Quantitative Aptitude Questions Daily Quiz Day – 49

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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.

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1) If (p2 + q2)/(r2 + s2) = (pq)/(rs), then what is the value of (p – q)/(p + q) in terms of r and s?

(a) (r + s) /(r – s)

(b) (r – s) /(r + s)

(c) (r + s) /(rs)

(d) (rs) /(r – s)

2) If the expression (px3 – 8x2 – qx + 1) is completely divisible by the expression (3x2 – 4x + 1), then what will be the value of p and q respectively?

(a) (21/4, 15/8)

(b) (6, 1)

(c) (33/4, 5/4)

(d) (1, 6)

3) If a = 2017, b = 2016 and c = 2015, then what is the value of a2 + b2 + c2 – ab – bc – ca?

(a) -2

(b) 0

(c) 3

(d) 4

4) 

(a) 3

(b) 7

(c) 11

(d) 18

5)

(a) 1

(b) 3

(c) 0

(d) 21

6) In ΔPQR, P :Q : R = 2 : 2 : 5. A line parallel to QR is drawn which touches PQ and PR at A and B respectively. What is the value of PBA – PAB?

(a) 60

(b) 30

(c) 24

(d) 36

7) In the given figure, O is the center of the circle, DAB = 110o and BEC = 100o. What is the value (in degrees) of OCB?  

(a) 5

(b) 10

(c) 15

(d) 20

8) If ΔDEF is right angled at E, DE = 15 and DFE = 60°, then what is the value of EF?

(a) 5√3

(b) 5

(c) 15

(d) 30

9) In the given figure, area of isosceles triangle PQT is 128 cm2 and QT = PQ and PQ = 4 PS, PTSR, then what is the area (in cm2) of the quadrilateral PTRS?

(a) 80

(b) 64

(c) 124

(d) 72

10) In the given figure, BD passes through centre O, AB = 12 and AC = 8. What is the radius of the circle?

(a) 3√2

(b) 4√3

(c) 3√5

(d) 3√3

Answers :

1) Answer: B

Put p = q = r = s = 2

So, option (b) satisfy the conditions.

2) Answer: C

3) Answer: C

4) Answer: D

5) Answer: D

6) Answer: A

The value of Angle P : Q : R = 40 & 40 and 100

So, ∠PRQ = ∠PBA

∠PAB =∠PQR parallel lines.

(∠PBA-∠PAB) = (100-40)

=60°

7) Answer: B

∠DAB=110°

∠BEC=100°

∠BOC (exterior)

= 2∠BEC

8) Answer: A

9) Answer: B

10) Answer: C

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