2023 CBSE (Class 10) Board Paper Solution Mathematics : 30/4/2

$\textbf{Series: WX1YZ/4} \ \ \ \ \ \textbf{SET } \sim 2 \ \ \ \ \ \textbf{Q.P. Code } 30/4/2$

$\textbf{MATHEMATICS (STANDARD) - Theory}$

Time: 3 Hours Max. Marks:80

General Instructions:

Read the following instructions carefully follow them:

(i) This question paper contains 38 questions. All questions are compulsory.

(ii) Question paper is divided into FIVE sections – Section A, B, C, D and E.

(iii) In section A, question number 1 to 18 are multiple choice questions (MCQs) and question number 19 and 20 are Assertion – Reason based questions of 1 mark each.

(iii) In section B, question number 21 to 25 are very short answer (VSA) type questions of 2 marks each.

(iv) In section C, question number 26 to 31 are short answer (SA) type questions carrying 3 marks each.

(v) In section D, question number 32 to 35 are long answer (LA) type questions carrying 5 marks each.

(vi) ln section E, question number 36 to 38 are case based integrated units of assessment questions carrying 4 marks each. Internal choice is provided in 2 marks question in each case study.

(viii) There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, 2 questions in Section D and 3 questions in Section E.

$\text{(ix) Draw neat figures wherever required. Take } \pi = \frac{22}{7} \text{ wherever required if not stated.}$

(x) Use of calculators is not allowed.

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SECTION A

Section-A consists of Multiple Choice

Type questions of 1 mark each

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$\displaystyle \text{Question 1: Which of the following is true for all values of } \theta ( 0^\circ \leq \theta \leq 90^\circ) \text{?}$

$\displaystyle \text{(a) } \cos^2 \theta - \sin^2 \theta=1 \hspace{1.0cm} \text{(b) } \mathrm{cosec}^2 \theta - \sec^2 \theta \hspace{1.0cm} \\ \\ \text{(c) } \sec^2 \theta - \tan^2 \theta = 1 \hspace{1.0cm} \text{(d) } \cot^2 \theta - \tan^2 \theta = 1$

$\displaystyle \text{Answer:}$

$\fbox{c}$

$\displaystyle \text{Explanation:} \sec^2 \theta = 1 + \tan^2 \theta$

$\displaystyle \Rightarrow \sec^2 \theta - \tan^2 \theta = 1$

$\\$

$\displaystyle \text{Question 2: If } k+2, 4k-6 \text{ and } 3k-2 \text{are three consecutive terms of an A.P., then the value of } \\ k \text{ is : }$

$\displaystyle \text{(a) } {3} \hspace{1.0cm} \text{(b) } {-3} \hspace{1.0cm} \text{(c) } {4} \hspace{1.0cm} \text{(d) } {-4}$

$\displaystyle \text{Answer:}$

$\fbox{a}$

$\displaystyle \text{Since} k+2, 4k-6, \text{ and } 3k-2 \text{ are consecutive terms of A.P. }$

$\displaystyle \Rightarrow (4k-6) - (k+2) = (3k-2) - (4k-6)$

$\displaystyle \Rightarrow 3k-8 = -k+4$

$\displaystyle 4k = 12$

$\displaystyle \therefore k = 3$

$\\$

$\displaystyle \text{Question 3: The next term of the A.P.} \sqrt{6}, \sqrt{24}, \sqrt{54} \text{ is}$

$\displaystyle \text{(a)} {\sqrt{60} } \hspace{1.0cm} \text{(b)} {\sqrt{96}} \hspace{1.0cm} \text{(c)} {\sqrt{72}} \hspace{1.0cm} \text{(d)} {\sqrt{216}}$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{First term, } a_1 = \sqrt{6}$

$\displaystyle \text{Second term, }a_2 = \sqrt{24} = 2\sqrt{6}$

$\displaystyle \text{Common difference } = 2\sqrt{6}-\sqrt{6} = \sqrt{6}$

$\displaystyle \text{Next term of A.P = Third term + common difference }$

$= \sqrt{54}+ \sqrt{6}$

$= 3\sqrt{6} + \sqrt{6}$

$= 4\sqrt{6} = \sqrt{96}$

$\\$

$\displaystyle \text{Question 4: The distance of the point} (-1, 7) \text{from x-axis is:}$

$\displaystyle \text{(a)} {-1} \hspace{1.0cm} \text{(b)} {7} \hspace{1.0cm} \text{(c)} {6} \hspace{1.0cm} \text{(d)} {\sqrt{50}}$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{The distance of } (-1, 7) \text{from the x-axis is } 7 \displaystyle \text{ units. }$

$\\$

$\displaystyle \text{Question 5: What is the area of semi-circle of diameter } 'd'$

$\displaystyle \text{(a)} \ {\frac{1}{16} \pi d^2 } \hspace{1.0cm} \text{(b)} \ {\frac{1}{4} \pi d^2 } \hspace{1.0cm} \text{(c)} \ {\frac{1}{8} \pi d^2 } \hspace{1.0cm} \text{(d)} \ {\frac{1}{2} \pi d^2 }$

$\displaystyle \text{Answer:}$

$\fbox{c }$

$\displaystyle \text{Given, diameter of semi-circle } = d$

$\displaystyle \therefore \text{radius of semi-circle }= \frac{d}{2}$

$\displaystyle \therefore \text{Area of semi-circle }= \frac{\pi \Big(\frac{d}{2}\Big)^2}{2} = \frac{\pi d^2}{8}$

$\\$

$\displaystyle \text{Question 6: The empirical relation between the mode, median and mean of a } \\ \text{distribution is:}$

$\displaystyle \text{(a) } \text{{Mode : 3Median - 2Mean }} \hspace{1.0cm} \text{(b) } \text{{Mode = 3Mean - 2Median }} \hspace{1.0cm} \\ \\ \text{(c) } \text{{Mode : 2Median - 3Mean}} \hspace{1.0cm} \text{(d) } \text{{Mode : 2Mean - 3Median }}$

$\displaystyle \text{Answer:}$

$\fbox{a }$

$\displaystyle \text{Empirical formula: Mode = 3Median - 2Mean }$

$\\$

$\displaystyle \text{Question 7: The pair of linear equations} 2x=5y+6 \text{ and } 15y = 6x-18 \\ \text{ represents two lines which are:}$

$\displaystyle \text{(a)} \text{{Intersecting}} \hspace{1.0cm} \text{(b)} \text{{parallel }} \hspace{1.0cm} \\ \\ \text{(c)} \text{{coincident }} \hspace{1.0cm} \text{(d)} \text{{either intersecting or parallel }}$

$\displaystyle \text{Answer:}$

$\fbox{c}$

Given equations can be re-written as:

$\displaystyle 2x-5y-6=0$

$\displaystyle 6x-15y-18=0$

$\displaystyle \frac{a_1}{a_2}= \frac{2}{6} = \frac{1}{3}$

$\displaystyle \frac{b_1}{b_2} =\frac{-5}{-15} = \frac{1}{3}$

$\displaystyle \frac{c_1}{c_2} = \frac{-6}{-18} = \frac{1}{3}$

Therefore the pair of equations has infinitely many solutions. Graphically pair of linear equations represent coincident.

