Mntop100_que_and_sol_maths_class10_fjk.pdf

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MATHEMATICS MN Top 100 Questions

Questions Real Numbers 93 1. Without performing actual division, state whether the rational number will have a 600 terminating or non-terminating decimal expansion. 2. Find the HCF of 1488 and 37888 using Euclid’s division algorithm. 3. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3p or 3p + 1, where p is an integer.





4. Prove that 4 5  9 is an irrational number. 5. On a circular track of length 1080 m, Srikant is walking at a speed of 90 m/min, Venkat at 72 m/min and Bala at 60 m/min. If all of them start walking at 5:30 a.m. in the same direction from same starting point, then at what time will they all meet each other at the starting point?

Polynomials 6. If two of the zeroes of the polynomial p(x) = 3x4 + 7x3 – 7x2 – 21x – 6 are find the other zeroes of p(x).

3 and – 3 , then

7. Find the zeroes of the quadratic polynomial, p(x) = 6x2 + 18x and verify the relationship between the zeroes and the coefficients. 8. If  and  are the zeroes of the polynomial f (x) = x2 – 5x + k such that  –  = 1, then find the value of k.

Pair of Linear Equations in Two Variables 9. For what value of k will the system of equations x + 2y = 3 and 5x + ky = –7 have a unique solution? 10. A group of students are seated in rows in such a way that the number of students in each row is the same. If the number of students in each row is increased by 15, then number of rows decreases by 5. If the number of students in each row is increased by 6, then the number of rows decreases by 3. How many students were there in the group? 11. What is the value of k for which the pair of linear equations kx + 5y – (k –5) = 0 and 20x + ky – k = 0 has infinite many solutions? 12. Three men and eight women can complete a piece of work together in 10 hours. Eleven men and thirteen women can complete the same piece of work together in 4 hours. How long will one man alone and one woman alone take to complete the same work? 1

MN Top 100 Questions

MATHEMATICS

13. Solve the following pair of equations by algebraic method. 4 1   5  2x  3y   2x  3y  2 1  8  2x  3y   2x  3y 

Quadratic Equations 14. For what value(s) of k does the quadratic equation 25x2 –10kx + 49 = 0 have equal roots? 15. Find the roots of the equation 2x2 –7x + 3 = 0 using the method of completing the square. 16. For an excursion, Rs 12,900 is collected from a group of students such that same amount of money was contributed by each student. Had there been 14 students more, each of them would have contributed Rs 21 less. How much amount of money is contributed by each student? 17. Determine the nature of the roots of the quadratic equation 5 x 2  2 3 x  7  0 . 18. The total cost of a certain length of a piece of cloth is Rs 200. If the piece was 5 m longer and each metre of cloth costs Rs 2 less, the cost of the piece would have remained changed. How long is the piece and what is its original rate per metre? 19. Solve the given quadratic equation for x : 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0. 20. Solve for x :

1

 x  1 x  2 



1

2  , x  1, 2,3  x  2  x  3 3

Arithmetic Progressions 21. Find the 25th term of the AP 5,

5 5 , 0, , ... . 2 2

22. The sum of 8th and 17th term of an AP is 36. Find the sum of first 24 terms of the AP. 23. The students from five schools in a locality thought of planting trees to reduce air pollution. It was decided that the number of trees that each class of each school will plant will be the same as the class in which they are studying. For e.g., class I will plant one tree, class II will plant two trees and so on till class XII. (i) How many trees will be planted by the students? (ii) Write values reflected in the question. 24. For what value of k will the consecutive terms 3k + 2, 4k + 5 and 6k – 4 form an AP? 25. The ratio of the 21st term and 29th term of an arithmetic progression is 2 : 1. What is the 37th term of the arithmetic progression? 26. A thief runs with a uniform speed of 100 m/min. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/min in the first minute and increases his speed by 10 m/min every succeeding minute. After how many minutes the policeman will catch the thief. 2

MN Top 100 Questions

MATHEMATICS

27. If the ratio of sum of the first m and n terms of an AP is m2 : n2, show that the ratio of its mth and nth terms is (2m –1) : (2n –1). 28. Find the sum of all three digit natural numbers, which are multiples of 11. 29. If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3 (S20 – S10).

Triangles 30. In the given  ABC, DE || BC and

AD 5  . If AC = 18 cm, then what is the length of EC? AB 12 A E

D C

B

31. The given figure shows an equilateral triangle ABC drawn on side BC of square BCED and another equilateral triangle FDC drawn on diagonal DC of the square BCED. A

B

C

D

E F

What is the ratio of the area of  ABC to the area of  FDC? 32. The given figure shows a right-angled triangle PQR. S and T are the points on PQ and QR respectively such that QS =

1 1 PQ and QT = QR. 3 3

Prove that 18(PT2 + RS2) = 20PR2.

P

S Q

T

R

33. State and prove the basic proportionality theorem. 34. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. 35. In two similar triangles ABC and PQR, the corresponding medians AD and PS are in the ratio 2 : 3. If the area of  ABC is 20 cm2, then what is the area of  PQR? 3

MN Top 100 Questions

MATHEMATICS

36. In the given figure, AC || QR and AB || PR. D

P A

Q

C

B

R

If DP = 8 cm, AP = 4 cm and CR = 3.9 cm, then what is the length of PC? 37. In  ABC, BD  AC and AC2 – AB2 = BC2. Prove that BD2 = AD × CD. 38. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. 39. State and prove the Pythagoras Theorem.

Coordinate Geometry 40. If the points (–1, –2), (p, q) and (4, 0) are collinear, then prove that 2p – 5q = 8. 41. The vertices of  ABC are A (2, 0), B (–1, 7) and C (4, 5). BX is the median of  ABC corresponding to the side AC. If a point P on BX divides BX in the ratio 2 : 5, then find the coordinates of the point P. 42. In what ratio does the y-axis divide the line joining the points (–4, 5) and (8, 3). Also, find the point of intersection. 43. Show that the triangle formed by vertices (–1, 1), (7, 5) and (9, 1) is a right triangle. Also, find the length of the altitude corresponding to the hypotenuse. 44. What are the coordinates of the point dividing the line joining the points (3, 4) and (4, 5) in the ratio 3 : 2? 45. What are the coordinates of the point (on the y-axis) that is equidistant from points A (6, 4) and B (3, –2)? 46. If point P(a, b) is equidistant from the points A (3, 2), B (–5, –6) and C (5, 0), then find the area of the triangle formed by the points P, B and C. 47. The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0,–3). The origin is the mid-point of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus. 48. Find the coordinates of a point P, which lies on the line segment joining the points A (–2, –2) and B (2, –4), such that AP =

3 AB . 7

49. In the given figure, ABC is a triangle coordinates of whose vertex A are (0, –1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the mid-point of BC, find the areas of  ABC and  DEF. 4

MN Top 100 Questions

MATHEMATICS A (0, –1)

E (0, 1)

(1, 0) D

B

C

F

Introduction to Trigonometry 50. If x = a sin  + b cos  and y = a cos  – b sin  , then what is the value of x2 + y2?

1 and tan(3B – A) = 1, where 0  A  90° and 0  B  90°, then what is 2 the value of sin(4A – 3B)?

51. If cos  2 A  B  

1 , then find the value of 52. If sinθ  3

2

2

1   1    tanθ     tanθ   . cosθ   cosθ  

53. Prove that

1 1   cosecθ  secθ . cosθ  sinθ  1 cosθ  sinθ- 1

54. If sinθ 

3 , then find the value of 3  3cot 2θ . 5

tan 2 28  cosec 2 59 cos 2 59  sin 59 cos 31  cos 31 cosec 59  55. Evaluate: sec 2 31  cot 2 62 sec 2 62  cot 2 28 56. If sin   cos   2 , then find value of 57. Prove that 58. Prove that:

sec 2   cosec 2 .

tan  cot    1  sec  cosec . 1  cot  1  tan  cos  cot   cosec  1 1  sin 

59. Prove that sec2 cosec 2  2  cot 2   tan 2 

Some Applications of Trigonometry 60. The angle of elevation of a jet fighter plane from a point P on the ground is 60°. The jet 3 3 km. After a flight of 15 seconds, the 2 angle of elevation changes to 30°. At what speed is the jet fighter plane flying?

fighter plane is flying at the constant height of

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MN Top 100 Questions

MATHEMATICS

61. From the top of a 10 m high building, the angle of depression of the foot of a tower is 30° and the angle of elevation of top of the tower from the foot of the building is 60°. Find the distance between the tops of the building and the tower. [Use 7 = 2.64] 62. A pole casts a shadow of length 2 3 m on the ground, when the sun’s elevation is 60°.Find the height of the pole. 63. 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. 64. The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 30° and 60° respectively. Find (a) the difference between the heights of the light-house and the building. (b) the distance between the light-house and the building. 65. The angles of elevation of the top of a tower from two points, distant a and b from its base in the same straight line, are complementary. Show that the height of the tower is ab .

Circles 66. If tangents XY and YZ from a point Y to a circle with centre O are inclined to each other at an angle of 64°, then find  YOZ. 67. From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ. 68. Prove that the lengths of tangents drawn from an external point to a circle are equal. 69. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. 70. Prove that a parallelogram circumscribing a circle is a rhombus. 71. In fig, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that  DOE = 90°. A

D

l

O C B

E

m

Construction 72. Draw  ABC with sides AB = 5 cm, BC = 6 cm and  ABC = 60°. Then, construct a triangle whose sides are

3 of the sides of  ABC. 4 6

MN Top 100 Questions

MATHEMATICS

73. Draw a circle of radius 3 cm. From a point P, 7 cm away from its centre draw two tangents to the circle. Measure the length of each tangent. 74. Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.

