The solution of mathematics question paper class X CBSE board exam 2020
Solutions of class 10 CBSE maths question paper 2020 board exam
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The solution of mathematics question paper set- 1 class X CBSE 2020
Series JBB/1 SET- 1
The time allowed: 3 hours Maximum marks: 80
(i) This question paper comprises four sections- A, B, C, and D. This question paper carries 40 questions. All questions are compulsory.
(ii) Section A: Q.No. 1 to 20 comprises of 20 questions of 1 mark each.
(iii) Section B: Q. No. 21 to 26 comprises of 6 questions of two marks each.
(iv) Section C: Q.No. 27 to 34 comprises of 8 questions of three marks each.
(v) Section D: Q.No. 35 to 40 comprises of 6 questions of four marks each.
(vi) There is no overall choice in the question paper. However, an internal choice has been provided in questions of one marks each, 2 questions of two marks each. 3 questions of three marks each and 3 questions of four marks each. You have to attempt only one of the choices in such questions.
(vii) In addition to this, separate instructions are given with each section and question, whenever necessary.
(viii) The use of calculators is not permitted.
Q. No. 1 to 10 are multiple-choice type questions of 1 mark each. Select the correct option.
Q1.If one of the zeroes of the quadratic polynomial x² + 3x + k is 2, then the value of k is
(a) 10 (b) –10 (c) –7 (d) –2
Q2. The total number of factors of a prime number is
(a) 1 (b) 0 (c) 2 (d) 3
Q3 The quadratic polynomial, the sum of whose zeroes is –5 and their product is 6, is
(a) x² + 5x + 6 (b) x² – 5x + 6 (c) x² – 5x –6 (d) –x² + 5x + 6
Ans.(a) x² + 5x + 6
Q4. The value of k for which the system of equations x + y – 4= 0 and 2x + ky = 3, has no solution, is
(a) –2 (b) ≠2 (c) 3 (d) 2
Ans. (d) 2
Q5. The HCF and the LCM of 12, 21, 15 respectively are
(a) 3, 140 (b) 12, 420 (c) 3, 420 (d) 420, 3
Ans. (c) 3, 420
Q6. The value of x for which 2x, (x +10) and (3x +2) are the three consecutive terms of an AP, is
(a) 6 (b) –6 (c) 18 (d) –18
Q7. The first term of an AP is p and the common difference is q, then its 10 th term is
(a) q + 9p (b) p – 9q (c) p+ 9q (d) 2p + 9q
Ans. (c) p+ 9q
Q8. The distance between the points (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ), is
(a) a² + b² (b) a² – b² (c) (d)
Q9. If the point P(k,0) divides the line segment joining the points A(2, –2) and B(–7, 4) in the ratio 1: 2, then the value of k is
(a) 1 (b) 2 (c) –2 (d) 1
Q10. The value of p, for which the points A(3, 1), B(5, p) and C(7, –5) are collinear, is
(a) –2 (b) 2 (c) –1 (d) 1
In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark.
Q11. In Fig.1, ΔABC is circumscribing a circle, the length of BC is — cm
Ans. From the fig.1 we have
AP = 4 cm, AP = AR = 4 cm (Tangents drawn from an external point of the circle)
AC = 11 cm
CR = AC – AR = 11 – 4 = 7 cm
CR = CQ = 7 cm (Tangents drawn from an external point of the circle)
BP = BQ = 3 cm (— do—)
BC = BQ + CQ = 3 + 7 = 10
Hence the length of BC is 10 cm.
Q12. Given Δ ABC ∼ ΔPQR, if , then
Ans. If Δ ABC ∼ ΔPQR then the ratio of the areas of both triangles is written as follows
Q13. ABC is an equilateral triangle of side 2a, then length of one of its altitude is ……….
