GATEWAY TO CGL MAINS 2015 : Advanced Quant Quiz

1.The diagonals AC and BD of a ∥ gm ABCD intersect each other at the point O such that ∠DAC = 50° and ∠AOB = 80°. Then ∠DBC =? 

(a) 50° 

(b) 40° 

(c) 45° 

(d) 30°

2.The perimeter of ∥ gm is 22 cm. If the longer side measures 6.5 cm. what is the measure of the shorter side? 

(a) 5.5 cm 

(b) 4.5 cm 

(c) 6.0 cm 

(d) 5.0 cm 

3.If an angle of a ∥ gm is tow- third of its adjacent angle, then the largest angle of ∥ gm :  

(a) 72° 

(b) 60° 

(c) 108° 

(d) 120°

4.PQRS is a square. The ∠SRP is equal to : 

(a) 90° 

(b) 45° 

(c) 100° 

(d) 60°

5.ABCD is a rhombus with ∠ABC = 50°, then ∠ACD is: 

(a) 50° 

(b) 90° 

(c) 65° 

(d) 70°

6.PQRS is a ∥ gm. PX and QY are respectively, the perpendicular from P and Q to SR and SR produced. Then PX is equal to : 

(a) QY 

(b) 2QY 

(c) ½ QY 

(d) XR 

7. ABCD is a ∥gm CL ⊥ AD and DM ⊥ BA. If CD = 16 units, DM = 12 units and CL = 15 units, then AD = ? 

(a) 12. 8 units 

(b) 13.6 units 

(c) 11.1 units 

(d) 12.4 units 

8.In a parallelogram ABCD, AO and BO are the bisectors of ∠A and ∠B respectively, then ∠AOB is equal to : 

(a) 60° 

(b) 120° 

(c) 100° 

(d) 90°

9.If one of the interior angles of a regular polygon is equal to 5/6 times of one of the interior angles of a regular pentagon, then the no. of sides of the polygon: 

(a) 3 

(b) 4 

(c) 6 

(d) 8 

10.Difference between the interior and exterior angles of regular polygon is 60°. The number of sides in the polygon is: 

(a) 5 

(b) 6 

(c) 8 

(d) 9

ANSWERS AND SOLUTION

1.(d) ∠ACB = ∠DAC = 50° (Alternate interior ∠s)
∠BOC = 180° – 80° = 100°
so Now, in ∆BOC
∠OBC = 180° – (100° + 50°) = 30°

2.(b)  Perimeter of ∥ gm = 22cm
=> 2(a + b) = 22cm => a + b = 11
=> b = 11 – a => 11 - 6.5 = 4.5cm
so shorter side, b = 4.5cm

3.(c) Since, adjacent angles of a ∥ gm are supplementary.
so x + 2/3 + x = 180°
=> 5x/3 = 180°
=> X = 108°
so 2/3x + 2/3 * 108° = 72°
so Agnles are = 108°, 72°, 108°, 72°
so largest angle  = 108°

4.(b) PQRS is square, SP = SR and ∠S = 90°
and ∠SRP= ∠SPR=1/2(90°)=45°
Hence, ∠SRP = 45°

5.(c) Since, AB = BC
 so ∠BAC=∠BCA= 1/2 (180°-50°)
 = 65°(AB = BC)
so ∠ACD = ∠BAC=65°
[AB ∥ DC]

6.(a) In ∆PSX and ∆QRY
∠x= ∠y=90° and SX=RY
(sx= SY – XY and RY = SY-SR= SY – PQ = SY – XY]
And PS = QR (sides of a ∥ gm)
 SO ∆PSX ≅ ∆QRY (R.H axion)
 SOPX = QY

7.(a) Area of  ∥ gm ABCD = Base *  height
=>AB * DM = AD * CL
=> 16 * 12 = AD * 15
=> AD = 12.8 units

8.(d)
9.(b) interior agnle of pentagon = 180° – 360°/5 = 108°
so Interior angle of required polygon = 5/6 * 108°= 90°
so Each exterior angle of the required polygon = 180°-90°=90°
so No.of sides = 360/90 = 4
10.(b)