CBSE Class 10 Mathematics Previous Year Question Paper 2015 All India Set-1

Q 1. If the quadratic equation   has two equal roots then find the value of p.

 

Q 2. In the below figure, a tower AB is 20 m high and BC, its shadow on the ground, is  m long. Find the Sun’s altitude.

 

Q 3. Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.            

 

Q 4. PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ

 

 

Q 5. Two tangents RQ and RP are drawn from an external point R to the circle with centre O, If ∠PRQ = 120°,

then prove that OR = PR + RQ.

 

 

Q 6. A ΔABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ∆ABC is 54 cm², then find the lengths of sides AB and AC.

 

 

Q 7. Solve the following quadratic equation for x:

4x² + 4bx – (a² – b²) = 0

 

Q 8. In an A.P., if S₆ + S₇ = 167 and S₁₀ = 235, then find the A.P., where S_n denotes the sum of its first n terms

 

Q 9. The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B, Find the values of P.

 

Q 10. Find the relation between x and y if the points A(x, y), B (-5, 7) and C (-4, 5) are collinear.

 

Q 11. The 14^th term of an A.P. is twice its 8^th term. If its 6^th term is -8, then find the sum of its first 20 terms.

 

Q 12. Solve for x:

 

Q 13. The angle of elevation of an aeroplane from point A on the ground is 60° After flight of 15 seconds, the angle of elevation changes to 30° If the aeroplane is flying at a constant height of  m, find the speed of the plane in km/hr.

 

Q 14. If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = , where P lies on the line segment AB.

 

Q 15. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls  is. The probability of selecting a blue ball at random from the same jar . If the jar contains 10 orange balls, find the total number of balls in the jar.

 

Q 16. Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also find the area of the corresponding major segment.

 

Q 17. Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but height 2.8 m, and the canvas to be used costs Rs. 100 per sq. m, find the amount, the associations will have to pay. What values are shown by these associations?

 

Q 18. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

 

Q 19. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. 5 per sq. cm. [Use ∏ = 3.14]

 

Q 20. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere, Find the diameter of the sphere and hence find its surface area.

 

Q 21. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

 

Q 22. Find the 60^th term of the AP 8, 10, 12… if it has a total of 60 terms and hence find the sum of its last 10 terms.

 

Q 23. A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

 

Q 24. Prove that the lengths of the tangents drawn from an external point to a circle are equal.

 

Q 25. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

 

Q 26. Construct a ∆ABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°, Construct another ∆AB’C’ similar to ∆ABC with base AB’ = 8 cm.

 

Q 27. At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30° The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A.

 

Q 28. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is

i. a card of spade or an ace.
ii. a black king.
iii. neither a jack nor a king
iv. either a king or a queen.

 

Q 29. Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

 

Q 30. PQRS is square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersections of its diagonals. Find the total area of the two flower beds (shaded parts).

 

Q 31. From each end of a solid metal cylinder, metal was scooped out in hemispherical from of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire.

Q 1. If the quadratic equation   has two equal roots then find the value of p.

Solution:

The given quadratic equation is,

Px² - px + 15 = 0

Here, a = p, b =  p, c = 15

For real equal roots, discriminant = 0

∴ b² – 4ac = 0

∴ (p)² – 4p(15) = 0

∴ 20p² – 60p = 0

∴ 20p(p – 3) = 0

∴ p = 3 or p = 0

But, p = 0 is not possible.

∴ p = 3

 

Q 2. In the below figure, a tower AB is 20 m high and BC, its shadow on the ground, is  m long. Find the Sun’s altitude.

 

Solution:

Let AB be the tower and BC be its shadow.

AB = 20, BC = 20

In ∆ABC,

 

Q 3. Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is 6.

Solution:

Two dice are tossed

S = [(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)]

Total number of outcomes when two dice are tossed = 6 x 6 = 36

Favourable events of getting the product as 6 are:

(1 ⨯ 6 = 6), (6 ⨯ 1 = 6), (2 ⨯ 3 = 6), (3 ⨯ 2 =6)

i.e.(1,6), (6,1), (2,3), (3,2)

Favourable events of getting product as 6 = 4

    

Q 4. PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ

 

 

Solution:

m∠OPT = 90^o (∵ radius is perpendicular to the tangent)

So, ∠OPQ = ∠OPT - ∠QPT

                  = 90° – 60°

                 = 30°

m∠POQ = 2∠QPT = 2 ⨯ 60° = 120°

reflex m∠POQ = 360° – 120° = 240°

 

Q 5. Two tangents RQ and RP are drawn from an external point R to the circle with centre O, If ∠PRQ = 120°,

then prove that OR = PR + RQ.

