CBSE Class 10 Previous Question Paper (2016) > Maths with Solution
CBSE Class 10 Maths Previous Year Question Paper 2016: All Set Previous Year Question Paper PDF Class 10 Maths 2016 with Solution (also name as Marking Scheme) Download from here.
(1) From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50º, then find ∠AOB.
Solution:
∠APB = 80º 1 2
∴ ∠AOB = 100°
(2) In Fig. 1, AB is a 6 m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m, find the length of the ladder. (use √3 = 1.73)
Solution:
DB = 3.46 m 1 2
∴ DC = 4 m
(3) Find the 9th term from the end (towards the first term) of the A.P. 5, 9, 13, …., 185.
Solution:
l = 185, d = –4 1 2
l9 = 153
(4) Cards marked with number 3, 4, 5, …., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Solution:
Possible outcomes are 4, 9, 16, 25, 36, 49, i.e. 6.
∴ P (perfect square number) = 6/48 or 1/8
(5) If x = 2/3 and x = -3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a and b.
Solution:
-7/a = 2/3 – 3
⇒ a = 3
and b/a = 2/3 × (-3)
⇒ b = -6
(6) Find the ratio in which y-axis divides the line segment joining the points A(5, -6) and B( -1, -4). Also find the coordinates of the point of division.
Solution:
Let the point on y-axis be (0, y) and AP: PB = k:1
Therefore 5-k/k+1 = 0 gives k = 5
Hence required ratio is 5:1.
y = -4(5)-6/6 = -13/3
Hence point on y-axis is (0, -13/3)
(7) In Fig. 2, a circle is inscribed in a ∆ABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.
Solution:
Let AD = AF = x
∴ DB = BE = 12 – x
and CF = CE = 10 – x
BC = BE + EC
⇒ 8 = 12 – x + 10 – x ⇒ x = 7 1
∴ AD = 7 cm, BE = 5 cm, CF = 3 cm
(8) The x-coordinate of a point P is twice its y-coordinate
Solution:
Let the point P be (2y, y)
PQ = PR ⇒ √(2y-2)2 + (y + 5)2 = √(2y + 3)2 + (y – 6)2
Solving to get y = 8
Hence coordinate of point P are (16, 8).
(9) How many terms of the A.P. 18, 16, 14, .…. be taken so that their sum is zero?
Solution:
Hence a = 18, d = -2, Sn = 0
Therefore n/2 [36 + (n – 1) (-2)] = 0
⇒ n = 19
(10) In Fig. 3, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB.
Solution:
PA = PB
⇒ ∠PAB = ∠PBA = 60°
∴ ∆PAB is an equilateral triangle.
Hence AB = PA = 5 cm
(11) In Fig. 4, ABCD is a square of side 14 cm. Semi-circles are drawn with each side of square as diameter. Find the area of the shaded region (Use π = 22/7)
Solution:
Area of square = 196 cm2
Area of semicircles AOB + DOC = 22/7 × 49 = 154 cm2
Hence area of four shaded parts = (X + Y = 196 – 154 = 42 cm2
Therefore area of four shaded parts = 84 cm2
(12) In Fig. 5, is a decorative block, made up of two solids – a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the total surface area of the block. (use π = 22/7)
Solution:
Surface area of block = 216 – 22/7 × 3.5/2 × 3.5/2 + 2×22/7 × 3.5/2 × 3.5/2
= 225.42 cm2
(13) In Fig. 6, ABC is a triangle coordinates of whose vertex A are (0, –1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF.
Solution:
Using Mid Point formula coordinates of point B are (2, 1) and coordinates of point C are (0, 3).
Area ∆ABC = 1/2 [0 + 2(3+1) + 0] = 4 sq u.
Coordinates of point F are (1, 2)
Area of ∆DEF = 1/2 |1 (1-2) + 0 + 1 (0-1)| = 1 sq u.
(14) In Fig. 7, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M. If OP = PQ = 10 cm show that area of shaded region is 25 (√3 – π/7) cm2
Solution:
∠POQ = 60°
Area of segment PAQM = (100π/6 – 100√3/4) cm2
Area of semicircle = 25π/2 cm2
Area of shaded region = 25π/2 – (50π/3 – 25√3)
= 25 (√3 – π/6) cm2
(15) If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.
Solution:
S7 = 49 ⇒ 2a + 6d = 14
S17 = 289 ⇒ 2a + 16d = 34
Solving equations to get a = 1 and d = 2
Hence Sn = n/2 [2 + (n-1) 2] = n2.
(16) Solve for x:
2x/x-3 + 1/2x+3 + 3x+9/(x-3)(2x+3) = 0, x≠3, -3/2
Solution:
2x (2x + 3) + (x – 3) + (3x + 9) = 0
⇒ 2x2 + 5x + 3 = 0
⇒ (x + 1) (2x + 3) = 0
⇒ x = -1, x = – 3/2
(17) A well of diameter 4 m is dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.