Areas Related to Circles 75. In the given figure, AD is the diameter of a circle. The length of AD is 42 cm. Three semicircles of equal radius are drawn inside this circle taking AB, BC and CD respectively as 22   diameters.  Use    7  

A

B

D

C

Find the area of the shaded region. 76. The given figure shows a circle with centre O. The points A, B and C are on the circle such that AB = 14 cm and  ACB = 45°. C

O

A

B X

Find the area of the major segment ACBA. 77. In the given figure, ABCD is a square of side 14 cm. Semi-circles are drawn with each side 22   of square as diameter. Find the area of the shaded region.  use    7   A B

D

C

7

MN Top 100 Questions

MATHEMATICS

78. In the given figure, ABDC is a quadrant of a circle of radius 28 cm and a semi circle BEC is drawn with BC as diameter. Find the area of the shaded region.  Use   22   7 

B

E D

A

C

79. In the given figure, AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. OB is the diameter of the smaller circle. If OA = 7 cm, find the 22 area of the shaded region.  Use    7  C

B

O

A

D

Surface Areas and Volumes 80. From a cone of radius 40 cm, a cone of radius 16 cm and height 12 cm is cut from the top so that the circular bases of these cones are parallel. Find the volume of the frustum so obtained. 81. Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe. 82. From a solid metallic cylinder of height 24 cm, a cone of same height and same base (as cylinder) is taken out. The density of the metal is 10 gm per cm3. If the mass of the remaining portion of the cylinder is 24.64 kg, then find the cost required to paint the remaining portion at a rate of Rs 3 per 440 cm2.

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MN Top 100 Questions

MATHEMATICS

83. A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 cm high embankment. Find the width of the embankment. 84. 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 the each bottle, if 10% liquid is wasted in this transfer. 85. Milk in a container, which is in the form of a frustum of a cone of height 35 cm and the radii of whose lower and upper circular ends are 15 cm and 30 cm respectively, is to be distributed in a camp for flood victims. If this milk is available at the rate of Rs 45 per litre and 1155 litres of milk is needed daily for a camp, find how many such containers of milk are needed for a camp and what cost will it put on the donor agency for this. What value is indicated through this by the donor agency? 86. In the figure given below is a right circular cone of height 30 cm. A small cone is cut off from the top by a plane parallel to the base. If the volume of the small cone is

1 of the volume of 27

cone, find at what height above the base is the section made.

30 cm

87. If the radius of the base of a right circular cylinder is halved, keeping the height the same, then find the ratio of the volume of the cylinder thus obtained to the volume of original cylinder. 88. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic 22 sphere. Find the diameter of the sphere and hence find its surface area.  Use    7 

Statistics 89. Find the mean and mode of the following data. Using empirical relationship find the median of the data.

Class Interval 0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 Frequency

6

10

9

18

7

9

MN Top 100 Questions

MATHEMATICS

90. Calculate the mode of the following data.

Class Interval

Frequency

More than equal to 10

68

More than equal to 20

64

More than equal to 30

58

More than equal to 40

45

More than equal to 50

25

More than equal to 60

12

More than equal to 70

4

91. Find median of the following data using graphical method:

Class Interval 50 - 60 Frequency

60 - 70

70 - 80

80 - 90

90 - 100

5

9

12

6

3

92. The given table shows the data of 230 observations whose median is 46. If the median class is 40 - 50, then what are the values of a and b? Class Interval Frequency 10 - 20

12

20 - 30

30

30 - 40

a

40 - 50

65

50 - 60

b

60 - 70

25

70 - 80

18

93. If the mode of the following distribution is 68, then find the missing frequency corresponding to the class interval 60 - 80.

Class Interval

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

Frequency

7

14

8

-

3

94. The following distribution gives the height of 100 students of a class. Height of Students (in cm) 120 - 130 130 - 140 140 - 150 150 - 160 160 - 170 170 - 180 Number of Students

5

15

25

30

20

5

Convert above frequency distribution into more than type distribution and find the median height of the students by drawing an ogive.

10

MN Top 100 Questions

MATHEMATICS

Probability 95. In a fruit basket, the oranges and bananas are in the ratio 2 : 3. 15% of oranges are rotten and 10% of bananas are rotten. If a fruit is selected at random, then what is the probability that it will be rotten? 96. 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. 97. There are twenty pens. Some of them are blue and some are red in colour. A pen is selected at random. The probability of getting a blue pen is one-third of the probability of getting a red pen. How many blue pens are there? 98. In a cupboard, there are eight shelves and every second shelf of the cupboard has 4 Math 3 books. A book is chosen at random. The probability of it being a non-Math book is . How 7 many books are there in the cupboard? 99. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is (a) a card of spade or an ace (b) a black king (c) neither a jack nor a king (c) either a king or a queen 100.A dice is rolled twice. Find the probability that (a) 5 will not come up either time. (b) 5 will come up exactly one time.

11

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MATHEMATICS MN Top 100 Questions

Real Numbers

1.

93 3  31 31  3  3 2 2 600 2  3  5 2 5

It can be observed that denominator of the given number is of the form 2n × 5m, where n and m are non-negative integers. Thus,

93 will have a terminating decimal expansion. 600

2. Here, 37888 > 1488. Applying Euclid’s division lemma, we obtain 37888 = 1488 × 25 + 688 Again applying Euclid’s division lemma for 1488 and 688, we obtain 1488 = 688 × 2 + 112 Similarly, 688 = 112 × 6 + 16 112 = 16 × 7 + 0 Thus, the HCF of 1488 and 37888 is 16. 3. Let a be a positive integer. Then, a will be of the form 3m, 3m + 1 or 3m + 2. Case I: a = 3m On squaring both sides, we get a2 = (3m)2 = 9m2 a2 = 3(3m2) = 3p, where p = 3m2 Case II: a = 3m + 1 On squaring both sides, we get a2 = (3m + 1)2 = 9m2 + 6m + 1 a2 = 3(3m2 + 2m) + 1 = 3p + 1, where p = 3m2 + 2m Case III: a = 3m + 2 On squaring both sides, we get a2 = (3m + 2)2 = 9m2 + 12m + 4 a2 = 3(3m2 + 4m + 1) + 1 = 3p + 1, where p = 3m2 + 4m + 1 Hence, a is the form of 3p or 3p + 1. 1

MN Top 100 Questions



MATHEMATICS



4. Let 4 5  9 be a rational number equal to

a , where a and b are positive co-prime numbers. b

a b 1a   5    9 4b 

 4 5 9 

It can be seen that RHS of the above expression is a rational number whereas LHS is an irrational number. This is a contradiction.





 4 5  9 cannot be a rational number..





Thus, 4 5  9 is an irrational number.. 5. Length of the track = 1080 m Speed of Srikant = 90 m/min

1080 m Time taken by Srikant to walk a complete round of track  90 m/min  12 min 1080 m Similarly, Time taken by Venkat to walk a complete round of track  72 m/min  15 min 1080 m Time taken by Bala to walk a complete round of track  60 m/min  18 min Therefore, the time after which they all meet at the starting point is equal to the LCM of 12, 15 and 18. LCM of 12, 15 and 18 = 180 Therefore, they all will meet each other at the starting point 180 minutes after 5:30 a.m. i.e. at 8:30 a.m.

Polynomials 6. It is given that

3 and – 3 are zeroes of p (x). Therefore,

 x  3  x  3   x

2

 3 is a factor of p(x).

p(x) can be divided by x2 – 3 as:

2

MN Top 100 Questions

MATHEMATICS

3x 2  7 x  2 x 2  3 3x 4  7 x3  7 x 2  21x  6 3x 4 

 9 x2  7 x 3  2 x 2  21x  6 7 x3 

 21x  2x2  6 2x2  6   0

 3x 4  7 x 3  7 x 2  21x  6   x 2  3 3x 2  7 x  2  3x2 + 7x + 2 can be factorised as: 3x2 + 7x + 2 = 3x2 + 6x + x + 2 [6x + x = 7x, 6x × x = 3x2 × 2] = 3x(x + 2) + 1(x + 2) = (x + 2)(3x + 1) The zeroes of (x + 2)(3x + 1) are x = –2 and x  

1 3

1 Thus, the other zeroes of p(x) are –2 and  . 3

7. The given quadratic polynomial is p(x) = 6x2 + 18x p(x) = 0  6x2 + 18x = 0  6x(x + 3) = 0 x(x + 3) = 0 x = 0 or (x + 3) = 0 x = 0 or x = –3 Therefore, 0 and –3 are the zeroes of the given quadratic polynomial. Sum of zeroes = 0 + (–3) = –3, product of zeroes = 0 × (–3) = 0 For a quadratic polynomial p(x) = ax2 + bx + c, sum of zeroes  

b c , product of zeroes = a a

For the quadratic polynomial p(x) = 6x2 + 18x, a = 6, b = 18, and c = 0 Sum of zeroes  

b 18    3 a 6

3

MN Top 100 Questions Product of zeroes 

MATHEMATICS

c 0  0 a 6

Thus, the relationship between the zeroes and the coefficients of the given quadratic polynomial is verified. b c , product of zeroes  a a k   5  Therefore, for the given polynomial,  +   = 5 and   = k. 1 1   5

8. For any quadratic polynomial p(x) = ax2 + bx + c, sum of zeroes 

   1 2  6

 Given   Adding the two equations 

  3 Putting  = 3 in  +  = 5: 3+=5   =5–3=2 k =  = 3 × 2 = 6

Pair of Linear Equations in Two Variables 9. The given system of equations can be written as x + 2y – 3 = 0 5x + ky + 7 = 0 Here, a1 = 1; b1 = 2; c1 = –3; a2 = 5; b2 = k; c2 = 7 For the pair of linear equations to have a unique solution,

a1 b1  a2 b2 1 2  5 k  k  10 Thus, the given system of equations will have a unique solution if k  10. 

10. Let the number of rows be x and number of students in each row be y.  Total number of students = xy According to given information:

 x  5  y  15   xy  xy  5 y  15 x  75  xy  15 x  5 y  75  3 x  y  15

... 1

Also,  x  3 y  6   xy  xy  3 y  6 x  18  xy 4

MN Top 100 Questions

MATHEMATICS

 6 x  3 y  18  2x  y  6

...