 

 

Solution:

 

 

Given that m∠PRQ = 120^o

We know that the line joining the centre and the external point is the angle bisector between the tangents.

Also we know that lengths of tangents from an external point are equal.

Thus, PR = RQ.

Join OP and OQ.

Since OP and OQ are the radii from the centre O,

OP ⏊ PR and OQ ⏊ RQ

Thus, ∆OPR and ∆OQR are right angled congruent triangles.

Hence, m ∠POR = 90°-m ∠PRO = 90°– 60°= 30°

m∠QOR = 90°- m∠QRO = 90° – 60° = 30°

 

Q 6. A ΔABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ∆ABC is 54 cm², then find the lengths of sides AB and AC.

 

 

Solution:

 

 

Let the given circle touch the sides AB and AC of the triangle at points F and E respectively and let the length of the line segment AF be x.

Now, it can be observed that:

BF = BD = 6 cm   (tangents from point B)

CE = CD = 9 cm   (tangents from point C)

AE = AF = x         (tangents from point A)

AB = AF + FB = x + 6

BC = BD + DC = 6 + 9 = 15

CA = CE + EA = 9 + x

2s = AB + BC + CA = x + 6 + 15 + 9 + x = 30 + 2x

s = 15 + x

s – a = 15 + x – 15 = x

s – b = 15 + x – (x + 9) = 6

s – c = 15 + x – (6 + x) = 9

 

324 = 6 (15x + x²)

54 = 15x = x²

x² + 15x - 54 = 0

x² + 18x - 3x - 54 = 0

x(x + 18) -3(x +18)

(x + 18) (x – 3) = 0

x = -18 and x = 3

As distance cannot be negative, x = 3

AC = 3 + 9 = 12

AB = AF + FB = 6 + x = 6 + 3 = 9

 

Q 7. Solve the following quadratic equation for x:

4x² + 4bx – (a² – b²) = 0

Solution:

 

Q 8. In an A.P., if S₆ + S₇ = 167 and S₁₀ = 235, then find the A.P., where S_n denotes the sum of its first n terms

Solution:

S₆ + S₇ = 167 and S₁₀ = 235

∴ S₆ + S₇ = 167

⇒ 5a + 10d + 7a + 21d = 167

⇒ 12a + 31d = 167                     ....... (1)

Also, S₁₀ = 235

⇒ 10a + 45d = 235

⇒ 2a + 9d = 47                       ……. (2)

Multiplying equation (2) by 6, we get

12a + 54d = 282              ……….(3)

Subtracting (1) from (3), we get

∴ d = 5

Substituting value of d in (2), we have

2a + 9 (5) = 47

⇒ 2a + 45 = 47

⇒ 2a = 2

⇒ a = 1

Thus, the given A.P. is 1, 6, 11, 16 …

 

Q 9. The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B, Find the values of P.

Solution:

∆ABC is right triangle at B.

∴ AC² = AB² + BC²                    ….(1)

Also, A ≡ (4,7), B ≡ (p,3) and C ≡ (7,3)

Now, AC² = (7 – 4)² + (3 – 7)² = (3)² + (-4)² = 9 + 16 = 25

        AB² = (p – 4)² + (3 – 7)² = p² – 8p + 16 + (-4)²

                                              =p² – 8p + 16 + 16

                                              =p² -8p + 32

         BC² = (7 – p)² + (3 – 3)² = 49 – 14p + p² + 0

                                             = p² – 14p + 49

From (1), we have

25 = (p² – 8p + 32) + (p² – 14p + 49)

⇒ 25 = 2p² – 22p + 81

⇒ 2p² – 22p + 56 = 0

⇒ p² – 11p + 28 = 0

⇒ p² – 7p – 4p + 28 = 0

⇒ p (p - 7) – 4p (p - 7) = 0

⇒ (p – 7) (p - 4) = 0

⇒ p = 7 and p = 4

 

Q 10. Find the relation between x and y if the points A(x, y), B (-5, 7) and C (-4, 5) are collinear.

Solution:

Given that the points A (x,y), B(-5,7) and C(-4,5) are collinear.

So, the area formed by these vertices is 0

 

Q 11. The 14^th term of an A.P. is twice its 8^th term. If its 6^th term is -8, then find the sum of its first 20 terms.

