O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS
Q uanti uantitati tativ ve Apti Aptitu tude de Solved Paper of Combined Defence Services Examination We are grateful to Sarvshri Aditya Prakash of New Delhi and Satish Kumar of Patna for sending the question booklet to us. 1. In the given figure, figure, AB and CD are par parallel allel and PQ is perpendicular to LM. If ∠ BNM = 50°, then ∠ PQD is:
(a) 30° (b) 40° (c) 50° (d) 65° 2. Sum of the the interior angles of of a regular regular polygon polygon having having ‘n’ sides is equal to: (a) (n + 2) π (b) (n + 1) π (c) (n – 1) π (d) (n – 2) π 3. Angl Anglee at a common common end point of of two opposit oppositee rays, rays, is equal to: (a) 135° (b) 180° (c) 270° (d) 360° 4. In a regular regular polygon, polygon, if an interior interior angle is equal to four times the exterior angle, then the number of sides in the polygon is: (a) 7 (b ) 8 ( c) 1 0 (d) 11 5. If one side of of a regular regular polygon polygon with with seven sides sides is produced, the exterior angle (in degrees), has the magnitude: (a) 60
3 (b) 51 7
(c ) 4 5
(d) 40
6. The angle BDE BDE in a regular regular hexagon hexagon ABCDEF ABCDEFA A is equal to: (a) 12 120° (b) 10 105° (c) 90° (d) 60° 7. Which one of the following represen represents ts the sides of of a triangle? ( b) 50 50,, 75, 75, 22 2211 (a)) 300, 400, 500 (a (c) 110, 221, 415 ( d) 50 50,, 150 150,, 750 750 8. If the the corresp corresponde ondence nce ABC ↔ RQP is a congruence, then which one of the following is n o t correct? (b) ∠A ≅ ∠R (a)) AB ≅ RQ (a ( c ) ∠C ≅ ∠P ( d ) A C ≅ RQ
9. In the the giv given en diagr diagram, am, ifif ∠ABC ABC = ∠DAC DAC = 90° and ∠ACB ACB = ∠DCA, DCA, then which one of the following statements is n o t correct?
(a) CD2 = DA2 + AB 2 + BC2 (b) AC2 = CD × BC (a) (c) AC × AD = AB × CD (d) AD × BC BC = AC × AB AB 10.. In the 10 the given given figu figure, re, ∆ ABC is an equilateral triangle. O is the point of intersection of the medians. If AB = 6 cm, then OB is equal to:
(a ) 3 3
( b) 2 3
(c )
3
(d )
3 2
11.. If a piece 11 piece of wire of length 15 cm is bent into the form 1 2
of a rectangle of area 13 cm2, then the length of the shorter side of the rectangle (in cms) is: ( a) 1
1 2
1 2
( b) 2
( c) 3
1 2
(d) 4
12. In a parallelogra parallelogram m ABCD, ABCD, bisectors bisectors of consecutive consecutive angles A and B intersect at P, then ∠APB is equal to: (a) 30 3 0° (b) 45° ( c) 60 (c 60° (d) 90° 13. If the straight straight line y = x + C is a tangent tangent to the circle 2 2 x + y = 1, then C is equal to:
APRI RIL L 20 2001 01 THE COMPETITION MASTER 860 AP
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS (a ) ± 2 (b) ± 2 (c) ± 1 (d ) ± 3 14.. The 14 The equa equati tion on 2 ax + 2hxy + by2 + 2gx + 2fy + c = 0 represents a circle if: (a) h2 = ab ( b) a = b 2 (c) h + ab = 0 (d) a = b an and d h= 0 15. Consider the following following statements statements relating relating to the the chords and arcs of a circle: 1. Equal chords are equidistant from the centre. centre. 2. Between two two arcs, the the one that subtends a larger larger angle angle at the centre is larger. 3. If two arcs arcs are of the same same length, so are are the chords chords joining their extremities. Of these statements: (a) 1 and 2 are correct (b) 2 and and 3 are are cor corre rect ct (c) 1 and 3 are correct (d) 1, 2 and 3 are are corr correct ect 16. The area area of the the given given triangle triangle is: is:
(a) 12 sq cm (b) 16 sq cm (c) 25 sq cm (d) 36 sq cm 17. If the perimet perimeter er of a square square is 24 cm, cm, then one of of the sides of the square is: (a) 12 cm (b) 8 cm (c) 6 cm (d) 2 cm 18. An isosceles isosceles right triangle triangle has has an area area of 200 200 sq cm. The area of a square drawn on its hypotenuse is: (a) 400 sq cm
(b) 40 400 2 sq cm
(c) 800 sq cm (d) 80 800 2 sq cm 19. The area area of the the shaded shaded portion portion is: is:
(a) 8 sq cm ( b ) 6 sq c m (c) 16 sq cm ( d ) 4 sq c m 20. The ar area ea of the shaded portion portion in the the given figure figure is
F π = 22 I G J : H 7 K