 2

Subtracting (2) from (1), we get

3 x  y  15 2x  y  6    x9 Substituting x = 9 in (1), we get 3 × 9 – y = 15  27 – y = 15 y = 12 xy = 12 × 9 = 108 Thus, 108 students were there in the group. 11. The given pair of linear equations is kx + 5y – (k – 5) = 0 and 20x + ky – k = 0. It is known that the pair of equations a1 x  b1 y  c1  0 and a2 x  b2 y  c2  0 has infinite

a1 b1 c1 many solutions, if a  b  c 2 2 2 For the given pair of equations, a1 = k, b1 = 5, c1 = – (k – 5), a2 = 20, b2 = k, c2 = – k

k 5   k  5   20 k k Equating the first two ratios, we have k2 = 100  k = ±10 Equating the last two ratios, we have

5k  k 2  5k  k 2  10k  0  k  k  10   0  k  0 or k  10 So, k = 10 satisfies both the conditions. Thus, the value of k is 10. 12. Let one man alone take x hours to complete the work and one woman alone take y hours to complete the same work. According to the given information:

3 8 1   x y 10

... 1

11 13 1   x y 4

...  2 

5

MN Top 100 Questions

MATHEMATICS

1 1  u and  v . y x Then, (1) and (2) can be re-written as Let

1 10 1 11u  13v  4 3u  8v 

...  3 ...  4 

Multiplying (3) by 11 and (4) by 3 and then subtracting, we get 11 10 3 33u  39v  4    33u  88v 

11 3  10 4 44  30  49v  40 14  49v  40 1 v 140  y  140 Substituting y = 140 in (1), we get 49v 

3 8 1   x 140 10 3 1 8   x 10 140 3 6  x 140 x  70 Thus, the time taken by a man alone to complete the work is 70 hours whereas the time taken by a woman alone to complete the work is 140 hours.

13. The given pair of linear equations are 4 1   5  2x  3 y   2x  3 y 2 1  8  2x  3 y   2x  3 y  Let

1 1  u and  v , then the given equations become  2x  3 y  2x  3 y 

6

MN Top 100 Questions 4u  v  5 2u  v  8

MATHEMATICS 1  2

Adding (1) and (2), we obtain 6u = 3 u 

1 2

On putting the value of u in equation (2), we obtain

1 2   v  8 2 1– v = 8 v = 1 – 8 v = – 7 Therefore, we obtain 1 1   2x  3 y  2  2x  3 y  2

 3

1  7  2x  3 y   – 7(2x – 3y) = 1  – 14x + 21y =1

(4) On multiplying equation (3) by 7 and adding to equation (4), we obtain

14 x  21 y  14 14 x  21 y  1 42 y  15 15 5  y  42 14 Putting the value of y in equation (3), we obtain 5 2x  3   2  14  15 2x   2 14 15 2x  2  14 28  15 2x  14 13 x 28 13 5 Thus, x  and y  is the required solution. 28 14 7

MN Top 100 Questions

MATHEMATICS

Quadratic Equations 14. The given equation is 25x2 – 10kx + 49 = 0. This equation has equal roots, if its discriminant D is 0. Here, a = 25, b = – 10k, c = 49

D0 2

  10k   4  25  49  0  100k 2  100  49  0  k 2  49  0  k 2  49  k  7 15. The equation is 2x2 – 7x + 3 = 0. 7 3  x2  x   0 2 2 7 3  x2  2  x      0 4 2 2

49 7 Adding and subtracting    to the LHS of the equation:  4  16 2 2  2 7 7  7 3  x  2 x           0 4  4    4  2  2

7  49 3  x    0 4  16 2  2

7  49 3  x    4  16 2  2

7  49  24  x   4 16  2

2

7  25  5   x    4  16  4   7  5   x      4   4 7 5 x  4 4 7 5 7 5  x   or x   4 4 4 4 12 2  x  or x  4 4 1  x  3or x  2 1 Thus, the roots of the equation 2x2 – 7x + 3 = 0 are 3and . 2

8

MN Top 100 Questions

MATHEMATICS

16. Let the amount of money contributed by each student be Rs x.  Number of students 

12, 900 x

According to the given information:  12,900   14   x  21  12,900   x  2,70,900  12,900   14 x  294  12,900 x  14 x 2  294 x  2, 70, 900  0  x 2  21x  19,350  0  x 2  150 x  129 x  19,350  0  x  x  150   129  x  150   0   x  129  x  150   0   x  129   0 or  x  150   0  x  129 or 150 Since the amount of money is non-negative, x = 150 Thus, an amount of Rs 150 was contributed by each student. 17. The given quadratic equation is 5 x 2  2 3x  7  0 . Here, a = 5, b = 2 3 , c = –7



 Discriminant, D  b 2  4ac  2 3



2

 4  5   7   12  140  152  0

Thus, the given quadratic equation has two distinct real roots. 18. Let the length of the piece of cloth be x m. Total cost of the piece of cloth = Rs 200 Then, cost per metre = Rs

200 x

New length = (x + 5) m Since, the cost of piece of cloth remains unchanged.  New cost per metre = Rs

200 x5

According to the question,

200 200  2 x x5 200  x  5  200 x  2 x  x  5 

200 x  1000  200 x 2 x2  5x 9

MN Top 100 Questions

MATHEMATICS

 2  x 2  5 x   1000  x 2  5 x  500  0  x 2  25 x  20 x  500  0  x  x  25   20  x  25   0   x  20  x  25   0  x  20 or x  25 As x cannot be negative, so x is 20.  Cost per metre = Rs

200 = Rs 10 20

Hence, the length of the piece of cloth is 20 m and original rate is Rs 10 per metre. 19. 9 x 2  9  a  b  x   2a 2  5ab  2b 2   0

 9 x 2  3 2a  b    a  2b  x   2a 2  4ab  ab  2b 2   0  9 x 2  3 2a  b    a  2b  x  2a  a  2b   b  a  2b   0  9 x 2  3 2a  b    a  2b  x   2a  b  a  2b   0  9 x 2  3  2a  b  x  3  a  2b  x   2a  b  a  2b   0  3 x 3x   2a  b    a  2b  3x   2a  b   0  3 x   a  2b 3x   2a  b   0  3 x   a  2b   0 or 3 x   2a  b   0  3 x  a  2b or 3x  2a  b a  2b 2a  b x or x  3 3 20.

1



1

2  , x  1, 2,3  x  2  x  3 3

 x  1 x  2   x  3   x  1  2  x  1 x  2  x  3 3 

2x  4 2   x  1 x  2  x  3 3



x2 1   x  1 x  2  x  3 3

 3   x  1 x  3  3  x2  4x  3  x2  4x  0  x  x  4  0  x  0, x  4

10

MN Top 100 Questions

MATHEMATICS

Arithmetic Progressions 21. Here, a  5 5 5 5 d   5   5 2 2 2 Now, an  a   n  1 d  a25  5   25  1   a25  5  24 

5 2

5 2

 a25  5  60  a25  55

Thus, the 25th term of the given AP is 55. 22. Let a be the first term and d be the common difference of the given AP. It is given that, 8th term + 17th term = 36 It is known that the nth term, an, of an AP with first term a and common difference d is given by, an = a + (n – 1) d a8 + a17 = 36   a  7 d    a  16d   36 ... 1

 2a  23d  36

 Sum of the first 24 terms of the given AP

24  2a   24  1 d  2   12  2a  23d  

 12  36

 Using 1 

 432 23. (a) Students of Class I of five schools will plant 1 × 5 = 5 trees Students of Class II of five schools will plant 2 × 5 = 10 trees Students of Class III of five schools will plant 3 × 5 = 15 trees . . . . . . . . . . . . . . . . . . Students of Class XII of five schools will plant 12 × 5 = 60 trees Number of trees planted by the students are 5 + 10 + 15 + …+ 60.

. .

Here, a = 5, d = 5, n = 12 We know that, S n 

n  2a   n  1 d  2

11

MN Top 100 Questions  Sn 

MATHEMATICS

12  2  5  12  1  5  390 2 

Thus, 390 trees will be planted by the students. (b) The values reflected in the question are teamwork, sharing, responsibility, commitment towards the society and importance of conservation of nature. 24. The given terms 3k + 2, 4k + 5 and 6k – 4 are in AP.

  4k  5    3k  2    6k  4    4k  5   k  3  2k  9  k  12 Hence, the value of k for which the given terms are in an AP is 12. 25. It is known that the nth term of an arithmetic progression is given by an = a + (n – 1)d a21 = a + (21 – 1)d = a + 20d a29 = a + (29 – 1)d = a + 28d a37 = a + (37 – 1)d = a + 36d It is given that: a21 : a29 = 2 : 1 a  20d 2  a  28d 1  a  20d  2a  56d



 2a  56d  a  20d  0  a  36d  0  a37  0

Thus, the 37th term of the arithmetic progression is 0. 26. Suppose the policeman catches the thief after t minutes. Uniform speed of the thief = 100 m/min  Distance covered by thief in (t + 1) min = 100 m/min × (t + 1) min = 100(t + 1) m Distance covered by policeman in t minutes = Sum of t terms of an AP with first term 100 and common difference 10 t  2  100   t  1  10  m 2  t 100  5  t  1  

 t  5t  95  5t 2  95t When the policeman catches the thief, 5t 2  95t  100  t  1  5t 2  95t  100t  100  5t 2  5t  100  0  t 2  t  20  0

12

MN Top 100 Questions

MATHEMATICS

 t 2  5t  4t  20  0  t  t  5  4  t  5   0   t  5  t  4   0  t  5 or t  4  t = 5 (As t cannot be negative)

Thus, the policeman catches the thief after 5 minutes. 27. Let the first term and the common difference of the AP be a and d, respectively. Therefore, Sum of the first m terms of the AP, S m  Sum of the first n terms of the AP, S n 

m  2a   m  1 d  2

n  2a   n  1 d  2

It is given that

m S m 2  2a   m  1 d  m 2   2 n Sn n  2a   n  1 d  2  2a   m  1 d  m    2a   n  1 d  n  2an  mnd  nd  2am  nmd  md  2an  2am  nd  md  2a  n  m   d  n  m   2a  d Now,

Tm a   m  1 d  Tn a   n  1 d 

Tm a   m  1  2 a  Tn a   n  1  2a

Tm a 1  2  m  1   Tn a 1  2  n  1  T 2m  1  m  Tn 2n  1 

Hence, the ratio of the mth term to the nth term is (2m – 1) : (2n – 1). 28. The smallest and the largest three digit natural numbers, which are divisible by 11 are 110 and 990 respectively. So, the sequence of three digit numbers which are divisible by 11 are 110, 121, 132, …, 990. Clearly, it is an A.P. with first term, a = 110 and common difference, d = 11. Let there be n terms in the sequence. So, an = 990 13

MN Top 100 Questions

MATHEMATICS

 a   n  1 d  990  110   n  111  990  110  11n  11  990  11n  99  990  11n  990  99  11n  891  n  81

Now, required sum 

n  2a   n  1 d  2

81  2  110   81  1 11 2 81   220  880  2 81   1100 2  44550 Hence, the sum of all three digit numbers which are multiples of 11 is 44550. 