Solution:

Here it is given that,

T₁₄ = 2 (T₈)

⇒ a + (14 – 1)d = 2[a + (8 – 1)d]

⇒ a + 13d = 2[ a + 7d]

⇒ a + 13d = 2a + 14d

⇒ 13d – 14d = 2a – a

⇒ -d = a                                 …….(1)

Now, it is given that its 6^th term is -8.

T₆ = -8

⇒ a + (6 – 1)d = -8

⇒ a + 5d = -8

⇒ -d + 5d = -8                [∵ Using (1)]

⇒ 4d = -8

⇒ d = -2

Subs. this in eq. (1), we get a  = 2

Now, the sum of 20 terms,

 

 

Q 12. Solve for x:

Solution:

Q 13. The angle of elevation of an aeroplane from point A on the ground is 60° After flight of 15 seconds, the angle of elevation changes to 30° If the aeroplane is flying at a constant height of  m, find the speed of the plane in km/hr.

Solution:

Let BC be the height at which the aeroplane is observed from point A.

Then, BC =  

In 15 seconds, the aeroplane moves from point A to D.

A and D are the points where the angles of elevation 60° and 30° are formed respectively.

Let BA = x metres and AD y metres

BC = x + y

 

∴ x + y = 1500 (3) = 4500

∴ 1500 + y = 4500

∴ y = 3000 m            …..(2)

We know that the aeroplane moves from point A to D in 15 seconds and the distance covered is 3000 metres. (by 2)

 

 

Q 14. If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = , where P lies on the line segment AB.

Solution:

Q 15. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls  is. The probability of selecting a blue ball at random from the same jar . If the jar contains 10 orange balls, find the total number of balls in the jar.

Solution:

Here the jar contains red, blue and orange balls.

Let the number of red balls be x.

Let the number of blue balls be y.

Number of orange balls = 10

∴ Total number of balls = x + y + 10

Now, let P be the probability of drawing a ball from the jar

Subs. x = 6 in eq. (i), we get y = 8

∴ Total number of balls = x + y + 10 = 6 + 8 + 10 = 24

Hence, total number of balls in the jar is 24.

 

Q 16. Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Also find the area of the corresponding major segment.

Solution:

Radius of the circle = 14 cm

Central Angle, θ = 60°

Area of the minor segment

 

Q 17. Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but height 2.8 m, and the canvas to be used costs Rs. 100 per sq. m, find the amount, the associations will have to pay. What values are shown by these associations?

Solution:

Diameter of the tent = 4.2 m

Radius of the tent, r = 2.1 m

Height of the cylindrical part of tent, h_cylinder = 4 m

Height of the conical part, h_cone = 2.8 m

Slant height of the conical part, l

= 22 ⨯ 0.3 ⨯ 8 = 52.8 m²

Curved surface area of the conical tent = πrl  ⨯ 2.1 ⨯ 3.5 = 23.1 m²

Total area of cloth required for building one tent

= Curved surface area of the cylinder + Curved surface area of the conical tent

= 52.8 + 23.1

= 75.9 m²

Cost of building one tent = 75.9 × 100 = Rs. 7590

Total cost of 100 tents = 7590 × 100 = Rs. 7, 59,000

It shows the helping nature, unity and cooperativeness of the associations.

 

Q 18. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

Solution:

Internal diameter of the bowl = 36 cm

Internal radius of the bowl, r = 18 cm

 

Let the height of the small bottle be ‘h’.

Diameter of a small cylindrical bottle = 6 cm

Radius of the small bottle, R = 3 cm

Volume of a single bottle = πR²h = 𝜋 × 3² × h

No. of small bottles, n = 72

 

Volume of liquid to be transferred in the bottles

Height of the small cylindrical bottle = 5.4 cm

 

Q 19. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. 5 per sq. cm. [Use ∏ = 3.14]

Solution:

Side of the cubical block, a = 10 cm

Longest diagonal of the cubical block = a  = 10 cm

Since the cube is surmounted by a hemisphere, therefore the side of the cube should

be equal to the diameter of the hemisphere.

Diameter of the sphere = 10 cm

Radius of the sphere, r = 5 cm

Total surface area of the solid = Total surface area of the cube – Inner cross-section

area of the hemisphere + Curved surface area of the hemisphere

= 6a² – πr² + 2πr²

= 6a² + πr²

= 6 ⨯ (10)² + 3.14 ⨯ 5²

= 600 + 78.5 = 678.5 cm²

Total surface area of the solid = 678.5 cm²

Cost of painting per sq. cm = Rs. 5

Cost of painting the total surface area of the solid = 678.5 × 5 = Rs. 3392.50

 

Q 20. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere, Find the diameter of the sphere and hence find its surface area. 