$\\$

$\displaystyle \text{Question 8: If } \alpha, \beta \text{ are zeros of the polynomial } x^2 - 1, \text{ then value of } (\alpha + \beta) \text{ is: }$

$\displaystyle \text{(a) } {2 } \hspace{1.0cm} \text{(b)} { 1} \hspace{1.0cm} \text{(c) } {-1 } \hspace{1.0cm} \text{(d) } {0 }$

$\displaystyle \text{Answer:}$

$\fbox{d }$

$\displaystyle \text{ Given polynomial:}$

$\displaystyle x^2- 1 = (x-1)(x+1)$

$\displaystyle \text{For zeros, } (x-1)(x+1) = 0$

$\displaystyle \Rightarrow x = 1 \text{ and } x = - 1$

$\displaystyle \text{ Let } \alpha = 1 \text{ and } \beta = - 1$

$\displaystyle \text{Sum of } (\alpha + \beta) = 1 + (-1) = 0$

$\\$

$\displaystyle \text{Question 9: If a pole 6 m high casts a shadow } 2\sqrt{3} \text{ m long on the ground, then sun's elevation is:}$

$\displaystyle \text{(a) } {60^\circ} \hspace{1.0cm} \text{(b) } {45^\circ } \hspace{1.0cm} \text{(c) } {30^\circ } \hspace{1.0cm} \text{(d) } {90^\circ }$

$\displaystyle \text{Answer:}$

$\fbox{ a}$

$\displaystyle \tan \theta = \frac{AB}{AC}$

$\displaystyle \Rightarrow \tan \theta = \frac{6}{2\sqrt{3}}$

$\displaystyle \Rightarrow \tan \theta = \frac{3}{\sqrt{3}}$

$\displaystyle \Rightarrow \tan \theta = \sqrt{3}$

$\displaystyle \Rightarrow \tan \theta = \tan 60^\circ$

$\displaystyle \therefore \theta = 60^\circ$

$\\$

$\displaystyle \text{Question 10:} \sec \theta \text{ when expressed in terms of } \cot \theta, \text{ is equal to: }$

$\displaystyle \text{(a) } { \frac{1+\cot^2 \theta}{\cot \theta} } \hspace{1.0cm} \text{(b) } {\sqrt{1+ \cot^2 \theta} } \hspace{1.0cm} \text{(c) } {\frac{\sqrt{1+\cot^2 \theta}}{\cot \theta} } \hspace{1.0cm} \text{(d)} {\frac{\sqrt{1-\cot^2 \theta}}{\cot \theta} }$

$\displaystyle \text{Answer:}$

$\fbox{c}$

We know that

$\displaystyle \sec \theta = \frac{1}{\cos \theta }= \frac{\sin \theta }{\cos \theta } \cdot \frac{1}{\sin \theta } =\frac{1}{\cot \theta } \cdot \mathrm{cosec} \theta =\frac{\sqrt{1+\cot^2 \theta}}{\cot \theta}$

$\\$

$\displaystyle \text{Question 11: For the following distribution:}$

$\begin{array}{|c|c|c|c|c|c|} \hline \text{Class } & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\ \hline \text{Frequency } & 10 & 15 & 12 & 20 & 9 \\ \hline \end{array}$

$\displaystyle \text{The sum of the lower limits of median class and modal class is:}$

$\displaystyle \text{(a) } {15} \hspace{1.0cm} \text{(b) } {25} \hspace{1.0cm} \text{(c) } {30} \hspace{1.0cm} \text{(d) } {35}$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{Modal class:} 15-20$

$\displaystyle \text{Since the highest frequency } = 20$

$\displaystyle \text{Lower limit of modal class is } 15$

$\displaystyle \text{Here, sum of frequencies, } N = 66$

$\displaystyle \therefore \frac{N}{2} = \frac{66}{2} = 33$

$\\$

$\displaystyle \text{Question 12: Length of a tangent drawn to a circle of radius 9 cm from a point} \\ \text{41 cm from the center of the circle is:}$

$\displaystyle \text{(a) } {40 \text{cm}} \hspace{1.0cm} \text{(b) } {9 \text{cm} } \hspace{1.0cm} \text{(c) } {41 \text{cm} } \hspace{1.0cm} \text{(d) } {50 \text{cm} }$

$\displaystyle \text{Answer:}$

$\fbox{a}$

$\displaystyle \text{In } \triangle PQO,$

$\displaystyle OP^2 = OQ^2 + PQ^2$

$\displaystyle \Rightarrow (41)^2 = (9)^2 + PQ^2$

$\displaystyle \Rightarrow 1681 = 81 + PQ^2$

$\displaystyle \Rightarrow PQ^2 = 1681 - 81 = 1600$

$\displaystyle \Rightarrow PQ = 40 \text{ cm }$

$\\$

$\displaystyle \text{Question 13: In the given figure, O is the center of the circle and PQ is the chord. If the } \\ \text{tangent PR at P makes an angle of } 50^\circ \text{with PQ, then the measure of } \angle POQ \text{ is: }$

$\displaystyle \text{(a) } {50^\circ } \hspace{1.0cm} \text{(b) } {40^\circ} \hspace{1.0cm} \text{(c) } {100^\circ} \hspace{1.0cm} \text{(d) } {130^\circ}$

$\displaystyle \text{Answer:}$

$\fbox{c}$

$\displaystyle \text{Here, } \angle OPQR= 90^\circ \text{ (angle between radius and tangent) }$