29. We know that the sum of the nth term is given by: n  2a   n  1 d  2 Here: Sn 

a = first term d = common difference n = number of terms in an A.P. We need to prove the following:

S30  3  S 20  S10  Considering R.H.S., we get: 3  S20  S10  10  20   3  2a   20  1 d   2a  10  1 d  2 2   3 10  2a  19d   5  2a  9d    3  20a  190d  10a  45d   3 10a  145d   15  2a  29d  Considering L.H.S., we get:

30  2a   30  1 d  2   15  2a  29d 

S30 

L.H.S. = R.H.S. Hence, proved. 14

MN Top 100 Questions

MATHEMATICS

Triangles 30. In  ABC, DE || BC. Using corollory of basic proportionality theorem, we have AD AE  AB AC 5 18  EC   12 18 5  18  EC   18  7.5 12  EC  18  7.5  10.5 cm Thus, the length of EC is 10.5 cm. 31. BCED is a square and DC is its diagonal which divides it into two congruent right triangles. Applying Pythagoras theorem in right  DBC, we obtain DC2 = BC2 + BD2 DC2 = BC2 + BC2 DC2 = 2BC2

(BD = BC) ...(1)

Since  ABC and  FDC are equilateral triangles, each of their angles measures 60°.  A = F = 60° B = D = 60°

 ABC   FDC (By AA similarity) We know that ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. 

ar  ABC  ar  FDC 



BC 2 DC 2

BC 2  From 1  2BC 2 1  2 Thus, the ratio of the area of  ABC to the area of  FDC is 1 : 2. 

32.  PQT is right-angled at Q. Applying Pythagoras theorem in  PQT, we have

PT 2  PQ 2  QT 2 2

1   PT 2  PQ2   QR  3  1  PT 2  PQ2  QR 2 9  QRS is right-angled at Q.

... 1

Applying Pythagoras theorem in  QRS, we have 15

MN Top 100 Questions

MATHEMATICS

RS2  QS2  QR 2 2

1   RS   PQ   QR 2 3  1  RS2  PQ 2  QR 2 9  PQR is right-angled at Q. 2

...  2 

Applying Pythagoras theorem in  PQR, we have PR2 = PQ2 + QR2 ...(3) Adding (1) and (2), we get

1   1  PT 2  RS2   PQ2  QR 2    PQ 2  QR 2  9   9  10 10  PT 2  RS2  PQ 2  QR 2 9 9 10  PT 2  RS2  PR 2  Using equation  3  9 10  18  PT 2  RS2   18  PR 2 9  18  PT 2  RS2   20 PR 2 33. Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. We can prove this theorem as follows: Suppose PQR is any triangle. ST is a line parallel to QR. Firstly, let us join QT and RS and then draw SV and TU perpendicular to PR and PQ respectively. To prove:

P V U S

Q

T

R

PS PT = SQ TR

1 Now, ar (  PST)   PS  TU ...(1) 2 1 Similarly, ar (  QST)   SQ  TU ...(2) 2 1 Also, ar (  PST)   PT  SV ...(3) 2 1 ar (  STR)   TR  SV ...(4) 2 On dividing equation (1) by (2), we obtain

1 ar  PST  2  PS  TU PS    ar  QST  1  SQ  TU SQ 2

...  5  16

MN Top 100 Questions

MATHEMATICS

Dividing (3) by (4), we obtain 1 ar  PST  2  PT  SV PT   ar  STR  1  TR  SV TR 2

...  6 

Observe that  QST and  STR are on the same base ST and lie between the same parallels ST and QR. Therefore, ar (  QST) = ar (  STR)

...(7)

From (5), (6), and (7), we obtain

PS PT  SQ TR Thus, the line ST divides the side PQ and PR in the same ratio. 34. Consider a  ABC in which AC2 = AB2 + BC2. We have to prove that B = 90°. Let us construct  PQR right-angled at Q such that PQ = AB and QR = BC. P

A

B

C

Q

R

From  PQR, PR2 = PQ2 + QR2 (By Pythagoras Theorem, as Q = 90°)  PR2 = AB2 + BC2 …(1) (By construction) However, AC2 = AB2 + BC2 …(2) (Given) From (1) and (2), we obtain AC = PR …(3) Now, in  ABC and  PQR, we obtain AB = PQ (By construction) BC = QR (By construction) AC = PR [From (3)] Therefore, by SSS congruency criterion,  ABC   PQR By CPCT, B = Q

However, Q = 90° (By construction) B = 90° Hence, in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

17

MN Top 100 Questions

MATHEMATICS

35. We know that the areas of two similar triangles are in the ratio of the squares of the corresponding medians.



ar  ABC   AD    ar  PQR   PS 

20 cm 2  2    ar  PQR   3 

2

2

20  9  45cm 2 4 Thus, the area of  PQR is 45 cm2.  ar  PQR  

36. In  DAB, AB || PC. Using basic proportionality theorem, we have DP DC  PA CB 8 cm DC   4 cm CB 2

DC CB

... 1

In  DQB, AC || QB. Using basic proportionality theorem, we have

DA DC  AQ CB DP  PA DC   AQ CB 8cm  4 cm DC  AQ CB 12 cm DC   ...  2  AQ CB From (1) and (2), we have 

12cm 2 AQ 12 cm  AQ   6cm 2 In  PQR, AC || QR. Using basic proportionality theorem, we have

PA PC  AQ CR 4cm PC   6 cm 3.9 cm 4  PC   3.9 cm  2.6 cm 6 Thus, the length of PC is 2.6 cm.

18

MN Top 100 Questions

MATHEMATICS B

37. Given: In  ABC, BD  AC and AC2 – AB2 = BC2. To prove: BD2 = AD × CD Proof: In  ABC, AC2  AB2  BC 2  AC 2  AB2  BC 2

... 1

A D  Converse of Pythagoras theorem 

 ABC  90

C

In right  ABD, .....(2)

(Pythagoras theorem)

.....(3) BC2  BD 2  CD 2 Adding (2) and (3), we get

(Pythagoras theorem)

AB2  BD2  AD 2 In right  BCD,

AB2  BC 2  2BD 2  AD 2  CD 2  AC 2  2BD 2  AD 2  CD 2

 Using 1 

2

  AD  CD   2BD 2  AD 2  CD 2  AD 2  CD 2  2AD  CD  2BD 2  AD 2  CD 2  BD 2  AD  CD 38.

D A

B

X

C

E

Y

F

Let  ABC and  DEF be such that  ABC   DEF.. 2

ar  ABC 

2

 AB   BC   CA   To prove:      ar  DEF   DE   EF   FD 

2

Construction: Draw AXBC and DYEF Proof: ar (  ABC) = ar (  DEF) =



1 × BC × AX 2

1 × EF × DY 2

ar  ABC  BC  AX  ar  DEF  EF  DY

... 1

19

MN Top 100 Questions

MATHEMATICS

In  ABX and  DEY::  ABC   DEF}

B = E {

X = Y = 90°   ABX   DEY {By AA similarity criterion}



AX AB  DY DE

...  2 

It is given that:  ABC   DEF 

AB BC CA   DE EF FD

...  3

Using (1) and (2), we get 

ar  ABC  BC  AB  ar  DEF  EF  DE 

BC BC  EF EF

 BC     EF  Using (3), we have ar  ABC 

2

 Using  3 

2

2

 AB   BC   CA        ar  DEF   DE   EF   FD 

2

Thus, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. 39. Statement of Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides of the triangle. The proof of Pythagoras Theorem is as follows: Let PQR be a triangle, right-angled at P.

P

Q

S

R

Draw PS  QR Now, we know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on the both sides of the perpendicular are similar to the whole triangle and to each other.   QSP   QPR

Therefore, QS  QP QP QR

(Since the sides of similar triangles are proportional)

QS × QR = QP2

…(1)

Also, we have  PSR   QPR

20

MN Top 100 Questions Therefore,

MATHEMATICS

RS RP  RP RQ

(Since the sides of similar triangles are proportional)

 RS × RQ = RP2 …(2)

Adding (1) and (2), we get QS × QR + RS × RQ = RP2 + QP2  QR × (QS + RS) = RP2 + QP2  QR × QR = RP2 + QP2  QR2 = RP2 + QP2 Thus, the square of the hypotenuse is equal to the sum of squares of the other two sides of the triangle.