Solution:

No. of cones = 504

Diameter of a cone = 3.5 cm

Radius of the cone, r = 1.75 cm

Height of the cone, h = 3 cm

Volume of a cone

 

Q 21. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Solution:

Let l be the length of the longer side and b be the length of the shorter side.

Given that the length of the diagonal of the rectangular field is 16 metres more than the shorter side.

Thus, diagonal = 16 + b

Since longer side is 14 metres more than shorter side, we have,

∴ l = 14 + b

Diagonal is the hypotenuse of the triangle.

Consider the following figure of the rectangular field.

 

By applying Pythagoras Theorem in ∆ABD, we have,

Diagonal² =  Length² + breadth²

⇒ (16 + b)² = (14+b)² + b²

⇒ 256  + b² + 32b = 196 + b² + 28b + b²

⇒ 256  + 32b = 196 + 28b + b²

⇒ 60 + 32b = 28b + b²

⇒ b² – 4b – 60 = 0

⇒ b² – 10b + 6b – 60 = 0

⇒ b(b - 10) + 6(b – 10) = 0

⇒ (b + 6) (b – 10) = 0

⇒ (b + 6) = 0 or (b – 10) = 0

⇒b =- 6 or b = 10

As breadth cannot be negative breadth = 10 m

Thus, length of the rectangular field = 14 + 10 = 24 m

 

Q 22. Find the 60^th term of the AP 8, 10, 12… if it has a total of 60 terms and hence find the sum of its last 10 terms.

Solution:

Consider the given A.P.8, 10,12, ……..

Here the initial term is 8 and the common difference is 10 – 8 = 2  and 12 – 10 = 2

General term of an A.P is t_n and formula to find out t_n is

t_n = a + (n – 1)d

⇒ t₆₀ = 8 + (60 – 1) ⨯ 2

⇒ t₆₀ = 8 + 59 ⨯ 2

⇒ t₆₀ = 8 + 118

⇒ t₆₀ = 126

We need to find the sum of last 10 terms.

Thus,

Sum of last 10 terms = Sum of first 60 terms – Sum of first 50 terms

 

Q 23. A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

Solution:

Let x be the first speed of the train

Thus, we have,

 

⇒ 54 (x + 6) + 63x = 3x (x + 6)

⇒ 54x + 324 + 63x = 3x² + 18x

⇒ 117x + 324 = 3x² + 18x

⇒ 3x² – 117x – 324 + 18x = 0

⇒ 3x² – 99x – 324 = 0

⇒ x² – 33x – 108 = 0

⇒ x² – 36x + 3x – 108 = 0

⇒ x (x – 36) + 3 (x - 36) = 0

⇒ (x + 3) (x – 36) = 0

⇒ (x + 3) = 0 or (x – 36) =0

⇒ x = -3 or x = 36

Speed cannot be negative and hence initial speed of the train is 36km/hour

 

Q 24. Prove that the lengths of the tangents drawn from an external point to a circle are equal.

Solution:

Let P be an external point and PA and PB are tangents to the circle.

We need to prove that PA = PB

Now consider the triangles ∆OAP and ∆OBP

m∠A = m∠B = 90°

OP = OP [common]

OA = OB = radii of the circle

Thus, by Right Angle-Hypotenuse-Side criterion of congruence we have,

∆OAP ≅ ∆OBP

The corresponding parts of the congruent triangles are congruent.

Thus,

PA = PB

Q 25. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Solution:

In the figure, C is the midpoint of the minor arc PQ, O is the centre of the circle and AB is tangent to the circle through point C.

We have to show the tangent drawn at the mid-point of the arc PQ of a circle is parallel to the chord joining the end points of the arc PQ.

We will show PQ || AB.

It is given that C is the midpoint point of the arc PQ.

So, arc PC = arc CQ.

⇒ PC = CQ

 

 

This shows that ∆PQC is an isosceles triangle.

Thus, the perpendicular bisector of the side PQ of ∆PQC passes through vertex C.

The perpendicular bisector of a chord passes through the centre of the circle.

So the perpendicular bisector of PQ passes through the centre O of the circle.

Thus the perpendicular bisector of PQ passes through the points O and C.

⇒ PQ ⏊ OC

AB is the tangent to the circle through the point C on the circle.

⇒ AB ⏊  OC

The chord PQ and the tangent PQ of the circle are perpendicular to the same line OC.

∴ PQ ∣∣ AB.

