CBSE Class 10 Mathematics Previous Year Question Paper 2016 All India Set-1

Q 1. PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

 

Q 2. For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P?

 

Q 3. A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.

 

Q 4. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.

 

Q 5. If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x)k = 0 has equal roots, find the value of k.

 

Q 6. Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q

 

Q 7. A quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.

 

Q 8. Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.

 

Q 9. The 4^th term of an A.P. is zero. Prove that the 25^th term of the A.P. is three times its 11^th term.

 

Q 10. From an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that OTS = ∠OST = 30°.

 

Q 11. O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)

 

Q 12. A tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs.500/sq. metre.

 

Q 13. If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.

 

Q 14. Find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. 

 

Q 15. If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their m^th terms.

 

Q 16. Solve for x:

 

Q 17. A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel.

Q 18. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by  cm. Find the diameter of the cylindrical vessel.

Q 19. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of a hill as 30°. Find the distance of the hill from the ship and the height of the hill.

 

Q 20. Three different coins are tossed together. Find the probability of getting

1. exactly two heads
2. at least two heads
3. at least two tails.

 

Q 21. Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the governments and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 cm and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? 

 Q 22. Prove that the lengths of the tangents drawn from an external point to a circle are equal.

 

Q 23. Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.

 

Q 24. Two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of 

 

Q 25. Solve for x:

 

Q 26. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. 

Q 27. The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.

 

Q 28. The vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that  . Calculate the area of ∆ADE and compare it with area of ∆ABC.

 

Q 29. A number x is selected at random from the numbers 1, 2, 3, and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.

 

Q 30.  Is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OQ and meets OP produced at B. Prove that the perimeter of shaded region is r 

 

Q 31. A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream.

 

Q 1. PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

Solution:

In the given figure,

In ∆ACO,

OA = OC … (Radii of the same circle)

∴ ∆ACO is an isosceles triangle.

∠CAB = 30° … (Given)

∴ ∠CAO = ∠ACO = 30°    …(angles opposite to equal sides of an isosceles triangle are equal)

∠PCO = 90°     … (radius drawn at the point of contact is perpendicular to the tangent)

Now ∠PCA = ∠PCO – ∠CAO

∴ ∠PCA = 90° – 30° = 60°

 

Q 2. For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P?.

Solution:

If k + 9, 2k – 1 and 2k + 7 are the consecutive terms of A.P., then the common difference will be the same.

∴ (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)

∴ k – 10 = 8

∴ k = 18

 

Q 3. A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.

Solution:

Let AB be the ladder and CA be the wall.

The ladder makes an angle of 60° with the horizontal.

⇒ ∠B = 60°

Here AC ⊥ CB ⇒ ∠C = 90°

∠B = 60°, ∠C = 90° ⇒ ∠A = 90°

∴ ΔABC is a 30° - 60° - 90° triangle.

∴ BC =  (Side Opposite to 30° angle is equal to the half of the hypotenuse)

⇒ AB = 2 × BC = 2 × 2.5 = 5 cm

Therefore the length of the ladder is 5cm.

∴ AB = 5 cm

 

Q 4. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.

Solution:

There are 26 red cards including 2 red queens.

Two more queens along with 26 red cards will be 26 + 2 = 28

∴ P(getting a red card or a queen)

∴ P(getting neither a red card nor a queen)

 

Q 5. If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x)k = 0 has equal roots, find the value of k.

Solution:

Given – 5 is a root of the quadratic equation 2x² + px – 15 = 0.

∴ -5 satisifies the given equation.

∴ 2 (-5)² + p (-5) – 15 = 0

∴ 50 – 5p – 15 = 0

∴ 35 – 5p = 0

∴ 5p = 35 ⇒ p = 7

Substituting p = 7 in p (x² + x) + k = 0, we get

7 (x² + x) + k = 0

∴ 7x² + 7x + k = 0

The roots of the equation are equal.

∴ Discriminant = b² – 4ac = 0

Here, a = 7, b = 7, c = k

b² – 4ac = 0

∴ (7)² – 4 (7)(k) = 0

∴ 49 – 28k = 0

∴ 28k = 49

 

Q 6. Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q

Solution:

Since P and Q are the points of trisection of AB, such the P is nearer  to A.

