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Time Allowed: 3 hours General Instructions: 1. This Question Paper has 5 Sections A, B, C, D and E. Maximum Marks: 80 234 ... Section A has 20 MCQs carrying 1 mark eachSection B has 5 questions carrying 02 marks each.Section C has 6 questions carrying 03 marks each. 56 .. Section D has 4 questions carrying 05 marks each.Section E has 3 case based integrated units of assessment (04 marks each) with sub- parts of the values of 1, 1 and 2 marks each respectively. 78 .. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2marks questions of Section EDraw neat figures wherever required. Take wherever required if not stated.
a) 360^ π^ =^ Section A^227 b) 90
1. If a = (2 c) 1802 × 33 × 54 ) and b = (2^3 × 32 ×5) then HCF (a, b) = ?d) 540 [1]
a) c) 24 b)d) (^31)
- The graph of y = f(x) is shown in the figure for some polynomial f(x). The number of zeroes of f(x) are [1]
- A system of linear equations is said to be inconsistent if it has [1]
a) c) one solutiontwo solutions b)d) at least one solutionno solution
4. If the equation x a) c) k −8 5 > < 85 k <^2 + 5 kx + 16 = 0 has no real roots then 85 b)d) kk <> −8−8 55 [1]
- If the sum of n terms of an A.P. be 3n a) c) 23 2 + n and its common difference is 6 then its first term isb)d) 14 [1]
6. The distance of point P(4, -5) from origin is a) c) 3 units √ − 40 −units b)d) 1 unit√ − 41 −units [1]
7. If (3, –6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is: a) c) (6, - 6) ( 32 , −3 ) b)d) (6, -12)(- 3, 6) [1]
a) c) 1 : 33 : 4 b)d) 1 : 42 : 3
8. In the given figure if BP || CF , DP ||E F ,then AD : DE is equal to [1]
a) b) cm
- Two concentric circles with centre O are of radii 6 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles as shown in the figure. If AP = 10 cm, then BP is equal to [1]
√^ − 91 − Page 2 of 20 √^ −− 119 −
Section B
Section C
c) all the three d) median a) c) Both A and R are true and R is the correct explanation of A.A is true but R is false. b)d) Both A and R are true but R is not the correct explanation of A.A is false but R is true.
- Assertion (A): resulting cuboid is 300 cm Reason (R): Total surface area of a cuboid is 2(lb + bh + lh) Two identical solid cubes of side 5 cm are joined end to end. The total surface area of the (^2). [1]
20. Assertion (A): Reason (R): a) c) Both A and R are true and R is the correct explanation of A.A is true but R is false. Given series is in A.P. and sum of n terms of an A.P. is S The sum of the series with the nth term. t b)d)n Both A and R are true but R is not the correct explanation of A.A is false but R is true.= (9 - 5n) is (465), when no. of terms n = 15.n = n 2 [2 a + ( n − 1) d ] [1]
2122 .. Prove thatIn Figure, ∠ACB√^27 is irrational. = 90o and CD ⊥ AB. Prove that BC AC 22 = BD AD. [2][2]
2324 .. From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm. Then what is the radius of the circle?Prove the trigonometric identity: [2][2]
25. If 3tanIn Figure, OACB is a quadrant of a circle with centre O and radius 7 cm. If OD = 3 cm, then find the area of the shaded region.1+cos^ sin^ θ (1+cos^ θ −^^ θ sin^^ = 4, evaluate^2 θ ) θ^ = cot^ θ^ 3 sin^ 3 sin^ θθ +2 cos−2 cos^ θθ.^ OR [2]
The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days. [Take π = 3.14.]^ OR
- In order to promote reading habits among students, a school organized a Library Week. As part of the celebration, three genres of books: Biography, Mystery, and Self-help books were bought. For optimum arrangement, the organizers have stacked the books in such a way that all the books are stored topic-wise and the [3]
Section D
height of each stack is the same. The number of Biography books is 96, the number of Mystery books is 240 and the number of Self-help books is 336. Assuming that the books are of the same thickness, determine the number of stacks of Biography, Mystery, and Self-help books.
