Download Sample Question Paper - 1 for Class X Mathematics (Standard) Session 2024-25 and more Exercises Mathematics in PDF only on Docsity!
Time Allowed: 3 hours General Instructions: Maximum Marks: 80 12 .. This Question Paper has 5 Sections A, B, C, D and E.Section A has 20 MCQs carrying 1 mark each 34 .. Section 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.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.
π = Section A
(^227)
1. (HCF a) c) 212100 × LCM) for the numbers 30 and 70 is: b)d) 70210 [1]
a) c) a > 0, b < 0 and c > 0a < 0, b > 0 and c > 0 b)d) a < 0, b < 0 and c < 0a > 0, b > 0 and c < 0
Figure show the graph of the polynomial f(x) = ax^2 + bx + c for which [1]
The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have [1]
a) c) a unique solutionno solution b)d) infinitely many solutionsexactly two solutions
a) (^) 2x^2 + 7x - 24 = 0 b) (^) x^2 + 4x + 14 = 0 c) (^) 3x^2 - 17x + 52 = 0 d) (^) x^2 - 14x + 48 = 0
- In the Maths Olympiad of 2020 at Animal Planet, two representatives from the donkey’s side, while solving a quadratic equation, committed the following mistakes.
ii^ i..^ One of them made a mistake in the constant term and got the roots as 5 and 9.Another one committed an error in the coefficient of x and he got the roots as 12 and 4. But in the meantime, they realised that they are wrong and they managed to get it right jointly. Find the quadratic equation.
[1]
a) c) -8-4 b)d) (^48)
What is the common difference of an AP in which a 18 - a 14 = 32? [1]
ABCD is a rectangle whose three vertices are B (4,0), C (4,3) and D (0,3). The length of one of its diagonals is a) c) 54 b)d) 325 [1]
In what ratio, does x-axis divide the line segment joining the points A(3, 6) and B(-12, -3)? a) c) 2 : 14 : 1 b)d) 1 : 21 : 4 [1]
a) 20 o, 30o. b) 50 o, 40o. c) 30 o, 20o. d) 40 o, 50o.
8. In the figures find the measures of ∠ P and ∠ R [1]
9. In the given figure, PQ is tangent to the circle centred at O. If ∠AOB = 95 o, then the measure of ∠ABQ will be [1]
Section B
a) c) 131 b)d)^56 6 12 a) c) b)d)
- 3 rotten eggs are mixed with 12 good ones. One egg is chosen at random. The probability of choosing a rotten egg is [1] 1511 45 5 25
a) c) 30 - 4050 - 60 b)d) 20 - 3010 - 20
- The distribution below gives the marks obtained by 80 students on a test: Marks Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60
The modal class of this distribution is:^ Number of Students^^3 12 27 57 75
[1]
a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true.
- Assertion (A): hemisphere and a cone surmounted on it. If the common radius is 7 m and height of the cone is 1 m, 463.39 cm A piece of cloth is required to completely cover a solid object. The solid object is composed of a 2 is the area of cloth required. Reason (R): Surface area of hemisphere = 2 r (^2).
[1]
a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true.