⇒ AP = PQ = QB

Thus, P divides AB internally in the ratio 1: 2 and Q divides AB internally in the ratio 2 : 1.

∴ By section formula,

 

 

Q 7. A quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.

Solution:

Since tangents drawn from an exterior point to a circle are equal in length,

AP = AS ….(1)

BP = BQ ….(2)

CR = CQ ….(3)

DR = DS ….(4)

Adding equations (1), (2), (3) and (4), we get

AP + BP + CR + DS = AS + BQ + CQ + DS

∴ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

∴ AB + CD = AD + BC  .....(Since A-P-B, C-R-D, A-S-D and B-Q-C.)

∴ AB + CD = BC + DA …..(Hence proved)

 

Q 8. Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.

Solution:

Let A (3,0), B(6,4) and C (-1,3) be the given points.

Now

∴ By converse of Pythagoras theorem,

ΔABC is a right-angled triangle.  ...(ii)

From (i) and (ii), we get

ΔABC is a right-angled isosceles triangle.

 

Q 9. The 4^th term of an A.P. is zero. Prove that the 25^th term of the A.P. is three times its 11^th term.

Solution:

n^th term of an A.P = a(n-1)d

Here 4^th term of an A.P. = a₄ = 0

∴ a + (4 – 1)d = 0

∴ a  + 3d = 0

∴ a = -3d                         …….(1)

25^th term of an A.P = a₂₅

= a + (25 – 1)d

= -3d + 24d               ………..[From (1)]

= 21d

3 times 11^th term of an A.P. = 3a₁₁

 = 3 [a + (11 – 1)d]

= 3 [a  + 10d]

= 3 [ -3d + 10d]

= 3 ⨯ 7d

= 21d

∴ a₂₅ = 3a₁₁

i.e., the 25^th term of an A.P. is three times its 11^th term

 

Q 10. From an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that OTS = ∠ OST = 30°.

Solution:

In the given figure,

OP = 2r             …….. (Given)

∠ OTP = 90° …… (radius drawn at the point of contact is perpendicular to the tangent)

In ∆OTP,

 

⇒ ∠OPT = 30°

∴ ∠TOP = 60°

∴ ∆OTP is 30° – 60° – 90°, right triangle.

In ∆OTS,

OT = OS   ……. (Radii of the same circle)

∴ ∆OTS is an isosceles triangle.

∴ ∠OTS = ∠OST ……. (Angles opposite to equal sides of an isosceles triangle are equal)

In ∆OTQ and ∆OSQ

OS = OT …… (Radii of the same circle)

OQ = OQ    ……. (side common to both triangles)

∠OTQ = ∠OSQ ….. (angles oppsite to equal sides of an isosceles triangle are equal)

∴ ∆OTQ ≅ ∆OSQ    ……(By S.A.S)

∴ ∠TOQ = ∠SOQ = 60°   ……. (C.A.C.T)

∴ ∠TOS = 120° ……. (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)

∴ ∠OTS + ∠OST= 180° – 120° = 60°

∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°    ...from (i)

 

Q 11. O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)

Solution:

Diameter, AB = 13 cm

∴ Radius of the circle, 

∠ACB is the angle in the semi-circle.

∴ ∠ACB = 90°

Now, in ΔACB using Pythagoras theorem, we have

AB² = AC² + BC²

∴ (13)² = (12)² + (BC)²

∴ (BC)² = (13)² – (12)² = 169 – 144 = 25

∴ BC =  = 5 cm

Now, area of shaded region = Area of semi–circle – Area of ∆ACB

Thus, the area of the shaded region is 36.33 cm²

 

Q 12. A tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs.500/sq. metre.

 

Solution:

For conical portion, we have

R = 1.5 m and l = 2.8 m

∴ S₁ =  Curved surface area of conical portion

∴ S₁ = πrl

 = π ⨯ 1.5 ⨯ 2.8

= 4.2π m²

For cylindrical portion, we have

R = 1.5 m and h = 2.1 m

∴ S₂ =  Curved surface area of cylindrical portion

∴ S₂ = 2πrh

= 2 ⨯ π ⨯ 1.5 ⨯ 2.1

= 6.3 π m²

Area of canvas used for making the tent = S₁ + S₂

= 4.2 π  + 6.3π

= 10.5 π

=

 = 33 m²

Total cost of the canvas at the rate of Rs. 500 per m²  = Rs. (500⨯33) =  Rs. 16500

 

Q 13. If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.