(a) 42 sq cm (b) 48 sq cm (c) 76 sq cm (d) 152 sq cm cm 21. A circle circle ‘A’ ‘A’ has a radius radius of 3 cm, two circles circles ‘B’ and ‘C’ have a radius each equal to the diameter of circle ‘A’. The radius of a circle ‘D’ which has an area equal to the total area of A, B and C is: ( a) 9 c m (b) 12 cm (c) 15 cm (d) 18 cm 22. Three cubes whose whose edges edges measure measure 3 cm, 4 cm and 5 cm respectively are melted to form a new cube. The surface area of the new cube is: (a) 50 cm2 (b) 216 cm2 (c) 250 cm2 (d) 300 cm2 23. 4 cubes each each of 6 sq cm total surface surface area area and another another cube of 24 sq cm total surface area are placed together in such a way to form a rectangular parallelopiped. The total surface area of the parallelopiped so formed is: (a) 48 sq cm (b) 40 sq cm (c) 36 sq cm (d) 32 sq cm 24. A rectangular rectangular cardboa cardboard rd sheet measures measures 48 48 cm × 36 cm. From each of its corners a square of 8 cm is cut off. An open box is made of the remaining sheet. The volume of the box is: (a) 8960 cm cm3 (b) 6400 cm cm3 (c) 5120 cm3 (d) 2560 cm cm3 25. The volume of a square square pyramid pyramid whose one side of of the base is 5 cm and height 6 cm is: (a) 25 cm3 (b) 30 cm3 (c) 36 cm3 (d) 50 cm3 26. If a right cone has a base base of radius radius 7 cm and and slant height height of 3 cm, then the total surface area of the cone is: ( a ) 9 sq c m (b) 21 sq cm (c) 49 sq cm (d) 220 sq cm cm 27. The slant height of a cone is l cm, and radius of its base is 7 cm. If the total surface area of the cone is 550 cm 2, then the value of l is is:: (a) 36 cm (b) 18 cm (c) 10 cm (d) 9 cm 28. The surface surface area area of a globe globe of radius radius r is: 2 2 (a) 2 π r (b) 3 π r 2 (c) 4 π r (d) 5 π r2 29. A sphere of radius 5 cm exactly fits into a cubical box. The ratio of the surface of box and the surface of the sphere is: (a) 19 : 9 (b) 21 : 11 (c) 23 : 13 (d) 25 : 13 30. A water water tank is is hemispherical hemispherical at the bottom bottom and cylindrical on top of it. The radius is 12 m. If the total capacity is 3312 π m3, then the capacities of the two portions are in the ratio: ( a) 8 : 9 (b) 8 : 11 (c) 8 : 13 (d) 8 : 15
APRI RIL L 20 2001 01 THE COMPETITION MASTER 861 AP
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS sin sin θ .tanθ + 1 3 31. If cos cos θ = , then the value of is: 5 2 tan2 θ
(a )
88 160
( b)
91 160
(c )
92 160
(d)
93 160
32. 32. (a) (c) 33.
If si sinn (30° (30° + θ ) = cos (60° – φ ), then: (b) θ – φ = 0 θ + φ = 0 (d) θ – φ = 90° θ + φ = 90° The value value of tan tan 40° tan tan 50° tan tan 60° is: is:
( a) 0
(b) 1
(c)
1 3
(d )
3
1 , where A and B are 2
34. If cos cos (A – B) = sin (A + B) =
positive, then smallest positive value of A + B (in degrees) is: (a ) 4 5 ( b) 60 ( c) 105 ( d) 1 5 0 35. Which of the following following pairs pairs is correctly matched? 1. tan x + cot x ...... ...... cosec cosec 2x 2. tan x +
1 cot x .... cosec 2x (1 + sin2 x) 2
3. cot x – tan tan x ..... ..... 2 cot 2 x
(a) 4 ( b) 3 (c) 2 ( d) 1 42. From a point point A due north of of the tower, tower, the elevation elevation of the top of the tower is 60°. 60° . From a point B due south, the elevation el evation is 45°. If AB = 100 meters, then the height of the tower is: (a) 50 3 ( 3 − 1) m (b) 50 3 ( 3 + 1) m (c) 50( 3 − 1)m (d ) 50( 3 + 1)m 43. The shadow shadow of a pole pole of height 10 10 meters, meters, when the angle of elevation of the sun is 45°, will be: (a) 10 meters (b) 20 met meter erss ( c ) 5 m e t e rs (d) 10 2 meters 44. A, B, C are three three points on a circle circle such that AB is the chord and CP is perpendicular to OP, where O is the centre and P is any point on AB. The radius r of the circle is: (a) r2 = OP2 + AP × CP (b) r2 = OP2 + PB × PC (c) r2 = OP2 + PB2 (d) r2 = OP2 + AP × PB 45. A balloon balloon is connected connected to a flying club by a cable of length 100 meters inclined at 30° to the vertical. The height of the balloon from the ground is: (a) 82.6 m (b) 83.6 m (c) 85.6 m (d) 86.6 m 46. The arithm arithmetic etic mean mean of k number numberss y1, y2 ..., yk is A. If y k is replaced by xk, then the new arithmetic mean will be:
cos 2x 4. ..... cos x − sin x cos 2 x + sin 2 x
(a) A – yk + xk
(b)
kA − y k + xk k
Select the correct answer using the codes given below:
(c) A – (yk – x k)
(d)
(k − 1)A − y k + xk k
cos x + sin x
Codes:
(a) 2, 3 and 4 (c) 1, 3 and 4 36.
(b) 1, 2 and 3 (d) 1 alone
1 3 – is equal to: sin 10° cos 10°
( a) 2
(b) 1
( c) 4
( d) 3
37.
