# Bank PO Questions

Q:

If log 2 = 0.30103, Find the number of digits in 256 is

 A) 17 B) 19 C) 23 D) 25

Answer & Explanation Answer: A) 17

Explanation:

${\color{Black}\log&space;(2^{56})=(56\times0.30103)&space;}$ =16.85768.

Its characteristics is 16.

Hence, the number of digits in ${\color{Black}2^{56}&space;}$ is 17.

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23 9213
Q:

The length of a rectangular hall is 5m more than its breadth. The area of the hall is 750 m. The length of the hall is

 A) 20 B) 25 C) 30 D) 35

Answer & Explanation Answer: B) 25

Explanation:

Let breadth = x m

Then, length = (x+5)m

x(x+5) = 750

x² + 5x - 750= 0

(x+30)(x-25)= 0

x = 25

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22 9154
Q:

Find the ratio of the areas of the incircle and circumcircle of a square.

 A) 1:1 B) 1:2 C) 1:3 D) 1:4

Answer & Explanation Answer: B) 1:2

Explanation:

Let the side of the square be x. Then, its diagonal = ${\color{Black}\sqrt{2x^{2}}=\sqrt{2}x}$

Radius of incircle = $\inline \fn_cm \frac{x}{2}$

Radius of circum circle= ${\color{Black}\sqrt{{2}}\times&space;\frac{x}{2}=\frac{x}{\sqrt{2}}}$

Required ratio = $\inline \fn_cm \frac{\prod x^{2}}{4}:\frac{\prod x^{2}}{2}=\frac{1}{4}:\frac{1}{2}=1:2$

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28 9019
Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

 A) 1 B) 2 C) 3 D) 4

Answer & Explanation Answer: C) 3

Explanation:

Area of the park = (60 x 40) = 2400$\inline m^{2}$

Area of the lawn = 2109$\inline m^{2}$

Area of the crossroads = (2400 - 2109) = 291$\inline m^{2}$

Let the width of the road be x metres. Then,

$\inline 60x+40x-x^{2}=291$

$\inline x^{2}-100x+291=0$

(x - 97)(x - 3) = 0
x = 3.

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9 8954
Q:

One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field?

 A) 110 B) 120 C) 130 D) 140

Answer & Explanation Answer: B) 120

Explanation:

Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. m

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32 8754