Solution:

P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b).

∴ AP = BP

 Hence proved.

 

Q 14. Find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. 

Solution:

Area of the region ABDC =  Area of sector AOC -  Area of sector BOD

 

Q 15. If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their m^th terms..

Solution:

Let a₁, a₂ be the first terms and d₁, d₂ the common differences of the two given A.P’s.

 Q 16. Solve for x:

Solution:

 

Q 17. A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel.

Solution:

Let the radius of the conical vessel = r₁ = 5 cm

Height of the conical vessel = h₁ = 24 cm

Radius of the cylindrical vessel = r₂ = 10 cm

Let the water rise upto the height of h2 cm in the cylindrical vessel.

Now, volume of water in conical vessel = volume of water in cylindrical vessel

 

Q 18. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by  cm. Find the diameter of the cylindrical vessel.

Solution:

 

Q 19. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of a hill as 30°. Find the distance of the hill from the ship and the height of the hill.

Solution:

Let CD be the hill and suppose the man is standing on the deck of a ship at point A.

The angle of depression of the base C of the hill CD observed from A is 30° and the angle of elevation of the top D of the hill CD observed from A is 60°.

∴ ∠EAD = 60° and ∠BCA = 30°

∴ DE = 30 m

∴ CD = CE + ED = 10 + 30 = 40 m

Thus, the distance of the hill from the ship is  m and the height of the hill is 40 m

 

Q 20. Three different coins are tossed together. Find the probability of getting

1. exactly two heads
2. at least two heads
3. at least two tails.

Solution:

When three coins are tossed together, the possible outcomes are

HHH, HTH, HHT, THH, THT, TTH, HTT, TTT

∴ Total number of possible outcomes = 8

(i) Favourable outcomes of exactly two heads are HTH, HHT, THH

    ∴ Total number of favourable outcomes 3

    

(ii) Favourable outcomes of at least two heads are HHH, HTH, HHT, THH

     ∴ Total number of favourable outcomes 4

    

(iii)   Favourable outcomes of at least two tails are THT, TTH, HTT, TTT

      ∴ Total number of favourable outcomes 4

      

 

Q 21. Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the governments and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 cm and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? 

Solution:

Height of conical upper part = 3.5 m, and radius = 2.8 m

(Slant height of cone)² = 2.1² +2.8² = 4.41 + 7.84

Slant height of cone = = 3.5 m

The canvas used for each tent

= Curved surface area of cylindrical base + curved surface area of conical upper part

= 2 πrh + πrl

= πr (2h + l)

 

= 92.4 m²

So, the canvas used for one tent is 92.4 m².

Thus, the canvas used for 1500 tents (92.4 ⨯ 1500) m².

Canvas used to make the tents cost Rs. 120 per sq. m

So, canvas used to make 1500 tents will cost Rs. 92.4 ⨯ 1500 ⨯ 120

The amount shared by each school to set up the tents

The amount shared by each school to set up the tents is Rs.332640.

The value to help others in times of troubles is generated from the problem.

 

Q 22. Prove that the lengths of the tangents drawn from an external point to a circle are equal.

Solution:

Consider a circle centered at O.

Let PR and QR are tangents drawn from an external point R to the circle touching at points P and Q respectively.

Join OR.

Proof:

In ∆OPR and ∆OQR,

OP = OQ            ... (Radii of the same circle)

∠OPR = ∠OQR     …. (Since PR and QR are tangents to the circle)

OR = OR             ... (Common side)

∴ ∆OPR ≅ ∆OQR    …. (By R.H.S)

∴ PR = QR             …. (c.p.c.t)

Thus, tangents drawn from an external point to a circle are equal.

 

Q 23. Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.

Solution:

Steps of construction:

(i)  Take a point O on the plane of the paper and draw a circle of radius OA = 4 cm.

(ii) Produce OA to B such that OA = AB = 4 cm.

(iii) Draw a circle with centre at A and radius AB.

(iv) Suppose it cuts the circle drawn in step (i) at P and Q.