cotA + cosecA − 1 is equal to: cot A − cosec A + 1
(a )
1 + cos A sin A
( b)
1 + sin A sin A
(c )
1 − co s A sin A
(d)
1 − sin A sin A
38. If sin sin θ = ( 2 − 1 ) cos θ , then cos θ – sin θ is equal to: (a) 2 cos θ (b) 2 sin θ (c )
(d)
3 cos θ
3 sin θ
39. Maximum value of cos4 θ – sin4 θ is: ( a) 2 (b) 0 ( c) 1 (d) –1 40. If (a)
5 3
6 sin 2 θ + tan 2 θ = 10 , then cos θ is equal to: 4 cos θ
(b)
4 3
(c)
2 3
(d)
1 3
41. If sin sin θ + sin2 θ = 1, then cos2 θ + cos4 θ is equal to:
47. Average age age of a class class is 16 years. years. If the class teacher teacher aged 40 years old is also included, the average age rises to 17 years. Then the number of students in the class are: (a ) 2 3 ( b) 3 3 (c) 40 (d) 16 48. Which one of the following frequency distribution distribution have area under the histogram equal to unity? (a) Rela Relative tive freque frequency ncy distri distributi bution on (b) Ord Ordinar inaryy frequency frequency distr distribut ibution ion (c) Perc Percenta entage ge frequen frequency cy distribu distribution tion (d) Cumu Cumulati lative ve frequency frequency distr distribut ibution ion 49. Consider the following frequency distributio distribution: n: Classes Frequency 0—4 1 4—8 5 8 — 12 9 12 — 16 12 16 — 20 8 20 — 24 4 24 — 28 3 If the ‘less than’ and ‘greater than’ ogives are drawn, then they will intersect each other at the coordinates: (a) (21, 14) (b) (1 (122, 14) (c) (14, 21) (d) (21, 21) 50. The class class intervals intervals in a frequency frequency distribu distribution tion are are (16 — 19), (20 — 23), (24 — 27), (28 — 31) etc. The correct specification for the class (24 — 27) is:
APRI RIL L 20 2001 01 THE COMPETITION MASTER 862 AP
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS (a) 24 < x < 27 ( b ) 24 ≤ x < 27 (c ) 2 4 < x ≤ 2 7 ( d ) 2 4 ≤ x ≤ 27 51. If for some values of x, 100 x + 99 99 is a positive positive integer, integer, then the next positive integer is: (a) 101x + 99 ( b) 10 1000x + 100 (c) 10 101x + 100 ( d) 10 1000 (x (x + 1) + 99 99 52. The number number of ration rational al numbers numbers between between 1 and 5 is: is: ( a) 3 ( b) 5 (c) finite (d) in infi fini nite tely ly man manyy 53. The number 152207 when multiplied multiplied with with x produce producess an 8-digit number with each digit 1. The number x is: (a) 63 (b ) 7 3 (c) 83 (d) 93 1+
54. If ( a) 1
27 169
x , then x equals: 13
(b) 2
55.. Th 55 Thee valu valuee of (a )
= 1+
21
(b)
( c) 3 10 4 4 −1 20
56. The rati rationa onall number number
( d) 4
is: (c) 20
(d) 21
7 expressed in decimal fraction, 1000
will be: (a) 0.07 (b) 0.007 (c) 0. 0.7 (d) 0.0007 57. Shiela’s house house is 10 km away from the school. She takes 30 minutes to reach the school by bus. If Ram travels from his house at the same speed as that of Shiela and takes only 12 minutes to reach the school, the distance between Ram’s house and his school (in km) is: ( a) 4 (b) 5 ( c) 6 ( d) 7 58. A train 100 meters meters long passes a platform 100 100 meters meters long in 10 seconds. The speed of the train is: (a) 36 km/hour (b) 45 km km/h /hou ourr (c) 63 km/hour (d) 72 km km/h /hou ourr 59. 16 men or or 24 women women can do a piece of of work in 20 days. days. The number of days needed to complete the job, if 20 men and 30 women are employed to do the same piece of work, is: (a) 16 (b ) 1 2 (c) 10 (d ) 8 60. In an examination, examination, the marks secured by Ram were were as as follows: Subject Total marks Marks obtained Sanskrit 100 60 Maths 100 80 Social Science 100 50 Hi n d i 100 60 S c i en c e 100 85 English 100 65 The percentage of total marks obtained is: (a) 56.66 (b) 65.66 (c) 66.66 (d) 75.66 61. If a shopkeeper shopkeeper sells an item for Rs Rs 141 his loss loss is 6%. To To earn a profit of 10% he should sell it for: (a) Rs 155 (b) Rs 160 ( c) Rs 165 (d) Rs 170 62. A sum sum invested at 5% 5% simple interest interest grows to to Rs 504 in 1 2
4 years. The same amount at 10% simple interest in 2 years
will grow to: (a) Rs 420 (b) Rs 525 ( c) Rs 450 (d) Rs 550 63. The compound interest on Rs 1500 1500 for 2 years years at 5% 5% is: (a) Rs 158.25 (b) Rs 153 53.7 .755 (c) Rs 143.75 (d) Rs 14 140.2 .255 64. In a 500 meters meters race, Q starts starts 45 meters ahead ahead of P. P. But P wins the race while Q is still 35 meters behind. The ratio of their speeds, assuming that both start at the same time, is: ( a) 5 : 7 (b) 5 : 3 (c) 5 : 6 (d) 25 : 21 65. In order to have a rate of Rs 9.20 per kg of sugar sugar, the sugar costing Rs 8.50 per kg and the sugar costing co sting Rs 9.50 per kilo must be mixed in the ratio of: ( a) 2 : 7 (b ) 3 : 7 (c) 3 : 8 (d) 2 : 3 66. A boy takes 20 20 minutes to reach the school school at an average average speed of 12 km/hour. If he wants to reach the school in 15 minutes, his average speed (in km/hour) must be: (a) 14 14 (b ) 1 6 (c) 18 (d) 20 67.. Th 67 Thee num numbe berr ‘l’ ‘l’ of positive integers less than 72, such that the HCF of ‘l‘l’’ and 72 is 3, is: ( a) 5 (b) 6 (c) 7 (d) 8 68. The number 23*7 is divisible by 3. The missing digit digit (*) is: ( a) 1 (b) 2 (c) 3 (d) 4 69. If p
