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8/12/2019 Maths Previous

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5. Which term of the sequence 114, 109, 104, .... is the first negative term?

2. The sum and product of the zeroes of a quadratic polynomial are - and -3 respectively. What is thequadratic polynomial?

3. For what values of k the quadratic equation x2- kx + 4 = 0 has equal roots?

6. A cylinder, a cone, and a hemisphere are of equal base and have the same height. What is the ratio in theirvolumes?8. In the figure given below, PA and PB are tangents to the circle; drawn from an external point P. CD is thethird tangent touching the circle at Q. If PB = 10 cm, and CQ = 2 cm, what is the length of PC?

9. Cards each marked with one of the numbers 4, 5, 6....20 are placed in a box and mixed thoroughly. Onecard is drawn at random from the box. What is the probability of getting an even prime number?

15 A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag, find theprobability of getting (i) a white ball or a green ball. (ii) neither a green ball not a red ball. OR One card is drawn

from a well shuffled deck of 52 playing cards. Find the probability of getting (i) a non-face card (ii) a black king ora red queen.21 Observe the graph given below, and state whether triangle ABC is scalene, isosceles or equilateral.Justify your answer. Also find its area.


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23 Construct a

ABC in which CA = 6 cm, AB = 5cm and BAC = 45, then construct a triangle similar to the given trianglewhose sides are

of the corresponding sides of the

ABC.

24 Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle atthe centre of the circle.

25 A square field and an equilateral triangular park have equal perimeters. If the cost of ploughing the field atrate of Rs 5/m2is Rs 720, find the cost of maintaining the park at the rate of Rs 10/m2. OR An iron solid sphere ofradius 3 cm is melted and recast into small spherical balls of radius 1 cm each. Assuming that there is nowastage in the process, find the number of small spherical balls made from the given sphere.26 Some students arranged a picnic. The budget for food was Rs 240. Because four students of the groupfailed to go, the cost of food to each student got increased by Rs 5. How many students went for the picnic? OR

A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, ithad to increase the speed by 250 km/h from the usual speed. Find its usual speed.

27 From the top of a building 100 m high, the angles of depression of the top and bottom of a tower areobserved to be 45 and 60 respectively. Find the height of the tower. Also find the distance between the foot ofthe building and bottom of the tower. OR The angle of elevation of the top a tower at a point on the level groundis 30. After walking a distance of 100 m towards the foot of the tower along the horizontal line through the foot ofthe tower on the same level ground, the angle of elevation of the top of the tower is 60. Find the height of thetower.

28 Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the othertwo sides. Using the above, solve the following: A ladder reaches a window which is 12 m above the ground onone side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to

reach a window 9 m high. Find the width of the street if the length of the ladder is 15 m.


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29. The interior of a building is in the form of a right circular cylinder of radius 7 m and height 6 m,surmounted by a right circular cone of same radius and of vertical angle 60. Find the cost of painting the buildingfrom inside at the rate of Rs 30/m2.

15(i) P (White or green ball) =

(ii) P (Neither green nor red) =

OR (i) P (non-face card) =

(ii) P (black king or red queen) =

18

19 A contract on a construction job specifies a penalty for delay of completion beyond a certain date as

follows: Rs 200 for first day, Rs. 250 for second day, Rs. 300 for third day and so on. If the contractor pays Rs

27750 as penalty, find the number of days for which the construction work is delayed.

19Let the delay in construction work be for n days. Here a = 200, d = 50, Sn = 27750.

n2+ 7n1110 = 0 (n + 37) (n 30) = 0 n =37 (Rejected)

or n = 30 \ Delay in construction work as for 30 days.

24


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Since tangent is perpendicular to the radius: SPO = SRO = OQT = 90 In right triangles OPS and ORS, OS= OS (Common) OP = OR (Radii of circle) \DOPS @ DORS (RHS Congruence) \ 1 = 2 Similarly 3 = 4

Now 1 + 2 + 3 + 4 = 180 (Sum of angles on the same side of Transversal) 2 + 3 = 90 \ SOT =

90

25Let the side of the square be a metres. 5 a2= 720 a = 12 m \ Perimeter of square = 48 m Perimeter of triangle =48 m Side of triangle = 16 m Now Area of triangle =

Cost of maintaining the park = Rs.

= Rs.

