof(x) = x \), we have shown that \( fog = gof = x \). This means f and g are inverse functions of each other.
14. Solution (Compulsory Question)

Representing the function \( f = \{ (1, 2), (2, 2), (3, 2), (4, 3), (5, 4)\} \):

i) Arrow Diagram:

Draw two ovals. The first represents the domain \( \{1, 2, 3, 4, 5\} \) and the second represents the codomain (which must include the range \( \{2, 3, 4\} \)). Draw arrows from each element in the domain to its corresponding element in the codomain.

Domain (A)             Codomain (B)

1   →   2

2   ↗

3   ↗

4   →   3

5   →   4

ii) Table Form:

Create a table with two columns, one for \( x \) and one for \( f(x) \).

x f(x)
1 2
2 2
3 2
4 3
5 4

iii) Graph:

Plot the ordered pairs as points on a Cartesian coordinate system.

The points to be plotted are: (1, 2), (2, 2), (3, 2), (4, 3), (5, 4).

(A simple sketch would show these discrete points on a graph with x and y axes).
PART-III Solutions
15. Solution

First, list the elements of the sets:

  • \( A = \{x \in N \mid 1 < x < 4 \} = \{2, 3\} \)
  • \( B = \{ x \in W \mid 0 \le x < 2 \} = \{0, 1\} \)
  • \( C = \{x \in N \mid x < 3 \} = \{1, 2\} \)

(i) Verify \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)

  • LHS: \( A \times (B \cup C) \)
    \( B \cup C = \{0, 1\} \cup \{1, 2\} = \{0, 1, 2\} \)
    \( A \times (B \cup C) = \{2, 3\} \times \{0, 1, 2\} = \{(2,0), (2,1), (2,2), (3,0), (3,1), (3,2)\} \)
  • RHS: \( (A \times B) \cup (A \times C) \)
    \( A \times B = \{2, 3\} \times \{0, 1\} = \{(2,0), (2,1), (3,0), (3,1)\} \)
    \( A \times C = \{2, 3\} \times \{1, 2\} = \{(2,1), (2,2), (3,1), (3,2)\} \)
    \( (A \times B) \cup (A \times C) = \{(2,0), (2,1), (3,0), (3,1), (2,2), (3,2)\} \)
  • Since LHS = RHS, the property is verified.

(ii) Verify \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)

  • LHS: \( A \times (B \cap C) \)
    \( B \cap C = \{0, 1\} \cap \{1, 2\} = \{1\} \)
    \( A \times (B \cap C) = \{2, 3\} \times \{1\} = \{(2,1), (3,1)\} \)
  • RHS: \( (A \times B) \cap (A \times C) \)
    From above, \( A \times B = \{(2,0), (2,1), (3,0), (3,1)\} \) and \( A \times C = \{(2,1), (2,2), (3,1), (3,2)\} \)
    \( (A \times B) \cap (A \times C) = \{(2,1), (3,1)\} \)
  • Since LHS = RHS, the property is verified.
16. Solution

The piecewise function is \( f(x) = \begin{cases} x+2 & \text{if } x > 1 \\ 2 & \text{if } -1 \le x \le 1 \\ x-1 & \text{if } -3 < x < -1 \end{cases} \)

  • (i) To find \( f(3) \):
    Since \( 3 > 1 \), we use the first rule: \( f(x) = x+2 \).
    \( f(3) = 3 + 2 = 5 \).
  • (ii) To find \( f(0) \):
    Since \( -1 \le 0 \le 1 \), we use the second rule: \( f(x) = 2 \).
    \( f(0) = 2 \).
  • (iii) To find \( f(-1.5) \):
    Since \( -3 < -1.5 < -1 \), we use the third rule: \( f(x) = x-1 \).
    \( f(-1.5) = -1.5 - 1 = -2.5 \).
  • (iv) To find \( f(2) + f(-2) \):
    First find \( f(2) \). Since \( 2 > 1 \), \( f(2) = 2 + 2 = 4 \).
    Next find \( f(-2) \). Since \( -3 < -2 < -1 \), \( f(-2) = -2 - 1 = -3 \).
    \( f(2) + f(-2) = 4 + (-3) = 1 \).
17. Solution

We will find the HCF of 396, 504, and 636 using prime factorization.