2728 .. IfSolve the following system of equation by elimination method: α , β are the zeroes of the x and^2 + 7x + 7, find the value of α^1 + 1 β − 2 αβ. [3] [3]
In the figure below ABCDE is a pentagon with BE of ABCDE is 21 cm, find the Values of x and y. x^2 −^^ y^5 = 4 x^7 +^15^ y = 3 ∥CD and BC || DE. BC is perpendicular to CD. If the perimeter^ OR
- In Fig. l and m are two parallel tangents at A and B. The tangent at C makes an intercept DE between l and m. Prove that = 90o. [3]
Prove that the tangents drawn at the ends of a chord of a circle make equal angles with chord.^ OR
∠ DF E
3031 .. Prove the following identity:Find the median of the following data. Class Interval 1−cossin0 - 10^ θ θ + 1+cos^ tan^ θ 10 - 20 θ = secθ ⋅ cos20 - 30 ecθ + cotθ 30 - 40 40 - 50 Total [3]
- Find the value of m for which the quadratic equation^ Frequency^^8 16 36 34 6 ,^100^ [3]
has equal roots. Hence find the roots of the equation. The sum of ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.^ ( m OR^ + 1)^ y^2 − 6( m^ + 1) y^ + 3( m^ + 9) = 0 m^ ≠ −1^ [5]
- In the given figure, DEFG is a square and Prove that i.. iii iv^ ii...
△ AGF ∼ △ DBG ∠ BAC^ =^90 ∘^ [5]
△ △ D EAGFDBG 2 = BD ∼∼ △△ × EF CEF C EC
faces and three rectangular faces. Also, the image of canvas on graph paper is shown in the adjacent figure.
- Read the text carefully and answer the questions: (a)(b)(c)^ What is the distance of point Q from y-axis?^ What are the coordinates of U?^ What is the distance between the points P and Q?^ What is the Perimeter of image of a rectangular face?^ OR The houses of Ajay and Sooraj are at 100 m distance and the height of their houses is the same as approx 150 m. One big tower was situated near their house. Once both friends decided to measure the height of the tower. They measure the angle of elevation of the top of the tower from the roof of their houses. The angle of elevation of ajay's house to the tower and sooraj's house to the tower are 45o^ and 30o^ respectively as shown in the figure.
[4]
(a)(b)(c) Find the height of the tower. What is the distance between the tower and the house of Sooraj? Find the distance between top of tower and top of Ajay's house? Find the distance between top of the tower and top of Sooraj's house? OR
(c) Explanation: (^180) It is given that: a = (2 (^2 33 54) ) and b Section A = (2 (^3 32) 5)
(c) Explanation:^ HCF (a, 4 b) = Product of smallest power of each common prime factor in the numbers = f(x) intersects the x -axis at 4 points. hence , f(x) has 4 zeroes.^22 32 5 = 180
(d) Explanation: and not cutting each other at any point. no solution A system of linear equations is said to be inconsistent if it has no solution means two lines are running parallel
(c) Explanation: For no real roots, we must have b (^2) - 4ac < 0..
