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

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Q 1. If the quadratic equation  begin mathsize 12px style px squared minus 2 square root of 5px plus 15 equals 0 end style 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 begin mathsize 12px style 20 square root of 3 end style 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 Sn 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 14th term of an A.P. is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 terms.

 

Q 12. Solve for x:

begin mathsize 12px style square root of 3 straight x squared end root minus 2 square root of 2 straight x end root minus 2 square root of 3 equals 0 end style

 

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 begin mathsize 12px style 1500 square root of 3 end style 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 = begin mathsize 12px style 3 over 7 end style, 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 begin mathsize 12px style 1 fourth end style is. The probability of selecting a blue ball at random from the same jar begin mathsize 12px style 1 third end style. 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.begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

 

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?begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

 

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.begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

 

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 60th 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.begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

Q 1. If the quadratic equation  begin mathsize 12px style px squared minus 2 square root of 5px plus 15 equals 0 end style has two equal roots then find the value of p.

Solution:

The given quadratic equation is,

Px2 - begin mathsize 12px style 2 square root of 5 end stylepx + 15 = 0

Here, a = p, b = begin mathsize 12px style negative 2 square root of 5 end style p, c = 15

For real equal roots, discriminant = 0

∴ b2 – 4ac = 0

∴ (begin mathsize 12px style negative 2 square root of 5 end stylep)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 begin mathsize 12px style 20 square root of 3 end style m long. Find the Sun’s altitude.

 

Solution:

Let AB be the tower and BC be its shadow.

AB = 20, BC = 20begin mathsize 12px style square root of 3 end style

In ∆ABC,

Error converting from MathML to accessible text.

 

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

Error converting from MathML to accessible text.

    

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 = 90o (∵ 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°

Error converting from MathML to accessible text.

 

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 = 120o

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

begin mathsize 12px style Thus ,  straight m angle PRO  =  straight m angle QRO equals 120 to the power of straight o over 2 equals 60 to the power of straight o end style

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°

Error converting from MathML to accessible text.

 

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

 begin mathsize 12px style table attributes columnalign left end attributes row cell Area space of space ΔABC equals square root of straight s left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis left parenthesis straight s minus straight c right parenthesis end root end cell row cell 54 equals square root of left parenthesis 15 plus straight x right parenthesis left parenthesis straight x right parenthesis left parenthesis 6 right parenthesis left parenthesis 9 right parenthesis end root end cell row cell 54 equals 3 square root of 6 open parentheses 15 straight x plus straight x squared close parentheses end root end cell row cell 18 equals square root of 6 open parentheses 15 straight x plus straight x squared close parentheses end root end cell end table end style

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:

begin mathsize 12px style table attributes columnalign left end attributes row cell 4 straight x squared plus 4 bx minus open parentheses straight a squared minus straight b squared close parentheses equals 0 end cell row cell rightwards double arrow straight x squared plus bx minus open parentheses fraction numerator straight a squared minus straight b squared over denominator 4 end fraction close parentheses equals 0 end cell row cell rightwards double arrow straight x squared plus 2 space open parentheses straight b over 2 close parentheses space straight x equals fraction numerator straight a squared minus straight b squared over denominator 4 end fraction end cell row cell rightwards double arrow straight x squared plus 2 space open parentheses straight b over 2 close parentheses space straight x space plus open parentheses straight b over 2 close parentheses squared equals fraction numerator straight a squared plus straight b squared over denominator 4 end fraction open parentheses straight b over 2 close parentheses squared end cell end table
table attributes columnalign left end attributes row cell rightwards double arrow open parentheses straight x plus straight b over 2 close parentheses squared equals straight a squared over 4 end cell row cell rightwards double arrow straight x plus straight b over 2 equals plus-or-minus straight a over 2 end cell end table
table attributes columnalign left end attributes row cell rightwards double arrow straight x equals fraction numerator negative straight b over denominator 2 end fraction plus-or-minus straight a over 2 end cell row cell rightwards double arrow straight x equals fraction numerator negative straight b minus straight a over denominator 2 end fraction comma space fraction numerator negative straight b plus straight a over denominator 2 end fraction end cell row cell Hence ,  the space roots space are  -  open parentheses fraction numerator straight a plus straight b over denominator 2 end fraction close parentheses space and space open parentheses fraction numerator straight a minus straight b over denominator 2 end fraction close parentheses. end cell end table end style

 

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

Solution:

S6 + S7 = 167 and S10 = 235

begin mathsize 12px style Now comma space straight S subscript straight n equals straight n over 2 left curly bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right curly bracket end style

∴ S6 + S7 = 167

begin mathsize 12px style text ⇒ end text 5 over 2 left curly bracket 2 straight a plus 4 straight d right curly bracket plus 7 over 2 left curly bracket 2 straight a plus 6 straight d right curly bracket equals 167 end style

⇒ 5a + 10d + 7a + 21d = 167

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

Also, S10 = 235

begin mathsize 12px style ∴  10 over 2 left curly bracket 2 straight a plus 9 straight d right curly bracket equals 235 end style

⇒ 10a + 45d = 235

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

Multiplying equation (2) by 6, we get

12a + 54d = 282              ……….(3)

Subtracting (1) from (3), we get

Error converting from MathML to accessible text.