OR

Radius of sphere = 3 cm Volume of sphere =

= 36 pcm3Radius of spherical ball = 1 cm Volume of one spherical ball =

Let the number of small spherical balls be N.

\ N = 27

1

1

1

26Section D

Let the number of students who arranged the picnic be x. \ Cost of food for one student =

New cost of food for one student =

x24x192 = 0


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(x16) (x + 12) = 0

x = 16 or x =12 (Rejected)

No of students who actually went for the picnic = 164 = 12

OR

Let the usual speed of plane be x km/hour.

Time taken =

hrs. with usual speed

Time taken after increasing speed =

hrs

x2+ 250x750000 = 0

(x + 1000 ) ( x750 ) = 0

x = 750 or1000 (Rejected)

\ Usual speed of the plane = 750 km/h.

27

In right DBED,

BE = DE \

. Height of the tower (CD) = AE = AB BE

= 42.27 m Distance between the foot the building and the bottom of the tower (AC)

= 57.73 m.

OR

1

1

1

1

1

1

1

1

1

1


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In right DBAC,

(i) In right BAD,

AB = AD

From (i) and (ii), we get (ii)

100 + AD = 3AD AD = 50 m From (ii) AB =

= 50 1.732 m Or, AB = 86.6 m

28

291

1

1

1

1


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Internal curved surface area of cylinder = 2prh = 2p 7 6 m2

= 264 m2 In right DOAB,

\ Slant height of cone (OB) = 14 m Internal curved surface area of cone = prl

= 308 m2 Total Area to be painted = 264 + 308 = 572 m2 Cost of painting = Rs (30 572) =

Rs 17160

26. Solve the following equation for x.9x29(a + b)x + (2a2+ 5ab + 2b2) = 0

Solution: Comparing the given equation with standard equation ax2+ bx + c = 0 we find a = 9, b = -9(a + b), c =2a2+ 5ab + 2b2

or

or


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orOR

If 5 is a root of the quadratic equation 2x2+ px15 = 0 and the quadratic equation p(x2+ x) + k = 0 has

equal roots, then find the values of p and k.Solution:

Given that 5 is a root of 2x2+ px15 = 02(5)2+ p(5)15 = 0

505p15 = 0

5p = 35

p = 7Putting p = 7 in the second quadratic equation, we get7x2+ 7x + k = 0Given that above equation has equal roots and henceb24ac = 0494(7)k = 0 ( a = 7, b = 7, c = k)

28k = 49

Question 28

28. An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at

an instant when the angles of elevation of the two planes from the same point on the ground are 30and

60respectively. Find the distance between the two planes at that instant.Solution:Let at some instant, A1be the 1

staero plane A2be the 2ndaeroplane A1A2is the vertical distance between these

aeroplanes at that instant.


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Let, BC = x m and A1A2= h mFrom the A1BC we have,

From A2BC we have,

h = 3 X31253125h = 2 X3125 = 6250

Hence the distance between the two planes at that instant is 6250 m

Question 29

29. A juice seller serves his customers using a glass as shown in figure. The inner diameter of the cylindricalglass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces the capacity of the

glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual capacity. (Use =3.14)

Solution: Since the inner diameter of the glass = 5 cmand height = 10 cmApparent capacity of the glass = r2h= 3.14 X2.5 X2.5 X10 cm3= 196.25 cm3But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass

It is less by= 32.71 cm3

Actual capacity of the glass = apparent capacity of glassvolume of the hemisphere= (196.2532.71) cm3= 163.54 cm3

OR

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of basediameter 7 cm and height 6 cm is completely immersed in water. Find the volume of(i) water displaced out of the cylindrical vessel(ii) water left in the cylindrical vessel

[Take ]Solution:

Diameter of cylindrical vessel = 10 cmSo, r= radius of cylindrical vessel = 5 cm

Height of cylindrical vessel = h = 10.5

Volume of cylindrical vessel = r2h

= X5 X5 X10.5 cm3

= 825 cm

3

Also given that diameter of the cone = 7 cm


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r = radius of cone =Height of the cone = 6 cm

Volume of the cone =

= 77 cm3Volume of water displaced by cone = volume of the cone = 76.93 cm3Water left in cylindrical vessel

= volume of water in cylindervolume of water in cone= 825cm377 cm3= 748 cm3

Question 3

3. If a, 2 are three consecutive terms of an A.P., then fine the value of a.Solution:

Since a, 2 are in AP, we have

Question 4

4. Two coins are tossed simultaneously. Find the probability of getting exactly one head.

Solution:When two coins are tossed simultaneously all the possible outcomes are (H,H),(H,T),(T,H)and (T,T) and the desired

outcomes of getting exactly one head are (H,T) and (T,H).

