#### SBI PO Exam Quantitative Aptitude Study Material

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b) 2cm

c) 3cm

d) 6cm

Volume of sphere= surfae area of sphere

(4/3)πr

⇒ r = 3cm diameter = 6 cm

b) 16m

c) 10.25m

d) 20m

b) 72cm

c) 36√3cm

d) 144√3cm

Let ABC is equilateral triangle.

Where sides is a cm.

⇒ Area of equilateral triangle =(?3/4)a

⇒ (1/2)*9*AD = (?3/4)a

⇒ (1/2)*12?3 =(?3/4)a

⇒ a=24cm.

Now, Area of triangle =((?3/4)a

=(?3/4)*24*24 = 148?3cm

b) 1 1/2sq.unit

c) 1sq.unit

d) 4 1/2sq.unit

2x + 3y = 6 ...(ii)

x = 0, y = 2

x = 3, y = 0

From eqn.

(iii) x+ y = 3

x = 0, y = 3

x = 3, y = 0

Area made by these three lines

= Area of triangle OBC – Area of OAC

=(1/2)*3*3-(1/2)*2*3 =(9/2)-3 =3/2 = 1 1/2 sq.

b) 1:2

c) 3:4

d) 2:3

b) 70

c) 72

d) 75

b) 3 sq. unit

c) 6 sq. unit

d) 12 sq. unit

❑(⇒┴ equation of a line parallel to y-axis. y=3)

❑(⇒┴ equation of a line parallel to x-axis)

Putting x=0 in the equation

3x+4y=12,

3×0+4y=12 ❑(⇒┴ y=12/4=3)

?co-ordinates of the point of

intersection on y-axis=(0,3)

Again putting y=0 in the

equation 3x+4y=12,

3x+4×0=12 ❑(⇒┴ )x=12/3=4

?co-ordinates of the point of

intersection on x-axis=(4,0) AC = 3units,BC= 4 units Therefore Area of ΔABC =1/2 ×BC×AC =1/2 ×4×3 = 6 sq.units

b) 3/2

c) 2/√3

d) 4

b) 9

c) 18

d) 15

b) 12 cm

c) 10 cm

d) 15, cm