20. Assertion (A): Reason (R): Sum of first 15 terms of 2 , 5 , 8 ... is 345. Sum of first n terms in an A.P. is given by the formula: Sn = 2n ×[2a + (n - 1)d] [1]
2122 .. Find the largest number which divides 320 and 457 leaving remainder 5 and 7 respectively.In the given figure XY ∥ BC .Find the length of XY. [2][2]
- Two concentric circles with centre O are of radii 3 cm and 5 cm. Find the length of chord AB of the larger circle which touches the smaller circle at P. [2]
Section C
- Show that: (^) OR [2] Prove that:
tan^4 θ + tan^2 θ = sec^4 θ − sec^2 θ for 0 ∘^ ≤ θ ≥ 90 ∘
25. Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre.^ 1−cottan^ θ^ θ^ +^ 1−tan^ cot^ θ^ θ = 1 + tan^ θ^ + cot^ θ [2]
A horse is tethered to one corner of a rectangular field of dimensions 70 m^ OR 52 m, by a rope of length 21 m. How
much area of the field can it graze?^ ×
2627 .. Prove that 3 +If α and β are the zeros of the polynomial f(x) = 6x √ 5 – is an irrational number. 2 + x - 2, find the value of ( α + ) [3][3]
- A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the^ β^ β^ α number. OR^ [3] A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
- In the given figure, a circle is inscribed in a quadrilateral ABCD in which cm and DS = 3 cm, then find the radius of the circle. B = 90o. If AD = 17 cm, AB = 20^ [3]
In figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If^ OR AB is a tangent to the circle at E, find the length of AB. where TP and TQ are two tangents to the circle.
30. Prove that:tan tan θθ +sec−sec θθ −1+1 = 1+sin cos θ θ [3]
- The percentage of marks obtained by 100 students in an examination are given below: Marks 30-35 35-40 40-45 45-50 50-55 55-60 60-
Determine the median percentage of marks.^ Frequency^^14 16 18 23 18 8
[3]
the figure.
OR
(a)(b) Find the coordinates of the point of intersection of diagonals AC and BD. Find the length of the diagonal AC. (c)^ Find the ratio of the length of side AB to the length of the diagonal AC. Find the area of the campaign Board ABCD.
- Read the text carefully and answer the questions: Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two Sections A and B. Tower is supported by wires from a point O. Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of the Section B is 30o (^) and the angle of elevation of the top of Section A is 45o.
[4]
OR
(a)(b) Find the length of the wire from the point O to the top of Section B. Find the distance AB.
(c) Find the height of the Section A from the base of the tower. Find the area of △OPB.
- Section A (c) Explanation: (^2100) As we know HCF LCM = Product of the Numbers
- Hence HCF (a) a > 0, b < 0 and c > 0^ LCM (30,70) = 30^ 70 = 2100 Explanation: Therefore, Clearly, represent a parabola opening upwards. The vertex of the parabola is in the fourth quadrant, therefore b < 0 cuts Y axis at P which lies on OY. Putting So the coordinates of P is (0, c). x = 0 in , we get y = c. Clearly, P lies on OY. Hence, a>0, b<0 and c>0 c>
- (c) no solution Explanation: x + 2y + 5 = 0, Given, equations are and
- 3x - 6y + 1 = 0. Comparing the equations with general form: a a 12 x + bx + b 12 y + cy + c 12 = 0= 0 Here, a And a 21 = - 3, b = 1, b 12 = 2, c= - 6, c 1 = 5 2 = 1 Taking the ratio of coefficients to compare So This represents a pair of parallel lines. 4.^ Hence, the pair of equations has no solution. (d) Explanation: x^2 - 14x + 48 = 0 For 1st (^) one, Let the equation be x Since roots are 5 and 9^2 + ax + b = 0 For 2^ a = -14 and b = 45nd^ one, Let the equation be x Since roots are 12 and 4.^2 + px + q = 0 Now, according to the question, b and p both are wrong.^ p = -16 and q = 48 Therefore, the correct equation would be x (^2) - 14x + 48 = 0
- (d) 8
× ××
a > 0^ f ( x ) =^ a^ x^2 +^ bx^ +^ c y = a x^2 + bx + y (^) c = a x (^2) + bx + c ⇒
a a (^12) = a 1 −1 = 3 , b b (^12) ≠= −1 3 , c c (^12) = (^51) a 2^ b b^12^ c c^12
Solution
- (c) 4 Explanation: OA = 2m, Suppose AB is the ladder of length x mOAB = 60o
In right 2 = AOB, sec 60= 4 m o^ =
- (c) 120.56 cm 2 Explanation: ar(segment) = =(205.33 =(205.33 - 84.77) cm -49 1.73) cm 2 2 15.^ = 120.56 cm^2 (d) Explanation: 16.8 cm Length of the pendulum = Radius of a sector of the circle Arc length = 8.8 = 8. r = 16.8 cm^ = 8.