$\displaystyle \therefore \angle OPQ = 90^\circ - 50^\circ = 40^\circ$

$\displaystyle \text{Also, } \angle OPQ = \angle OQP = 40^\circ \text{ (being of equal radius) }$

$\displaystyle \text{In } \triangle POQ, \angle OPQ + \angle OQP + \angle POQ = 180^\circ$

$\displaystyle \Rightarrow \angle POQ = 180^\circ - 80^\circ = 100^\circ$

$\\$

$\displaystyle \text{Question 14: A bag contains 5 red balls and n green balls. If the probability of } \\ \text{drawing a green balls is three times that of a red ball, then the value of } n \text{ is: }$

$\displaystyle \text{(a) } {18} \hspace{1.0cm} \text{(b) } {15} \hspace{1.0cm} \text{(c) } {10} \hspace{1.0cm} \text{(d) } {20}$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{Total Balls } = 5 + n$

$\displaystyle \text{Probability of drawing Red ball P(R) }= \frac{5}{5+n}$

$\displaystyle \text{Probability of drawing red green ball P(G) }= \frac{n}{5+n}$

$\displaystyle \text{Given } P(G) = 3 P(R)$

$\displaystyle \Rightarrow \frac{n}{5+n} = \frac{5}{5+n}$

$\displaystyle \Rightarrow n = 15$

$\\$

$\displaystyle \text{Question 15: In the given figure, TA is a tangent to the circle with center O such that OT= 4 cm, } \\ \angle OTA = 30^\circ, \text{ then length of TA is:}$

$\displaystyle \text{(a) } {2\sqrt{3} \text{ cm}} \hspace{1.0cm} \text{(b) } {2 \text{ cm}} \hspace{1.0cm} \text{(c) } {2\sqrt{2} \text{ cm} } \hspace{1.0cm} \text{(d) } {\sqrt{3} \text{ cm} }$

$\displaystyle \text{Answer:}$

$\fbox{a}$

$\displaystyle \text{Here, } \angle OAT = 90^\circ \text{ (angle between radius and tangent)}$

$\displaystyle \text{In } \triangle OAT,$

$\displaystyle \cos 30^\circ = \frac{TA}{OT}$

$\displaystyle \Rightarrow \frac{\sqrt{3}}{2} = \frac{TA}{4}$

$\displaystyle \Rightarrow TA = \frac{4\sqrt{3}}{2} = 2\sqrt{3} \text{cm}$

$\\$

$\displaystyle \text{Question 16: In} \triangle ABC, PQ \parallel BC. \\ \text{ If, PB=6 cm, AP = 4 cm, AQ = 8 cm, find the length of AC. }$

$\displaystyle \text{(a) } {12\text{ cm} } \hspace{1.0cm} \text{(b) } {20\text{ cm} } \hspace{1.0cm} \text{(c) } {6\text{ cm} } \hspace{1.0cm} \text{(d) } {14\text{ cm} }$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{If } PQ \parallel BC \text{ by using basic proportionality theorem, }$

$\displaystyle \frac{AP}{PB} = \frac{AQ}{QC}$

$\displaystyle \Rightarrow \frac{4}{6} = \frac{8}{QC}$

$\displaystyle \Rightarrow QC = \frac{8 \times 6}{4}$

$\displaystyle \Rightarrow QC = 12 \text{ cm }$

$\displaystyle \text{Now, } AC = AQ + QC =8 + 12 = 20 \text{ cm }$

$\\$

$\displaystyle \text{Question 17: If } \alpha , \beta \text{ are zeros of the polynomial } p(x) = 4x^2 - 3x - 7, \\ \text{ then } \Big(\frac{1}{\alpha}+\frac{1}{\beta} \Big) \text{ is equal to:}$

$\displaystyle \text{(a) } {\frac{7}{3}} \hspace{1.0cm} \text{(b) } {\frac{-7}{3} } \hspace{1.0cm} \text{(c) } {\frac{3}{7}} \hspace{1.0cm} \text{(d) } {\frac{-3}{7}}$

$\displaystyle \text{Answer:}$

$\fbox{d}$

$\displaystyle \text{For zeros of polynomial, put } p(x) = 0$

$\displaystyle 4x^2-3x-7 = 0$

$\displaystyle 4x^2 - 7x + 4x - 7 = 0$

$\displaystyle x(4x-7) + ( 4x-7) = 0$

$\displaystyle (4x-7)(x+1) = 0$

$\displaystyle \therefore x = \frac{7}{4} \text{ and } x = -1$

$\displaystyle \text{Let } \alpha = \frac{7}{4} \text{ and } \beta = - 1$

$\displaystyle \therefore \frac{1}{\alpha}+\frac{1}{\beta} = \frac{4}{7} + \frac{1}{-1} =\frac{4}{7} - 1 = \frac{-3}{7}$

$\\$

$\displaystyle \text{Question 18: A card is drawn at random from a well-shuffled pack of 52 cards.} \\ \text{The probability that the card drawn is not an ace is: }$

$\displaystyle \text{(a) } {\frac{1}{13} } \hspace{1.0cm} \text{(b) } {\frac{9}{13} } \hspace{1.0cm} \text{(c) } {\frac{4}{13} } \hspace{1.0cm} \text{(d) } {\frac{12}{13} }$

$\displaystyle \text{Answer:}$

$\fbox{d}$

$\displaystyle \text{Number of ace cards in a pack of 52 = 4 }$

$\displaystyle \text{Number of non-ace cards in a pack of 52 = 48 }$

$\displaystyle \text{Required probability }= \frac{48}{52} = \frac{12}{13}$

$\\$

DIRECTIONS: In the question number 19 and 20 a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option out of the following:

(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).

(d) Assertion (A) is false but Reason (R) is true.