Thus the required probability =

6. If the diameter of a semicircular protractor is 14 cm, then find its perimeter.Solution:Diameter = 14 cm

RadiusPerimeter of semicircular protractor = r+ 2r

= 36 cm

Question 7

7. If 1 is a zero of the polynomial p(x) = ax23(a1)x1, then find the value of a.Solution:

Given that 1 is a zero of a polynomialTherefore,

P(1) = 0

a(1)23 (a1) (1)1 = 0


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a3(a1)1 = 0

a3a + 31 = 0

2a + 2 = 0

2a = 2

a = 1

Hence value of a is 1

10. Find the discriminant of the quadratic equation

.

Solution:

Discriminant= 10036 = 64

11. Find all the zeroes of the polynomial 2x

3

+ x

2

6x3,

if two of its zeroes are and .Solution:

The given equation is cubic and hence has 3 zeroes.We find 3rdzero.

Given that are zeroes of the polynomial and

hence are factors of the given polynomial.

is also a factor of the given polynomial.

Now we apply the division algorithm between x23 and 2x3+ x26x3 to find the third zero.

Thus we have,Hence the third zero of the given polynomial p(x) is obtained by

making

Therefore, the third zero of p(x) is .13. If the points A(4, 3) and B(x, 5) are on the circle with the centre O(2, 3), find the value of x.

Solution:OA = OB = r

OA2= OB2

(42)2+ (33)2= (x2)2+ (53)2

4 + 0 = x24x + 4 + 4

x24x + 4 = 0

(x2)2= 0

x = 2

Question 14

14. Which term of the A.P. 3, 15, 27, 39, will be 120 more than its 21stterm?Solution:Given A.P. is 3, 15, 27, 39,


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Here a = 3, d = t2t1= 153 = 12nthterm of A.P. = tn= a + (n1)dPut n = 21

t21= 3 + (211)12

t21= 3 + (20)12

t21= 243

Let tmbe the term which is 120 more than 21sttermtm= t21+ 120

tm = 243 + 120

a + (m1)d = 363

3+ (m1)12 = 363

(m1)12 = 3633= 360

m1 = 30

m = 31

Hence, the 31stterm of the A.P. is 120 more than its 21stterm.Question 18

18. The area of an equilateral triangle is . Taking each angular point as centre, circles are drawn withradius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles.

[Take

Solution:

Given area of equilateral triangle =

(a is side of equilateral triangle)

a = 14 cm

Radius of circles

Area of sector =

Area of 3 sectors= 77 cm2

Area of the triangle not included in the circles = Area of equilateral triangle Area of 3 sectors of equal area

= 84.7777= 7.77 cm2

OR


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The figure below shows a decorative block which is made of two solidsa cube and a hemisphere. The base ofthe block is a cube with edge 5 cm and the hemisphere, fixed on the top, has a diameter of 4.2 cm. Find the total

surface area of the block. [Take ]Solution:

Given side of cube, a = 5 cmTSA of cube = 6a2= 6 X 52= 150 cm2Diameter of hemisphere = 4.2 cmRadius, r = 2.1 cmSurface area of hemisphere = 2r2

= 27.72 cm2

Area of base of hemisphere = r2


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= 13.86 cm2

Total surface area of the block = Total surface area of cube + Surface area of hemisphereArea of baseof hemisphere

= 177.7213.86

= 163.86 cm

2

Question 19

19. Two dice are thrown simultaneously. What is the probability that(i) 5 will not come up on either of them?(ii) 5 will come up on at least one?(iii) 5 will come up at both dice?

Solution:Total number of possible outcomes of throwing two dice = 6 X 6 = 36

(i) Number of outcomes that 5 will not come up on either of them = 25i.e. [on both the dice, any one of the numbers 1, 2 3, 4, 6 appear]

P [5 will not come up on either of them](ii) 5 will come up on at least one = 5 on the 1stdie and any one of {1, 2, 3, 4, 6} on the 2nddie (or) 5 on

the 2nddice and any one of {1, 2, 3, 4, 6} on the 1 stdice (or) 5 on both dice= 5 + 5 + 1 = 11 ways.