Ans. Altitude of equilateral triangle bisect the side of equilateral triagle
Let one of the altitude of triangle is AO, which bisects BC
So, BO = BC/2 = 2a/2 =a, AB = 2a
Altitude AO =
= 1 + 1 = 2
Q15. The value of
= sin² θ + cos² θ = 1
The value of (1 + tan² θ)(1 –sin θ)( 1 + sin θ) = ………..
Ans. (1 + tan² θ)(1 –sin θ)( 1 + sin θ)
= sec² θ ( 1 – sin² θ )
= sec² θ ( cos² θ )
Q.Nos.16 to 20 are short answer type questions of 1 marks each.
Q16. The ratio of the length of a vertical rod and the length of its shadow is 1 : √3. Find the angle of elevation of the sun at that moment?
Ans. Let the length of vertical rod = x and the length of its shadow is = √3 x, let the angle of elevation of the sun is θ
θ = 30º
Therefore the angle of elevation of the sun is 30º.
Q17.Two cones of have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes?
Ans. Let the heights of both cones are x and 3x and their radii are 3y and y
Volume of cone is =
The ratio of their volume is given by
= 3 : 1
Therefore the ratio of the volume of given cones is 3: 1
Q18. A letter of the English alphabet is chosen at random. What is the probability that the chosen letter is a consonant?
Ans. The total number of English alphabets are = 26
Total possible outcome = 26
Number of consonants in English alphabets are = 21
Number of favourable outcomes =21
Hence the probability that the letter is chosen is consonant = 21/26.
Q19. A die is thrown once. What is the probability of getting a number less than 3 ?
Ans. If die is thrown once then the total possible outcomes are 6 → 1,2,3,4,5, and 6
The number less than 3 are 2→ 1 and 2
Number of favourable outcomes = 2
Therefore the probability of getting a number less than 3 is = 1/3
If the probability of winning a game is 0.07, what is the probability of losing it?
Ans. As we know the sum of the probabilities of happening and not happening of the event is =1
Where P(E) is the probability of happening(winning the game) the event and is the probability of not happening (losing the game ) the event
P(E) = 0.07
Therefore the probability of losing the game is 0.93
Q20. If the mean of the first n natural number is 15, then find n.
Ans. The first n natural number are 1,2,3,4,5…….n
Mean of first n natural numbers = 15
n + 1 = 30
n = 29
Q.Nos. 21 to 26 carry 2 marks each.
Q21. Show that (a– b)², (a² + b²) and (a + b)² are in AP.
Ans.Condition of AP series is, 2nd term – 1 st term = 3 rd term – 2 nd term
2nd term – 1 st term = (a² + b²) – (a– b)² = a² + b² –( a² + b² – 2ab)= 2ab
3 rd term – 2 nd term = (a + b)² – (a² + b²) = a² + b² + 2ab – a² – b² = 2ab
Therefore condition of AP is satisfied because common difference is = 2ab, so given terms are in AP.
Q22. In Fig.2, DE ∥ AC and DC ∥ AP. Prove that
Ans. From fig.2 In ΔABC , DE ∥ AC
………..(i) ( BPT theorem)
In ΔABP DC ∥ AP
………….(ii) (BPT theorem)
From (i) and (ii)
In Fig. 3, two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that
Ans. From the Fig.3, OP is the radius
∠OPT = 90° ( tangent is perpendicular to radius)
∠OPQ = ∠OPT – ∠QPT
∠OPQ = 90 ° – ∠QPT
∠QPT = 90°– ∠OPQ ………(i)
TP = TQ ( tangents drawn to the circle from an external point)
∠QPT = ∠PQT………………(ii)
∠PTQ + ∠QPT + ∠PQT = 180° (angle sum property of the triangle)
∠PTQ + ∠QPT + ∠QPT = 180°(from ii equation)
∠PTQ + 2QPT = 180°………(iii)
Substituting the value of ∠QPT from (i) in the equation no.(iii)
∠PTQ + 2(90°– ∠OPQ) = 180°
∠PTQ + 180° – 2∠OPQ = 180°
∠PTQ = 2∠OPQ Hence proved
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Q23.The rod AC of a TV disc antenna is fixed at right angles to the wall AB and a rod CD is supporting the disc as shown in Fig.4. If AC = 1.5 m long and CD = 3 m, find (i) tanθ (ii) secθ + cosecθ.