$\displaystyle \text{Question 19: } \\ \text{Assertion (A): The probability that a leap year has 53 Sundays is } \frac{5}{7} \\ \text{Reason (R): The probability that a non-leap year has 53 Sundays is } \frac{5}{7}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{A week has 7 days and total days are 366. }$

$\displaystyle \text{Number of Sundays is a leap year = 52 Sundays + 2 days }$

$\displaystyle \text{Therefore, probability of leap year with 53 Sundays } = \frac{2}{7}$

$\displaystyle \text{Reason: There are 52 Sundays in a non-leap year. But one left over days apart} \\ \text{from those 52 weeks can be either a Monday. Tuesday, Wednesday, Thursday,} \\ \text{Friday, Saturday or Sunday. }$

$\displaystyle \text{Therefore Required probability } = \frac{1}{7}$

$\\$

$\displaystyle \text{Question 20: Assertion (A): a, b, c are in A.P if only if 2b =a+c. } \\ \text{Reason (R): The sum of first n odd natural numbers is } n^2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Assertion is true because }$

$\displaystyle b-a= c-b \ \ \ \ (a, b, c \text{ are in A.P)}$

$\displaystyle \Rightarrow 2b=a+c$

$\displaystyle \text{Reason: Let } 1+ 3 + 5 + 7 + 9 + \ldots + n, \text{ are sum of n odd natural numbers. }$

$\displaystyle S_n = \frac{n}{2} [ 2a + (n-1)d]$

$\displaystyle S_n = \frac{n}{2} [ 2(1) + (n-1)2]$

$\displaystyle S_n = \frac{n}{2} (2n)$

$\displaystyle S_n = n^2$

$\displaystyle \text{Hence, the sum of the first } n \text{ odd natural number is } n^2$

$\\$

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SECTION – B

Section – B consists of Very Short Answer (VSA) type questions of 2 marks each.

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$\displaystyle \text{Question 21: }$

$\displaystyle \text{(A) Evaluate: }$

$\displaystyle \frac{5}{\cot^2 30^\circ} + \frac{1}{\sin^2 60^\circ} - \cot^2 45^\circ + 2 \sin^2 90^\circ$

$\displaystyle \text{(B) If }\theta \text{ is an acute angle and } \sin \theta = \cos \theta, \text{ find value of } \tan^2 \theta + \cot^2 \theta - 2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) We have, }$

$\displaystyle \frac{5}{\cot^2 30^\circ} + \frac{1}{\sin^2 60^\circ} - \cot^2 45^\circ + 2 \sin^2 90^\circ$

$\displaystyle = \frac{5}{(\sqrt{3} )^2} + \frac{1}{\Big(\frac{\sqrt{3}}{2}\Big)^2} - (1)^2 + 2 ( 1)^2$

$\displaystyle = \frac{5}{3}+ \frac{4}{3} - 1 + 2$

$\displaystyle = \frac{9}{3}+ 1$

$\displaystyle = 3+1$

$\displaystyle = 4$

$\displaystyle \text{Given } \sin \theta = \cos \theta$

$\displaystyle \therefore \frac{\sin \theta}{\cos \theta}=1$

$\displaystyle \Rightarrow \tan \theta = 1$

$\displaystyle \Rightarrow \tan \theta = \tan 45^\circ$

$\displaystyle \Rightarrow \theta = 45^\circ$

$\displaystyle \text{Now, } \tan^2 \theta + \cot^2 \theta - 2$

$\displaystyle = \tan^2 45^\circ + \cot^2 45^\circ - 2$

$\displaystyle = 1^1 + 1^2 - 2 = 1 + 1 -2 = 0$

$\\$

$\displaystyle \text{Question 22: If one zero of the polynomial } p(x) = 6x^2+37x-(k-2) \\ \text{is reciprocal of the other, then find the value of } k$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the zeros of the polynomial are } \alpha \text{ and } \frac{1}{\alpha}$

$\displaystyle \text{Product of the zeros } = \frac{-(k-2)}{6}$

$\displaystyle \Rightarrow \alpha \times \frac{1}{\alpha} = \frac{-(k-2)}{6}$

$\displaystyle \Rightarrow 6 = - (k-2)$

$\displaystyle \Rightarrow k = 2-6$

$\displaystyle \Rightarrow k = -4$

$\displaystyle \text{Therefore, value of } k \text{ is } -4$

$\\$

$\displaystyle \text{Question 23:} \\ \text{(A) Find the sum and product of the roots of the quadratic equation } \\ 2x^2-9x + 4=0 \\ {\hspace{5.0cm} \text{OR}} \\ \text{(B) Find the discriminant of the quadratic equation } 4x^2-5=0 \\ \text{ and hence comment on the nature of roots of the equation. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) Given quadratic equation is } 2x^2 - 9x + 4 = 0$

$\displaystyle \text{Sum of the roots } = \frac{-(-9)}{2} = \frac{9}{2}$

$\displaystyle \text{Product of the roots } = \frac{4}{2} = 2$

$\displaystyle \text{Note: For quadratic equation } ax^2 + bx + c = 0, \text{sum of the roots is } \frac{-b}{2} \\ \text{ and product of the roots} = \frac{c}{a}$

$\displaystyle {\hspace{5.0cm} \text{OR}}$

$\displaystyle \text{(B) Given quadratic equation is } 4x^2 - 5=0$

$\displaystyle Q \text{ discriminant, } D = b^2 - 4ac$

$\displaystyle \Rightarrow D = 0 - 4(4) (- 5)$

$\displaystyle \Rightarrow D = 80$

$\displaystyle \text{Thus determinant } D = 80$

$\displaystyle \text{Since } D> 0, \text{ then the roots are real and distinct. }$

$\\$

$\displaystyle \text{Question 24: If a fair coin is tossed twice, find the probability of getting atmost one head.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When a coin is tossed two times.}$

$\displaystyle \text{The possible outcomes are \{TT, HH, TH, HT\} }$

$\displaystyle \therefore n(S) = 4$

$\displaystyle \text{Favourable outcomes : \{HH, HT TH \} }$

$\displaystyle \therefore n(E) = 3$

$\displaystyle \text{Required probability }= \frac{n(E)}{n(s)} = \frac{3}{4}$

$\\$

$\displaystyle \text{Question 25:} \\ \text{(A) Evaluate: } \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^ 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \\ \\ {\hspace{5.0cm} \text{OR}} \\ \\ \text{(B) If A and B are acute angles such that } \sin(A-B) = 0 \text{ and } \\ 2 \cos (A+B) - 1 = 0, \text{ then find the angles A and B. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) We have } \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^ 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

$\displaystyle = \frac{5\Big( \frac{1}{2} \Big)^2 + 4 \Big( \frac{2}{\sqrt{3}} \Big) - (1)^2}{\Big( \frac{1}{2} \Big)^2 + \Big( \frac{\sqrt{3}}{2} \Big)^2}$

