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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, we have shown that both 𝑝  and  q ar

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) SECTION B -  Question 21 ABCD is a parallelogram. Point P divides AB in the ratio 2:3 and point Q divides DC in the ratio 4:1. Prove that OC is half of OA. Explanation : ABCD is a parallelogram. we know that AB = DC let AB = l then AB = DC = l given that Point P divides AB in the ratio 2:3 \[\Rightarrow AP = \frac{2}{5}\times l;and\;BP \;=\frac{3}{5}\times l \] \[Since\; we \;considered\; AP + BP = AB = l \] \[\Rightarrow DQ = \frac{4}{5}\times l;and\;CQ \;=\frac{1}{5}\times l \] given that point Q divides DC in the ratio 4:1 \[Since \; DQ + CQ = DC= l \] As we know that [AA similarity] \[\therefore ∆ APO \thicksim ∆ CQO \] \[\therefore \frac{AP}{CQ} = \frac{PO}{QO} = \frac{AO}{CO} \] \[\therefore \frac{AO}{CO} = \frac{\frac{2}{5}\times l}{\frac{1}{5}\times l} = \frac{2}{1} \] After simplifying the above equation, we get \[\Rightarrow OC = \frac{1}{2} \times AO \]

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 20 \begin{flalign} & Statement A (Assertion )\; 5,\; \frac{-5}{2},\; 0,\; \frac{5}{2} …. \;is\; in \;Arithmetic\; Progression.&\\ \end{flalign} Statement R (Reason) : The terms of an Arithmetic Progression cannot have both positive and negative rational numbers. (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:   let's understand the AP where AP is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. The general form of an arithmetic progression is a , a+d,a+2d,a+3d,… where  a is the first term . d is the common difference between terms. a+nd r

 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 because the total surface

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 18 The upper limit of the modal class of the given distribution is: (a) 165  (b) 160  (c) 155  (d) 150 Explanation :  The class interval  Below 165 has the highest frequency of 51 girls. To find the upper limit of this class interval, we consider that the class interval "Below 165" starts from 160 and ends at the upper limit of the class interval before it. So, the upper limit of the modal class Below 165 is Upper limit = 165  Upper limit=165. Guess the Option and comment below   👇

 MATH SAMPLE QUESTION PAPER Class X Session 2023-24 MATHEMATICS STANDARD (Code No.041) Question 17 2 cards of hearts and 4 cards of spades are missing from a pack of 52 cards. A card is drawn at random from the remaining pack. What is the probability of getting a black card? \begin{flalign} (a)\;\;& \frac{22}{52}\\ (b)\;\;& \frac{22}{46}&\\ (c)\;\;& \frac{24}{52}&\\ (d)\;\;& \frac{24}{46}&\\ \end{flalign} Explanation :  let's first determine the number of black cards in the original pack, and then subtract the number of hearts and spades removed. In a standard deck of 52 cards, there are 26 black cards (13 spades and 13 clubs), 13 hearts, and 13 diamonds. \[total\; number\; of\; remaining\; cards\; is\; = \; 11 + 9 + 13 + 13 \] After removing 2 hearts and 4 spades, we are left with 11 hearts (13 - 2) 9 spades (13 - 4) 13 clubs (unchanged) 13 diamonds (unchanged) \[total\; number\; of\; remaining\; cards\; is\; = \; 11 + 9 + 13 + 13 \] there are 22 black

 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 \[\therefore Area \; of\; shaded\; region\; = \; 4a^2 - \pi(a)^2 \;square \; units\] The probability that