 

Q 26. Construct a ∆ABC in which AB = 6 cm, ∠A = 30° and ∠B = 60°, Construct another ∆AB’C’ similar to ∆ABC with base AB’ = 8 cm.

Solution:

i.  Construct the ∆ABC as per given measurements.

ii.  In the half plane of  which does not contain C, draw   Such that ∠BAX is an acute angle.

iii. With some appropriate radius and centre A, Draw an arc to intersect  at B₁.

    Similarly, with center B₁ and the same radius, draw an arc to intersect  at B₂ 

     such that B₁B₂ = B₃B₄= B₄B₅= B₅B₆= B₆B₇= B₇B₈ 

iv.  Draw  

v. Through B₈ draw a ray parallel to  to intersect  a B’

vi. Through B’ draw a ray parallel to  BC to intersect  at C’.

Thus, ∆AB’C’ is the required triangle

 

Q 27. At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30° The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A.

Solution:

 

Let PB be the surface of the lake and A be the point of observation such that

AP = 20 metres. Let C be the position of the cloud and C’ be its reflection in the lake.

Then CB = C’B. Let AM be perpendicular from A on CB.

Then m ∠CAM = 30° and m∠C'AM = 60°

Let CM = h. Then CB = h + 20 and C’B = h + 20. (CB=CB’ since refection about PB)

⇒ 3h = h + 40

⇒ 2h = 40

⇒ h = 20 m

⇒ AM = 

Now, to find AC using Pythagoras theorem

AC² = AM² + MC²               

 

AC = 40

Hence, the height of the cloud from the surface of the lake is 40 metres.

 

Q 28. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is

i. a card of spade or an ace.
ii. a black king.
iii. neither a jack nor a king
iv. either a king or a queen.

Solution:

Let S be the sample space of drawing a card from a well – shuffled deck

n (S) = ⁵²C₁ = 52

i. There are 13 spade cards and 4 ace’s in a deck

As ace of spade is included in 13 spade cards,

so there are 13 spade cards and 3 ace’s

a card of spade or an ace can be drawn in ¹³C₁+ ³C₁ = 13 + 3 = 16

Probablity of drawing card of spade or an ace = 

ii. There are 2 black king cards in a deck

A card of black king can be drawn in ²C₁ = 2

Probablity of drawing a black king = 

iii. There are 4 jack and 4 king cards in a deck

So there are 52 - 8 = 44 cards which are neither jack or nor king

A card which neither a jack nor a king can be drawn in ⁴⁴C₁ = 44

Probablity of drawing card which is neither a jack or nor a king = 

iv. There are 4 king and 4 queen cards in a deck

So there are 4 + 4 = 8 cards which are either king or queen

A card which is neither a king or a queen can be drawn in ⁸C₁ = 8

Probability of drawing a card which is either a king or a queen = 

 

Q 29. Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

Solution:

Take (x₁, y₁) = (1, -1), (-4, 2k) and (-k, -5)

It is given that the area of the triangle is 24 sq.unit

Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

given by

48 = [(2k + 5) + 16 + (k + 2k²)]

∴ 2k² + 3k – 27 = 0

 

Q 30. PQRS is square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersections of its diagonals. Find the total area of the two flower beds (shaded parts).

 

Solution:

PQRS is a square.

So each side is equal and angle between the adjacent sides is a right angle.

Also the diagonals perpendicularly bisect each other.

In ∆PQR using Pythagoras theorem,

PR² = PQ² + QR²

PR² = (42)² + (42)²

PR =  (42)

From the figure we can see that the radius of the flower bed ORQ is OR.

Area of the flower bed ORQ = Area of the flower bed OPS

= 251.37 cm²

Total area of the two flower beds

= area of the flower bed ORQ + Area of the flower bed OPS

= 251.37 + 251.37

= 502.74 cm²

 

Q 31. From each end of a solid metal cylinder, metal was scooped out in hemispherical from of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire.

Solution:

Height of the cylinder (h) = 10 cm

Radius of the base of the cylinder = 4.2 cm

Volume of original cylinder = πr²h

Volume of the remaining cylinder after scooping out the hemisphere from each end

= Volume of original cylinder - 2 ⨯ Volume of hemisphere

= 554.4 – 2 ⨯  155.232

= 243.936 cm³

The remaining cylinder is melted and converted to a new cylindrical wire of 1.4 cm thickness.

So they have the same volume and radius of the new cylindrical wire, i.e. 0.7 cm.

Volume of the remaining cylinder =  Volume of the new cylindrical wire

243.936 = πr²h