- (d) Explanation: Even numbers are 2,4, 6. Their number is 3. Number of all possible outcomes = 6. 17.^ P (getting an even number) = (c) Explanation: Number of possible outcomes = 3 Number of Total outcomes = 15 Required Probability =
- (a) Explanation: 30 - 40 30 - 40
∴ tan θ = BC AB =^34
⇒ x 2 △⇒^ x^2 = ( 227 × 14 × 14 × 120360 ) − (14 × 14 × sin^ (^ π^360^ r^2 θ^ −^ r^2 sin^2 θ cos 60 ∘^^ θ^2 cos) 60 ∘) = ( 6163 − √ (^23) ×× 12 × 14 × 14)cm^2
36036030 θ^ (2 × 2 × πr ) 227 × r (^12) ∴ 36 =^12 (^15) ∴ 153 =^15
- (c) A is true but R is false. 20.^ Explanation:(a) Both A and R are true and R is the correct explanation of A.^ A is true but R is false. Explanation: Both A and R are true and R is the correct explanation of A. Section B
- The given numbers are 320 and 457 Now as 5 and 7 are remainders on division of 320 and 457 by said number On subtracting the reminders 5 and 7 from 320 and 457 respectively we get: 320 - 5 = 315, 457 - 7 = 450 The prime factorizations of 315 and 405 are
Hence the said number = 45^ H.C.F. of 315 and 450 =
- Given XY AX = 1 cm, XB = 3 cm, and BC = 6 cm BC AB = AX + XB = 1 + 3 = 4 cm In AXY and [Common] ABC Then, AXY ~ ABC [By AA similarity]^ [Corresponding angles] [Corresponding parts of similar are proportional]
- Join OA and OP OP AB (radius tangent at the point of contact)
OP is the radius of smaller circle and AB is tangent at P. AB is chord of larger circle and OP AB In right^ AP = PB( AOP, AP^ from centre bisects the chord)^2 = OA^2 - OP^2 = (5) AP = 4 cm = PB^2 - (3)^2 = 16 24.^ AB = 8 cm
= R.H.S. OR
315 = 3 × 3 × 5 × 7 = 32 × 5 × 7
450 = 2 × 3 × 3 × 5 × 5 = 2 × 32 × 52
∴ 32 × 5 = 9 × 5 = 45
∠ A^ Δ = ∠^ A Δ
∠ AXY △ = ∠ ABC △
∴ ⇒^ AX AB 14 == XY^ XY BC 6 △
⇒ XY = 64 = 1.5 cm ⊥ ⊥
∴ ⊥^ ⊥
∴ L. HS = tan (^4) θ + tan (^2) θ = = tantan^22 θθ ( tansec^22 θθ + 1) = = (sec sec (^4 2) θ^ θ − − 1 sec) 2 sec θ^2 θ [∵ tan^2 θ = sec^2 θ − 1]
- Let us suppose that the digit at unit place be x Suppose the digit at tens place be y. Thus, the number is 10y + x. According to question it is given that the number is 4 times the sum of the two digits. Therefore, 10y + x = 4(x + y) we have 10y + x = 4x + 4y 4x + 4y - 10y - x = 0 3x - 6y = 0 3(x - 2y) = 0 After interchanging the digits, the number becomes 10x + y.^ x - 2y = 0 Again according to question If 18 is added to the number, the digits are reversed. Thus, we have (10y + x) + 18 = 10x + y 10x + y - 10y - x = 18 9x - 9y = 18 9(x - y) = 18 Therefore, we have the following^ x - y = 2 systems of equations x - 2y = 0 ...........