(v) Join BP and BQ to get the desired tangents.

Justification:

In ∆OAP, OA = OP = 4 cm ...(radii of the same circle)

Also, AP = 4 cm ….(Radius of the circle with centre A)

∴ ∆OAP is equilateral.

∴ ∠PAO = 60°

∴ ∠BAP = 120° Exterior angle of a triangle is equal to the sum of the interior opposite angles.

In ∆BAP, we have BA = AP and ∠BAP = 120°

∴ ∠ABP = ∠APB = 30°

Similarly we can get ∠ABQ = 30°

∴ ∠PBQ = 60°

 

Q 24. Two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of 

Solution:

AO’ = O’X = XO = OC      ……… (Since the two circles are equal)

So, OA = AO’ + O’X + XO      ………. (A-O’-X-O)

∴ OA = 30’A

In ∆ AO’D and ∆AOC,

∠DAO’ = ∠CAO   …….. (Common angle)

∠ADO’ = ∠ACO   ……..(both measure 90°)

∴ ∆ADO’ ∼ ∆ACO  ……..(Both AA test of similarity)

 

Q 25. Solve for x:

Solution:

L.C.M. of all the denominators is (x + 1) (x + 2) (x + 4)

Multiply throughout by the L.C.M., we get

(x + 2) (x + 4) + 2 (x + 1) (x + 4) = 4(x +1) (x + 2)

∴ (x + 4) (x + 2 + 2x + 2) = 4(x² + 3x + 2)

∴ (x + 4) (3x + 4) = 4x² + 12x + 8

∴ 3x² + 16x + 16 = 4x² + 12x + 8

∴ x² – 4x – 8 = 0

Now, a = 1, b = -4, c = -8

 

Q 26. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. 

Solution:

MP = YX = 40 m

∴ QM = h – 40

In right angled ∆QMY,

∴ PX = h – 40         ………..(1)

In right angled ∆QPX,

∴ 1.73 h – h = 40 (1.73) ⇒ h = 94.79 m

Thus, PQ is 94.79 m

 

Q 27. The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.

Solution:

Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.

That is, 1 + 2 + 3 + ……………… + (x – 1) = (x + 1) + (x + 2) + …………. + 49

∴ 1 + 2 + 3 + ………….. + (x – 1)

     = [1 + 2 + ………. + x + (x + 1) + ……+ 49] – (1 + 2 + 3 + ………. + x)

∴ x (x – 1) = 49 ⨯ 50 – x (1 + x)

∴ x (x – 1) + x (1 + x) = 49 ⨯ 50

∴ x² – x + x + x² = 49 ⨯ 50

∴ x² = 49 ⨯ 25

∴ x = 7 ⨯ 5 = 35

Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.

 

Q 28. The vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that  . Calculate the area of ∆ADE and compare it with area of ∆ABC.

Solution:

∴ AD: DB = AE : EC = 1: 2

So, D and E divide AB and AC respectively in the ratio 1:2.

So the coordinates of D and E are  

 

 

 Q 29. A number x is selected at random from the numbers 1, 2, 3, and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.

Solution:

x is selected from 1,2,3 and 4

1,2,3,4

y is selected from 1,4,9 and 16

Let A {1,4,9,16,2,8,18,32,3,12,27,48,36,64} which consists of elements that are product of x and y

 

 

Q 30. Is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OQ and meets OP produced at B. Prove that the perimeter of shaded region is r 

Solution:

Perimeter of shaded region = AB = PB +arc length AP ……. (1)

 

In right angled ∆OAB,

 

OB = OP + PB

∴ r sec θ = r + PB

∴ PB = r sec θ – r ………. (4)

Substitute (2), (3) and (4) in (1), we get

Perimeter of shaded region = AB + PB + arc length AP

 

Q 31. A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:

Let the speed of the stream be s km/h.

Speed of the motor boat = 24 km / h

Speed of the motor boat upstream = 24 - s

Speed of the motor boat downstream = 24 + s

According to the given condition,

∴ 32 ⨯ 2s = 576 – s²

∴ s² + 64s – 576 = 0

∴ (s + 72) (s – 8) = 8

∴ s = -72 or s = 8

Since, speed of the stream cannot be negative, the speed of the stream is 8 km / h.