(d) Explanation: and common difference (d) = 6 (^4) Sum of n terms of an A.P = 3n (^2) +n Let the first term be a, then
(d) Explanation: units OP = 7.^ = (b) Explanation:^ (6, -12)units If (a, b) and (c, d) be the coordinates of any two points, then the coordinates of the mid-point joining those points be The line segment is formed by points are (0, 0) and (x, y), whose mid-point is (3, -6). Then, (^) = 3 and. = -
or,^ or,^ Therefore the required point is (6, -12). (a)^ x = 61 : 3= 3 or,^ or, y = -12^ = - Explanation: Since BP CF,
∴ ×^ ×^ ×^ × × ×
∴ ( 25 −8^5 k^ <^2 − 4 × 16^ k^ <^85 ) < 0 ⇒ 25 k^2 < 64 ⇒ k^2 < 6425 ⇒ −8 5 < k <^85 S ⇒ (^2) ⇒ 2 an (^) (^) + 6 = n (^2) a [2 (^) n + 6 (^2) an [2 + (− 6 = an + ( − 6 = (3 n − 1)6] = 3 n ( (^) 3 − 1) n (^2) n + d + 1)2 = 6] = 3 n n ) 2 × n + (^22) n n += (^) nn n + 2 (3 n +1)×2 n ⇒ 2 a = √ a (^82) − 41 = 6−= 4 n + 2 − 6 n + 6 = 8 = (^) √ √ − 41 −−(16 + 25)−^ −−−−−−^ √^ −^ (4 − 0−^ −−−−^ )^2 −^ + (0 − (−5)−−−−−−−−−−)−^2 (0+ 2 ( x^ )^ ( a +^2 c ),^ ( b +^2 d (0+)) 2 y ) x 2 2^ y ∥
Solution
- (c) Explanation: So, area 130.95 cm (^2) Here the angle swept is 150o. We need to find the area of this sector which subtends 150o (^) at the centre. 15.^ = 130.95 cm (c) Explanation: 52 cm (^22) We know that perimeter of a sector of radius, r …(1) Therefore, substituting the corresponding values of perimeter and radius in equation (1), we get, …(2) We know that area of the sector From equation (3), we get Area of the sector^ ………(3) Substituting r = 6.5 Area of the sector we get,
16.^ = 52^ Therefore, area of the sector is 52 cm (b)^2. Explanation: Number of king cards in the pack = 4 So, favourable number of outcomes = 48 Number of cards that are not king = 52 - 4 = 48 Total number of possible outcomes = 52
- (a) Explanation: No. of Mondays = 52^ Required probability = No. of days in a leap year = 366^ = Extra days = 366 - 52 = 366 - 364 = 2 Remaining days in the week = 7 - 2 = 5 Probability of being 52 Mondays in the leap 7 18.^ year = (b) Explanation: mode The most frequent value in the data is known as the Mode. e.g let us consider the following data set: 19.^ 3,5,7,5,9,5,8,4 the mode is 5 since it occurs most often in data set. (d) Explanation: A is false but R is true. A is false but R is true.
= 227 × = 10 π^2^ r ×^2 ×^150360360 θ ∘ 29 = 2 × 6.5 + 29 = 2 × 6.5 (1 + 360 θ 360 × 2 θ × π (^) π × 6.5)^ = 2 r^ +^360 θ × 2 πr 2×6.5 2×6.5^2929 = − 1 =^ (1 + 360 θ 360^ θ ×× π^ π ) = 360 θ × πr 2 = ( 29×6.5^2 − =^ =)^ (^ (^ 2×6.5^ 2×6.5^2929 − 1− 1)) r 6.5^22 = = = (94.25 − 42.25) (( 29×6.529×6.5^ 2×6.5 22 −− 6.56.56.5^222 )) ∴^1213 ∴ (^5748521213) ∴ ∴ 5 × 7
- (d) Explanation: A is false but R is true. A is false but R is true. Section B
- Let us assume that As 2b and a are rational numbers. is rational. Then, there exist positive co-primes a and b such that Then But Since a rational number cannot be equal to an irrational number. Our assumption that Hence is not a rational number.is rational number.is an irrational number is rational number is wrong.
- (^) Similarly CB (^2) = AB BD ...(i)(By AA similarity) By (i) and (ii)^ AC = 2 = AB^ =AD ...(ii)
- OAP = 90° in By applying Pythagoras theorem , we get , = we should not take negative value because length cannot be negative.
