## Very important Maths questions for SSC CGL-13 Series#3

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61. If O is the circumcentre of ?ABC and ?OBC = 350, then the ?BAC is equal to-

a. 550

b. 1100

c. 700

d. 350

Ans : We know ?BOC = 1800 – (350 + 350) = 1100

We know ?BAC = 1/2 * 1100 = 550

62.. If I is the incentre of ? ABC and angle BIC = 1350, then &Delta ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans :

angle BIC = 1350

=> B/2 + C/2 = 1800 – 1350 = 450

=> angle B + angle C = 900

We know angle A = 1800 – ( angle B + angle C) = 900

i.e., ? ABC is a right angled.

63. The length of a shadow of a vertical tower is 1//?3 times its height. The angle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans :

We know that tan ? = h/1?3 h = ?3 = tan 600

64. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 => [0, y1 =3/2]

[3x – 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 – y2)

= 3/2 – (- ½) = 2 units

65. If x+1/16x = 1, then the value of 64×3 + 1/64×3 is-

a. 4

b. 52

c. 64

d. 76

We know that x + 1/16x = 1

=> 16×2 – 16x + 1 =0

=> 16×2 – 16x + 4 = 3

=> (4x – 2)2 = 3

=> 4x = 2 + ?3

=> 64×3 = (2±?3)3

=8 + 3?3 + 6?3 (2 + ?3)

= 26 + 15?3

We know 64×3 + 1/64×3 = (26 + 15?3) + 1/ (26 + 15?3)

= (26 + 15?3) + 26 – 15?3/676 -675

=52

66. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and b2 ? ca, then the value of a2 + b2 + c2/ b2 –ca is-

a. 3

b. 2

c. 0

d. 1

We know that a + b + c = 0

=> a + c = -b

=> a2 + c2 = b2 -2ac

=> a2 + b2 + c2 = 2b2 – 2ac

We know a2 + b2 + c2/ b2 ac = 2

67. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

We know that a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

=> (a2 + b2)2 – (ab) 2/ (a2 + b2 + ab) = 2

=> a2 – ab + b2 = 2 ….(1)

and a2 + ab + b2 = 4 …..(2)

=> 2ab = 2

=> ab = 1

68. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 – 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

We know that a3 + b3 + c3 – 3abc

= (25)3 + (15)3 + (-10)3 – 3*25*15*(-10)

=15625 + 3375 – 1000 + 11250 = 29250

and (a – b)2 + (b – c)2 + (c – a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

We know Required value = 29250/1950 =15

69. A, B, C are three points on a circle. The tangent at A meets BC produced at T, angle BTA = 400, angle CAT = 440. The angle subtended by BC at the centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans :

angle ACB = 400 + 440 = 840

We know angle ACO = 900 – 440 = 460 = angle OAC

=> angle OCB = angle ACB – angle ACO

= 840 – 460 = 380 = angle OBC

We know angle BOC = 1800 – ( angle OCB + angle OBC)

= 1800 – (380 + 380) = 1040

70. If the length of a chord of a circle at a distance of 12 cm from the Centre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans :

We know that OA = ?OM2 + AM2

= ?122 + 52 = 13

We know Diameter of the circle = 2*OA

= 2*13 = 26cm

71. In &Delta ABC, P and Q are the middle points of the sides AB and AC respectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR = 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

We know that PR/RQ = ½

But, PR = 2cm

RQ = 2*PR

= 4cm

We know PQ = PR + RQ

= 2 + 4 = 6 cm

We know BC = 2*PQ = 12CM

72. The value of cot ?/20 cot 3? /20 cot 5? /20 cot 7? /20 cot 9? /20 is-

a. -1

b. ½

c. 0

d. 1

Ans : Given Exp.

= cot ? /20. cot 3? /20. cot 5? /20. cot 7? /20. cot 9?/20

= cot 90. cot 270. cot 450. cot 630. cot 810

= cot 90. cot 270*1*tan 270. tan 90.

= cot 90. cot 270*1*1/ cot 270*1/ cot 90

=1

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