$\displaystyle = \frac{\frac{5}{4} + \frac{16}{3} - 1}{\frac{1}{4}+ \frac{3}{4}}$

$\displaystyle = \frac{5}{4}+ \frac{16}{3} - 1 = \frac{15+64-12}{12} = \frac{67}{12}$

$\displaystyle {\hspace{5.0cm} \text{OR}}$

$\displaystyle \text{(B) Given } \sin(A-B) = 0 \text{ and } 2 \cos (A+B) - 1 = 0$

$\displaystyle \Rightarrow \sin(A-B) = \sin 0^\circ$

$\displaystyle \text{and } \cos(A+B) = \frac{1}{2}$

$\displaystyle \Rightarrow A-B = 0^\circ \ldots \ldots (i)$

$\displaystyle \text{and } \cos (A+B) = \cos 60^\circ$

$\displaystyle \text{and } A+B = 60^\circ \ldots \ldots (ii)$

$\displaystyle \text{On solving equations (i) and (ii) we get }$

$\displaystyle A = 30^\circ \text{ and } B = 30^\circ$

$\\$

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SECTION – C

Section – C consists of Short Answer (SA) type questions of 3 marks each.

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$\displaystyle \text{Question 26:} \\ \text{(A) How many terms are there in an A.P, whose first and fifth terms are -14 and 2, } \\ \text{respectively and the last term is 62. }\\ {\hspace{5.0cm} \text{OR}} \\ \text{(B) Which term of the A.P: 65, 61,57,53, ... is the first negative term? }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) Given first term } (a) = - 14, \text{ fifth term } (a_5) = 2 \\ \text{ and the last term } (a_n) = 62$

$\displaystyle \text{Let common difference be } d.$

$\displaystyle \text{Since } a_5 = a + 4d$

$\displaystyle \Rightarrow 2 = - 14 + 4d$

$\displaystyle \Rightarrow d = 4$

$\displaystyle \Rightarrow a_n = a + (n-1) 4$

$\displaystyle \Rightarrow 62 = -14+(n-14) 4$

$\displaystyle \Rightarrow n-1 = 19$

$\displaystyle \Rightarrow n=20$

$\displaystyle \text{Thus the number of terms in the A.P. is 20. }$

$\displaystyle {\hspace{5.0cm} \text{OR}}$

$\displaystyle \text{Given A.P. is 65, 61 57, 53, ... }$

$\displaystyle \text{Here, first term, } a= 65$

$\displaystyle \text{Common difference, } d = -4$

$\displaystyle \text{Let the } n^{th} \text{ term of the given A.P. be the first negative term.}$

$\displaystyle \because a+n < 0$

$\displaystyle \Rightarrow a+ (n-1) d < 0$

$\displaystyle \Rightarrow 65 + (n-1)(-4) < 0$

$\displaystyle \Rightarrow 69-4n < 0$

$\displaystyle \Rightarrow -4n < -69$

$\displaystyle \Rightarrow n > \frac{69}{4}$

$\displaystyle \Rightarrow n > 17\frac{1}{4}$

$\displaystyle \text{Since, 18 is the natural number just greater that } 17\frac{1}{4}$

$\displaystyle \text{Hence, } n = 18$

$\displaystyle \text{Hence, }18^{th} \text{ term is the first negative term.}$

$\\$

$\displaystyle \text{Question 27: Prove that } \sqrt{5} \text{ is an irrational number.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Assume that } \sqrt{5} \text{ is a rational number. }$

$\displaystyle \text{Then } \sqrt{5} = \frac{a}{b} \text{ where HCF of } a, b \text{ is } 1. \ldots (i)$

$\displaystyle \Rightarrow a = \sqrt{5} b$

$\displaystyle \Rightarrow a^2 = 5b^2 \text{ where } b \neq 0$

$\displaystyle \text{Since, } a^2 \text{ is a multiple of 5, so } a \text{ is also a multiple of 5. }$

$\displaystyle \text{Let } a = 5m$

$\displaystyle \Rightarrow (5m)^2 = 5b^2$

$\displaystyle \Rightarrow 25m^2 = 5b^2$

$\displaystyle \Rightarrow b^2 = 5m^2$

$\displaystyle \text{Since } b^2 \text{ is a multiple of 5, so, } b \text{ is also a multiple of 5. }$

$\displaystyle \text{Let } b = 5n$

$\displaystyle \text{Thus, HCF of } (a, b) = 5 \ldots (ii)$

$\displaystyle \text{From equation (i) and (ii), we get that our assumption is wrong. }$

$\displaystyle \text{Therefore } \sqrt{5} \text{ is not a rational number. It is a irrational number. }$

$\\$

$\displaystyle \text{Question 28: Prove that the angle between the two tangents drawn from an external to} \\ \text{circle is supplementary to the angle subtended by the line joining the points of } \\\text{contact at the center.}$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{Given: } \text{PA and PB are the tangent drawn from a point P to a circle with center O. }$

$\displaystyle \text{Also, the line segments OA and OB are drawn. }$

$\displaystyle \textbf{To Prove: }\angle APB+ \angle AOB= 180^\circ$

$\displaystyle \textbf{Proof: } \text{We know that the tangents to a circle is perpendicular to the radius } \\ \text{through the points of contact. }$

$\displaystyle \therefore PA \perp OA \Rightarrow \angle OAP = 90^\circ$

$\displaystyle \text{and } PB \perp OB \Rightarrow \angle OBP = 90^\circ$

$\displaystyle \text{Therefore, } \angle OAP + \angle OBP = 180^\circ$

$\displaystyle \text{Hence, }\angle APB + \angle AOB = 180^\circ$

$\displaystyle \text{Note: Sum of all the angles of a quadrilateral is } 360^\circ$

$\\$

$\displaystyle \text{Question 29:} \\ \text{(A) The sum of first 15 terms of an A.P is 750 and its first term is 15. Find its 20th term. }$

$\displaystyle \text{(B) Rohan repays his total loan of Rs. 1,18,000 by paying every month starting } \\ \text{with the first instalment by Rs 1000. If he increases his instalment by Rs. 100 every } \\ \text{month, what amount will be paid by him in the 30th instalment? What amount of } \\ \text{loan has he paid after 30th instalment? }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) Given, } S_{15}= 750 \text{ and the first term } a = 15$