P [5 will come up on either of them) =

(iii) 5 will come up on both the dice in only one way

P [5 will come up at both dice]

Question 20

20. The sum of first six terms of an arithmetic progression is 42. The ratio of its 10thterm to its 30thterm is 1 :3. Calculate the first and the thirteenth terms of the A.P.

Solution:Let a be the first term and d be the common difference of the AP

S6= 42

.(1)T10: T30= 1 : 3 (given)

3[a + 9d] = 1(a + 29d)

3a + 27d = a + 29d

2a = 2d

a = d ..(2)Using the result of (2) in (1)3(2a + 5a) = 423(7a) = 42a = 2

d = 2 (from (2))

First term = a = 213thterm = T13= a + (131)d = a + 12d = a + 12a = 13a = 13(2) = 26T13= 26

Question 22

22. Find the ratio in which the point (x, 2) divides the line segment joining the points (3, 4) and (3, 5). Also find thevalue of x.

Solution:

Let the ratio be k : 1 in which the point (x, 2) divides thejoin of (3, 4) and (3, 5)

x1= 3, y1= 4 and m1= k


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x2= 3, y2= 5 and m2= 1

By using section formula we get

2(k + 1) = 5k4 (By cross multiplication)

2k + 2 = 5k4

5k2k = 6

3k = 6

k = 2

Thus the ratio is 2 : 1

Also

x = 1

Question 23

23. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices

are (0,1), (2, 1) and (0, 3).Solution:

Let A(0, 1), B(2, 1) and C(0, 3) are the vertices of the triangle.Let D, E, F be the mid points of the sides AB, BC, CA respectively. Then

We use the formulaTo find the area of triangle DEF, we havex1= 1, y1= 0x2= 1, y2= 2

x3= 0, y3= 1


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Area of DEF =

= 1 sq unit.

Question 24

24. Solve for x and y:

axby = 2abSolution:

---------(1)

axby = 2ab -----------(2)Multiply (1) with ab, we geta2xb2y = a2b + ab2-------(3)

Multiply (2) with a, we geta2xaby = 2a2b --- (4)

Solve (3) and (4)a2xb2y = a2b + ab2

a2xaby = 2a2b___________________________y(abb2) = ab2a2b

y = a

Put y = a in (2)ax = 2abab = abx = b

Hence, x = b and y = aOR

The sum of two numbers is 8. Determine the numbers if the sum of their reciprocals isSolution:Given sum of two numbers = 8

Sum of their reciprocals =Let the numbers be x and yx + y = 8 ..(1)

And

xy = 15

Substitute x value in equation (1)

y28y + 15 = 0y25y3y + 15 = 0

y(y5)3(y5) = 0

(y3) (y5) = 0

y = 3 or 5

Case (i) If y = 3,


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x = 5 and y = 3

Case (ii): If y = 5,

x = 3 and y = 5

Question 25

25. Draw a right triangle in which sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then

construct another triangle whose sides are times the corresponding sides of the first triangle.Solution:

Steps of construction:1. Draw a right triangle using given conditions. Consider the triangle as ABC in which BC = 8 cm and AB =

6 cm, and B = 902. Make any acute angle CBX below the line BC

3. Locate 4 points (greater of 3 and 4 in )B1, B2, B3, B4on BX such thatBB1= B1B2= B2B3= B3B44 . Join B4and C5. Through B3, draw a line parallel to B4C meeting BC in C.6. Again draw a line through C parallel to CA meeting BA in A on being produced. ABC is the required

triangle whole sides are times the sides of ABC.

26. A juice seller serves his customers using a glass as shown in figure 6. The inner diameter of thecylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces thecapacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actualcapacity. (Use = 3.14)Solution:


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Let x m be the horizontal distance and h be the height of the lamp post

(i) In ACB

Thus the horizontal distance between the building and the lamp post is

(ii) from EDB

From problem (i) x =

Thus

[By cross multiplication]

h = 6020h = 40 m

Thus height of the lamp post is 40 m30. Solve the following equation for x.