∴ 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

begin mathsize 12px style table attributes columnalign left end attributes row cell therefore 1 half left square bracket straight x left parenthesis 7 minus 5 right parenthesis plus left parenthesis negative 5 right parenthesis left parenthesis 5 minus straight y right parenthesis plus left parenthesis negative 4 right parenthesis left parenthesis straight y minus 7 right parenthesis right square bracket equals 0 end cell row cell rightwards double arrow 1 half left square bracket 2 straight x minus 25 plus 5 straight y minus 4 straight y plus 28 right square bracket equals 0 end cell row cell rightwards double arrow 1 half left square bracket 2 straight x plus straight y plus 3 right square bracket equals 0 end cell row cell rightwards double arrow 2 straight x plus straight y plus 3 equals 0 end cell row cell rightwards double arrow straight y equals negative 2 straight x minus 3 end cell end table end style

 

Q 11. The 14th term of an A.P. is twice its 8th term. If its 6th 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 6th 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,

 begin mathsize 12px style table attributes columnalign left end attributes row cell straight S subscript straight n equals straight n over 2 open square brackets 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d close square brackets end cell row cell straight S subscript 20 equals 20 over 2 left square bracket 2 straight a plus left parenthesis 20 minus 1 right parenthesis straight d right square bracket end cell row = 10 [ 2 ( 2 ) +  19 (- 2 )] row =  10 [ 4  -  38 ] row =- 340 end table end style

 

Q 12. Solve for x:

begin mathsize 12px style square root of 3 straight x squared end root minus 2 square root of 2 straight x end root minus 2 square root of 3 equals 0 end style

Solution:

Error converting from MathML to accessible text.


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 begin mathsize 12px style 1500 square root of 3 end style 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 = begin mathsize 12px style 1500 square root of 3 end style 

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

 Error converting from MathML to accessible text.

∴ 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)

 Error converting from MathML to accessible text.

 

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 = begin mathsize 12px style 3 over 7 end style, where P lies on the line segment AB.

Solution:

Error converting from MathML to accessible text.


Q 15. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls begin mathsize 12px style 1 fourth end style is. The probability of selecting a blue ball at random from the same jar begin mathsize 12px style 1 third end style. 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

Error converting from MathML to accessible text.

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.begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

Solution:

Radius of the circle = 14 cm

Central Angle, θ = 60°

Area of the minor segment

begin mathsize 12px style table attributes columnalign left end attributes row cell equals straight theta over 360 to the power of straight o cross times πr squared minus 1 half straight r squared sinθ end cell row cell straight equals 60 to the power of straight o over 360 to the power of straight o cross times straight pi cross times 14 squared minus 1 half cross times 14 squared cross times sin 60 to the power of straight o end cell row cell equals 1 half cross times 22 over 7 cross times 14 cross times 14 minus 1 half cross times 14 cross times 14 cross times fraction numerator square root of 3 over denominator 2 end fraction end cell row cell equals fraction numerator 22 cross times 14 over denominator 3 end fraction minus 49 square root of 3 end cell row cell equals fraction numerator 22 cross times 14 over denominator 3 end fraction minus fraction numerator 147 square root of 3 over denominator 3 end fraction end cell row cell equals fraction numerator 308 minus 147 square root of 3 over denominator 3 end fraction space cm to the power of straight 2 end cell row cell Area space of space the space minor space segment  =  fraction numerator 308 minus 147 square root of 3 over denominator 3 end fraction space cm to the power of straight 2 end cell row cell Area space of space major space segment space
=  πr to the power of straight 2 minus fraction numerator 308 minus 147 square root of 3 over denominator 3 end fraction space cm to the power of straight 2 end cell row cell equals 22 over 7 cross times 14 cross times 14 minus fraction numerator 308 minus 147 square root of 3 over denominator 3 end fraction space cm to the power of straight 2 end cell row cell equals 616 minus fraction numerator 308 minus 147 square root of 3 over denominator 3 end fraction space cm to the power of straight 2 end cell end table end style

 

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?begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

Solution:

Diameter of the tent = 4.2 m

Radius of the tent, r = 2.1 m

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

Height of the conical part, hcone = 2.8 m

Slant height of the conical part, l

Error converting from MathML to accessible text.