$\displaystyle \because S_n = \frac{n}{2} [2a + (n-1) d ]$

$\displaystyle \therefore S_{15} = \frac{15}{2} [2a + (15-1) d ]$

$\displaystyle \Rightarrow 750 = \frac{15}{2} [2 \times 15 + 14 d ]$

$\displaystyle \Rightarrow 50 \times 2 = 30 + 14 d$

$\displaystyle \Rightarrow 14d = 70$

$\displaystyle \Rightarrow d = \frac{70}{14} = 5$

$\displaystyle \because a_n = a + ( n-1) d$

$\displaystyle \therefore a_{20} = 15 + (20-1) (5)$

$\displaystyle \Rightarrow a_{20} = 15+95 = 110$

$\displaystyle \text{(B) First Instalment, } a = 1000 \text{ Rs. }$

$\displaystyle \text{Common difference } d = 100 \text{ Rs. }$

$\displaystyle \because a_n = a + (n-1) d$

$\displaystyle \therefore a_{30} = a + (30-1) d$

$\displaystyle = 1000+ 29 \times 100$

$\displaystyle = 1000 + 2900$

$\displaystyle = 3900$

$\displaystyle \text{Thus 3900 Rs. will be paid by Rohan on the 30th instalment.}$

$\displaystyle \text{Amount of loan paid by Rohan after 30 instalments }$

$\displaystyle = \text{Total Loan } - \text{Amount paid in 30 instalments}$

$\displaystyle = 118000- \frac{30}{2} [ 2 \times 1000 + (30-1) \times 100]$

$\displaystyle = 118000- 15(2000+2900)$

$\displaystyle = 118000 - 15 \times 4900$

$\displaystyle = 118000 - 73500$

$\displaystyle = 44500 \text{ Rs. }$

$\\$

$\displaystyle \text{Question 30: Prove that } \sqrt{3} \text{ is an irrational number}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } \sqrt{3} \ \text{is a rational number.}$

$\displaystyle \therefore \sqrt{3} = \frac{p}{q} \ \text{where } \ \ \ \ \ [ p \text{ and } q \text{ are co-prime integers and } q \neq 0]$

$\displaystyle \Rightarrow 3 = \frac{p^2}{q^2}$

$\displaystyle \Rightarrow p^2 = 3q^2 \ldots \ldots (i)$

$\displaystyle 3 \text{ is a factor of } p^2$

$\displaystyle \Rightarrow 3 \text{ is a factor of } p$

$\displaystyle \text{Hence, } p = 3 \times m \ \ \ \ \ [m \text{ is any integer } ] \ldots \ldots (ii)$

$\displaystyle \text{From equation } (i) \ \ \ \ 9m^2 = 3q^2$

$\displaystyle \Rightarrow q^2 = 3m^2$

$\displaystyle \therefore 3 \text{ is a factor of } q^2$

$\displaystyle \Rightarrow 3 \text{ is a factor of } q$

$\displaystyle \text{From equation (i) and (ii), } 3 \text{ is a common factor of } p \text{ and } q.$

$\displaystyle \text{It contradicts our assumption that } p \text{ and } q \text{ are co-prime integers. } \\ \text{Hence our assumption is wrong. }$

$\displaystyle \therefore \sqrt{3} \text{ is irrational. }$

$\\$

$\displaystyle \text{Question 31: Find the value of } 'p' \text{ for which the quadratic equation } \\ px(x-2) + 6 =0 \text{ has two equal roots.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{For equal roots determinant is } = 0$

$\displaystyle \therefore b^2 - 4ac = 0$

$\displaystyle \text{Given equation is } px(x-2)+6 = 0$

$\displaystyle \Rightarrow px^2 - 2px + 6 = 0$

$\displaystyle \text{Here, } a = p, b= - 2p \text{ and } c = 6$

$\displaystyle \text{Therefore } (-2p)^2 - 4(p)(6) = 0$

$\displaystyle \Rightarrow 4p^2 - 24 p = 0$

$\displaystyle \Rightarrow 4p^2 = 24p$

$\displaystyle \Rightarrow p^2 = 6 p$

$\displaystyle \Rightarrow p^2 - 6p = 0$

$\displaystyle \Rightarrow p ( p-6) = 0$

$\displaystyle \therefore p = 0 \text{ or } p = 6$

$\displaystyle \therefore p = 6$

$\displaystyle \text{Note: If } p = 0, \text{ the the given equation is not a quadratic equation. }$

$\\$

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SECTION. D

Section – D consists of Long Answer (LA) type questions of 4 marks each.

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$\displaystyle \text{Question 32: From a solid cylinder of height 20 cm and a diameter of 12 cm, } \\ \text{a conical cavity of height 8 cam and a radius 6 cm is hollowed out. Find the } \\ \text{total surface are of the remaining solid.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The remaining solid, after removing the conical cavity can be drawn as: }$

$\displaystyle \text{Height of the cylinder, } h_1 = 20 \text{ cm}$

$\displaystyle \text{Radius of the cylinder, } r = \frac{12}{2} = 6 \text{ cm}$

$\displaystyle \text{Height of the cone, } h_2 = 8 \text{ cm}$

$\displaystyle \text{Radius of the cone, } r = 6 \text{ cm}$

$\displaystyle \text{Total surface area of remaining solid } \\ \text{= Areas of the top face of the cylinder + curved surface area of the cylinder } \\ \text{+ curved surface area of cone }$

$\displaystyle \text{Now, slant height of cone,}$
$\displaystyle l = \sqrt{r^2 + {h_2}^2} = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10 \text{ cm}$

$\displaystyle \text{Curved surface area of the cone}$
$\displaystyle = \pi r l = \frac{22}{7} \times 6 \times 10 = \frac{1320}{7} \text{ cm}^2$

$\displaystyle \text{Curved surface area of the cylinder}$
$\displaystyle = 2 \pi r {h_1} = 2 \times \frac{22}{7} \times 6 \times 20 = \frac{5280}{7} \text{ cm}^2$

$\displaystyle \text{Area of the top face of the cylinder}$
$\displaystyle = \pi r^2 = \frac{22}{7} \times 6^2 = \frac{792}{7} \text{ cm}^2$