Coordinate Geometry 40. The given points are collinear; therefore, the area of the triangle formed by these points is zero.  x1  y2  y3   x2  y3  y1   x3  y1  y2   0  1 q  0   p  0  2   4  2  q   0   q  2 p  8  4q  0  2 p  5q  8

41. Since BX is the median of  ABC, X is the mid-point of the side AC. Therefore, by mid-point formula, coordinates

A(2, 0)

 24 05  5 , of point X are     3,  2   2  2 Let the coordinates of point P be (a, b). Then, by section formula, we obtain   2  3  5  1 a , b  ,    25   

 5 2    57  1 40  2    ,  25  7 7   

X 5 P 2 B(–1, 7)

C(4, 5)

 1 40  Thus, the coordinates of point P are  , . 7 7  42. Let the y-axis divide the line segment joining A (–4, 5) and B (8, 3) in the ratio 1 :  . Therefore, by section formula, the point of intersection is given by

 1 8   4   1 3  5  ,   1  1     8  4 3  5   ,   1  1   The x-coordinate of any point on the y-axis is 0. 21

MN Top 100 Questions

MATHEMATICS

8  4 0 1   8  4  0



2 Thus, y-axis divides AB in the ratio 1 : 2.

 3  5  2   13   Point of intersection =  0,    0,  3   3  Thus, the point of intersection of y-axis with the line joining the points (–4, 5) and (8, 3)  13  is  0,  .  3 43. Let the triangle be ABC such that its vertices are A (–1, 1), B (7, 5) and C (9, 1) The lengths of the sides of  ABC can be calculated as: 2

2

2

2

AB 

 7  1   5  1

BC 

 9  7   1  5 

CA 

 9  1  1  1

2

2

 64  16  80 units  4  16  20 units  100 units

It can be seen that AB2 + BC2 = CA2. Therefore,  ABC is right-angled at B and AC is the hypotenuse. Let BD be the altitude corresponding to AC. C Now, Area of  ABC =

1 1 × AB × BC = × AC × BD 2 2 20 units

D

1 1 10 units   80 units  20 units  10 units  BD 2 2 1600  BD  units  4 units B A 80 units 10 Thus, the length of the altitude corresponding to the hypotenuse is 4 units. 44. Let P(x, y) be the point dividing the line joining the points A(3, 4) and B(4, 5) in the ratio 3 : 2. By section formula, the coordinates of P are given by  3  4  2  3 3  5  2  4   18 23  ,   ,  3 2   5 5   3 2

 18 23  Thus, the coordinates of the required point are  ,  . 5 5  45. Let the required point on the y-axis be R (0, p). Point R (0, p) is equidistant from points A and B.  RA = RB



2

 6  0   4  p 

2



2

 3  0    2  p 

2

22

MN Top 100 Questions

MATHEMATICS

 36  16  p 2  8 p  9  4  p 2  4 p  12 p  39 39 13  p  12 4  13  Thus, the coordinates of the required point are  0,  .  4 46. It is given that point P is equidistant from the points A, B and C. PA = PB = PC PA2 = PB2 (a – 3)2 + (b – 2)2 = (a + 5)2 + (b + 6)2 a2 + b2 + 13 – 6a – 4b = a2 + b2 + 61 + 10a + 12b  16a + 16b + 48 = 0 a + b + 3 = 0 ...(1) PA2 = PC2  (a – 3)2 + (b – 2)2 = (a – 5)2 + (b – 0)2 a2 + b2 + 13 – 6a – 4b = a2 + b2 + 25 – 10a  4a – 4b = 12 a – b = 3 ...(2) Solving (1) and (2), we get a = 0 and b = – 3 Therefore, the coordinates of point P are (0, – 3). It is known that area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by

1  x1  y2  y3   x2  y3  y1   x3  y1  y2  2 1 Area  PBC   0  6  0   (5)  0  3  5  3  6  2 1   15  15  2 0 Thus, the area formed by the points P, B, and C is zero i.e., they are collinear. 47. Let the coordinates of B be (0, y). If origin is the midpoint of BC, then y   3 0 2  y 3

Thus, the coordinates of B are (0, 3). If triangle ABC is an equilateral triangle and the origin is the midpoint of the base, then A must lie on x-axis. Let the coordinates of A be (x, 0). BC 

2

 0  0    3  3

2

 62  6 units

23

MN Top 100 Questions AB = 6 units

MATHEMATICS

[  ABC is an equilateral triangle.]

Y

In right triangle ABO, B

AB2  OB2  OA 2  62  32  OA 2  36  9  OA 2

X’

 OA 2  36  9

D

X

A

O

 OA 2  27  OA  27

C(0, –3)

 OA  3 3



 



So, the coordinates of A are 3 3, 0 or 3 3, 0 .

Y’

Since BACD is a rhombus, so the diagonals BC and AD bisect each other at 90°. Therefore, the Point D must lie on x-axis and opposite to point A.

  the coordinates of A are  3 3, 0  , then the coordinates of D will be  3

 3, 0  .



Thus, if the coordinates of A are 3 3, 0 , then the coordinates of D will be 3 3, 0 or if

48. It is given that, AP 

3 AB , where A, P and B are three points on line segment AB. 7

AB 7  AP 3 AB 7  1   1 AP 3 AB  AP 7  3   AP 3 PB 4   AP 3 

Thus, AP : PB = 3 : 4 It is given that, the coordinates of points A and B are (– 2, – 2) and (2, – 4). Using section formula,  3  2  4   2  3   4   4   2   Coordinates of P are  ,  3 4 3 4    6  8 12  8   2 20   ,    ,  7   7 7   7

 2 20  Hence, the coordinates of point P are   ,   . 7   7 49. Let the coordinates of B and C be

 x2 , y2  and  x3 , y3  , respectively.. D is the midpoint of AB. 24

MN Top 100 Questions

MATHEMATICS A (0, –1)

So,

x  0 y 1 1, 0   2 , 2  2   2 x y 1  1  2 and 0  2 2 2  x2  2 and y2  1 Thus, the coordinates of B are (2, 1). Similarly, E is the midpoint of AC. So, x  0 y 1  0,1   3 , 3  2   2 x y 1  0  3 and 1  3 2 2  x3  0 and y3  3

E (0, 1)

(1, 0) D

B

F

C

Thus, the coordinates of C are (0, 3).

 2  0 1 3  , Also, F is the midpoint of BC. So, its coordinates are    1, 2  2   2 Now, 1  x1  y2  y3   x2  y3  y1   x3  y1  y2   2 Thus, the area of  ABC is 1  0 1  3  2  3  1  0  1  1  2 1  8 2 = 4 square units Area of a triangle 

And the area of  DEF is 1 11  2   0  2  0   1 0  1  2 1    2  2 = 1 square units (Taking the numerical value, as the area cannot be negative)

Introduction to Trigonometry 50. It is given that: x = a sin+ b cos and y = a cos– b sin 2

 x 2  y 2   a sin   b cos     a cos   b sin  

2

  a 2 sin 2   b 2 cos 2   2ab sin  cos     a 2 cos2   b 2 sin 2   2ab sin  cos    a 2  sin 2   cos 2    b 2  sin 2   cos     a2 1  b2 1

sin 2   cos2   1 Thus, the value of x2 + y2 is a2 + b2. 25

MN Top 100 Questions

MATHEMATICS

1 2  cos  2 A  B   cos 60

51. cos  2 A  B  

 2 A  B  60 ... 1

 B  2 A  60

tan(3B – A) = 1 tan(3B – A) = tan 45° 3B – A = 45°  3(2A – 60°) – A = 45° [Using (1)]  6A – 180° – A = 45°  5A = 180° + 45° = 225°  A

225  45 5

Substituting the value of A in (1), we get B = 2 × 45° – 60º = 90º – 60º = 30º  sin(4A – 3B) = sin(4 × 45° – 3 × 30º) = sin(180° – 90º) = sin 90° =1 Thus, the value of the expression sin(4A – 3B) is 1. 2

1   1  52.  tan     tan    cos    cos    2

  tan   sec     tan   sec  

2

2

 tan 2   sec2   2 tan  sec   tan 2   sec 2   2 tan  sec   2  tan 2   sec 2    sin 2  1   2   2 2  cos  cos    1  sin 2    2  2  cos    1  sin 2    2  2  1  sin    1  1   2 3   1 1   3 4  2  4 2

1    sin    3 

26

MN Top 100 Questions 53.

MATHEMATICS

1 1  cos   sin   1 cos   sin   1 cos   sin   1  cos   sin   1   cos   sin   1 cos   sin   1 

2cos   2sin  2

 cos   sin    1 2  cos   sin   

cos 2   sin    2sin  cos   1 2  cos   sin    2sin  cos  cos  sin    sin  cos  sin  cos   cosec  sec  54. 3  3cot 2   3 1  cot 2    3cos ec2  

3 sin 2  3  3    5

2

5

2 2 2 55. tan 28  cos ec 59  cos 59  sin 59 cos 31  cos 31cosec59 sec 2 31  cot 2 62 sec 2 62  cot  28 cos2 59  sin 59 cos  90  59   cos 31 cos ec  90  31  tan 2 28  co sec2 59   sec 2  90  59   cot 2  90  28  sec 2 62  cot 2  90  62 

tan 2 28  cosec 2 59 cos2 59  sin 59 sin 59  cos 31 sec 31  cos ec2 59  tan 2 28 sec 2 62  tan 2 62 1 cos2 59  sin 2 59  cos 31 cos 31  1 2 2 sec 62  tan 62 1  1  1 1  1 2 3 

27

MN Top 100 Questions 56.

MATHEMATICS

sec 2   cosec2 

1 1  2 2 cos  sin 

sin 2   cos 2   sin 2  cos 2  1 sin  cos 2  1  sin  cos  2  2sin  cos  

  

2

2 sin   cos   2sin  cos   1 2 2

2

 sin   cos  

2

1

2

  2

2

1

sin   cos  2 

2 2 1 2 

57.

tan  cot   1  cot  1  tan  1 tan    tan  1 1  tan  1 tan  tan 2  1   tan   1 tan   tan   1 tan 3   1  tan   tan   1

 tan   1  tan 2   tan   1  tan   tan   1 tan 2   tan   1 tan   tan   1  cot  sin  cos   1  cos  sin  sin 2   cos 2   1 sin  cos  

28

MN Top 100 Questions

MATHEMATICS

1 sin cos  1 sec cosec  1

cos  cot  1  sin  cos  cos  cos 2  sin    1  sin  sin  1  sin  

58. LHS 

 

1  sin   cos 2   sin  1  sin   1  sin   cos 2  1  sin   sin  1  sin 2  

cos 2  1  sin   sin  .cos 2  1  sin   sin  1 sin    sin  sin   cosec  1  RHS 

 cos 2   sin 2   1

Hence proved 2 2 2 59. sec  cosec   2  cot 

 sec2  cosec2  1  1  cot 2   sec2  cosec2  1  1  cot 2    sec2  cosec2  1  cosec 2

1  cot 2 A  cosec 2 A

 cosec 2  sec 2   1  1  cosec 2 tan 2   1 1 sin 2    1 sin 2  cos 2  1  1 cos 2 

1  tan 2 A  sec 2 A 1    cosec A  sin A   

 sec 2   1

1   sec A  cos A   

 tan 2 

1  tan 2 A  sec 2 A

29

MN Top 100 Questions

MATHEMATICS

Some Applications of Trigonometry B

60. The given information can be represented diagrammatically as: Here, B and D are the positions of the jet fighter plane, P is the observation point, and BPA = 60° and DPC = 30°, where A and C are points on the ground such that BA  PC and DC  PC.