=
n(n + 1) is prime for some natural number n, 2
then p equals: ( a) 5 (b) 3 (c) 7 (d) 11 70. The number of composite composite numbers between 110 110 and 120 is: ( a) 5 (b) 6 (c) 7 (d) 8 71. The HCF of 608, 544; 544; 638, 783 783 and 425, 425, 476 respectively respectively is: (a) 32, 29, 17 (b) 17, 32, 29 29 (c) 29, 32, 17 (d) 32 32,, 17 17, 29 29 72.. The val 72 value ue of log10 .00001 is: ( a) – 4 (b) – 5 ( c) 4 ( d ) 5 73. If ( a)
1 log (11 + 4 7 ) = log (2 + x), then the value of x is: 2 7
(b) 11
(c ) 4
(d) 2
12 log10 10 74.. Th 74 Thee val value ue of 2 log 100 is: 10
( a) 2 (b) 3 (c) 4 (d) 5 2 75.. If log 75 log (k (k – 4k + 5) = 0, then the value of k is: ( a) 0 (b) 1 (c) 2 (d) 3 76. The smallest integral value of x, for which
5 is an integer, is: x
( a) – 1 (b) 1 ( c) – 5 ( d ) 5 77. The missing digit (*) in 8276*84 8276*8455 so as to make a multiple of 11 is: ( a) 1 (b) 2 (c) 3 (d) 4 78. The sum sum of the coeffici coefficients ents of of even powers powers of of x of a polynomial f(x) is:
APRI RIL L 20 2001 01 THE COMPETITION MASTER 863 AP
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS ( a) f ( 1 ) (c )
(b) f(0)
f(1) f(1) + f(−1) 2
(d)
f(1) − f(−1) 2
79. A factor factor of of the polyn polynomia omiall 3 (x2 + x)2 – 10 (x 2 + x) + 3 is: (a) 3x2 + 3x – 1 (b) 3x2 – 3x – 1 (c) x2 – x + 3 (d) x2 + x + 3 80. Which one of of the following following is the the product product of two identi identical cal factors? (a) x2 + 5x + 25 (b) x2 + 10x + 25 (c) x2 + 15x + 25 (d) x2 + 20x + 25 81.. Fa 81 Facto ctors rs of (2x2 – 3x – 2) (2x 2 – 3x) – 63 are: (a) (x – 3) 3) (2x + 3) (x (x – 1) 1) (x – 7) (b) (x + 3) 3) (2x – 3) (x – 1) 1) (x – 7) 2 (c) (x + 3) (2x (2x + 3) (2 (2xx – 3x + 7) (d) (x – 3) (2x + 3) (2x2 – 3x + 7) 82.. If the 82 the polyn polynomi omial al 3x 3x4 – 5x2 + x + 8 is divided by x – 3, then the remainder is: ( a) 2 0 0 (b) 204 (c) 207 (d) 209 3 83. If x + ax – 28 is exactly divisible by x – 4, then the value of ‘a’ will be: ( a) 9 (b) – 9 (c) – 8 ( d) 8 3 2 2 84.. G. 84 G.C. C.D. D. of x – x – 4x – 6 and x – 2x – 3 is: (a) x + 3 (b) – 2x + 3 (c) x – 3 ( d) – x – 3 85. If x = 2a – 1, 1, y = 2a – 2, z = 3 – 4a, then then the value of of 3 3 3 x + y + z will be: (a) 6( 6(33 – 13 13aa + 18 18aa 2 – 8a3) (b) 6( 6(33 + 13 13aa – 18 18aa 2 + 8a3) (c) 6( 6(33 + 13 13aa + 18 18aa2 – 8a3) (d) 6( 6(33 – 13 13aa – 18 18aa2 – 8a3) (a 2 − b2 )3 + (b2 − c2 )3 + (c2 − a2 )3 86.. Th 86 Thee valu valuee of is: (a − b)3 + (b − c) c )3 + (c − a) a )3
(a) (b) (c) (d)
( a)
87. If x + y = a and xy = b, then then the valu valuee of (a) a – 3ab (c )
96. The value value of x in the equa equation tion 16x 16x +
(a + b)3 + (b + c)3 + (c + a)3 (a + b) + (b + c) + (c + a) (a + b)3 (b + c)3 (c + a) 3 (a + b) (b + c) (c (c + a) a)
3
( b)
a 3 − 3ab b
3
1 x3
+
1 is: y3
a 3 + 3ab b 3
(d) a3 + 3ab
89. The solut solution ion of the equa equatio tions ns =
x+4 4
2 is:
(a) x = 4, y = 3 (c) x = – 4, y = – 3
1 1 , 4 4 1 b I bc
F x J H x c K
97. G
(b)
1 ,2 4 1 c I ca
F x J H x a K
× G
(c)
1 1 , 4 2
1 = 8, is: x
(d )
1 1 , 2 2
1 a I ab
F x J simplifies to: H x b K
×G
1 x
(a ) x
(b )
(c ) 1
(d) no none ne of of the the abo above ve
x a+ b × x b + c × xc + a 98.. Th 98 Thee val value ue of is: (x a × x b × xc )2
88. The value of x and y respectively respectively in the simultaneous simultaneous equations 2x + 3y = 15, 2x – 3y = 9 is: (a) 3 and 3 (b) 6 and 1 (c) 9 and 1 (d) 9 and 3
2x + 1 y + 5 − 3 2
90. A man’s age age is six times that of his son’s age. In six six years the father’s age will be three times the son’s age. The ages of the father and the son are respectively: (a) 24, 4 (b) 18, 3 (c) 30 3 0, 5 (d) 42, 7 91. The cost of of 5 tables is 10 more more than the cost cost of 11 11 chairs. The total cost of 9 chairs and 9 tables is 450. The costs of one table and one chair are respectively: (a) 35 and 15 (b) 15 an and 35 35 (c) 25 and 25 (d) 30 and 20 92. The speed of of a boat in still still water is is 10 km/hour. km/hour. If it can travel 26 km downstream and 14 km upstream at the same time, the speed of the stream is: (a) 2 km/hour (b) 2. 2.55 km/ km/ho hour ur (c) 3 km/hour (d) 4 km/ km/ho hou ur 93. The value of x, y and z respectively respectively in the equations equations 2x + 3y = 0 3y + 4z = 14 2x + 4z = 26, is: (a) – 3, – 2 and 5 (b) – 3, 3, 2 and 5 (c) 3, 3, – 2 and 5 (d) 3, 2 and 5 94. The value value of x and y respect respectively ively in the equation equation x + y = 4, 3x – 8y = 1, is: ( a) 1 a n d 3 (b) 3 and 1 (c) 1 and 2 (d) 2 and – 1 95. The equati equation on whose whose roots roots are 4 and and 5, is: 2 2 (a) x + 9x – 20 = 0 (b) x + 9x + 20 = 0 2 (c) x – 9x + 20 = 0 (d) x2 – 9x – 20 = 0
