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MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) SECTION B -  Question 22 Prove that √2 is an irrational number.  Proof: Let us assume  √2  is a rational number. Then, there exist positive integers a and b such that \[\sqrt{2} =\frac{p}{q} where,\; p\; and\; q,\; are \;co-prime\; i.e. \;their\; HCF \;is \;1\] \[ OR \; (\sqrt{2})^2 =\frac{p^2}{q^2} \] \[OR\; 2 =\frac{p^2}{q^2} \] \[OR\; 2 \times q^2 = p^2 \] This implies that 𝑝^2 is even because it is equal to 2𝑞^2. From this, we can conclude that 𝑝 must also be even because the square of an odd number is odd, and the square of an even number is even. Let 𝑝 = 2k, where k is arbitrary integer , Substituting 𝑝 = 2k into the equation 2q^2 = p^2 we get  Substituting 𝑝 = 2k into the equation 2q^2 = p^2 \[\Rightarrow 2q^2 = (2k)^2 \] \[\Rightarrow 2q^2 = 4(k)^2 \] \[\Rightarrow q^2 = 2(k)^2 \] This implies that 𝑞^2  is even. Therefore, 𝑞 must also be even. Now...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 19 DIRECTION: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct option Statement A (Assertion): Total Surface area of the top is the sum of the curved surface area of the hemisphere and the curved surface area of the cone. Statement R( Reason) : Top is obtained by joining the plane surfaces of the hemisphere and cone together (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)  (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A) (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true.  Explanation :  (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A) Statement A (Assertion) is true becaus...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 16 There is a square board of side ‘2a’ units circumscribing a red circle. Jayadev is asked to  keep a dot on the above said board. The probability that he keeps the dot on the shaded  region is. \begin{flalign} (a)\;\;& \frac{\pi}{4}\\ (b)\;\;& \frac{4-\pi}{4}&\\ (c)\;\;& \frac{\pi - 4}{4}&\\ (d)\;\;& \frac{4}{\pi}&\\ \end{flalign} Explanation :  Given that , The square board has side length 2a  units and the board circumscribes a red circle. The shaded region represents the area within the square board but outside the red circle. so we have to find the ratio of the area of the shaded region to the total area of the square board. we know that area of the square is side^2  => 4a^2 square units. similarly  area of the red circle   => π(radius)^2 = π(a)^2 \[\therefore Area \; of\; shaded\; r...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041)  Question 13 If a pole 6 m high casts a shadow 2 √3m long on the ground, then the Sun’s elevation is (a) 60°  (b) 45°  (c) 30°  (d) 90° Explanation :  let's assume  h as the height of the pole (6 meters) s as the length of the shadow (2√3 meters) θ as the Sun's elevation angle The triangles formed by the pole, its shadow, and the ground are similar. Therefore, the ratio of corresponding sides is equal. In this case, the ratio of the height of the pole to the length of its shadow is  the same as the ratio of the height of the observer to the distance from the observer to the tip of the shadow. \[\therefore \frac{h}{s} \; = \tan\theta\] \[\Rightarrow \frac{6}{2\sqrt{3}} \; = \tan\theta\] \[\therefore \sqrt{3} \; = \tan\theta\] \[\theta\; =\; \sqrt{3}\] The above question is illustrated in the picture below as well  : Where X =  2√3 and θ =...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041)  Question 12 (sec A + tan A) (1 – sin A) equals : (a) sec A  (b) sin A  (c) cosec A  (d) cos A Explanation :  let's first express  sec A and tan A in terms of sinA and cos A We know that  \[sec A \; =\; \frac{1}{cos A }\]  \[tan A \; =\; \frac{sin A}{cos A }\]  \[(secA+tanA)(1−sinA) \; = (\frac{1}{cosA} + \frac{sinA}{cosA}) \;(1−sinA) \] \[\Rightarrow (\frac{1+sinA}{cosA})(1−sinA) \] Now, let's expand this expression \[\Rightarrow \frac{(1+sinA)(1-sinA)}{cosA} \] \[\Rightarrow \frac{(1-sin^2A)}{cosA} \] \[we\; know \;that \; { 1-cos^2A} \;=\; sin^2A, 💡  \] \[\therefore \frac{(cos^2A)}{cosA} \; =\; cosA \] Guess the Option and comment below   👇

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041)  Question 11 \begin{flalign} & Given \; that \; \sin\theta \; = \; \frac{a}{b},then \; \cos\theta\ is\;? &\\ \end{flalign} \begin{flalign} (a)\;\;& \frac{b}{\sqrt{b^2 - a^2}}\\ (b)\;\;& \frac{b}{a} \\ (c)\;\;& \frac{\sqrt{b^2 - a^2}}{b}\\ (d)\;\;& \frac{a}{\sqrt{b^2 - a^2}}&\\ \end{flalign} Explanation :  \begin{flalign} & Given \; that \; \sin\theta \; = \; \frac{a}{b},to \; find \;the \; \cos\theta &\\ \end{flalign} The Pythagorean identity states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (𝑎 and 𝑏).  Mathematically, it is represented as  \[ \sin^2\Theta + \cos^2\Theta = 1, -----(a) \]  \begin{flalign} & Since \; we \; know\; that\; \sin\theta \; = \; \frac{a}{b} &\\ \end{flalign} we can substitute this value into the Pyth...

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041)  Question 10   A quadrilateral PQRS is drawn to circumscribe a circle the tangent  If PQ = 12 cm, QR = 15 cm and RS = 14 cm, then find the length of SP is (a) 15 cm  (b) 14 cm (c) 12 cm  (d) 11 cm Explanation :  A circle is inscribed inside the quadrilateral, such that each side of the quadrilateral is tangent to the circle.  Now, in a cyclic quadrilateral, the opposite angles are supplementary Given that PQRS is a cyclic quadrilateral,  \[ \therefore PQ + RS = QR + SP,-----(a) \] Given that  PQ=12 cm QR=15 cm RS=14 cm  and putting eq(a) \[ \therefore 12+14=15+SP \] \[ \therefore SP=26−15 \] \[ \therefore SP=11 \] Guess the Option and comment below   👇

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041)  Question 9  If O is centre of a circle and Chord PQ makes an angle 50° with the tangent PR at the point of contact  P, then the angle subtended by the chord at the centre is (a) 130°  (b) 100°                                                               (c) 50°    (d) 30° Explanation :  OP ⊥ PR   [.. it indicates that the line segment 𝑂𝑃 is perpendicular to the line 𝑃𝑅 ] ∠OPQ = 90° – 50° = 40° OP = OQ   ...[Radii] ∴ ∠OPQ = ∠OQP = 40° In ΔOPQ,  ∠POQ + ∠OPQ + ∠OQP = 180° [.. applied the angle sum property ]  ∠POQ + 40° + 40° = 180°  ∠POQ = 180° – 80° = 100° Guess the Option and comment below   👇

 CBSE MATHEMATICS (IX-X):