3 3 AB  CD  km 2

D

3 3 km 2 60° 30° P

A

C

In  ABP:

AB  tan 60 AP 3 3   km 2     3 AP 3 3 1  3  AP     km  km 2 3  2 In  DCP: CD  tan 30 PC 3 3   km 2  1    PC 3 3 3  9  PC    3  km  km  2  2   Distance covered by the jet fighter in 15 seconds = BD = AC

= PC – AP 

9 3 km  km 2 2

= 3 km = 3000 m Thus, speed at which the jet fighter plane is flying =

Distance 3000 m  = 200 m/s Time 15s

30

MN Top 100 Questions

MATHEMATICS

61. The given information can be represented diagrammatically as: Here, AB and CD represent the building and the tower respectively. According to the given information: AB = 10 m, CBD = 60°, and EAD = 30° so that ADB = 30°

C

In  ABD:

AB  tan 30 BD 10 m 1   BD 3

A

E

30°

 BD  10 3 m AE  10 3 m In  BCD:

60°

CD  tan 60 BD CD   3 10 3 m



30°

B

D



 CD  10 3  3 m  30 m  CE = CD – DE = CD – AB = 30 m – 10 m = 20 m

Applying Pythagoras theorem in  ACE: AC2 = AE2 + CE2  AC2 = (10 3 m)2 + (20 m)2  AC2 = (300 + 400) m2 = 700 m2  AC = 10 7 m  AC = 10 × 2.64 m  AC = 26.4 m

Thus, the distance between the tops of the building and the tower is 26.4 m. 62. Let AB be the height of the pole and BC be the shadow of the pole. Given:

A

BC  2 3 m ACB = 60°

In  ABC, AB BC AB  3 2 3  AB  6 m Hence, the height of the pole is 6 m. tan 60 

60° C

2 3m

B

31

MN Top 100 Questions

MATHEMATICS

63. Let BE be the surface of the lake. Suppose C be the position of the cloud and C’ be its reflection in the lake. Then, AB = 20 m A Let CD = x m. Then, 20 m CE = C’E = (x + 20) m B In  ACD,

C x 30° 60°

D 20 m E

CD AD 1 x   3 AD tan 30 

 AD  x 3

(x + 20)

... 1

Also,

C’

CD sin 30  AC 1 x   2 AC  AC  2 x

...  2 

Now, in  ADC’, CD AD  x  20   20

tan 60   3  AD 

AD  x  40  3

... 3

From (1) and (3), we get x 3

 x  40 

3  3 x  x  40  2 x  40  x  20 From (2), we have

AC = 2 × 20 = 40 m Thus, the distance of the cloud from A is 40 m. 64.

Let AB be the building and CD be the light house. Suppose the height of the light house be h m. Given: AB = 60 m,  EAD = 30° and  CAE = 60°. CE = AB = 60 m  DE = CD – CE = (h – 60) m 32

MN Top 100 Questions

MATHEMATICS

In  EAD,

D

DE AE 1 h  60   AE 3 tan 30 

30°

A

E

60°

 AE  3  h  60 

... 1

h

60 m

In  ACE, CE AE 60  3 AE 60  AE  ...  2  3 From (1) and (2), we get tan 30 

B

C

60 3  3  h  60   60 3  h  60  

 3h  180  60  3h  240  h  80  Difference between the height of light house and building = CD – AB = 80 m – 60 m = 20 m Distance between the light house and building = BC = AE 

60  20 3 m 3

65. Let the two discussed points be A and B. We have BC = a and AC = b Here, CD represents the tower. Let  DBC =    DAC = 90°– 

(  the angles are given to be complementary) D

In  DBC, tan  

DC DC  BC a

1

In  DAC,

tan  90    

DC DC  AC b

DC  2  tan  90     cot   b Multiplying (1) and (2), we obtain  cot  

90 – A

B

a

C

b

33

MN Top 100 Questions

MATHEMATICS

DC DC  a b 2 DC 1 ab 2  DC  ab tan   cot  

 DC  ab

Thus, the height of the tower is ab . Hence, proved

Circles 66. We are given that XY and YZ are two tangents such that  XYZ = 64°. Join OX, OY, and OZ. We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

X

64°

O

Y

 XO  XY and OZ  YZ

  OXY =  OZY = 90° In quadrilateral XYZO,

Z

 OXY +  XYZ +  OZY +  XOZ = 360° (Sum of angles of a quadrilateral is 360°)  90° + 64° + 90° +  XOZ = 360°  XOZ = 116° Now, in  XYO and  ZYO,  OXY =  OZY (Each equal to 90°) XY = YZ (Tangents drawn from an external point to a circle are equal) XO = OZ (Radii of the same circle)   XYO  ZYO

(By SAS congruency criterion)

  XOY =  ZOY Also,  XOZ =  XOY +  ZOY = 116° 16°   XOY =  ZOY =

116  58 2

Thus,  YOZ = 58° 67. TP and TQ are tangents drawn from an external point T to the circle. O is the centre of the circle. Suppose OT intersect PQ at point R. In  OPT and  OQT,, OP = OQ (Radii of the circle) TP = TQ (Lengths of tangents drawn from an external point to a circle are equal.) 34

MN Top 100 Questions OT = OT

MATHEMATICS

(Common sides) P

O

T

R

Q

  OPT   OQT (By SSS congruence rule)

So,  PTO =  QTO (By CPCT)

.....(1)

Now, in  PRT and  QRT, T, TP = TQ (Lengths of tangents drawn from an external point to a circle are equal.)  PTO =  QTO O RT = RT   PRT T  QRT

So, PR = QR

[From (1)] (Common sides) (By SAS congruence rule)

.....(2)

(By CPCT)

And,  PRT =  QRT (By CPCT) Now,  PRT +  QRT = 180°

(Linear pair)

 2  PRT = 180°   PRT = 90°   PRT T =  QRT = 90 .....(3) From (2) and (3), we can conclude that OT is the right bisector of the line segment PQ. Q

68. We have a circle with centre O. A point P lies outside the circle. Also, PQ and PR are two tangents to the circle. P To prove: PQ = PR Construction: Join OP, OQ and OR. Proof:

O

R

 OQP =  ORP = 90º (Tangent is perpendicular to the radius through the point of contact.) OQ = OR (Radii) OP = OP (Common)   OQP  ORP

 PQ = PR Hence proved.

(By the RHS congruency criterion) (Corresponding parts of congruent triangles)

69. Given: A circle C (0, r) and a tangent l at point A To prove: OA  l

35

MN Top 100 Questions

MATHEMATICS

Construction: Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle at C. Proof: We know that among all line segments O joining the point O to a point on l, the perpendicular is shortest to l. OA = OC (Radius of the same circle) C Now, A B OB = OC + BC  OB > OC  OB > OA  OA < OB B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l. Here,

l

OA  l Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact. D

R

C

70. Given: ABCD be a parallelogram circumscribing a circle with centre O. To prove: ABCD is a rhombus. We know that the tangents drawn to a circle from an exterior point are equal in length.  AP = AS, BP = BQ, CR = CQ and DR = DS.

S O Q

AP + BP + CR + DR = AS + BQ + CQ + DS A P B (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)  AB + CD = AD + BC or 2AB = 2BC (since AB = DC and AD = BC)  AB = BC = DC = AD.

Therefore, ABCD is a rhombus. Hence, proved. 71. Given: l and m at are two parallel tangents to the circle with centre O touching the circle at A and B respectively. DE is a tangent at the point C, which intersects l at D and m at E. To prove:  DOE = 90° Construction: Join OC. Proof:

A

D

l

O C B

E

m

In  ODA and  ODC, OA = OC (Radii of the same circle) 36

MN Top 100 Questions

MATHEMATICS

AD = DC (Length of tangents drawn from an external point to a circle are equal) DO = OD (Common side)  ODA  ODC (SSS congruence criterion)   DOA =  COD … (1) (C.P.C.T)

Similarly,  OEB   OEC  EOB =  COE … (2) AOB is a diameter of the circle. Hence, it is a straight line.   DOA +  COD +  COE +  EOB = 180º

From (1) and (2), we have 2  COD + 2  COE = 180º   COD +  COE = 90º   DOE = 90° Hence, proved.

Construction 72. Firstly, draw  ABC with AB = 5 cm, BC = 6 cm and ABC = 60°. 3 of the sides of  ABC: 4 (a) Draw a ray BD making an acute angle with BC on the side opposite to vertex A.

Following are the steps to construct a triangle whose sides are

(b) Mark four points B1, B2, B3, and B4 on BD such that BB1 = B1B2 = B2B3 = B3B4. (c) Join B4 with C, and then draw a line through B3 parallel to B4C intersecting BC at C’. (d) Draw a line through C’parallel to AC. Let it intersect AB at A’

A A’ 5 cm

60° B

6 cm

C’

C

B1 B2 B3 B4 D

Now,  A’BC’ is the required triangle whose sides are

3 of the sides of  ABC. 4

37

MN Top 100 Questions

MATHEMATICS

73. Steps of Construction (a) Taking any point O as centre, draw a circle of 3 cm radius. Locate a point P, 7 cm away Q from O. Join OP. (b) Bisect OP. Let M be the mid-point of OP. 7 cm (c) Taking M as centre and MO as radius, O M draw a circle. 3 cm (d) Let this circle intersect the previous circle at points Q and R. R (e) Join PQ and PR. Thus, PQ and PR are the required tangents. The length of each of the tangents PQ and PR is approximately 6.32 cm. 74. Follow the given steps to construct the tangents. Step 1: Draw a circle of radius 4 cm, with O as centre. Step 2: Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at A. Step 3: Draw a radius OB, making an angle of 120° (180° – 60º ) with OA. Step 4: Draw a perpendicular to OB at B. Let these

P

P

B 120° O

A

perpendiculars intersect at P. Here, PA and PB are two tangents drawn to the circle inclined at an angle of 60° to each other. Justification The construction can be justified by proving that  APB = 60°.  OAP = 90° (Construction)  OBP = 90° (Construction)  AOB = 120° (Construction) We know that the sum of all interior angles of a quadrilateral is 360°.  OAP +  AOB +  OBP +  APB = 360°  90° + 120° + 90° +  APB = 360°   APB = 60° This justifies the construction.