(b) x = 4, y = – 3 (d) x = – 4, y = 3
+
y+3 5
=
2,
(a ) 1 (b) x2 (c) xa + b + c (d) xabc 99. For any thre threee sets sets A, A, B, C A – (B ( B ∩ C) is equal to: (a) (A − B) ∪ (A − C) (b) (A ∩ B) − C (c) (A − B) ∩ (A − C) (d) A ∪ (B − C) 100.. Let A, B, C be three 100 three finite finite sets sets with with k, l, m elements respectively. If B ∩ C contains n elements, then the number of elements in the set A × (B ∪ C) is: (a) klm – n (b) k + l + m – n (c) k(l + m − n) (d ) k(lm − n)
APRI RIL L 20 2001 01 THE COMPETITION MASTER 864 AP
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS ANSWERS AND EXPLANATIONS
1. (b)
= ∠NQD = 50° (corresp. LM ∠PQM = 90° Q PQ⊥ LM ∠BNM
∴ ∠PQD = ∠PQM – ∠NQD =
∠s )
12. (d)
1 1 1 ∠DAB + ∠ABC = × 180° 2 2 2 ∠ 1 + ∠ 2 = 90 ° ∠1 + ∠ 2 + ∠APB = 180 ° or 90° +
9 0 ° – 50 ° = 4 0 °
2. (d) 3. ( b) 4. (c) Let Let the the ex extt ∠ be x° ∴ Int angle = 4x° Sum of ext. ∠s of a regular polygon = 360° Let the no. of sides be n ∴ nx = 360
⇒
x
=
360 n
Sum of int ∠s of a regular polygon = n × 4x = (n – 2) 180 n×4×
360 n
5. (b) Each ext 6. (c)
∠C =
= 180 (n
– 2)
⇒ n
= 10
∠ =
360° 7
=
3 = 51 ° 7
30 30 ° 120 – 30 = 90 ° ∴ ∠BDE = 120
7. (a) Q Sum of 2 sides of a 8. (d) 9. All ar aree cor corre rect. ct. As (a) True CD D2 = DA2 + AC2 Q C
2
2
ABC ABC ~ ∆DAC DAC AB BC AC ∴ = = AD AC CD
15. (d)
Q∆
(Q AC = AB + BC )
10. (b) In an an equil equilat atera erall of sides of a
=
BC AC
=
AC = AB × CD = AC × AD CD AC = BC × CD = AC 2 CD
∆ medians
∆ ∴ AE =
are also
BE ⊥ AC In rt ∆ AEB, AB2 = AE2 + BE2 62 = 32 + BE2 or BE = 3 3 BO : OE = 2 : 1 BO
=
2 2 , BE = × 3 3 = 2 3 3 3
11. (c) 2(L + B) = 15 15 .... .... ((ii) LB = Solving (i) and (ii (ii)) B = 3, Shorter side = 3
27 .... (ii (ii)) 2
9 2 9 2
[Q If B = , L = 3 ]
P
⊥ distance
from (x1, y1) to the line Ax1 + By1 + C A 2 + B2
The eqn represents a circle of coefft of x2 = coeff of y2 2nd degree eqn is x and y There is no term involving the product xy
E
D
∆ =
1 1 × B × H = × 6 × 4 = 12 sq cm 2 2
17. (c) Sid Sidee of of a squ squar aree
=
Perimeter
4
=
24 4
=
6 cm
18. (c) a2 + a2 = h2 or h2 = 2a2 Area of
∆
=
1 × a × a = 200 2
h
a
1 22 2 × ×7 ) 2 7
= 196 – 154 = 42 sq cm 21. (a) π R = π (3)2 + π (6)2 + π (6)2 ⇒ R = 9 cm 22. (b) l3 = 33 + 43 + 53 ⇒ l = 6 Surface area of new cube = 6l 6l2 = 6 × 62 = 216 2
O B
1 80 °
2 ⇒c=± 2
=
20. (a) Req Reqd d are areaa = (7 + 77))2 – (2 ×
A
F
B
∠A P B =
or a2 = 400 ∴ h2 = 2 × 400 = 800 a ∴ Reqd area = 800 sq cm 19. (c) Are Areaa of the shade shaded d portion portion = 4 × 2 – (1 × 2) = 6
⊥ bisectors
1 × 6 = 3 cm 2
2
Q
16. (a) ar of
AB × AC = AD × BC
AB AD
c
ax + by + c = 0 is 14. (d)
2
1 A
B
BC ⇒ AC
C
+ ( − 1)
[Q
> the third side
=
= 1 or
C
A
AB AD
2
D
F
∆ is
CD2 = DA2 + (AB2 + BC2)
∴
1 E
∴ ∠ CD B = ∠ CBD =
0−0+c 2
120°, DC = BC
D
or ∠APB = 180 – 90 = 90° 13. (a) If a line line is tangen tangentt to a circle circle,, the ⊥ distance of this line from the centre of the circle = radius. Centre of the circle x2 + y2 = 12 is (0, 0) Eqn of line x – y + c = 0 O ∴
360° n
(Q DA D BC)
1 80 ° ∠D A B + ∠A BC = 18
C
23. (d) 6l 12 = 24 ∴ l1 = 2 6l 22 = 6 ∴ l2 = 1 l = 2 + 1 = 3, b = 2, h = 2 2 1 TSA of parallelopiped 1 1 = 2 (l b b + bh + hl) 2 = 2(3 × 2 + 2 × 2 + 2 × 3) 1 = 32 sq cm 1 2 24. (c) Vol of of open open box box = l × b × h = (48 – 2 × 8) × (36 – 2 × 8) × 8 = 32 × 20 × 8 = 5120 cm 3 25. (d) Vol of squa square re pyrami pyramid d=
APRI RIL L 20 2001 01 THE COMPETITION MASTER 865 AP
1 × area of base × height 3
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS 1 × 52 × 6 = 50 cm3 3
=
26. (d) T.S. .S.A. A. of of cone cone = π r (r + l) 27. (b)
22 × 7 (7 + l) = 550 7
=
22 × 7(7 + 3) = 220 cm 2 7
l = 18 cm
⇒
2 3 πr + π r 2 h 3
=
3312π ⇒
2 3 π r Reqd ratio = 3 π r 2h
cos 2 x + sin 2 x
MP = 5
2
−
3
=
= 15
=
2 r 3 h
=
2 3
×
12 15
=
8 15
=
8 :15
5
36. (c)
sin θ . ta tanθ + 1 2 tan 2θ
=
4
=
O
4 4 × +1 5 3 = 4 2 2×( ) 3
3
= sin 2x
1 3 2[ cos 10°− sin 10° ] 2 2 sin 10° cos 10°
=
4 sin (30°−10° ) 4 sin 20° = sin 20° sin 20°
=
4
[sin (A – B) = (sin A cos B – cos A sin B)
= sin (90°− 60 − φ ) = sin (30° + φ ) sin (30° + φ ) = sin (30° + φ ) ⇒ 30° + θ = 30° + φ ⇒ θ = φ or θ – φ = 0 33. (c) tan 40°. tan 50°. tan 60° = tan (90° (90° – 50°) 50°) . tan 50° . 3 =
1 + tan2 x
L sin sin 30° cos cos 10°−cos cos 30° sin sin 10° O 4M P 2 sin sin 10 10° cos 10° N Q
93 160
2tanx
=
M
3
32. (b) si sinn (30 (30°° + θ ) = cos (60 – φ )
= cot 50° . tan 50° .
