**Dear Aspirants,** **Our SSC Crackers team** is providing a new series of Quantitative Aptitude Questions for **Upcoming Exam **so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

**1) If (1/x) + (1/y) + (1/z) = 0 and x + y + z = 11, then what is the value of x ^{3} + y^{3} + z^{3} – 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 ****Δ****ABC?**

(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 cm ^{2} and BE = AB and AB = 2 AD, AE||DC, then what is the area (in cm^{2}) 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

**Answers :**

**1) Answer: A**

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

**2) Answer: C**

**3) Answer: A**

By the given condition in question x < 0, only option (a) satisfy the condition,

**4) Answer: B**

**5) Answer: B**

**6) Answer: B**

**7) Answer: C**

∵ OA = OB = radius

∠A = ∠B = 60°

So, ∆AOB is equilateral triangle

So, R = 7

**8) Answer: A**

**9) Answer: D**

**10) Answer: A**