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# IBPS RRB PO Most Expected Questions PDF

Download Top-20 IBPS RRB PO Most Expected Questions PDF. Most Expected questions based on asked questions in previous year exam papers very important for the IBPS RRB PO (Officer Scale-I, II & III) exam

Question 1:Â The amount that the banking ombudsman charge when an individual files a complaint regarding any bank to it?

a)Â 25

b)Â 50

c)Â 100

d)Â No fees involved

e)Â 100rs and fees will be reimbursed if the individual wins the case.

Question 2:Â â€˜Letâ€™s Make Money Simpleâ€™ is the tagline of which bank?

a)Â ICICI Bank

b)Â Kotak Mahindra Bank

c)Â RBL Bank

d)Â Karnataka Bank

e)Â Citi Bank

Question 3:Â Which of the following is the first bank to introduce internet banking?

a)Â Punjab National Bank

b)Â ICICI Bank

c)Â Corporation Bank

d)Â State Bank of India

e)Â HDFC Bank

Question 4:Â An asset is classified as a substandard asset if it has remained as an NPA (Non Performing Asset) for a period ____________

a)Â less than 12 months

b)Â more than or equal to 12 months

c)Â less than or equal to 12 months

d)Â less than or equal to 6 months

e)Â equal to 6 months

Question 5:Â What is the interest rate of Public Provident Fund (PPF) at present (July 2019 to September 2019)?

a)Â 8.5%

b)Â 7.8%

c)Â 8.3%

d)Â 8.0%

e)Â 7.9%

Question 6:Â A person lent same amount of Rs 10000 at 30% Simple interest for time period of 2 years and 20% Compound interest for time period of 3 years then what is the difference between the interest received ?

a)Â 1200

b)Â 1250

c)Â 1280

d)Â 1240

e)Â 1260

Question 7:Â Compound interest on a certain sum for 3 years is Rs 2880 and for 4 years it is Rs 3456. What is the value of sum taken ?

a)Â Rs 5000

b)Â Rs10000

c)Â Rs 8000

d)Â Rs 9000

e)Â Rs 7000

Instructions

Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.

Question 8:Â Quantity 1: Simple interest charged by a bank on a sum of Rs. 1000 at the rate of 22% annum for 2 years.
Quantity 2: Compound interest charged by another bank on a sum of Rs. 1000 at the rate of 20% annum for 2 years compounded annually.

a)Â Quantity 1 > Quantity 2

b)Â Quantity 1 $\geq$ Quantity 2

c)Â Quantity 1 < Quantity 2

d)Â Quantity 1 $\leq$ Quantity 2

e)Â Quantity 1 = Quantity 2

Question 9:Â The coordinates of the vertices of a right-angled triangle are A (6, 2), B(8, 0) and C (2, -2). The coordinates of the orthocentre of triangle PQR are

a)Â (2, -2)

b)Â (2, 1)

c)Â (6, 2)

d)Â (8, 0)

Question 10:Â find the area of an equilateral triangle if the height of the triangle is 24 cm.

a)Â $192\sqrt{3} cm^2$

b)Â $175\sqrt{3} cm^2$

c)Â $178\sqrt{3} cm^2$

d)Â $164\sqrt{3} cm^2$

Question 11:Â If P, Q and R are the three points on a circle such that the angle subtended by the chords PQ and PR at the centre O are 80Â° and 130Â° respectively then LQPR is equal to

a)Â 80Â°

b)Â 75Â°

c)Â 90Â°

d)Â 70Â°

Question 12:Â â€‹If a metallic cone of radius 30 cm and height 40 cm is melted and recast into metallic spheres of radius 10 cm, find the number of spheres formed.

a)Â 5

b)Â 9

c)Â 6

d)Â 12

Instructions

Find the missing number (?) in the given series from the options provided.

Question 13:Â 2, 4, 7, 11, 16, ?

a)Â 22

b)Â 21

c)Â 20

d)Â 23

e)Â 24

Question 14:Â 9, 27, 45, 63, ?

a)Â 79

b)Â 80

c)Â 81

d)Â 82

e)Â 83

Question 15:Â 3, 8, 15, 24, 35, ?

a)Â 63

b)Â 49

c)Â 80

d)Â 48

e)Â 56

Question 16:Â If two sides of a triangle are 6 cm and 4 cm then how many integral values are possible for the third side ?

a)Â 6

b)Â 7

c)Â 8

d)Â 9

e)Â 10

Question 17:Â A person standing on the top of a building of height 20m observes the angle of elevation of the top of other building as 60 degrees and angle of depression of the bottom the same building as 30 degrees. What is the difference between the heights of two building ?

a)Â 10m

b)Â 40m

c)Â 50m

d)Â 30m

e)Â 60m

Question 18:Â A cuboid has its length, breadth and height in the ratio of 4:5:6 If the length is decreased b 25% ,breadth is increased by 20% and height is increased by 33.33 % then what is the percentage change in total surface area of the cuboid ?

a)Â 21.1%

b)Â 23.8%

c)Â 22.3%

d)Â 21.6%

e)Â 22.6%

Question 19:Â Three sides of a triangle are 12 cm,14 cm and 16 cm. What is the value of inradius of the triangle ?

a)Â $\frac{32}{\sqrt{15}}$

b)Â $\frac{12}{\sqrt{15}}$

c)Â $\frac{8}{\sqrt{15}}$

d)Â $\frac{24}{\sqrt{15}}$

e)Â $\frac{16}{\sqrt{15}}$

Question 20:Â Area of a rectangle is 9600 square meters. Length of the rectangle is twice the length of the diagonal of a square having area as 1800 square meters.If a lady travels whole perimeter of the rectangle in 8 seconds then what is the speed of the lady ?