(4) is correct
cos 10°− 3 sin 10° 1 3 – = sin 10° cos 10° sin 10° cos 10°
4
sin θ = , tan θ = 5 3 ∴
cos x − sin x cosx + sinx × cosx − sinx 1
cos2 x – sin2 x = cos 2x,
4
4
=
cos2x cos 2 x − sin 2 x = = cosx − sin x cosx − sinx
P
2
−
sinx cos 2 x − sin 2 x cosx = sin sin x cosx cosx
[sin2 x + cos2 x = 1, 2 sin x cos x = sin 2x
3 5
31. (d) cos θ =
(2) is correct
2 cos2x
cosx + sinx
2 3 2 × 12 + 12 × h 3
⇒ h
2 sin 2 x + cos 2 x 2 sin x cos x
= 2sin x cos x × 2 = sin sin 2x = 2 cot 2x (3) is correct
21 = 21 : 11 11
=
(1) is incorrect
1 sin sin 2x + cos cos 2 x + sin sin 2x = sin 2 x (1 + sin2 x) sin 2x
cos2x
6 × 102 22 2 ×5 4× 7
= 3312
=
cosx cot x – tan x = sinx
30. (d) Vol of hemispherical portion + cylindrical cylindrical portion portion =
1 cos x sin x tan x + cot x = + 2 cos x 2 sin x
=
= cosec 2x (1 + sin2 x)
S.A. of box box 6l 2 S.A. of sphere = 4π r 2 =
∴
=
28. (c) 29. (b) If a sphere sphere exactly fits into a cubical box, then edge of cubical box = diameter of a sphere ∴ l = 2 × 5 = 10 cm
2 sin sin 2 x
2 cose cosecc 2x
=
1 tan 50° × tan 50° .
3 =
3
1 = cos 60° ∴ A – B = 60° ... (i ( i) 2 1 sin (A + B) = = sin 30° ∴ A + B = 150° ... (ii ( ii)) 2
34. (d) co coss (A (A – B) B) =
= sin (180 – 30°) [As A and B are +ve and = sin 150° A – B = 60° ∴ A + B ≠ 30°
tan 2 x + 1 1 35. (a) tan x + cot x = tan x + = tan x tan x
=
2 2tan x 1 + tan 2x
cot A + cosec A − 1
37. (a) cot A − cosec A + 1 cot cot A + cos cosec ec A − (cos cosec2A − cot 2 A) = cot A − cose osec A + 1
=
(cosec A + cot A) (1 − cosecA + cot A) cot A − cosec A + 1
= cosec A + cot A =
1 sinA
38. (b) sinθ = ( 2 − 1 ) cos θ cos θ = ⇒ cos
2 sin θ + sin θ = 2−1
θ – sin θ =
APRI RIL L 20 2001 01 THE COMPETITION MASTER 866 AP
⇒
1 + cos A sinA sin θ 2 cosθ = × 2 −1 2
+
cosA sinA
=
2 sin θ + sin θ
2 sin θ
+1 +1
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS 39. (c) cos4 θ – sin4 θ = (cos2 θ + sin2 θ ) (cos2 θ – sin2 θ ) = 1.cos 2θ = cos 2θ (as max value of cosine of any angle = 1) Max. value is 1 40. (d)
6 sin 2 θ + tan 2 θ = 10 4 cos θ
New A.M. =
8
=x
3
h
h = 3 (100 – x) h = 100 3 – 3 x
∴
= 100 3 –
3h
h ( 3 + 1) = 100 =
⇒
3
⇒h
100 3 ( 3 3−1
45
60
A
(100-x
B
x
3 h + h = 100
3
100 3 × 3 +1
3 –1 3 –1
=
− 1)
10 43. (a) = tan 45° = 1 x 45 x
O
r
C
2
P
OC2 = OP2 + PC2 2
= OP +
2
2
2
r – OP
or r – OP = PA × PB
9
15
= 12 +
27 35 39 42
∴ Reqd
= OP2 +
PA × PB PQ 2
⇒ (r
⇒r
2
2
– O P 2 )2 2
=
54. (a)
1+
27 = 169
= OP + AP × PB
196 14 1 = =1+ 169 13 13
55. (c)
10 4 10 × 2 = 2−1 4 −1
56. (b)
7 = .007 1000
57. (a) Speed =
2
PA 2
D T
=
20
10 30
=
2
×h
2 1 − 15 × 4 = 14 12
pt is (14, 21)
x 12
⇒
x=1
⇒
x = 4 km km
100 + 100 200 = = 20 m/sec 10 10
59. (d) 16 me menn = 24 24 wom women en
× PB
f
11111111 = 73 152207
=
18 5
= 20 ×
2
−c
52. (d)
58. (d) Sp Spee eed d of of trai trainn =
PA 2 × PB 2
PA 2 × PB 2 r2 – OP2 = 2 r – OP 2 2
8—12
2