- We have, L.H.S
= R.H.S Hence proved. OR Given,
√√^277 – = =^ a^ b^2 ab √^27 √^27 –^ ab 2 √ (^27) △ ⇒AB CBABC ∼ = √ BC BD^7 △=CBD AC CD× ⇒AB^ AC =^ BC^ CD^ △=ABC ∼^ AD^ AC×^ △ACD CB AC^22 AB×BD AB×AD BD AD ∠ ⇒ △ 17 OAP (^2) = r (^2) + 152 ⇒ ⇒ 2 × 32 rr^22 == 64 ⇒ 172 − r (^15) = ±8cm^2 = (17 − 15)(17 + 15) ⇒ = 1− r = 8 sin = (^2) θcm 1+cos+cos sin θ (1+cos θ θ − sin θ (^2) θ ) = = = (^) sincoscos^ sin sincos θθθθθ^2 (1+cos(1+cos(1+cos(1+cos θ +cos^ θ θθθθ )))) [∵ 1 − sin^2 θ = cos^2 θ ] = cot^ sin^ θ θ [∵ cos sin^ θθ = cotθ ] (^3) = tanθ 3 sin 3 sin = 4 θθ +2 cos−2 cos ⇒ tan θθ θ = (^43)
So, the solution of the given system of equations is x = 14, y = 15 Verification ; Substituting x = 14, y = 15. We find that both the equations (1) and (2) are satisfied as shown below; Hence, the solution is correct. OR Since BC | | DE and BE | | CD with BC Since, BE = CD Also, DE = BC = x - y x + y = 5 ..(i) CD, BCDE is a rectangle. Since, perimeter of ABCDE is 21 AB + BC + CD+ DE + EA = 21 3 + x - y + x + y + x - y + 3 = 21 6 + 3x - y = 21 Adding eqns. (i) and (ii), we get 4x = 20^ 3x - y = 15.....(ii) x = 5 On substituting the value of x in (i), we get y = 0 x = 5 and y = 0.
- Since tangents drawn from an external point to a circle are equal. Therefore, DA = DC. Thus, in triangles ADF and DFC, we have DA = DC DF = DF Common] AF = CF (radii of the circle) So, by SSS-criterion of congruence, we obtain ...(i) Similarly, we can prove that Now, = 180° (Sum of the interior angles on the same side of transversal is 180 .....(ii) o) = 90°^ = 180° [ Using equations (i) and (ii) ]
⇒ 2 + ⇒ 15 y = 1 ⇒ 15^ y = 3 ⇒ y = 15 15^ y = 3 − 2 x (^2) x 7 − +^ y (^515) y == 142147 −+ 15 51515 = 7 − 3 = 4= 2 + 1 = 3
∴^ ⊥ ∴ ⇒ ⇒ ⇒ ⇒ ∴
Δ ⇒ ⇒ ADF ∠∠ ADFADC ≅Δ DF C = ∠= 2∠ CDFCDF
∠ ⇒ ⇒ BEF 2∠∠ ∠ CDF CEBADC = ∠ CEF + 2∠ = 2∠+ ∠ CEBCEFCEF
⇒ ∠ CDF + ∠ CEF Page 13 of 20
Let NM be chord of circle with centre C. Let tangents at M and N meet at the point O. Since OM is a tangent OR ∴ ∵ ∴ Again in ΔCMN , CM = CN = r MOON is a tangentON ⊥⊥ CN i.e.CM i.e. ∠ ∠ONC = 90°OMC = 90° ∴ ∴ ⇒ Thus, tangents make equal angle with the chord. ∠∠ ∠CMN =OMC –OML ∠∠∠CMN =CNMONL ∠ONC – ∠CNM
- LHS
. [Since, Sin^2 A + Cos^ [ taking LCM ]^ [Since,(a-b)(a+b) = a^2 A = 1 ]^2 - b^2 ] = cot = sec = RHS. Hence, Proved. + sec. cosec. cosec+ cot
