Maths questions series #4

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Very important Maths questions for SSC

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Q. 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


Q. 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.



Q. 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



Q. 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


Q. 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


Q. 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

Q. 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

Q. 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


Q. 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



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