$\displaystyle \text{Thus, total surface area of remaining solid}$
$\displaystyle = \frac{1320}{7} + \frac{5280}{7} + \frac{792}{7} = \frac{7392}{7} = 1056 \text{ cm}^2$

$\\$

$\displaystyle \text{Question 33:} \\ \text{(A) D is a point on the side BC of a triangle ABC such that } \angle ADC = \angle BAC, \\ \text{ prove that } CA^2 = CB \cdot CD \\ {\hspace{5.0cm} OR} \\ \text{(B) If AD and PM are medians of triangle ABC and PQR respectively where } \\ \triangle ABC \sim \triangle PQR, \text{ prove that } \frac{AB}{PQ} = \frac{AD}{PM}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) } \textbf{Given: } \text{D is the point on the side BC of } \triangle ABC \text{ such that } \angle ADC = \angle BAC$

$\displaystyle \textbf{To Prove: } CA^2 = CB \cdot CD$

$\displaystyle \textbf{Proof: } \text{From } \triangle ADC \text{ and } \triangle BAC$

$\displaystyle \angle ADC = \angle BAC \text{ (given) }$

$\displaystyle \angle ACD = \angle BCA \text{ (common angle) }$

$\displaystyle \therefore \triangle ADC \sim \triangle BAC \text{ (By AA similarity criterion) }$

$\displaystyle \text{We know that, the corresponding sides of similar triangles are in proportion. }$

$\displaystyle \therefore \frac{CA}{CB} = \frac{CD}{CA}$

$\displaystyle \Rightarrow CA^2 = CB \cdot CD$

$\displaystyle \text{Hence Proved.}$

$\displaystyle \text{(B) Given, } \triangle ABC \sim \triangle PQR$

$\displaystyle \text{We know that, the corresponding sides of similar triangles are in proportion. }$

$\displaystyle \therefore \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} \ldots \ldots (i)$

$\displaystyle \text{Also, } \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R \ldots \ldots (ii)$

$\displaystyle \text{Since AD and PM are medians, they will divide opposite sides. }$

$\displaystyle \therefore BD = \frac{BC}{2} \text{ and } QM = \frac{QR}{2} \ldots \ldots (iii)$

$\displaystyle \text{From equation (i) and (ii) we get }$

$\displaystyle \frac{AB}{PQ} = \frac{BD}{QM} \ldots \ldots (iv)$

$\displaystyle \text{In } \triangle ABD \text{ and } \triangle PQM$

$\displaystyle \angle B = \angle Q$

$\displaystyle \frac{AB}{PQ} = \frac{BD}{QM}$

$\displaystyle \therefore \triangle ABD \sim \triangle PQM \text{ (By SAS similarity criterion) }$

$\displaystyle \text{Thus, } \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$

$\displaystyle \text{Hence, } \frac{AB}{PQ} = \frac{AD}{PM}$

$\displaystyle \text{Hence proved. }$

$\\$

$\displaystyle \text{Question 34: A student was asked to make a model shaped like a cylinder with two} \\ \text{cones attached to its ends by using a thin aluminium sheet. The diameter of the model}\\ \text{is 3 cm and its total length is 12 cm. If each cone has a height of 2 cm, find the}\\ \text{volume of air contained in the model. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have (the figure) }$

$\displaystyle \text{Height of the cylinder:} = 12 - 4 = 8 \text{ cm }$

$\displaystyle \text{Radius of cone / cylinder } = \frac{3}{2} = 1.5 \text{ cm }$

$\displaystyle \text{Height of the cone } = 2 \text{ cm }$

$\displaystyle \text{Volume of the cylinder } = \pi r^2 h = \pi (1.5)^2 \times 8 = 18 \pi \ cm^3$

$\displaystyle \text{Volume of the cone } = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1.5)^2 \times 2 = 1.5 \pi \ cm^3$

$\displaystyle \text{Total volume = Volume of cylinder + (volume of cone) } \times 2 \\ {\hspace{3.0cm} = 18\pi + 3\pi = 21\pi = 21 \times \frac{22}{7} = 66 \ cm^3 }$

$\\$

$\displaystyle \text{Question 35: }$

$\displaystyle \text{(A) In the given figure, } \angle ADC = \angle BCA; \text{ prove that }\triangle ACB \sim \triangle ADC. \\ \text{ Hence find } BD \text{ if } AC = 8 \text{ cm and } AD=3 \text{ cm.}$

$\displaystyle \text{(B) If a line is drawn parallel to one side of a triangle to intersect the other} \\ \text{two sides in distinct points, then prove that the other two sides are divided in} \\ \text{the same ratio. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) }$

$\displaystyle \text{In } \triangle ACB \text{ and } \triangle ADC,$

$\displaystyle \angle ADC = \angle BCA \text{ (given) }$
$\displaystyle \angle A= \angle A \text{ (common) }$
$\displaystyle \therefore \triangle ACB \sim \triangle ADC$
$\displaystyle \text{(By AA similarity criterion) Hence Proved. }$

$\displaystyle \text{Since } \triangle ACB \sim \triangle ADC$
$\displaystyle \frac{AC}{AD} = \frac {BC}{CD} = \frac{AB}{AC}$
$\displaystyle \frac{AC}{AD} = \frac{AB}{AC}$
$\displaystyle \text{(on equating first and last term) }$

$\displaystyle AC^2 = AD \times AB$
$\displaystyle 8^2 = 3 \times AB$
$\displaystyle \text{[Given AC = 8 cm and AD = 3 cm] }$

$\displaystyle \Rightarrow AB = \frac{64}{3} \text{ cm}$
$\displaystyle \text{Thus, } BD = AB - AD = \frac{64}{3} - 3 = \frac{64-9}{3} = \frac{55}{3} = 18.3 \text{ cm}$

$\displaystyle \text{(B) }$

$\displaystyle \text{(B) Let } \triangle ABC \text{ in which a line DE parallel to BC intersects AB at D and AC at E. }$

$\displaystyle \textbf{To prove: } \text{DE divides the two sides in the same ratio. }$
$\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$