- Class Interval^ 10 - 20 0 - 10 Frequency^ 168 Cumulative Frequency 248 20 - 30 30 - 40 40 - 50 (^36346 ) Here, N = 100 The cumulative frequency just greater than 50 is 60. Hence, median class is 20 - 30. l = 20, h = 10, f = 36, cf = cf of preceding class is 24 Now, Median = = 20 + 7.22 = 27.2 Thus, the median of the data is 27.2. Section D
- In equation
⇒ ⇒ (^180) ∠∘ DF E^ − ∠ DF E = 90 (^) ∘= 90 ∘^ [ ∵ ∴^ ∠∠ CDFCDF^ , ∠+ ∠ CEFCEF^ and ∠ + ∠ DF EDF E are angles of a triangle= 180 ∘
= (^) 1−cossin^ θ θ +^ ≅ 1+cos^ tan^ θ θ = = = (^) 1−cos 1−cossinsinsin θ^ ⋅cos θθ (1−cos θθ θ ++(1+cos 1+cos cos θ ) cos^ sin^ cos θ^ θ^ sin(1+cos θθ )+sin θθ (1+cos θ (^) θθ (1−cos) θ ) θ ) = = = sinsinsin cos^ cos θθθθ ⋅cos⋅⋅⋅ cos cos sin θ ⋅^2 sin 22 θ +sin θθθ +sin (^2) + cos θ^ θ θ (^) cos⋅ θ^ cos(1− sin θ^2 ⋅ cos θ sin θ +sin (^22) θθ )^ θ −sin^ θ ⋅cos^ θ = cos sin^ θθ θθ + cos (^) θθ^1 ⋅sin θ θ θθ
∴^ ⇒^ N^2 = 50
= 20 + = 20 + {{10 ×10 ×^ l^^ (50−24) 2636 + (^36) }^ {h ×}^ (^^ N^2 −^ fcf )} ( m + 1) y^2 − 6( m + 1) y + 3( m + 9) = 0Page 14 of 20
iv. Hence, [ DG = DE and FE = DE)..
- According to question Diameterdiameter of the well = 7m Radius of the well (r) = Volume of the earth dugout = m = 3.5m and, height of the well (h) (3.5) (^2) 22.5 m (^3) = = 22.5 m m 3 Let the width of the embankment be r metres. Clearly, embankment forms a cylindrical shell whose inner and outer radii are 3.5 m and (r + 3.5) m respectively and height 1.5 m. = Volume of the embankment = Area of ring at top{(r + 3.5) (^2) - (3.5) (^2) } 1.5 m (^3) = (r + 7) r m 3 height of the embankment But, Volume of the embankment = Volume of the well
4r^2 r (r + 7) =^ 4r^ 4r+ 70x - 42x - 735 = 0^22 + 28r = 735+ 28r - 735 = 0^15 2r(2r + 35) - 21(2r + 35) = 0 (2r + 35)(2r - 21) = 0 2r + 35 = 0 or 2r - 21 = 0 r = or x = Hence, the width of the embankment is 10.5 m^ is negative, hence neglect this value^ = 10.5 m OR We have; A Cube, Cube's A Cylinder: , a = 7 cm Cylinder's Radius, r = 2.1 cm or r = Cylinder's Height, h = 7 cm A cylinder is scooped out from a cube, TSA of the resulting cuboid: cm = TSA of whole Cube - 2 = 6a = 6 = 6 2 + 2(7)49 + (44 2 rh - 2+ 2 (22 7 ( r (^2) 14.7) - (44) 7 (Area of upper circle or Area of lower circle) + CSA of the scooped out Cylinder2.1 7) - 2 7 4.41)[22 7 (2.1) (^2) ] = 294 + 92.4 - 27.72 = 294 + 64.68 = 358.68 cm Hence, the total surface area of the remaining solid is 358.68 cm (^22)
△ ⇒ ⇒ DBG BD FEBD DE == D ∼ E DG EC DE EC △ 2 = EF C ∵ BD × EC
∴ 72 π × × π × 72 × 72 ×^452 ∴ (^) π × π × 32 × ⇒ ⇒ ⇒ ⇒ πr ( r + 7) × (^494) ×^32 = π × 72 × 72 ×^452 ⇒ ⇒ ⇒ ⇒ −35 2 212 ⇒^ −35 2 x = (^212) length Edge (^2110) ∵ ∴ π × π × × × ×÷ (^) ×÷ × ×÷ (^) ×× ÷ ×