Ans.(i) In the fig.4, in ΔADC we are given AC = 1.5 m and CD = 3 m
(ii) secθ + cosecθ
Q24. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.[Use π = 22/7]
We are given length of arc (l) = 22 cm
Applying the following for the length of arc(l)
l = 22 cm, θ = 60°, r is radius of the circle
r = 21
Therefore the radius of the circle is = 21 cm
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Q25. If a number x is chosen at random from the numbers –3 , –2, –1, 0, 1, 2, 3. What is the probability that x² ≤ 4?
Ans. Total numbers are = 7 (–3 , –2, –1, 0, 1, 2, 3), so possible O/C = 7
We have to calculate the probability of the number x such that x² ≤ 4
The given condition is satisfied by the number (x) when its value is –2, –1, 0, 1 and 2
Favourable O/C = 5
Therefore the probability of the number x is chosen when x² ≤ 4 is 5/7
If the point C (–1, 2) divides internally the line segment joining A(2, 5) and B(x,y) in the ratio 3: 4, find the coordinates of B.
Ans. The given line segment is given to us as follows
Let the coordinates of C are represented by (p,q) which divides line segment AB in the ratio of 3: 4
Applying the section formula
–7 = 8 + 3x, 14 = 20 + 3y
x = – 5, y = –2
Hence the coordinates of B are (– 5,–2)
Q26. Find the mean of the following distribution:
|Class:||3- 5||5 – 7||7 – 9||9 – 11||11 – 13|
|Class||Class mark (x)||Frequency (f)||fx|
|3 – 5||4||5||20|
|5 – 7||6||10||60|
|7 – 9||8||10||80|
|9 – 11||10||7||70|
|11 – 13||12||8||96|
N= 40 ∑fx=326
Hence the mean of the above data is = 8.15
Find the mode of the following data:
|Class:||0 – 20||20 – 40||40 – 60||60 – 80||80 – 100||100 – 120||120 – 140|
|0 – 20||6|
|20 – 40||8|
|40 – 60||10 (F0)|
|60(l) – 80||12(f1)|
|80 – 100||6(f2)|
|100 – 120||5|
|120 – 140||3|
Where l is the lower limit of modal class = 60
f1 =12, f0 =10, f2 =6 and h = 80 – 60 = 20
Mode of the data is 65
SECTION – C
Q. Nos. 27 to 34 carry 3 marks each.
Q27. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax² + bx +c, a ≠ 0, c ≠ 0.
Ans. We are given the polynomial f(x) = ax² + bx +c, a ≠ 0, c ≠ 0.
We have to find out a polynomial whose zeroes are the reciprocal of the zeroes of the given polynomial
Let the zeroes of the given polynomial are α and β and of the polynomial we have to find out are 1/α and 1/β
Applying the relationship between zeroes and coefficient of the polynomial
Sum of the zeroes of the polynomial we are going to find out
Substituting the values from eq.(i)
The product of the zeroes of the polynomial
So, the required polynomial is
x²–(sum of zeroes)x + product of zeroes
The required polynomial can be written as cx² + bx + a
Divide the polynomial by the polynomial and verify the division algorithm.
According to division algorithm, the relationship between f(x)(polynomial) , g(x)(divisor) and r(x) (remiender)is following
f(x) = g(x) q(x) + r(x)
LHS = RHS, verified
Q28. Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are given by 2y – x = 8, 5y – x = 14 and y – 2x = 1.
