Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
Reference: Ganita Manjari Part 1 (Syllabus: Chapters 1-7)
Section A: Multiple Choice Questions (Part 1 | Q.1 to Q.5)
Section A: Multiple Choice Questions (Part 1 | Q.1 to Q.5)
Q.1 Coordinate Geometry
If the coordinates of two points A and B are \((2, -4)\) and \((-6, 7)\), then the value of (x-coordinate of B) - (x-coordinate of A) is:
Concept:
1
Every point in the Cartesian plane is represented by an ordered pair \((x, y)\). The first number is the x-coordinate (abscissa), and the second is the y-coordinate (ordinate).
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Hint:
1
Identify the first number in point A and the first number in point B. Then evaluate: \( (\text{Value from B}) - (\text{Value from A}) \).
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Mistakes to Avoid:
!
Reversing the order: Calculating \((\text{x of A}) - (\text{x of B})\) which gives \(2 - (-6) = 8\). Always follow the exact wording given in the question!
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Given points:
\( A(2, -4) \implies x_1 = 2 \)
\( B(-6, 7) \implies x_2 = -6 \)
Expression to evaluate: (x-coordinate of B) - (x-coordinate of A)
\(= x_2 - x_1 \)
\(= -6 - 2 \)
\(= -8 \) [1 M]
Correct Option: (D) -8
Q.2 Number Systems
Which of the following is an irrational number?
Concept:
1
Rational Numbers: Terminating decimals (e.g., 0.5) and non-terminating repeating decimals (e.g., \(0.\overline{3}\)).
2
Irrational Numbers: Non-terminating, non-repeating decimals. The square roots of prime numbers or non-perfect squares fall strictly into this category.
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Hint:
1
Check the square roots. Is 225 a perfect square? Yes (\(15 \times 15\)). Is 23 a perfect square? No.
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Mistakes to Avoid:
!
Bar Notation Confusion: Many students see the bar in \(7.\overline{478}\) and think it means irrational. The bar means it repeats endlessly. Repeating decimals can always be converted to fractions (\(p/q\)), so they are completely rational!
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Let's check each option:
(A) \(0.3796\) is a terminating decimal \(\rightarrow\) Rational.
(B) \(\sqrt{225} = 15\) (Integer) \(\rightarrow\) Rational.
(D) \(7.\overline{478}\) is a non-terminating repeating decimal \(\rightarrow\) Rational.
(C) 23 is a prime number, not a perfect square.
Therefore, \(\sqrt{23}\) yields a non-terminating, non-repeating decimal. [1 M]
Correct Option: (C) \(\sqrt{23}\)
Q.3 Evaluating Polynomials
The value of the polynomial \(4x^3 - 5x^4 + 2x^2 + 3\) at \(x = -1\) is:
Concept:
1
Evaluating polynomials involves substituting the given variable value and strictly following the order of operations (BODMAS/PEMDAS).
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Hint:
1
Be careful with negative signs raised to even and odd powers. A negative number raised to an odd power remains negative (\((-1)^3 = -1\)). A negative number raised to an even power becomes positive (\((-1)^4 = 1\)).
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Mistakes to Avoid:
!
Incorrect Powers: Calculating \(-5(-1)^4\) as \(-5(-1) = +5\). Remember that \((-1)^4\) is positive \(1\), so the term remains \(-5(1) = -5\).
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Let \( P(x) = 4x^3 - 5x^4 + 2x^2 + 3 \)
Substitute \( x = -1 \):
\( P(-1) = 4(-1)^3 - 5(-1)^4 + 2(-1)^2 + 3 \)
Evaluate the powers first:
\( P(-1) = 4(-1) - 5(1) + 2(1) + 3 \)
Multiply:
\( P(-1) = -4 - 5 + 2 + 3 \)
Combine terms:
\( P(-1) = -9 + 5 \)
\( P(-1) = -4 \) [1 M]
Correct Option: (C) -4
Q.4 Algebraic Identities
If \(a - b = 7\) and \(ab = 9\) then \(a^2 + b^2 =\)
Concept:
1
To relate \((a-b)\), \(ab\), and \((a^2+b^2)\), we use the binomial square expansion identity: \((a - b)^2 = a^2 - 2ab + b^2\).
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Hint:
1
Take the equation \(a - b = 7\) and square both sides entirely: \((a - b)^2 = 7^2\). Then replace \(ab\) with 9.
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Mistakes to Avoid:
!
Sign Errors during Transposition: When you have \(-2ab\) on the left side, you must ADD it to the other side to isolate \(a^2 + b^2\). A common mistake is calculating \(49 - 18\) instead of \(49 + 18\).
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Given: \( a - b = 7 \) and \( ab = 9 \)
We know the identity:
\( (a - b)^2 = a^2 + b^2 - 2ab \)
Substitute the given values into the identity:
\( (7)^2 = a^2 + b^2 - 2(9) \)
\( 49 = a^2 + b^2 - 18 \)
Transpose \(-18\) to the LHS:
\( 49 + 18 = a^2 + b^2 \)
\( a^2 + b^2 = 67 \) [1 M]
Correct Option: (C) 67
Q.5 Properties of Circles
In the given figure, O is the centre of the circle. \(\angle AOB = \angle COD = 55^\circ\) and \(CD = 5 \text{ cm}\) then \(AB\) is equal to:
Concept:
1
Central Angle Theorem: In a circle, if two central angles are equal, then the corresponding chords they subtend are also equal in length. This is easily proven using Side-Angle-Side (SAS) congruence for triangles.
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Hint:
1
Look at triangles \(\Delta AOB\) and \(\Delta COD\). The sides \(OA, OB, OC, OD\) are all radii of the same circle. What congruence rule applies here when the angle trapped between them is exactly \(55^\circ\)?
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Mistakes to Avoid:
!
Visual Guessing: Never guess chord lengths purely by how the drawing looks. Always rely on geometric congruence proofs. If the central angles were different, the chords would be strictly different lengths.
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In \(\Delta AOB\) and \(\Delta COD\):
\( OA = OC \) (Radii of the same circle)
\( OB = OD \) (Radii of the same circle)
\( \angle AOB = \angle COD = 55^\circ \) (Given)
Therefore, by SAS congruence criterion:
\( \Delta AOB \cong \Delta COD \)
By CPCT (Corresponding Parts of Congruent Triangles):
\( AB = CD \)
Since \( CD = 5 \text{ cm} \), it must be true that \( AB = 5 \text{ cm} \). [1 M]
Correct Option: (C) 5 cm
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