- Let the missing frequencies are a and b. Class Interval 0 - 5 Frequency f 12 i Cumulative frequency 12 10 - 15 15 - 20^ 5 - 10^1215 a^ 12 + a24 + a39 + a 20 - 25 25 - 30 30 - 35 b 66 39 + a + b45 + a + b51 + a + b Then, 55 + a + b = 70 a + b = 15 Median is 16, which lies in 15 - 20 ......(1)^ 35 - 40^4 55 + a + b = 70 So, The median class is 15 - 20 Therefore, l = 15, h = 5, N = 70, f = 15 and cf = 24 + a Median is 16, which lies in the class 15 - 20. Hence, median class is 15 - 20. Now, Median = Now,
- Hence, the missing frequencies are a = 8 and b = 7. Read the text carefully and answer the questions: The students of a school decided to beautify the school on an annual day by fixing colourful flags on the straight passage of the^ Section E school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the position of the middlemost flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time.
∴ ∴ (^) 16 = 15 + l = 15, h = 5, {5 × l f + = 15, { (35−24− h × 15 c. ( f (^) a N (^2) ). = 24 +}− fcf )} a ⇒ 16 = 15 + ⇒ 1 = ⇒ ⇒3 = 11 − a = 8 11− 3 a a { 11− 3 a } ⇒ ⇒ ⇒ 55 + 8 +63 + b = 7^ 55 + b = 70^ a^ b + = 70^ b^ = 70
Distance covered in placing 6 flags on either side of center point is 84 + 84 = 168 m Sn = S (^6) = 3[8 + 20][2a + (n - 1)d] S 6 = 84
(i) (^) ⇒ ⇒ S 6 n (^2) = 62 [2 × 4 + (6 − 1) × 4] (ii)^ ⇒ Let A be the position of the middle-most flag.
of the top of the tower from the roof of their houses. The angle of elevation of ajay's house to the tower and sooraj's house to the tower are 45o^ and 30o^ respectively as shown in the figure.
OR
The above figure can be redrawn as shown below: Let PQ = y In tan 45 = 1 = PQA, = x = y ...(i) In tan 30 = PQB, x = From the figure, height of tower h = PQ + QR = x + 150 = 136.61 + 150 = x + 100= 136.61 m h = 286.61 m
(i)
△ y x (^) QAPQ y x x√^1 √^3 △ 3 – =^ x+100^ PQ^ QB^ x^ =^ x+100^ PQ =^ x+100^ y =^ x+100^ x √^1003 − The above figure can be redrawn as shown below: Distance of Sooraj's house from tower = QA + AB = x + 100 = 136.61 + 100 = 236.61 m
(ii)
The above figure can be redrawn as shown below: Distance between top of the tower and top of Ajay's house is PA In sin 45 PQAo (^) = ⇒ ⇒ ⇒ The above figure can be redrawn as shown below: △ PA = PA = 193.20 m P A = (^) √^ PQ^ PA y 1 sin 2 PQ (^45) = o √ 2 – × 136.
Distance between top of tower and Top of Sooraj's house is PB In sin 30 PQBo (^) = PB = PB = 273.20 m
(iii)
⇒ ⇒ △ P B = (^) y 1^ PQ^ PB sin PQ = 2 × 136.61 30 o ⇒^2