  • Prime factorization of 396:
    \( 396 = 2 \times 198 = 2 \times 2 \times 99 = 2^2 \times 9 \times 11 = 2^2 \times 3^2 \times 11 \)
  • Prime factorization of 504:
    \( 504 = 2 \times 252 = 2 \times 2 \times 126 = 2^3 \times 63 = 2^3 \times 9 \times 7 = 2^3 \times 3^2 \times 7 \)
  • Prime factorization of 636:
    \( 636 = 2 \times 318 = 2 \times 2 \times 159 = 2^2 \times 3 \times 53 \)
  • The HCF is the product of the lowest powers of the common prime factors.
    Common prime factors are 2 and 3.
    Lowest power of 2 is \( 2^2 \).
    Lowest power of 3 is \( 3^1 \).
    HCF = \( 2^2 \times 3^1 = 4 \times 3 = 12 \).
  • The HCF of 396, 504, and 636 is 12.
18. Solution

We are given \( p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} = 113400 \).

  • First, perform the prime factorization of 113400.
    \( 113400 = 1134 \times 100 \)
    \( = 1134 \times 10^2 = 1134 \times (2 \times 5)^2 = 1134 \times 2^2 \times 5^2 \)
    Now factorize 1134:
    \( 1134 = 2 \times 567 \)
    The sum of digits of 567 is \( 5+6+7=18 \), so it's divisible by 9.
    \( 567 = 9 \times 63 = 9 \times 9 \times 7 = 3^2 \times 3^2 \times 7 = 3^4 \times 7 \)
    So, \( 1134 = 2 \times 3^4 \times 7 \).
  • Combine all factors:
    \( 113400 = (2 \times 3^4 \times 7) \times 2^2 \times 5^2 = 2^{1+2} \times 3^4 \times 5^2 \times 7^1 \)
    \( 113400 = 2^3 \times 3^4 \times 5^2 \times 7^1 \)
  • The primes \( p_1, p_2, p_3, p_4 \) are in ascending order, so:
    \( p_1 = 2, p_2 = 3, p_3 = 5, p_4 = 7 \)
  • The corresponding integer exponents are:
    \( x_1 = 3, x_2 = 4, x_3 = 2, x_4 = 1 \)
19. Solution

Let the three consecutive terms in A.P. be \( a-d, a, a+d \).

  • Sum of the terms:
    Given sum is 27.
    \( (a-d) + a + (a+d) = 27 \)
    \( 3a = 27 \implies a = 9 \).
    The middle term is 9.
  • Product of the terms:
    Given product is 288.
    \( (a-d) \times a \times (a+d) = 288 \)
    Substitute \( a=9 \):
    \( (9-d) \times 9 \times (9+d) = 288 \)
    \( 9 \times (9^2 - d^2) = 288 \)
    \( 81 - d^2 = \frac{288}{9} = 32 \)
    \( d^2 = 81 - 32 = 49 \)
    \( d = \pm\sqrt{49} \implies d = \pm 7 \).
  • Finding the terms:
    If \( d=7 \), the terms are \( 9-7, 9, 9+7 \), which are \( 2, 9, 16 \).
    If \( d=-7 \), the terms are \( 9-(-7), 9, 9-7 \), which are \( 16, 9, 2 \).
  • In both cases, the three terms are 2, 9, and 16.
20. Solution

We need to find the sum of natural numbers between 300 and 600 that are divisible by 7.

  • Find the first term (a): The first number greater than 300 divisible by 7.
    \( 300 \div 7 = 42 \) with a remainder of 6.
    So, the first term is \( 300 + (7-6) = 301 \). So, \( a=301 \).
  • Find the last term (l): The last number less than 600 divisible by 7.
    \( 600 \div 7 = 85 \) with a remainder of 5.
    So, the last term is \( 600 - 5 = 595 \). So, \( l=595 \).
  • The sequence is an A.P: 301, 308, ..., 595 with common difference \( d=7 \).
  • Find the number of terms (n):
    Using \( l = a + (n-1)d \).
    \( 595 = 301 + (n-1)7 \)
    \( 595 - 301 = (n-1)7 \implies 294 = (n-1)7 \)
    \( n-1 = \frac{294}{7} = 42 \implies n = 43 \).
  • Find the sum (S_n):
    Using \( S_n = \frac{n}{2}(a+l) \).
    \( S_{43} = \frac{43}{2}(301 + 595) = \frac{43}{2}(896) \)
    \( S_{43} = 43 \times 448 = 19264 \).
  • The sum is 19264.
21. Solution (Compulsory Question)

Given \( f(x) = x^2 \), \( g(x) = 2x \), and \( h(x) = x + 4 \). We need to prove \( (f \circ g) \circ h = f \circ (g \circ h) \).