(1) x - y = 2...............(2) Here x and y are unknowns. Now let us solve the above systems of equations for x and y. Subtracting the equation (1) from the (2), we get (x - y) - (x - 2y) = 2 - 0 x - y - x + 2y = 2 Now, substitute^ y = 2 the value of y in equation (1), we get x - 2 x - 4 = 0 2 = 0 Therefore the number is 10^ x = 4 2 + 4 = 24 Thus the number is 24 OR Let the usual speed of the plane = x km/hr. Distance to the destination = 1500 km Case (i): So, in case(i) Time = Case (iI) Hrs Distance to the destination = 1500 km Increased speed = 100 km/hr So, speed = x+100 So, in case(ii) Time = Hrs So, according to the question - =
= − = − 25253613 ×
⇒ ⇒ x − y = (^189) ⇒
⇒ ⇒ ⇒^ × ⇒ ×
we know that, ⇒ T ime = Distance speed Speed = Distance T ime (^1500) x
x^1500 + ∴ (^1500) x x^1500 +100^3060
x x^22 + 100x - 300000 = 0+ 600x - 500x - 300000 = 0 x = 500 or x = -600^ (x + 600)(x - 500) = 0 Since, speed can not be negative, x = 500 Therefore, Speed of plane = 500 km/hr.
DR = DS = 3 cm AR = AD - DR = 17 - 3 = 14 cm QB = AB - AQ = 20 - 14 = 6 cm^ AQ = AR = 14 cm Since QB = OP = r radius = 6 cm OR According to the question, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E.
OP OPT = 90° TP [Radius from point of contact of the tangent] In right OT (^2) = OP 2 OPT * (^) + PT 2 Let AP = x cm AE = AP^ (13)^2 = (5)^2 + PT^2 PT = 12 cmAE = x cm and AT = (12 - x)cm TE = OT - OE = 13 - 5 = 8 cm OE AEO = 90° AB [Radius from the point of contact] AET = 90° In right AET, Also BE = AE = cm^ cm
- We have,^ cm
[ sec^2 - tan^2 = 1]
AT^2 = △AE^2 +ET^2
(12 − ⇒ 144 + x )^2 x = 2 x − 24^2 + x (^8)^2 = x (^2) + 64 ⇒ 24 x = 80 ⇒ 103 x = 8024 =^103 ⇒ AB = 103 + 103 =^203 LHS = ⇒ LHS = tan tan^ θθ +sec−sec (tan (tan^ θθ −1+1 θθ +sec−sec θθ )−1)+ ⇒ LHS = (sec^ θ +tan tan^ θ θ )−−sec(^ sec θ^2 +1^ θ − tan^2 θ ) ∵ θ θ
1125 = x x (^2) + 20x - 1125 = 0^2 + 20x x x(x + 45) - 25(x + 45) = 0^2 + 45x - 25x - 1125 = 0 (x + 45)(x - 25) = 0 x - 25 = 0 [ The number of persons cannot be negative. x + 45 0] Hence, the original number of persons is 25.^ x = 25
- GF || DE (DEFG is square) = (Corresponding angles) = = 90o (By AA similarity) Again DEFG being a square = (each 90o) = (corresponding angles) (By AA similarity)