9x29(a + b)x + (2a2+ 5ab + 2b2) = 0Solution:Comparing the given equation with standard equation ax2+ bx + c = 0 we find a = 9, b = -9(a + b), c = 2a2+ 5ab+ 2b2


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or

or

or

ORIf (5) is a root of the quadratic equation 2x2+ px15 = 0 and the quadratic equation p(x2+ x) + k = 0 has equal

roots, then find thevalues of p and k.

Solution:

Given that 5 is a root of 2x2+ px15 = 0

2(5)2+ p(5)15 = 0

505p15 = 0

5p = 35

p = 7

Putting p = 7 in the second quadratic equation we get7x2+ 7x + k = 0

Given that above equation has equal roots and henceb24ac = 0494(7)k = 0 ( a = 7, b = 7, c = k)

28k = 49

1. Find the discriminant of the quadratic

equation .Solution:

Here

Discriminant = 10036 = 64

Question 2

2. If a, 2 are three consecutive terms of an A.P., then fine the value of a.Solution:

Since a, 2 are in AP, we have


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Question 5

5. Two coins are tossed simultaneously. Find the probability of getting exactly one head.Solution:When two coins are tossed simultaneously all the possible outcomes are (H,H),(H,T),(T,H)and (T,T) and thedesired outcomes of getting exactly one head are (H,T) and (T,H).

Thus the required probability =

Question 7

7. If 1 is a zero of the polynomial p(x) = ax23(a1)x1, then find the value of a.Solution:

Given that 1 is a zero of a polynomialTherefore, p(1) = 0

a(1)23 (a1) (1)1 = 0

a3(a1)1 = 0

a3a + 31 = 0

2a + 2 = 0

2a = 2

a = 1Hence value of a is 1

Question 10

10. Find the number of solutions of the following pair of linear equations:x + 2y8 = 02x + 4y = 16

Solution:Infinitely many solutions

[ .

Question 11

11. If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), find the value of x.Solution:Given points on circle are A (4, 3) and B (x, 5)Centre is at O (2, 3)

OA = OB ( Radii of same circle) ------- (1)Distance between two points (x1, y1) and (x2, y2)

is


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From equation (1),

4 = (x2)2+ 444 = (x2)2x2 = 0x = 2

Question 12

12. Which term of the A.P. 4, 12, 20, 28, will be 120 more than its 21stterm?Solution:Given A.P. is 4, 12, 20, 28, ..

Here a = 4, d = t2t1= 124 = 8nthterm of A.P. = tn= a + (n1)dPut n = 21t21= 4 + (211)8t21= 4 + (20)8t21= 164Let tmbe the term which is 120 more than 21

sttermtm= t21+ 120tm = 164 + 120a + (m1)d = 2844 + (m1)8 = 284 (m1)8 = 2844 = 280

m1 = 35m = 36Hence, the 36thterm of the A.P. is 120 more than its 21stterm.

Question 14

14. Find all the zeroes of the polynomial x3+ 3x22x6, if two

of its zeroes are and .Solution:Let P(x) = x3+ 3x22x6

Given two zeroes of the given polynomial

P(x) = x3+ 3x22x6 are and

is a factor of the given polynomialx22 is a factor of the given polynomial.Now, we apply the division algorithm to P(x) and x22


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P(x) = x3+ 3x22x6 = (x22) (x + 3)x + 3 is a factor of P(x)

3 is the other zero of the polynomial.

Question 17

17. Draw a right triangle in which sides (other than hypotenuse) are of lengths 8 cm and 6 cm. Then

construct another triangle whose sides are times the corresponding sides of the first triangle.

Solution:Steps of construction:

(1) Construct ABC such that BC = 8 cm, BCA = 90and CA = 6 cm(2) Draw a ray BX, making an acute angle with BC on the side opposite to BA(3) Mark points B1, B2, B3, B4on BX such that BB1= B1B2= B2B3= B3B4

(4) Join B4and C(5) Draw B3C || B4C(6) Draw CA || CA

Now ABC is the required triangle, whose sides are times to the corresponding sides of the ABC.

Question 18

18. Two dice are thrown simultaneously. What is the probability that(i) 5 will not come up on either of them?(ii) 5 will come up on at least one?(iii) 5 will come up at both dice?