  • LHS: \( (f \circ g) \circ h \)
    First, find \( f \circ g \):
    \( (f \circ g)(x) = f(g(x)) = f(2x) = (2x)^2 = 4x^2 \).
    Now, apply \( h(x) \) to this result:
    \( ((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = (f \circ g)(x+4) \).
    Substitute \( x+4 \) into \( 4x^2 \):
    \( = 4(x+4)^2 = 4(x^2 + 8x + 16) = 4x^2 + 32x + 64 \).
  • RHS: \( f \circ (g \circ h) \)
    First, find \( g \circ h \):
    \( (g \circ h)(x) = g(h(x)) = g(x+4) = 2(x+4) = 2x+8 \).
    Now, apply \( f(x) \) to this result:
    \( (f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(2x+8) \).
    Substitute \( 2x+8 \) into \( f(x) = x^2 \):
    \( = (2x+8)^2 = (2(x+4))^2 = 4(x+4)^2 = 4(x^2 + 8x + 16) = 4x^2 + 32x + 64 \).
  • Since LHS = RHS, it is proved that \( (f \circ g) \circ h = f \circ (g \circ h) \).
PART-IV Solutions
22. Solution: Construction (Scale factor \( \frac{2}{3} < 1 \))

To construct a triangle P'Q'R' similar to a given triangle PQR such that its sides are \( \frac{2}{3} \) of the corresponding sides of triangle PQR.

Steps of Construction:
  1. Draw any triangle PQR.
  2. Draw a ray QX starting from Q, making an acute angle with QR on the side opposite to vertex P.
  3. Since the ratio is \( \frac{2}{3} \), the greater number is 3. Locate 3 points \(Q_1, Q_2, Q_3\) on the ray QX such that \(QQ_1 = Q_1Q_2 = Q_2Q_3\).
  4. Join the last point \(Q_3\) to R. This forms the line segment \(Q_3R\).
  5. The numerator of the scale factor is 2. From the point \(Q_2\), draw a line parallel to \(Q_3R\). This line will intersect QR at a point. Name this point R'. (To draw a parallel line, you can construct equal corresponding angles at \(Q_2\) and \(Q_3\)).
  6. From the point R', draw a line parallel to the side PR. This line will intersect PQ at a point. Name this point P'.
  7. The triangle P'Q'R' is the required similar triangle, with each side being \( \frac{2}{3} \) of the corresponding side of triangle PQR.
23. Solution: Construction (Scale factor \( \frac{7}{4} > 1 \))

To construct a triangle P'Q'R' similar to a given triangle PQR such that its sides are \( \frac{7}{4} \) of the corresponding sides of triangle PQR.

Steps of Construction:
  1. Draw any triangle PQR.
  2. Extend the sides QP and QR.
  3. Draw a ray QX starting from Q, making an acute angle with QR on the side opposite to vertex P.
  4. Since the ratio is \( \frac{7}{4} \), the greater number is 7. Locate 7 points \(Q_1, Q_2, ..., Q_7\) on the ray QX such that all segments \(QQ_1, Q_1Q_2, ...\) are equal.
  5. The denominator of the scale factor is 4. Join the point \(Q_4\) to R. This forms the line segment \(Q_4R\).
  6. The numerator of the scale factor is 7. From the point \(Q_7\), draw a line parallel to \(Q_4R\). This line will intersect the extended line segment QR at a point. Name this point R'.
  7. From the point R', draw a line parallel to the side PR. This line will intersect the extended line segment QP at a point. Name this point P'.
  8. The triangle P'Q'R' is the required similar triangle, with each side being \( \frac{7}{4} \) of the corresponding side of triangle PQR.