# 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 doesnâ€™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.

Start Quiz

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)Â

(a) 1

(b) (x – 3)/(x + 3)

(c) (x + 4)/(x – 3)

(d) (x – 3)/(x + 4)

3)Â

(a) (3âˆš3)/5

(b) (3âˆš15)/5

(c) (3âˆš15)/8

(d) (3âˆš5)/8

4)Â

(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 Î”ABC, âˆ BPC = 30Â°, then what is the value (in degrees) of âˆ BAC?

(a) 30

(b) 60

(c) 75

(d) 15

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

(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 = 10âˆš2, then what is the value of PR2?

(a) 200(1 + âˆš3)

(b) 300(2 + âˆš3)

(c) 200(2 + âˆš3)

(d) 100(3 + 2âˆš3)

10) If Î”ABC is right angled at B, AB = 30 and âˆ ACB = 60Â°, then what is the value of AC?

(a) 20

(b) 20âˆš3

(c) 40

(d) 60

As base are same

âˆ´ 5a â€“ 3 = -17 + 3a

2a = â€“ 14

a = â€“ 7

Put x = 0

â‡’ 1

As P is circumcenter

âˆ´ âˆ  BAC = Â½ Ã— âˆ  BPC

=15Â°

âˆ´ Radius of the circum â€“ circle = 13/2=6.5

QO = OR

âˆ´ âˆ  ORQ = âˆ  OQR = 45Â°

And âˆ  PQR = 30Â° + 45Â° = 75Â°

âˆ´ âˆ  POR = 2 Ã— âˆ  PQR

=Â  2 Ã— 75

= 150Â°