Solution:
Volume of earth dug out = π × 2 × 2 × 21 = 264 m3
Volume of embankment = π (25 – 4) × h = 66 h m3
∴ 66h = 264 1 2
⇒ h = 4 m
(18) The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of the cylinder. (use π = 22/7)
Solution:
Here r + h = 37 and 2πr (r + h) = 1628
⇒ 2πr = 1628/37
⇒ r = 7 cm and h = 30 cm
Hence volume of cylinder = 22/7 × 7 × 30 = 4620 cm3
(19) The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building. (use 3 = √1.73)
Solution:
Correct Figure
tan45° = h-50/x ⇒ x = h – 50
tan60° = h/x ⇒ x = h/√3
Hence h-50 = h/√3
⇒ h= 75 + 25√3 = 118.25 m.
(20) In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice ? (ii) a total of 9 or 11?
Solution:
(i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5) (5, 2) (5, 3) (5, 5) i.e. 9 outcomes.
P (a prime number on each die) = 9/36 or 1/4
(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e. 6 outcomes
P (a total of 9 or 11) = 6/36 or 1/6
(21) A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for treatment. Due to this, the plane got delayed by half an hour. To reach the destination 1500 km away in time, so that the passengers could catch the connecting flight, the speed of the plane was increased by 250 km/hour than the usual speed. Find the usual speed of the plane.
What value is depicted in this question?
Solution:
Let the usual speed of plane be x km/h.
∴ 1500/x – 1500/x+250 = 1/2
(x + 1000) (x – 750) = 0 ⇒ 750
Speed of plane = 750 km/h
For writing value
(22) Prove that the lengths of tangents drawn from an external point to a circle are equal.
Solution:
For correct Given, To prove, construction and figure correct proof
(23) Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length.
Solution:
Construction of tangent length of tangent
(24) In Fig. 8, 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 AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
Solution:
PT = √169-25 = 12 cm and TE = 8cm
Let PA = AE = x
= TE2 + EA2
⇒ (12 – x)2 = 64 + x2
⇒ x = 3.3 cm
Thus AB = 6.6 cm
(25) Find x in terms of a, b and c :
a/x-a + b/x-b = 2c/x-c, x ≠ a, b, c
Solution:
a (x – b) (x – c) + b (x – a) (x – c) = 2c (x – a) (x – b)
x2 (a + b – 2c) + x (-ab – ac – ab – bc + 2ac + 2bc) = 0
x2 (a + b – 2c) + x (-2ab + ac + bc) = 0
x = ac+bc-2ab/a+b-2c
(26) A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird. (Take √3 = 1.732)
Solution:
Correct Figure
tan45° = 80/y ⇒ y = 80
tan30° = 80/x+y ⇒ x + y = 80√3
∴ x = 80 (√3 – 1) = 58.4 m
Hence speed of bird = 58.4/2 = 29.2 m/s
(27) A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief.
Solution:
Let total time be n minutes
Total distance convered by thief = (100 n) metres 1 2
Total distance covered by policeman = 100 + 110 + 120 + … + (n – 1) terms
∴ 100n = n-1/2 [200 + (n-2) 10]
n2 – 3n – 18 = 0
(n – 6) (n + 3) = 0
⇒ n = 6
Policeman took 5 minutes to catch the thief
(28) Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t.
Solution:
Area of the triangle = 1/2 |t (t + 2 – t) + (t + 2) (t – t + 2) + (t + 3) (t – 2 – t – 2)|
= 1/2 [2t + 2t + 4 – 4t – 12]
Which is independent of t.
(29) A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, …, 8 (Fig. 9), which are equally likely outcomes. What is the probability that the arrow will point at
(i) an odd number
(ii) a number greater than 3
(iii) a number less than 9.
Solution:
(i) Favourable outcomes are 1, 3, 5, 7 i.e, 4 outcomes.
∴ P (an odd number) = 4/8 or 1/2
(ii) Favourable outcomes are 4, 5, 6, 7, 8, i.e, 5 outcomes P (a number greater than 3) = 5/8
(iii) Favourable outcomes are 1, 2, 3 …. 8 P (a number less than 9) = 8/8 = 1
(30) An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 10) From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length of the belt that is still in contact with the pulley. Also find the shaded area. (use π = 3.14 and √3 = 1.73)
Solution:
cosθ = 1/2 ⇒ θ = 60°
Reflex ∠AOB = 240°
∴ ADB = 2×3.14×5×240/360 = 20.93 cm
Hence length of elastic in contact = 20.93 cm
Now, AP = 5√3 cm
Area (∆OAP + ∆OBP) = 25√3 = 43.25 cm2
Area of sector OACB = 25×3.14×120/360 = 26.16 cm2
Shaded Area = 43.25 – 26.16 = 17.09 cm2
(31) A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (use π = 3.14)
Solution:
Here R = 20, r = 12, V = 12308.8
Therefore 12308.8 = 1/3 × 3.14 (400+240+144) h
⇒ h = 15 cm
l = √(20-12)2 + 152 = 17cm
Total area of metal sheet used = CSA + base area
= π [(20 + 12) × 17 + 12 × 12]
= 2160.32 cm2
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