$\displaystyle \textbf{Construction: } \text{Join BE and CD.}$

$\displaystyle \textbf{Draw } EF \perp AB \text{ and } DG \perp AC$

$\displaystyle \text{Proof: We known that, }$
$\displaystyle \text{Area of triangle } = \frac{1}{2} \times \text{Base} \times \text{Height}$
$\displaystyle \text{Then, }$
$\displaystyle \frac{area(\triangle ADE)}{area(\triangle DEC)} = \frac{\frac{1}{2} \times AD \times EF}{\frac{1}{2} \times DB \times EF} = \frac{AD}{DB} \ldots \ldots (i)$
$\displaystyle \text{and, }$
$\displaystyle \frac{area(\triangle ADE)}{area(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times GD}{\frac{1}{2} \times EC \times GD} = \frac{AE}{EC} \ldots \ldots (ii)$

$\displaystyle \text{Since, } \triangle BDE \text{ and } \triangle DEC \\ \text{ lie between the same parallel DE and BE, and are on the same base DE.}$

$\displaystyle \text{We have, } area(\triangle BDE) = area (\triangle DEC) \ldots \ldots (iii)$

$\displaystyle \text{From equations (i), (ii) and (iii), we get }$

$\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$

$\\$

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SECTION. E

Section – E consists of three Case Studies. Based questions of 4 marks each.

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$\displaystyle \text{Question 36:}$

Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey Rs. $x$ per student and Cricket Rs. $y$ per student. School ‘P’ decided to award a total of Rs. 9,500 for the two games to 5 and 4 students respectively; while school ‘Q’ decided to award Rs. 7,370 for the two games to 4 and 3 students respectively.

$\displaystyle \text{Based on the above information answer the following questions: }$

$\displaystyle \text{(i) Represent the following information algebraically (in terms of } x \text{ and } y ).$

$\displaystyle \text{(ii) (a) What is the prize amount for hockey? }$

$\displaystyle \text{ (b) Prize amount on which game is more and by how much? }$

$\displaystyle \text{(iii) What will be the total prize amount if there are 2 students each from two games ? }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Given Rs. } x \text{ and Rs.} y \text{ are the price money per student for Hockey and Cricket respectively. }$

$\displaystyle \therefore 5x + 4y = 9500 \ldots \ldots (i)$

$\displaystyle \text{and } 4x + 3y = 7370 \ldots \ldots (ii)$

(ii)

(a) On multiplying equation (i) by 4 and equation (ii) by 5 we get

$\begin{array}{ccccc} 20x & + & 16y & = & 38000 \\ 20x & + & 15y & = & 36850 \\ - & - & - & & \\ \hline & & y & = & 1150 \end{array}$

On substituting value of $y$ in equation (ii) we get

$5x + 4( 1150) = 9500$

$\Rightarrow 5x + 4600 = 9500$

$\Rightarrow 5x = 4900$

$\Rightarrow x = 980$

$\displaystyle \text{Thus, prize money for Hockey is Rs. 980. }$

$\displaystyle \text{(b) From part (a),}$

$\displaystyle \text{Prize money for Hockey = Rs. 980 }$

$\displaystyle \text{Price money for Cricket = Rs. 1150 }$

$\displaystyle \text{Difference between price money = 1150 - 980 = Rs. 170 }$

$\displaystyle \text{Thus the price money is Rs. 170 more for cricket in comparison to hockey }$

$\displaystyle \text{(iii) Total prize money = 2 ( Prize money for Hockey) + (Prize money for Cricket) }$

$= 2 (980 + 1150) = 2 \times 2130 = 4260 \ \text{Rs. }$

$\\$

$\displaystyle \text{Question 37:}$

Jagdhish has a field which is in the shape of a right angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field from growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O.

$\displaystyle \text{(i) Taking O as origin, coordinates of P are (-200, 0) and of Q are (200, 0).} \\ \text{PQRS being a square, what are the coordinates of R and S? }$

$\displaystyle \text{(ii) (a) What is the area of square PQRS ? }$

$\displaystyle \text{(b) What is the length of diagonal PR in square PQRS? }$

$\displaystyle \text{(iii) If S divides CA in the ratio K : 1, what is the value of K where point } \\ \text{A is (200, 800)? }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Coordinates of R = (200,400) }$

$\displaystyle \text{Coordinates of S = (- 200,400) }$

$\displaystyle \text{(ii) Since, side of square PQRS = 400 }$

$\displaystyle \text{Thus, area of square PQRS } = (400)^2 = 160000 \text{ units}^2$

$\displaystyle \text{We know that, diagonal of square } = \sqrt{2} \times \text{side }$

$\displaystyle \text{Therefore diagonal PR of square PQRS } = \sqrt{2} \times 400 = 400\sqrt{2} \text{ unit }$

$\displaystyle \text{(iii) Using section formula }$

$\displaystyle 200 = \frac{200K + 1(-600)}{K+1}$

$\displaystyle \Rightarrow - 200 K - 200 = 200 K - 60$

$\displaystyle \Rightarrow - 400 K = - 400$

$\displaystyle \Rightarrow K =$

$\displaystyle \text{Note: Here S is the mid-point of CA, hence S divides CA in ratio 1:1 }$

$\\$

$\displaystyle \text{Question 38:}$

Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking.

After survey, it was decided to build a rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats.

$\displaystyle \text{Based on the above information, answer the following questions: }$

$\displaystyle \text{(i) What is the total perimeter of the parking area? }$

$\displaystyle \text{(ii) (a) What is the total area of parking and the two quadrants? }$

$\displaystyle \text{(b) What is the ratio of area of playground to the area of parking area? }$

$\displaystyle \text{(iii) Find the cost of fencing the playground and parking area at the rate of } \\ \text{Rs. 2 per unit. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Radius of semi-circle } (r) = \frac{7}{2} =3.5 \text{ units}$

$\displaystyle \text{Circumference of semi-circle } = \pi r = \frac{22}{7} \times 3.5 = 11 \text{ units}$

$\displaystyle \therefore \text{Perimeter of parking area = circumference of semi-circle + diameter of semi-circle } = 11 + 7 = 18 \text{ units}$

$\displaystyle \text{(ii) (a) Area of parking } = \frac{\pi r^2}{2} = \frac{22}{7} \times \frac{1}{2} \times (3.5)^2 = 11\times 0.5 \times 3.5 = 19.25 \text{ unit}^2$

$\displaystyle \text{Area of quadrants } = 2 \times \text{(area of one quadrant)} \\ \\ = 2 \times \frac{\pi {r_1}^2 } {4} = 2 \times \frac{22}{7} \times \frac{1}{4} \times 2^2 = 6.285 \text{ unit}^2$