Areas Related to Circles 75. It is given that the length of AD is 42 cm.  AB = BC = CD =

42 cm  14 cm 3

38

MN Top 100 Questions

MATHEMATICS

Area of shaded region = Area of semi-circle with diameter AD + Area of semi-circle with diameter AB and CD – Area of semi-circle with diameter BC. 2 2 2 2 1   42   14   14   14     π    π    π    π    cm 2 2   2   2 2  2  

1  π  21 21  π  7  7 cm 2 2 1  22 22     21 21   7  7  cm 2 2 7 7  22   63  7  cm 2 2  11 70  cm 2 

 770cm 2 Thus, the area of the shaded region is 770 cm2. 76. Join OA, OB. It is known that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

C

  AOB = 2  ACB = 90°

O

Therefore,  OAB is a right-angled triangle. Let the radius of the circle be r.  OA = OB = r

A

Using Pythagoras theorem in  OAB: AB2 = OA2 + OB2

B X

 (14 cm)2 = r2 + r2  2r2 = 196 cm2  r2 = 98 cm2  r  98 cm  7 2 cm  OA = OB = 7 2 cm

Area of major segment ACBA = Area of circle – Area of minor segment AXBA Area of circle =  r 2 

22  7 2 cm 7





2



22  98 cm 2  308 cm 2 7

Area of minor segment AXBA = Area of sector OAXBO – Area of  OAB 90 22 1   7 2 cm  7 2 cm   7 2 cm  7 2 cm 360 7 2 1  =   22 14  49  cm 2 4  

39

MN Top 100 Questions

MATHEMATICS

 1  πr 2  OA×OB 360 2 2   77  49  cm 

 28cm 2 Thus, area of major segment ACBA = (308 – 28) cm2 = 280 cm2

1  14  14  2   π ×7 2 2 22  196  49  7  196  154

A

P

B

I

S

IV

O

II Q

14 cm

77. Let the four shaded regions be I, II, III and IV and the centres of the semicircles be P, Q, R and S, as shown in the figure. It is given that the side of the square is 14 cm. Now, Area of region I + Area of region III = Area of the square – Areas of the semicircles with centres S and Q.

III D

R

C

Radius of the semicircle  7 cm 

 42cm 2 Similarly, Area of region II + Area of region IV = Area of the square – Areas of the semicircles with centres P and R. 1  14  14  2   π ×7 2 2 22  196  49  7  196  154

Radius of the semicircle  7 cm 

 42cm 2 Thus, Area of the shaded region = Area of region I + Area of region III + Area of region II + Area of region IV = 42 cm2 + 42 cm2 = 84 cm2 78. Given: Radius (r) of the circle = AB = AC = 28 cm Area of quadrant ABDC: 1   π  r2 4  1 22      28  28  cm 2 4 7   616cm 2 40

MN Top 100 Questions

MATHEMATICS

Area of  ABC: 1   AC  AB 2 1     28  28  cm 2 2  2  392 cm

Area of segment BDC = Area of quadrant ABDC – Area of  ABC = (616 – 392) cm2 = 224 cm2 ....(i) Now, in right-angled triangle BAC: BC2 = BA2 + AC2 (By Pythagoras theorem)  BC2 = (282 + 282) cm2  BC2 = 28 × 28 × 2 cm2  BC = 28 2 cm Since BC is the diameter of the semi-circle BEC, its radius will be as follows:

28 2 cm  14 2 cm 2 Area of semi-circle BEC: 1   π  r2 2  1 22      14 2  14 2  cm 2 2 7   1 22      14  14  2  cm 2 2 7  2  616 cm Area of the shaded portion = Area of semi-circle BEC – Area of segment BDC = 616 cm2 – 224 cm2 = 392 cm2 79. AB and CD are the diameters of a circle with centre O.  OA = OB = OC = OD = 7 cm (Radius of the circle) Area of the shaded region = Area of the circle with diameter OB + (Area of the semi-circle ACDA – Area of  ACD) 2

1 2 7 1   π      π   7    CD  OA  2 2 2  22 49 1 22 1   cm 2    49 cm 2   14 cm  7 cm 7 4 2 7 2

77 2 cm  77 cm 2  49 cm 2 2  66.5cm 2 Thus, the area of the shaded region is 66.5 cm2. 

41

MN Top 100 Questions

MATHEMATICS

Surface Areas and Volumes 80. The given information can be represented with the help of a figure as It is given that the bases of the cones are parallel to each other.

A 12 cm D

In  ADE and  ABC,

E 16 cm

 ADE =  ABC (Corresponding angles)  AED =  ACB (Corresponding angles)

30 cm

  ADE ~ ABC (By AA similarity criterion)

AD DE  AB BC 12 cm 16cm   AB 40cm

B



C 40 cm

 40  12   AB    cm  30 cm  16   BD = AB –AD = 30 cm –12 cm = 18 cm In the frustum of the cone, r1 = 40 cm, r2 = 16 cm, and h = 18 cm The volume of the frustum of cone is given by

1 πh  r12  r22  r1r2  . 3

1 V  π  18  402  16 2  40  16  cm3  14976 π cm3 3 Thus, the volume of the frustum of the cone is 14976 π cm3.

81. Increase in the level of water in half an hour, h = 3.15 m = 315 cm Radius of the water tank, r = 40 cm Volume of water that falls in the tank in half an hour = π r2h = π × 402 × 315 = 504000 π cm3 Rate of flow of water = 2.52 km/h Length of water column in half an hour =

2.52 = 1.26 km = 126000 cm 2

Let the internal diameter of the cylindrical pipe be d. 2

d  3 Volume of the water that flows through the pipe in half an hour = π     126000 cm 2  

We know Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour 2

d  π     126000  504000π 2

42

MN Top 100 Questions

MATHEMATICS

2

d   4 2  d 2  16  d  4cm Thus, the internal diameter of the pipe is 4 cm. 82. Height of the cone and cylinder = 24 cm Let radii of the cone and cylinder be r cm. Therefore, Volume of the remaining portion = Volume of cylinder – Volume of cone 1 22  22     r 2  24    r 2  24  cm3 3 7 7  352 2  r cm3 7 Now, volume of the remaining portion

Mass Density 24.64 kg 24640 3   cm  2464 cm3 3 10g per cm 10 352 2  r  2464 7 2464  7  r2   49 352 r 7 Therefore,radii of the cone and cylinder = 7 cm Now, slant height of the cone, 

l  r 2  h2 

2

 7 cm    24 cm 

2

 49  576 cm  25 cm

Surface area of the remaining portion = Lateral surface area of cone + Lateral surface area of cylinder + Area of circular base 22 22  22     7  25  2   7  24   7  7  cm 2 7 7 7  2  550  1056  154 cm  1760 cm 2

Therefore, cost of painting the remaining portion

3   = Rs 1760  440   = Rs 12 43

MN Top 100 Questions

MATHEMATICS

83. Depth (h1) of the well = 14 m Radius (r1) of the circular end of the well =

40 cm

4 m=2m 2

Height (h2) of embankment = 40 cm = 0.4 m Let the width of embankment be x. From the figure, it can be observed that the embankment will be cylindrical in shape having outer radius (r2) as (2 + x) m and inner radius (r1) as 2 m. 4m Volume of earth dug from the well = Volume of earth used to form embankment

14 m

πr12 h1  π  r22  r12  h2 2 2    2  14    2  x   22  0.4   x  x  4 4  4  14  10 2  x  4 x  140  0

 x 2  14 x  10 x  140  0   x  14  x  10   0  x  10

 Because x cannot be negative 

Therefore, the width of the embankment will be 10 m. 84. Radius of the hemispherical bowl, r  Volume of the hemispherical bowl =

36 = 18 cm 2 2 3 πr 3

2   π 18  18  18 3 = 3888 π cm3

Volume of the liquid transferred = 3888 π – 10% of 3888 π = 3888 π – 388.8 π = 3499.2 π cm3 Radius of each cylindrical bottle, R =

6 = 3 cm 2

Let the height of each cylindrical bottle be h. Volume of each cylindrical bottle = π R2h = π×3×3×h  Total volume of 72 such cylindrical bottles = π × 3 × 3 × h × 72

= 648 π h cm3 According to the question, Total volume of 72 such cylindrical bottles = Volume of the liquid transferred 44

MN Top 100 Questions

MATHEMATICS

 648πh  3499.2π 3499.2 h 648  h  5.4cm Hence, the height of each cylindrical bottle is 5.4 cm. 85. Height of the container, h = 35 cm Bottom radius, r1 = 15 cm Top radius, r2 = 30 cm Volume of the container 1  π  r 21  r 22  r1r2  h 3 1 22   152  302  15  30   35 3 7  57750 cm3  57.75 L

1L  1000 cm  3

Amount of milk needed daily for the camp = 1155 L  Number of containers required 

1155 L  20 57.75 L

Thus, 20 such containers of milk are needed for a camp. Also, milk is available at the rate of Rs 45 per litre. Cost of milk = Rs 45 × 1155 = Rs 51, 975 Charity, sympathy, humanity are some of the values shown by the donor agency. 86. Let the radius and height of the bigger cone be R and H, respectively. Given: H = 30 cm Let the radius and height of the smaller cone be r and h, respectively. Now, in  AFE and  AGC, AEF = ACG (Corresponding angles) AFE = AGC (90° each)