⇒r
N c = 15 Md = L + 2
6
B
Q
2
60
5
r A
PA 2 × PB2 = PC × PQ ⇒ PC = PQ 2 2
h
4—8
53. (b) x
10
= 10 m 44. (d) Pro Produce duce CP CP to meet meet the circ circle le at Q AB and CQ are the two chords intersecting at P 2 2 ∴PA × PB = PC × PQ ⇒ PA × PB
30 100
kA − y k + x k k
12—16 12 16—20 8 20—24 4 24—28 3 N = 42 50. (b) 51. (b)
h = 50 3 ( 3 – 1)
⇒ x
3 × 100 2
h =
⇒
47. (a) Let the no. of stud students ents be n ATS 16n + 40 = 17(n + 1) ⇒ n = 23 48. (a) 49. (c) f cf l = 12 (median class is 12–16 0—4 1 1 f = 12
1
h = tan 60 = 100 – x
3 2
sin 60° =
y1 + y 2 +. .. .+ yk k
46. (b) A =
1
cos θ = satisfies the given equation 3 As – 1 ≤ cosθ ≤ 1 ∴ ( (aa) and (b (b) are false 2 41. (d) sin θ + sin θ = 1 ⇒ sin θ = 1 – sin 2 θ = cos2 θ 2 4 ⇒ sin θ = cos θ or 1 – cos2 θ = cos4 θ ⇒ cos4 θ + cos2 θ = 1 ⇒ h
=
or kA = y 1 + y2 + ... + yk New total = (y1 + y2 + ... + yk) – yk + xk = kA – y k + xk
By inspection we can see
h = tan 45° = 1 x
h 100
= 1.732 × 50 = 86.600 m
3
1 8 If cos θ = , sin θ = , tan 8 = 3 3
42. (a)
45. (d)
= 72 km/hr
∴ 20
men =
24 × 20 = 30 women 16
20 men + 30 women = 30 + 30 = 60 women Women Days 24 20 Inverse variation 60 x 24 : 60 = x : 20
APRI RIL L 20 2001 01 THE COMPETITION MASTER 867 AP
⇒
x=
24 × 20 60
=
8 days
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS 400 200 × 100 = = 66.66 60. (c) Re Reqd qd %a %age ge = 600 3
[( 2 +
A × 100 504 × 100 = Rs 420 = 100 + R × T 100 + 4 × 5
SI =
420 × 10 5 × 100 2
= 105 ∴ A
= 420 + 105 = Rs 525
64. (d) T =
41 = 153.75 400
500 500 – (45 (45 + 35) S1 D 500 ⇒ ATS S = = S2 S2 S 420 1
= 65. (b)
8.50
25 = 25 : 21 21
9.50
0.30
0.70
0.30
:
0.70
=
30 70 : 100 100
=
3:7
Reqd ratio = 3 : 7 66. (b) It is a case of of inverse inverse va variat riation ion Time Speed 20 hrr h 60
12 km/hr
15 hr 60
68. (c)
x
82.
20 15 : = x : 12 60 60 20 60 × 12 × 60 15
x=
=
16 km/hr
HCF is a factor of both l and 72 ∴ l must be a multiple of 3 ∴ l = 6 But HCF of 6 and 72 = 6 Question is wrong 2+3+x+7 3
= + ve
integer
12 + x 3
=
4+
x 3
∴ x = 3
The no. is divisible by 3 if the sum of digits of a no. is divisible by 3. 69. (b) 70. (d) Composite nos. between between 110 and 120 120 are 11 111, 112, 112, 114, 114, 115, 116, 117, 118, 119 i.e. 8 (113 is a prime no.) 71. (a) −5 72. (b) log10 .00001 = log10 10 = –5 log1010 = – 5 × 1 = – 5 [log mn = n log m, log ee = 1
73. (a)
1 log (11 + 4 7 ) = log (2 + x) 2
log (11 + 4 (4 + 7
+
4
1 7 ) 2 =
1 7 )2
78. 79.
80. 81.
9.20
67.
75. 76. 77.
5 2 r T ) − 1] 63. (b) CI = P[(1 + ) − 1] = 1500 [(1 + 100 100
= 1500 ×
log (2 + x)
∴ (1 1 +
4
1 2 7)
= 2+ x
= 2 + x ⇒ [(2)2 + ( 7 )2 + 2 × 2 =2+x
2+x
⇒ 2
+ 7=2+x
⇒
x=
7
12 12 log 10 10 12 × 1 74. (b) = = 2 × 2 log 10 2 log10 100 2 log10 102 10 12 = = 3 2×2×1
100 (100 + 10) × = Rs 165 61. (c) Req Reqd d S.P S.P. = 141 141 × (100 − 6) 100
62. (b) P =
1 2 2 7 ) ] =
1 7 ]2
83.
84. 85.