Ans. Let the triangle is ABC the equations of whose sides are AB→2y – x = 8, BC→5y – x = 14,AC→y – 2x = 1
Possible solutions of these equations
2y – x = 8 ⇒(0,4), (2,5),(4,6), 5y – x = 14 ⇒(1,3), (6,4),(–4,2), y – 2x = 1 ⇒(0,1), (2,5), (1,3)
Drawing the graph of these equations
The vertices of the triangle are A(–4,2), B(1,3) and C(2,5)
If 4 is a zero of the cubic polynomial , find its other two zeroes.
Ans. If 4 is a zero of the given polynomial , then its one of the factor is (x – 4)
So, dividing the given polynomial by (x – 4)
Factorizing the quotient x² + x – 6
x² + 3x – 2x – 6
x(x + 3) – 2(x + 3)
(x + 3)(x – 2)
x =–3, 2
Therefore other two zeroes of the given polynomial are –3 and 2
Q29. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the original duration of the flight.
Ans. Let the original speed of the flight is x km/h
The distance covered by aircraft = 600 km
The original time duration of the flight =
Modified speed = x – 200
The time taken with reduced speed =
30 minutes = 30/60 = 0.5 hrs
According to question
600x – 600x + 120000 = 0.5(x²– 200x)
0.5x² – 100x – 120000 = 0
x² – 200x – 240000 = 0
x² – 600x + 400x – 240000= 0
x(x – 600) + 400(x – 600) = 0
(x – 600)(x + 400) = 0
x = 600, – 400
Speed can’t be negative ,so the original speed of aircraft is = 600 km/h
∴ Original time duration of aircraft = Distance/original speed = 600/600 = 1
Therefore the original time duration of aircraft = 1 hour
Q30.Find the area of triangle PQR formed by the points P(– 5, 7), Q(– 4, – 5) and R(4, 5).
Ans. Area of the triangle is given as follows
ar ΔPQR = 1/2[X1(Y2 – Y3) + X2(Y3 – Y1) + X3(Y1 – Y2)]
X1 = –5, X2 = –4, X3 = 4, Y1 = 7, Y2 = – 5, Y3 = 5
arΔPQR = 1/2[ –5(– 5– 5) –4(5 – 7) + 4(7 + 5)] = 1/2[50 + 8 + 48] = 1/2 = 53
Therefore the area of the triangle = 53 sq. unit
If the point C(–1, 2) divides internally the line segment joining A( 2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.
Ans. Applying the section formula
We are given in the question m : n = 3 : 4, p = –1, q = 2,
4x +6 = –7 , 14 = 3y + 20
x = 13/4, y = –2
Hence the coordinates of B are = (13/4, –2)
Q31. In Fig.5, ∠D = ∠E and , prove that BAC is an isoscelles triangle.
Given that: ∠D = ∠E
To prove: ΔBAC is an isosceles triangle
Inverting both sides
Adding 1 both sides
∠D = ∠E (given)
AD = AE ……….(ii) ( opposite sides of similar angles in a Δ)
From equation (i) and (ii)
AB = AE
Hence ΔBAC is an isosceles triangle, Hence proved
Q32. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.
Given that: ΔABC in which AC² = AB² + BC²
To prove: ∠B = 90º
Construction: Drawing a right ΔDEF such that AB = DE and BC = EF
Proof : In ΔABC
AC² = AB² + BC² ……….(i) (Given)
DF² = DE² + EF²………..(ii) ( Pythogorus theorem)
AB = DE (constructed)
BC = EF (constructed)
Putting AB = DE and BC = EF in eq.(i)
AC² = DE² + EF²…….(iii)
Comparing (ii) and (iii)
AC² = DF²
AC = DF
In triagle ABC and triagle DEF, we have
AB = DE, BC = EF and AC = DF (proved above)
In ΔDEF ∼ ΔABC
∠E = ∠B (By CPCT)
∠E = 90
∴∠B = 90°, Hence proved
Q33.If sinθ + cosθ = √3, then prove that tanθ + cotθ = 1.
Ans. tanθ + cotθ = 1.