a)Â 20 m/s

b)Â 30 m/s

c)Â 50 m/s

d)Â 25 m/s

e)Â 45 m/s

Given P=10000
R=30%
T=2 years
I=(10000*2*30)/100
=Rs 6000
R%=r
CI=$P(1+(r/100))^{3}-P$
CI=10000(216/125) -10000
CI=10000(91/125)
CI=7280
Difference=7280-6000
=Rs 1280

Let R%=r
Given $Pr(1+r)^{2}$=2880
$Pr(1+r)^{3}$=3456
Dividing both the equations we have
1/(1+r)=5/6
6=5+5r
r=1/5
R=(1/5)*100
R=20%
Therefore $P(1/5)(1+(1/5))^{2}$=2880
P(36/125)=2880
P=2880*125/36
P=Rs 10000

Simple interest paid to bank = 1000*.22*2 = Rs. 440
Compound interest paid to bank = $1000(1 + \dfrac{20}{100})^2 – 1000$ = Rs. 440
Hence, we can say that Quantity 1 = Quantity 2. Option E is the correct answer.

Given,
AD = 15 cm and ABC is an equilateral triangle

Centre of the circle is O

Given, a metallic cone of radius 30 cm and height 40 cm
Volume of metallic cone = $\frac{1}{3} \pi r^2 h$

$\Rightarrow$ $T_{2} – T _{1} = 4 – 2 = 2$
$\Rightarrow$ $T_{3} – T _{2} = 7 – 4 = 3$
$\Rightarrow$ $T_{4} – T _{3} = 11 – 7 = 4$
$\Rightarrow$ $T_{5} – T _{4} = 16 – 11 = 5$
Here we can see the pattern that difference between consecutive terms are in A.P. and it is increasing by 1. Therefore,
$\Rightarrow$ $T_{6} = 6 + T _{5} = 6+16 = 22$
Hence, option A is the correct answer.

$\Rightarrow$ $T_{1} = 9 = 5^2 – 4^2$
$\Rightarrow$ $T_{2} = 27 = 6^2 – 3^2$
$\Rightarrow$ $T_{3} = 45 = 7^2 – 2^2$
$\Rightarrow$ $T_{4} = 63 = 8^2 – 1^2$
Hence, we can say that $\Rightarrow$ $T_{n} = (n+4)^2 – (5-n)^2$
$\Rightarrow$ $T_{5} = (5+4)^2 – (5-4)^2 = 81$
Hence, option C is the correct answer.

$\Rightarrow$ $T_{1} = 3 = 2^2 – 1$
$\Rightarrow$ $T_{2} = 8 = 3^2 – 1$
$\Rightarrow$ $T_{3} = 15 = 4^2 – 1$
$\Rightarrow$ $T_{4} = 24 = 5^2 – 1$
$\Rightarrow$ $T_{5} = 35 = 6^2 – 1$
Hence, we can say that $\Rightarrow$ $T_{n} = (n+1)^2 – 1$
$\Rightarrow$ $T_{6} = (6+1)^2 – 1 = 48$
Hence, option D is the correct answer.

For a triangle to form the sum of two sides should be greater than the third side and the difference of two sides should be less than the third side
Therefore (6+4)>s>(6-4)
10>s>2
So all the values between 2 and 10 are possible as third side
i.e 3,4,5,6,7,8 and 9
Therefore 7 values are possible

Let the height of other building be â€˜hâ€™
Distance between the buildings be â€˜dâ€™
From the figure we have $\tan60$=d/20
d=$\sqrt{3}20$
From the figure we have in the upper triangle $\tan60$=x/d
$\sqrt{3}$=x/$\sqrt{3}20$
x=60 cm
Therefore h=60+20
h=80 cm
Difference=80-20
=60 cm

Let the length,breadth and height of the cuboid be 4x,5x and 6x respectively
Total surface area of the cube=2lb+2bh+2lh
=$2*(20*x^{2})+2*(24*x^{2})+2*(30*x^{2})$
=$(40+48+60)x^{2}$
=$148x^{2}$
New length of cuboid=4x*3/4
=3x
=6x
New height of cuboid=6x*(4/3)
=8x
Total surface area of cuboid with new dimensions=$2*(18*x^{2})+2*(48*x^{2})+2*(24*x^{2})$
=$(36+96+48)x^{2}$
=$180x^{2}$
Percentage change in the total surface area=((180-148)/148)*100
=800/37
=21.6%

We know that Area of a triangle=abc/(4R)
S=(a+b+c)/2
=(12+14+16)/2
=42/2
=21
Area of a triangle=$\sqrt{s(s-a)(s-b)(s-c)}$
=$\sqrt{21(21-12)(21-14)(21-16)}$
=$\sqrt{21(9)(7)(5)}$
=$21\sqrt{15}$
R=abc/(4*area)
R=(12*14*16)/(4*$21\sqrt{15}$)
R=$\frac{32}{\sqrt{15}}$

Given that l*b=9600 sq mtrs
Area of a square $s^{2}$=1800
$s^{2}$=30*30*2
s=30$\sqrt{2}$
Given that l=2d
d=$\sqrt{2}$s
l=2d
l=2$\sqrt{2}$s
l=$2\sqrt{2}*30\sqrt{2}$
l=120 meters
Aslo 120*b=9600
l=80 meters
Perimeter of the rectangle=2(l+b)
=2*(120+80)
=2*200
=400 meters
Given time=8 sec
Speed=400/8
=50 m/s