[logee = 1, log mn = n log m] (c) log (k2 – 4k + 5) = 0 = log 1 ⇒ k 2 – 4k + 5 = 1 k2 – 4k + 4 = 0 ⇒ (k – 2) 2 = 0 ⇒ k – 2 = 0 ⇒ k = 2 ( c) (b) Sum of digit digitss at odd odd places places = 5 + 8 + 6 + 2 = 21 Sum of digits at even places = 4 + x + 7 + 8 = 19 + x Diff = 21 – 19 – x = 2 – x The no. is divisible by 11 if this diff is 0 or divisible by 11 ∴ Clearly x = 2 ( c) (a) 3(x2 + x)2 – 9(x2 + x) – (x2 + x) + 3 = 3 (x2 + x) (x2 + x – 3) – (x2 + x – 3) = (x2 + x – 3) (3x2 + 3x – 1) (b) x2 + 10x + 25 = (x + 5) (x + 5) (d) Put 2x2 – 3x = y (y – 2) y – 63 = y2 – 2y – 63 = y 2 – 9y + 7y – 63 = y(y – 9) + 7(y – 9) = (y – 9) (y + 7) = (2x2 – 3x – 9) (2x 2 – 3x + 7) = (2x2 – 6x + 3x – 9) (2x2 – 3x + 7) = [2x (x – 3) + 3 (x – 3)] [2x2 – 3x + 7] = (x – 3) (2x + 3) (2x 2 – 3x + 7) (d) Le Lett p(x p(x)) = 3x4 – 5x2 + x + 8 If p(x) is divided by x – 3, R = p(3) p(3) = 3(3)4 – 5(3)2 + 3 + 8 = 209 ∴ R = 209 ( b ) I f x 3 + ax – 28 is exactly divisible by x – 4 then R = 0 ⇒ 43 + a × 4 – 28 = 0 or 64 + 4a – 28 = 0 4a = – 36 ⇒ a = – 9 (c) If we we put put x = 3 in in both both polyno polynomials mials,, we get get = 0 ∴ x – 3 is a factor of both ∴ G.C.D. = x – 3 (a) x + y + z = 2a – 1 + 2a – 2 + 3 – 4a = 0 3 3 3 ∴ x + y + z = 3xyz = 3 (2a – 1) (2a – 2) (3 – 4a) = 6 (3 – 13a + 18a2 – 8a3)
3(a 2 − b2 )( b2 − c2 )( c2 − a 2 ) 86. (d) Give Givenn exp exp = 3(a − b) b )( b − c)( c − a)
= (a + b) (b + c) (c – a) Let a2 – b2 = x, b2 – c2 = y, c2 – a2 = z 3 3 3 ∴ x + y + z = 0 ∴ x + y + z = 3xyz 2 2 3 2 2 3 2 2 3 ∴ (a – b ) + (b – c ) + (c – a ) = 3 (a2 – b2) (b2 – c2) (c2 – a2) Also a – b + b – c + c – a = 0 3 3 3 ∴ (a – b) + (b – c) + (c – a) = 3 (a – b) (b – c) (c – a) 1 87. (c) 3 x
+
1 y3
APRI RIL L 20 2001 01 THE COMPETITION MASTER 868 AP
=
y3 + x3 x3 y 3
= =
(y + x)3 − 3xy (x (x + y) (xy)3 a 3 − 3ab b 3
=
a 3 − 3ba b 3
O BJ ECT ECTIVE-T IVE-TYPE YPE QUESTIONS 88. (b) 2x + 3y 3y = 15 15 ... ... (i) 2x – 3y = 9 Adding 4x = 24 ⇒ x = 6 Putting the value of x in (i) 2 × 6 + 3y = 15 ⇒ y = 1 89. (b) x = 4, y = –3 –3 satisfy satisfy both both the the equation equationss 90. (a) Let son’s present present age age x years years and that that of father = 6x years years ATS 6x + 6 = 3 (x + 6) ⇒ x = 4 ∴ 6x = 24 years 91. (a) Let the cost cost of 1 table table and 1 chair chair be Rs x and y resp respective ectively ly ATS 5x = 11y + 10 or 5x – 11y = 10 ... (i ( i) 9x + 9y = 450 ⇒ x + y = 50 ... (ii (ii)) Solving (i(i) and (ii (ii)) x = Rs 35, y = Rs 15 92. (c) Let the the speed speed of of stream stream be be x km/hr km/hr Speed downstream = 10 + x km/hr Speed upstream = 10 – x km/hr T=
D 26 14 ATS = or x = 3 km/hr S 10 + x 10 − x
93. (c) 2x + 3y = 0 ...... (i), 3y + 4z = 14 ... (ii ( ii)) Subtracting (ii (ii)) from (i(i), 2x – 4z = – 14 ... ( iii iii)) 2x + 4z = 26 ... (iv iv)) Add (iii (iv)) 4x = 12 ⇒ x = 3, 2 × 3 + 3y = 0 iii)) and (iv 3y + 4z = 14 3y = – 6 ⇒ y = – 2 3 × – 2 + 4z = 14 or 4z = 14 + 6 = 20 ⇒ z = 5 x = 3 , y = – 2, z = 5 94. (b) x = 3, y = 1 satis satisfy fy both both the equati equations ons
95. (c) The reqd reqd.. equati equation on is x2 – (α + β ) x + αβ = 0 α = 4, β = 5 ∴ We get x2 – 9x + 20 = 0 96. (a)
16 x 2 + 1 = 8 or 16x 2 – 8x + 1 = 0 x
(4x–1) (4x–1) = 0 or 4x–1 = 0 1 b I bc
F x J H xc K
97. (c) G
=
1 b × x bc 1 c× x bc
×
×
1 c I ca
F x G J H x a K x x
c× a×
1 ca 1 ac
98. (a) Th Thee give givenn exp. exp. =
×
×
⇒ (4x
x=
⇒
1 4
– 1)2 = 0 1 1 , 4 4
x=
1 a I ab
F x G J H x b K a×
1 ab
b ×
1 ab
x x
=
1 xc 1 x b
x a+b+b+c+c+a
x 2a × x 2b × x 2c 99. (a) A – (B ∩ C) = (A – B) ∪ (A – C)
×
=
100. (c) O( O(A) A) = k, k, O(B O(B)) = l, O(C) = m O (B ∪ C) = O(B) + O(C) – O (B ∩ C)
1 xa 1 xc
×
1 x b 1 xa
x 2(a+b+c) x 2(a+b+c)
=1
=1
O (B ∩ C) = n = l + m – n
A × (B ∪ C) = k( k(ll + m – n) [O(A) means no. of elements in the set A]
APRI RIL L 20 2001 01 THE COMPETITION MASTER 869 AP