L.H.S . tanθ + cotθ
sinθ/cosθ + cosθ/sinθ
(sin²θ + cos² θ )/sinθ cosθ = 1/sinθ cosθ…….(i)
We are given, sinθ + cosθ = √3
sin²θ + cos²θ + 2sinθ cosθ = 3
1 + 2sinθ cosθ = 3
sinθ cosθ = 1……….(ii)
From (i) and (ii)
tanθ + cotθ = 1, Hence proved
Q34.A cone of base radius 4 cm is divided into two parts by drawing a plane through the mid-point of its height and parallel to its base. Compare the volume of the two parts.
Let the height of the cone = h and its radius of base = r
ΔABF ∼ ΔAED, From the property of similar triangle
ED = r/2
Radius of the smaller cone = r/2 and height = h/2
So, volume of smaller cone (V1) = ( 1/3)πR²H where R =r/2 and H = h/2
Volume of the frustum =
r1 = r/2, r2 = r, H = h/2
V2 = 7 V1
The volume of the frustum is 7 times the volume of the smaller cone
Q.Nos 35 to 40 carry 4 marks each.
Q35. Show that the square of any positive integer cannot be of the form (5q + 2) or (5q +3) for any integer q.
Ans. According to Euclid’s algorithm any positive integer can be written in the form of bQ + r where 0≤ r < b
Let b = 5, then possible values of r = 0,1,2,3 and 4
Corresponding values of positive integer are = 5Q, 5Q +1, 5Q +2, 5Q +3, 5Q +4
Squaring the positive integers = (5Q)² = 5(5Q²) =5q where q = 5Q²
(5Q +1)² = 25Q² + 10Q + 1 = 5(5Q² +2Q) + 1 = 5q + 1(where q = 5Q² +2Q, is another positive integer)
(5Q +2)² = 25Q² +20Q +4 = 5(5Q² + 4Q) + 4 = 5q + 4(where q =5Q² + 4Q. is another positive integer)
It has been proved that the square of any positive integer cannot be written in the form of (5q+2) or (5q +3)
Q36.The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7:15. Find the numbers.
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Solve : 1 + 4 + 7 + 10 + ……+ ………+ x = 287
Ans. Applying the n th term formula of AP
Where a = 1, d = 4 –1 = 3,
x = 1 + (n – 1) 3
x = 1 + 3n –3
The sum of the n th term of an AP is written as
Where l is final term of the given AP, and Sn = 287, substituting the value of n
1722 = (x + 2)(1 + x) = x² + x + 2 + 2x
x² + 3x –1720 = 0
x² + 43x – 40x –1720 = 0
x(x + 43) – 40(x + 43) = 0
(x + 43)(x – 40) = 0
x = – 43, 40
All the terms of AP are positive so x can not be negative, therefore value of x = 40
Q37. Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.
Steps of construction.
(i) Drawing a line segment AB of the length 7 cm.
(ii) Taking A and B as centres drawing two circles of radii 3 cm and 2 cm respectively.
(iii) Drawing perpendicular bisection of the line segments AB we get a midpoint C of AB and taking C as a centre drawing a circle of the diameter of the length AB. This circle intersects the circle of 3 cm at T and R and to the circle of 2 cm at P and Q.
(iv) Joining B to T and B to R, we get two tangents on the circle BT and BR drawn from the centre B of the circle of radius 2 cm and joining A to P and A to Q, we get two tangents AP and AQ drawn from the centre A of the circle of radius 3 cm.