Solution:Total number of possible outcomes of throwing two dice = 6 X6 = 36

(i) Number of outcomes that 5 will not come up on either of them = 25

i.e. [on both the dice, any one of the numbers 1, 2 3, 4, 6 appear]


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P [5 will not come up on either of them](ii) 5 will come up on at least one = 5 on the 1stdie and any one of {1, 2, 3, 4, 6} on the 2nddie (or) 5 on

the 2nddice andany one of {1, 2, 3, 4, 6} on the 1 stdice (or) 5 on both dice

= 5 + 5 + 1 = 11 ways.

P [5 will come up on either of them) =(iii) 5 will come up on both the dice in only one way

P [5 will come up at both dice]

Question 20

20. The sum of first six terms of an arithmetic progression is 42. The ratio of its 10thterm to its 30thterm is 1 :3. Calculate the first and the thirteenth term of the A.P.

Solution:Given sum of 1stsix terms of an A.P. = 42

And t10: t30= 1 : 3

Sum of n terms of AP,Put n = 6

42 = 3(2a + 5d)2a + 5d = 14 --- (1)t10: t30= 1 : 3

3a + 27d = a + 29d2a = 2da = d --- (2)Substituting (2) in (1), we get

2a + 5a = 147a = 14a = 21stterm = 2d = 2 [From (2)]t13= a + 12d= 2 + 12(2)t13= 26

Question 22

22. Find the ratio in which the point (x, 1) divides the line segment joining the points (3, 5) and

(2, 5). Also find the value of x.Solution:

Given (x, 1) divides the line segment joining the points (3, 5) and (2, 5)

By section formula,

5m + 5n = mn


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4m = 6n

m : n = 3 : 2

x = 0

Question 26

26. A juice seller serves his customers using a glass as shown in figure 6. The inner diameter of thecylindrical glass is 5 cm, but the bottom of the glass has a hemispherical portion raised which reduces thecapacity of the glass. If the height of the glass is 10 cm, find the apparent capacity of the glass and its actual

capacity. (Use = 3.14)

Solution:Since the inner diameter of the glass = 5 cm

Height = 10 cmApparent capacity of the glass = r2h

= 3.14 X2.5 X2.5 X10 cm3= 196.25 cm3

But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass

It is less by= 32.71 cm3

Actual capacity of the glass = apparent capacity of glassvolume of hemisphere= (196.2532.71) cm3= 163.54 cm3

OrA cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base

diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of(i) water displaced out of the cylindrical vessel

(ii) water left in the cylindrical vessel

[Take ]Solution:Diameter of cylindrical vessel = 10 cmr= radius of cylindrical vessel = 5 cmheight of cylindrical vessel = h = 10.5Volume of cylindrical vessel = r2h

= X5 X5 X10.5 cm3= 825 cm3

Also given that diameter of the cone = 7 cm

r = radius of cone =height of the cone = 6 cm

volume of the cone =


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= 77 cm3

Volume of water displaced by cone= volume of the cone = 76.93 cm3

Water left in cylindrical vessel

= volume of water in cylindervolume of water incone

= 825cm377 cm3= 748 cm3

Question 28

28. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the

ground. The angle of elevation of the balloon from the eyes of the girl at that instant is 60. After some time, the

angle of elevation reduces to 30. Find the distance travelled by the balloon during the interval.Solution:

Given height of girl, AB = 1.2 mHeight of balloon from ground, CG = 88.2 mCF = 88.21.2 = 87 mDE = 87 mIn right AFC,

In right AED,

Distance travelled by the balloon = CD = FECD = AEAF

CD = 100.46 m (nearly)Hence, the distance travelled by the balloon during the interval is 100.46 m.

Question 29

29. Solve the following equation for x.

9x

2

9(a + b)x + (2a

2

+ 5ab + 2b

2

) = 0Solution:


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Comparing the given equation with standard equation ax2+ bx + c = 0 we find a = 9, b = -9(a + b), c = 2a2+ 5ab+ 2b2

or

or

orOr

If (5) is a root of the quadratic equation 2x2+ px15 = 0 and the quadratic equation p(x2+ x) + k = 0

has equal roots, then find the values of p and k.Solution:

Given that 5 is a root of 2x2+ px15 = 0

2(5)2+ p(5)15 = 0

505p15 = 05p = 35p = 7Putting p = 7 in the second quadratic equation we get7x2+ 7x + k = 0Given that above equation has equal roots and hence

B2

4ac = 0

494(7)k = 0 ( a = 7, b = 7, c = k)28k = 49

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