  AFE ~  AGC AF FE   AG GC h r   H R

(AA similarity) A h

... 1

It is given that Volume of the smaller cone =

1 × Volume of the bigger cone 27

D

F r

E

H

B

G

R

C

45

MN Top 100 Questions

MATHEMATICS

1 2 1 1 πr h   πR 2 H 3 27 3 2

1 r h      R  H 27 2

1  h h      H  H 27

 Using 1 

3

1  h     H  27 h 1   H 3 1 1 h   H   30  10 cm 3 3 Now, FG = AG – AF = 30 cm – 10 cm = 20 cm Hence, the section is made 20 cm above the base. 87. Let the radius and height of the original cylinder be r and h respectively. 2 Volume of the original cylinder = r h

According to the question, radius of the new cylinder is halved keeping the height same.  Radius of the new cylinder 

r 2

Also, height of the new cylinder = h 2

πr 2 h r  Volume of the new cylinder  π   h  4  2

πr 2 h Therefore, Volume of the new cylinder  42  1 Volume of the original cylinder πr h 4 88. Diameter of the cone = 3.5 cm  Radius of the cone, r 

3.5 cm 2

Height of the cone, h = 3 cm Now, Volume of each cone =

1 2 πr h 3

2

1 22  3.5   × ×  ×3 3 7  2  = 9.625 cm3  Volume of 504 cone = 504 × 9.625

= 4851 cm3 46

MN Top 100 Questions

MATHEMATICS

Let the radius of metallic sphere be R. It is given that cones are melted and recast into a metallic sphere.  Volume of the sphere = Volume of 504 cones 4 3 πR  4851 3 4 22    R 3  4851 3 7 4851 7  3  R3  4  22 3  R  1157.625 

 R  10.5 cm Diameter of the sphere = 2R = 2 × 10.5 cm = 21 cm  Surface area of the sphere  4πR 2  4 

22 2  10.5  1386 cm 2 7

Statistics 89. To find the mean, we construct the following table: Class marks, Class marks, xi fixi

Class Interval

Frequency

0 - 100

6

50

300

100 - 200

10

150

1500

200 - 300

9

250

2250

300 - 400

18

350

6300

400 - 500

7

450

3150

Total

50

13500

We know that:

   f

Mean x 

f i xi i



13500  270 50

In the above table, the class interval 300 - 400 has the maximum frequency i.e., 18. Therefore, modal class = 300 - 400 Here, l = Lower limit of modal class = 300 h = Class size = 100 f1 = Frequency of modal class = 18 f0 = Frequency of class preceding modal class = 9 f2 = Frequency of class succeeding modal class = 7 Now, 47

MN Top 100 Questions

MATHEMATICS

 f1  f 0  Mode  l   h  2 f1  f 0  f 2   18  9   300    100  2  18  9  7  900  300  20  345 We know that:

3 Median = Mode + 2 Mean = 345 + 2 × 270 = 345 + 540 = 885  Median =

885  295 3

90. The frequency distribution table for the given data can be drawn as:

Class Interval Frequency 10 - 20

4

20 - 30

6

30 - 40

13

40 - 50

20

50 - 60

13

60 - 70

8

70 - 80

4

Total

68

Here, the maximum class frequency is 20, which lies in the class interval 40 - 50. Therefore, the modal class is 40 - 50. l = 40 h = 10 f1 = 20 f0 = 13 f2 = 13  f1  f 0  Mode  l   h  2 f1  f 0  f 2  20  13    40     10  2  20  13  13   7   40     10  40  26 

70 14  40  5  45 Thus, the mode of the given data is 45.  40 

48

MN Top 100 Questions

MATHEMATICS

91. Firstly, the given data is converted into less than equal to type.

Class Cumulative frequency 

Less than Less than Less than Less than Less than 60 80 90 100 70 3

8

17

35

N 35   17.5 2 2

The points on the graph are plotted and joined to obtain the ogive. A perpendicular is drawn from point 17.5 on y-axis to the ogive and from that point to the xaxis. The point where the perpendicular line intersects the x-axis is the median of the data. Y 50

Cumulative frequency

45 40

Less than ogive

35 30 25 20 15 10 5 Median = 80 X O

60 70 80 90 100 Upper class tlimit

It can be observed that the perpendicular line intersects the x-axis at the point 80. Thus, the median is 80.14(approx 80). 92. The cumulative frequency table for the given data can be drawn as: Class Interval

Frequency

c.f.

10 - 20

12

12

20 - 30

30

42

30 - 40

a

42 + a

40 - 50

65

107 + a

50 - 60

b

107 + a + b

60 - 70

25

132 + a + b

70 - 80

18

150 + a + b

Σ fi = 230 49

MN Top 100 Questions

f

i

MATHEMATICS

 230

Now, 150 + a + b = 230 a + b = 230 – 150 a + b= 80 …(1)

f

i

 n  230

n 230    115 2 2 It is given that median is 46 and median class is 40 - 50. Here, l = 40 f = 65 h = 10 c.f. = 42 + a n   2  cf  Median  l    h  f    115   42  a    46  40     10 65  

6

115  42  a  10

65  65  6   73  a  10  390  730  10a  10a  730  390 340 a 10  a  34

From (1) we have b = 80 – a = 80 – 34 = 46 Thus, the values of a and b are 34 and 46 respectively. 93. Let the missing frequency corresponding to the class interval 60 - 80 be f. Since the mode is 68, which lies in the class interval 60 - 80, the modal class of the given data is 60 - 80. From the table, we have Lower class limit, l = 60 Class size, h = 20 Frequency (f1) of modal class = f Frequency (f0) of class preceding the modal class = 8 Frequency (f2) of class succeeding the modal class = 3

50

MN Top 100 Questions

MATHEMATICS

 f1  f 0  Mode  l   h  2 f1  f 0  f 2   f 8   60     20  2 f  8  3 20 f  160  68  60  2 f  11 20 f  160  8 2 f  11  20 f  160  16 f  88  4 f  160  88  72  f  18 Thus, the missing frequency of the class interval 60 - 80 is 18.

94. We can write the given distribution in more than type as follows:

Height (in cm)

Number of students (cf)

More than or equal to 120

100

More than or equal to 130

100 – 5 = 95

More than or equal to 140

95 – 15 = 80

More than or equal to 150

80 – 25 = 55

More than or equal to 160

55 – 30 = 25

More than or equal to 170

25 – 20 = 5

(Lower limit, Corresponding cumulative frequency) is the ordered pair of the points to be plotted. Thus, the points are (120, 100), (130, 95), (140, 80), (150, 55), (160, 25), (170, 5). Now, by choosing an appropriate scale (along x-axis, 1 unit = 10 cm height and along y-axis, 1 unit = 10 students), we can draw the required ogive. Y

Now, n  100  50 2 2

(approx). Therefore, the median height of the students is 152.

100 90 Cumulative frequency

We mark the point 50 on vertical line and then draw a horizontal line through this point. Let the horizontal line intersect the ogive at point A. Now, through A, draw a vertical line that intersects the x-axis at point B. Point B represents the point 152

80 70 60 A

50 40 30 20 10

B O

X

110 120 130 140 150 160 170 180 190 Lower class limit

51

MN Top 100 Questions

MATHEMATICS

Probability 95. Let the number of oranges and bananas in the basket be 2a and 3a respectively.  Total number of fruits = 2a + 3a = 5a  Number of rotten oranges 

Number of rotten bananas 

15 30a 3a  2a   100 100 10

10 3a  3a  100 10

 Total number of rotten fruits 

3 3 3a a a 10 10 5

Let A be the event of selecting a rotten fruit. 3a 3a 1 3 P  A   5    5a 5 5a 25

Thus, the probability selecting a rotten fruit is

3 . 25

96. The numbers x and y can be chosen in the following ways: (1, 1), (1, 4), (1, 9), (1, 16), (2, 1), (2, 4), (2, 9), (2, 16), (3, 1), (3, 4), (3, 9), (3, 16), (4, 1), (4, 4), (4, 9), (4, 16)  Total number of possible outcomes = 16 The product xy will be less than 16 if x and y are chosen in one of the following ways: (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4), (4, 1)  Favourable number of outcomes = 8 Probability that the product of x and y is less than 16  Thus, the required probability is

8 1  16 2

1 . 2

97. Let the number of blue pens be b. Total number of outcomes = 20 Let the probability of getting a blue pen be P(B) and of getting a red pen be P(R). 1 P  B    P  R  3 It is known that:

P  B   P  R   1 1   PR  P R  1 3 4   PR  1 3

52

MN Top 100 Questions

MATHEMATICS

3 4 1 3 1  P  B    3 4 4 b 1   20 4 b5 Thus, there are five blue pens.  PR 

98. Let the total number of books in the cupboard be x. It is given that every second shelf of the cupboard has 4 Math books.  Total number of Math books in the cupboard = 4 × 4 = 16 The probability of choosing a non-Math book is Probability of choosing a Math book  1 

3 . 7

3 4  7 7

16 4   x 7 16  7 x  28 4 Thus, there are 28 books in the cupboard. 99. Total number of cards = 52 (i) Number of cards that are spades or aces = 13 + 3 = 16 Probability that the card drawn is a card of spade or an ace =

Favourable number of outcomes 16 4   Total number of outcomes 52 13 (ii) Number of black kings = 2 Probability that the card drawn is a black king = Favourable number of outcomes 2 1   Total number of outcomes 52 26

(iii) Number of cards that are jacks or kings = 4 + 4 = 8 Number of cards that are neither jacks nor kings = 52 – 8 = 44 Probability that the card drawn is neither a jack nor a king =

Favourable number of outcomes 44 11   Total number of outcomes 52 13 (iv) Number of cards that are kings or queens = 4 + 4 = 8 Probability that the card drawn is either a king or a queen =

Favourable number of outcomes 8 2   Total number of outcomes 52 13

53

MN Top 100 Questions

MATHEMATICS

100. Possible outcomes on rolling the dice twice are given below: {(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 = 36 (a) Favourable outcomes for the event that 5 will not come up either time are given below: {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)} Total number of favourable outcomes = 25  Probability of not getting 5 either time =

Number of favourable outcomes 25  Total number of outcomes 36

(b) Favourable outcomes for the event that 5 will come up exactly one time are given below: {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 5)} Total number of favourable outcomes = 10  Probability of getting 5 such that it will come up exactly one time =

Number of favourable outcomes 10 5   Total number of outcomes 36 18

54