Radius of each hemispherical end = 7 cm. Height of each hemispherical part = its radius = 7 cm. Height of the cylindrical part = (104 - 2 Area of surface to be polished = 2(curved surface area of the hemisphere) + (curved surface area of the cylinder) 7) cm = 90 cm. = [2 ( ) + 2 ] sq units = (616 + 3960) cm = = 45.76 dm^2 = 4576 cm (^2) [^2 10 cm = 1 dm]. = ₹(45.76^ cost of polishing the surface of the solid 10) = ₹ 457.60. Diameter of the cylinder = 7 cm^ OR Therefore radius of the cylinder = Total height of the solid = 19 cm Therefore, Height of the cylinder portion = 19 - 7 = 12 cm Also, radius of hemisphere = cm
⇒ ⇒ (^9000180000) x (^2) +20 x +180000−9000 x = 160 (^2) +20 x = 160 ⇒ ⇒ 180000160 = x^2 + 20 ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ∵ ∴ ≠ ∴ ∴ ∠AGF∠A ∠GDB ∠ABC ∴ ∠AGF ∼ ∠DBG ∠AFG ∠ACB ∴ ∠AGF ∠A (^) ∼∠CEF ∠EFC
×
= [(^2 4 × πr 2 227 × 7 × 7 πrh ) + (2 × 227 × 7 × 90)]cm^2 ∴^ (^ 10×10^4576 )dm^2 ∵ × (^72) cm (^72)
Let V be the volume and S be the surface area of the solid. Then, V = Volume of the cylinder + Volume of two hemispheres
and, S = Curved surface area of cylinder + Surface area of two hemispheres
- = 418 cmLet the assumed mean be A = 25 and h = 5.^2 marks (x 5 1 ): no. of students (f 15 1 ): d-20 1 = x 1 = A = x 1 - 25 u-4 1 = (d 1 ) f-60 1 u 1 1015 5080 -15-10 -3-2 -150- 2025 7672 -5 0 -1 0 -76 0 (^3035 4539 510 12 ) (^3045 98 1520 34 ) (^50 6) f 1 = 400 25 5 30 f 1 u 1 = - We know that mean, Now, we have N = f 1 = A + h = 400, = -234, h = 5 and A = - 234, h = 5 and A = 25. Putting the values in the above formula, we get = A + h = 25 + 5 = 25 -
⇒ ⇒ VV == { ππ r 2 r (^) (^2 hh + 2+ 43 ( r ) (^23) cm π r^33 )}cm^3 ⇒ V = { 227 × ( 72 )^2 × (12 + 43 × 72 )}cm^3 = 227 × 72 × 72 × 503 cm^3 = 641.66cm^3 ⇒ ⇒ SS == 2 ( (^2) πrπrh ( h (^) + 2+ 2 × 2 r )cm π 2 r^2 )cm^2 ⇒ = (2 × S = 2 × 227 × 72227 × 19× (^72) )×cm (12 + 2 × 2 72 )cm^2 (^1) h
∑^ X^ ¯ ¯¯¯^ (^^ N^1^ ∑^ i =1^ n^ fi^ ui ) X ¯^ ( ( 1 N^1 × (−234) ∑ i =1^ n fi ui )) 23480400
Sections A and B. Tower is supported by wires from a point O. Distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of the Section B is 30 o^ and the angle of elevation of the top of Section A is 45o.
In APO, tan 45o^ =^ OR 1 = A = 36 cm Height of section A from the base of the tower = AP = 36 cm.
Let the length of wire BO = x cm
cos 30o^ = = 12^ x = = cm
(i)
∴ ⇒ (^) √ 23 = (^36) xPO BO ⇒ (^) × 36×2 2 √ √ (^33) – ×√√^33 In^24 √APO, tan 45^3 – o^ = 1 = AP = 36 cm ...(i) Now, In tan 30o (^) = PBO, BP = = (^) AB = AP - BPcm = 36 - cm
(ii) (^) ⇒ △ (^) AP 36 AP PO ⇒ △ ⇒ 1 =^ BP^ PO ⇒ √^3 √^36^ BP^363 ×√√^33 = (^12363) √ √ 3 – 3 – ∴ 12 √ 3 – △ AP 36 AP 36 In tan 30 OPBo (^) = BP =
(iii) △ BP PO ⇒ (^) √^13 √ 36 = 3^ BP 36 = (^) √^363 ×√√^33
= Now, Area of cm OPB = height base = = BP OP 36 = cm^2
121 √ 3 – △ 12 × ×
212 ××^12 √× 3 – ×
