GATEWAY TO CGL MAINS 2015 : Advanced Quant Quiz

1.The radius of a circle is 13 cm and the length of one of its chords is 10 cm. Find the distance of chord from the centre. 
(a) 8 cm 
(b) 10 cm 
(c) 9 cm 
(d) 12 cm


2.In the given figure PQ = 12, BQ = 8cm, then the length of chord AB : - 

(a) 10 cm 
(b) 4√5 cm 
(c) 4 cm 
(d) 18 cm




3.The value of x will be : 

(a) 70° 
(b) 90° 
(c) 60°  
(d) 75°







4.The value of x : - 

(a) 10 cm
(b) 9 cm
(c) 7.5 cm
(d) 11 cm





5.In the given figure ABCD is a cyclic quadrilateral and O is the centre of the circle. If ∠BOC = 138, then ∠BDC will be 

(a) 112° 
(b) 111° 
(c) 109°  
(d) None of these






6.In the given figure, O is the centre of the circle, then the value of x will be : - 

(a) 40°
(b) 90°
(c) 45° 
(d) 30°






7.In the given figure, PA and PB are tangents from a point P to a circle such that PA = 6 cm and ∠APB = 60°. What is the length of the chord AB? 

(a) 12cm  
(b) 8cm  
(c) 9cm  
(d) 6cm






8.In the given figure, AB = AC and ∠ABC = 50°, ∠BDC : - 

(a) 60° 
(b) 80° 
(c) 100° 
(d) 90°









9.In the given figure, ∆ABC is an equilateral triangle. Find ∠BEC. 

(a) 60° 
(b) 120° 
(c) 80° 
(d) 90°









10.In the given figure, AD ∥ BC, if ∠ABC =72°, then ∠BCD = ? 

(a) 108° 
(b) 36° 
(c) 90° 
(d) 72°






ANSWERS AND SOLUTION:
1.(d)

The perpendicular from the centre of a circle to chord bisects the chord.
so AL = LB = 1/2 AB = 5 cm
OA^2 = OL^2 + AL^2
=> 13^2 = OL^2 + 5^2
=> OL^2 = 13^2 – 5^2
=> OL = 12 cm

2.(a) PQ^2 = BQ * AQ
=> (12)^2 = AQ * 8 =>AQ = 18 cm
so AB = AQ – BQ = 18 – 8 = 10 cm

3.(d) ∠ACB = ∠ADB = 20°
(made by same are AB)
so In triangle ACB, ∠x° = 180°-85° – 20° = 75°

4.(d) PA * PC = PB * PD
=> 14 * 9 = (7 + x) * 7
=> 18 = 7 + x => x = 11 m 

5.(b) ∠BAC = 1/2 * 138° =  69°
so ∠BDC = 180° – 69° = 111°

6.(c) In triangle OBC
OB = OC  so ∠B= ∠C=45°
so  ∠D= ∠C
(made by same arc AB)
so ∠D=x°=45°

7.(d) PA = PB
so ∠PAB= ∠PBA 
Also, ∠PAB+ ∠PBA+ ∠APB=180°
=> ∠PAB+ ∠PBA=120°
 so ∠PAB= ∠PBA=60°
i.e ∆PAB is an equilateral triangle.
AB = 6 cm

8.(b)
AB = AC => ∠ACB=∠ABC50°
so ∠BAC=180°-(50+50)= 80°
so  ∠BDC= ∠BAC=80°
(angle by same are BC)

9.(b) ∠BAC=60° 
so ∠BEC=180°-60°=120°

10.(d) ∠D=180° 72°=108°
so ∠BCD=180°-108°
= 72°(∵AD ∥ BC)