Q38.A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff 6m. At a point on the plane, the angles of elevation of the bottom and top of the flag-staff are 30° and 45° respectively. Find the height of the tower.(Take √3 = 1.73)
Let height of vertical tower = h and the distance of the point of observation at the ground from the foot of the tower =x, Height of given flag-staff = 6 m
Considering both of the triangles, we have
x = h√3…..(i), x = h + 6……(ii)
From (i) and (ii), we have
h√3 = h + 6
h(√3 –1) = 6
Rationalizing the denominator
h = 3(1.73 +1) = 3 ×2.73 = 8.19
Hence height of the tower is 8.19 m
Q39. A bucket in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm, respectively. Find the capacity of the bucket. Also, find the cost of milk which can completely fill the bucket at the rate of Rs 40 per litre.(π/7)
Ans. The height of bucket (h) = 30 cm, radii of upper and lower radii, r1 = 10 cm and r2 =20 cm
The volume (V) of the frustum is given by the formula is as follows
V = 22000
Volume of the bucket is 22000 cub.cm
22000 cub.cm = 22000/1000 = 22 litre (1000 cub.cm = 1 litre)
Rate of milk = Rs 4o/litre
The cost of milk in the bucket is = 40 ×22 = 880
Hence the total cost of the milk in the bucket is = Rs 880
Q40. The following table gives production yield per hectare (in quintals) of wheat of 100 farms of a village:
|Production yield/hect||40 – 45||45 – 50||50 – 55||55 – 60||60 – 65||65 – 70|
|No. of farms||4||6||16||20||30||24|
Change the distribution to ‘a more than’ type distribution and draw its ogive.
|Production yield/hect(more than)||Number of farms (f)||Cumulative frequency (Cf)|
The median of the following data is 525. Find the values of x and y, if the total frequency is 100:
|0 – 100|
N/2 = 100/2 = 50, median = 525
We have 76+x+y = 100⇒x + y = 24…….(i)
Therefore the median group class lies in cf 56+ x
The lower limit of the median group (L) = 500, Cumulative frequency preceding the median group(CF) =36+x
Frequency of median group (F) = 20, C = 100(Class interval of median group)
25 × 20 = (50 –36 –x)100 = 1400 – 100x
100x = 900
x = 9
Putting this value in eq.(i) 9+ y = 24⇒ y = 15
Therefore the value of x and y are 9 and 15 respectively
NCERT Solutions of class 9 maths
|Chapter 1- Number System||Chapter 9-Areas of parallelogram and triangles|
|Chapter 2-Polynomial||Chapter 10-Circles|
|Chapter 3- Coordinate Geometry||Chapter 11-Construction|
|Chapter 4- Linear equations in two variables||Chapter 12-Heron’s Formula|
|Chapter 5- Introduction to Euclid’s Geometry||Chapter 13-Surface Areas and Volumes|
|Chapter 6-Lines and Angles||Chapter 14-Statistics|
|Chapter 7-Triangles||Chapter 15-Probability|
|Chapter 8- Quadrilateral|
NCERT Solutions of class 9 science
NCERT solutions of class 11 maths
|Chapter 1-Sets||Chapter 9-Sequences and Series|
|Chapter 2- Relations and functions||Chapter 10- Straight Lines|
|Chapter 3- Trigonometry||Chapter 11-Conic Sections|
|Chapter 4-Principle of mathematical induction||Chapter 12-Introduction to three Dimensional Geometry|
|Chapter 5-Complex numbers||Chapter 13- Limits and Derivatives|
|Chapter 6- Linear Inequalities||Chapter 14-Mathematical Reasoning|
|Chapter 7- Permutations and Combinations||Chapter 15- Statistics|
|Chapter 8- Binomial Theorem||Chapter 16- Probability|
NCERT solutions of class 12 maths
|Chapter 1-Relations and Functions||Chapter 9-Differential Equations|
|Chapter 2-Inverse Trigonometric Functions||Chapter 10-Vector Algebra|
|Chapter 3-Matrices||Chapter 11 – Three Dimensional Geometry|
|Chapter 4-Determinants||Chapter 12-Linear Programming|
|Chapter 5- Continuity and Differentiability||Chapter 13-Probability|
|Chapter 6- Application of Derivation||CBSE Class 12- Question paper of maths 2021 with solutions|
|Chapter 7- Integrals|
|Chapter 8-Application of Integrals|