= 22 ⨯ 0.3 ⨯ 8 = 52.8 m2

Curved surface area of the conical tent = πrl begin mathsize 12px style equals 22 over 7 end style ⨯ 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

begin mathsize 12px style Cost space to space be space borne space by space the space associations  =  759000 over 2 equals Rs.3 , 79 , 500 end style

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

 begin mathsize 12px style Volume space of space the space liquid ,  straight V  =  2 over 3 πr cubed equals 2 over 3 cross times straight pi cross times 18 cubed end style

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

 begin mathsize 12px style Volume space wasted space in space the space transfer  =  10 over 100 cross times 2 over 3 cross times straight pi cross times 18 cubed end style

Volume of liquid to be transferred in the bottles

Error converting from MathML to accessible text.

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 begin mathsize 12px style square root of 3 end style = 10begin mathsize 12px style square root of 3 end style 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. begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

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

Error converting from MathML to accessible text.

 

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,

Diagonal=  Length2 + breadth2

⇒ (16 + b)= (14+b)+ 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 60th 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 tand formula to find out tis

tn = 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

Error converting from MathML to accessible text.

 

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

begin mathsize 12px style We space know space that space Distance over Speed equals time end style

Thus, we have,

 begin mathsize 12px style table attributes columnalign left end attributes row cell 54 over straight x plus fraction numerator 63 over denominator straight x plus 6 end fraction equals 3 space hours end cell row cell rightwards double arrow fraction numerator 54 left parenthesis straight x plus 6 right parenthesis 63 straight x over denominator straight x left parenthesis straight x plus 6 right parenthesis end fraction equals 3 end cell end table end style

⇒ 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 begin mathsize 12px style AB with bar on top end style which does not contain C, draw  begin mathsize 12px style AX with bar on top end style Such that ∠BAX is an acute angle.

iii. With some appropriate radius and centre A, Draw an arc to intersect begin mathsize 12px style AX with bar on top end style at B1.

    Similarly, with center Band the same radius, draw an arc to intersect begin mathsize 12px style BX with bar on top end style at B

     such that B1B2 = B3B4= B4B5= B5B6= B6B7= B7B

iv.  Draw  begin mathsize 12px style stack straight B subscript 6 straight B with bar on top end style

v. Through Bdraw a ray parallel to begin mathsize 12px style stack straight B subscript 6 straight B with bar on top end style to intersect  begin mathsize 12px style AY with bar on top end stylea B’

vi. Through B’ draw a ray parallel to  BC to intersect begin mathsize 12px style AZ with bar on top end style 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)

Error converting from MathML to accessible text.

⇒ 3h = h + 40

⇒ 2h = 40

⇒ h = 20 m

⇒ AM = begin mathsize 12px style 20 square root of 3 end style

Now, to find AC using Pythagoras theorem

AC2 = AM2 + MC2               

 begin mathsize 12px style equals square root of 1600 end style

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 13C13C1 = 13 + 3 = 16

Probablity of drawing card of spade or an ace = begin mathsize 12px style 16 over 52 equals 4 over 13 end style

ii. There are 2 black king cards in a deck

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

Probablity of drawing a black king = begin mathsize 12px style 2 over 52 equals 1 over 26 end style

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 = begin mathsize 12px style 44 over 52 equals 11 over 13 end style

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 = begin mathsize 12px style 8 over 52 equals 2 over 13 end style

 

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

begin mathsize 12px style table attributes columnalign left end attributes row cell equals 1 half left square bracket straight x subscript 1 left parenthesis straight y subscript 2 minus straight y subscript 3 right parenthesis plus straight x subscript 2 left parenthesis straight y subscript 3 minus straight y subscript 1 right parenthesis plus straight x subscript 3 left parenthesis straight y subscript 1 minus straight y subscript 2 right parenthesis right square bracket end cell row cell therefore space 24 equals 1 half left square bracket 1 left parenthesis 2 straight k minus left parenthesis negative 5 right parenthesis right parenthesis left parenthesis left parenthesis negative 5 right parenthesis minus left parenthesis negative 1 right parenthesis right parenthesis plus left parenthesis negative straight k right parenthesis left parenthesis left parenthesis negative 1 right parenthesis minus 2 straight k right parenthesis right square bracket end cell end table end style

48 = [(2k + 5) + 16 + (k + 2k2)]

∴ 2k2 + 3k – 27 = 0

Error converting from MathML to accessible text.

 

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 = begin mathsize 12px style square root of 2 end style (42)

begin mathsize 12px style OR equals 1 half PR equals fraction numerator 42 over denominator square root of 2 end fraction equals OQ end style

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

Error converting from MathML to accessible text.

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.begin mathsize 12px style open square brackets Use space straight pi equals fraction numerator begin display style 22 end style over denominator begin display style 7 end style end fraction close square brackets end style

Solution:

Height of the cylinder (h) = 10 cm

Radius of the base of the cylinder = 4.2 cm

Volume of original cylinder = πr2h

Error converting from MathML to accessible text.

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

Error converting from MathML to accessible text.

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