# SSC Quantitative Aptitude Questions (Day-41)

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 (x/y)5a-3 = (y/x)17-3a, then what is the value of a?

(a) -7

(b) -5

(c) 0

(d) 3

2) (a) 1

(b) (x – 3)/(x + 3)

(c) (x + 4)/(x – 3)

(d) (x – 3)/(x + 4)

3) (a) (3√3)/5

(b) (3√15)/5

(c) (3√15)/8

(d) (3√5)/8

4) (a) 18

(b) 34

(c) 40

(d) 62

5) What is the simplified value of ? (a) (b) (c) (d) 6) If P is the circum-center in ΔABC, ∠BPC = 30°, then what is the value (in degrees) of ∠BAC?

(a) 30

(b) 60

(c) 75

(d) 15

7) In ΔPQR, ∠PQR = 90°, PQ = 5 cm and QR = 12. What is the radius (in cm) of the circum-circle of ΔPQR?

(a) 6.5

(b) 7.5

(c) 13

(d) 15

8) In the given figure, O is the centre of the circle,
PQO = 30o and QRO = 45o. What is the value (in degrees) of POR? (a) 150

(b) 110

(c) 160

(d) 130

9) In the given figure, O is the centre of the circle, OQ is perpendicular to RS and
SRT = 30o. If RS = 10√2, then what is the value of PR2? (a) 200(1 + √3)

(b) 300(2 + √3)

(c) 200(2 + √3)

(d) 100(3 + 2√3)

10) If ΔABC is right angled at B, AB = 30 and ∠ACB = 60°, then what is the value of AC?

(a) 20

(b) 20√3

(c) 40

(d) 60 As base are same

∴ 5a – 3 = -17 + 3a

2a = – 14

a = – 7

Put x = 0 ⇒ 1    As P is circumcenter

∴ ∠ BAC = ½ × ∠ BPC

=15° ∴ Radius of the circum – circle = 13/2=6.5 QO = OR

∴ ∠ ORQ = ∠ OQR = 45°

And ∠ PQR = 30° + 45° = 75°

∴ ∠ POR = 2 × ∠ PQR

=  2 × 75

= 150°    