# SSC Quantitative Aptitude Questions (Day-37)

Dear Aspirants, you can find the Quantitative Aptitude questions with detailed explanations for the SSC exams. Nowadays the competitive level of the exam has been increasing consistently. Due to the great demand for the government job, the level of the toughness reached greater. Candidates have to enhance the preparation process in order to drive in the right path. It doesn’t need to clear the prescribed cutoff. You must have to score good marks more than the cut off marks to get into the final provisional list. Here we have updating the Quantitative Aptitude questions with detailed explanations on a daily basis. You can practice with us and measure your level of preparation. According to that you can sculpt yourself in a proper way. SSC aspirants kindly make use of it and grab your success in your career.

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1) If (1/x) + (1/y) + (1/z) = 0 and x + y + z = 11, then what is the value of x3 + y3 + z3 – 3xyz?

(a) 1331

(b) 2662

(c) 3993

(d) 14641

2) (a) 1

(b) x

(c) x – 1

(d) 1/(x – 1)

3) (a) -4

(b) 0

(c) 3

(d) 16

4) (a) 3

(b) 7

(c) 9

(d) 11

5) (a) 2

(b) 2√2

(c) √5

(d) √3

6) In ΔABC,
BCA = 90°, AC = 24 cm and BC = 10 cm. What is the radius (in cm) of the circum-circle of ΔABC?

(a) 12.5

(b) 13

(c) 25

(d) 26

7) A chord of length 7 cm subtends an angle of 60° at the centre of a circle. What is the radius (in cm) of the circle?

(a) 7√2

(b) 7√3

(c) 7

(d) 14

8) If ΔPQR is right angled at Q, PQ = 12 and
PRQ = 30°, then what is the value of QR?

(a) 12√3

(b) 12√2

(c) 12

(d) 24

9) In the given figure, area of isosceles triangle ABE is 72 cm2 and BE = AB and AB = 2 AD, AE||DC, then what is the area (in cm2) of the trapezium ABCD? (a) 108

(b) 124

(c) 136

(d) 144

10) In the given figure, AC and DE are perpendicular to tangent CB. AB passes through centre O of the circle whose radius is 20 cm. If AC= 36 cm, what is the length (in cm) of DE ? (a) 4

(b) 6

(c) 2

(d) 8

1/x + 1/y + 1/z = 0

Xy + yz + zx = 0,         x + y + z  = 11

x² + y² + z² = (x + y+ z)² – 2(xy + yz + zx)

= 121 – 2 × 0 = 121

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)

= 11 × (121 – 0)

= 1331  By the given condition in question x < 0, only option (a) satisfy the condition,    ∵ OA = OB = radius

∠A = ∠B = 60°

So, ∆AOB is equilateral triangle

So, R = 7    