Banking Exams Quant: Essential Concepts, Practice Questions & Preparation Tips Free PDF

In this exam era in which all competitive examinations are being conducted online, especially the banking and government exams, it’s time to get your quantitative aptitude section strong. A huge number of aspirants sit for these banking exams every year and they face difficulties in the quant section majorly. Quantitative Aptitude is one section that involves a lot of calculation and practice failing which you will not be able to clear the exam. Even many banking exams like RBI, SEBI, SBI, include tough quantitative aptitude questions. Quan is an integral part of the competitive bank examinations and one must clear both the prelims and mains stage to crack the exam. As far as prelims and mains is concerned the difficulty level of the questions is high as compared to the prelims exam. Below we have categorized the important quant concepts and their explanations.

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Important Quant Concepts & Their Explanations

It is important to understand these critical concepts or topics for quant preparation especially for banks. Thus, it’s important for the candidates to prepare the below mentioned topics thoroughly. 

• Partnership
• Time, Speed & Distance
• Simple Interest Compound Interest
• Mensuration
• Permutation & Combination
• Ratio & Proportions
• Time & Work
• Probability
• Profit & Loss
• Average
• Income & Expenditure
• Ages
• Simplification & Approximation

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Banking Exams Reasoning: Essential Concepts, Practice Questions & Preparation Tips Free PDF

Partnership

Basics:

In a partnership, the ratio of profit shares of two different persons ‘A’

and ‘B’ is the respective ratio of the product of their investment and

time period of investment.

Suppose two different persons ‘A’ and ‘B’ started a business by

investing their capitals in the ratio p:q, respectively

And, the ratio of time period of investment of ‘A’ and ‘B’ is m:n,

respectively,

Then, ratio of profit shares of ‘A’ and ‘B’, respectively = (p × m):(q ×

n)

Practice Questions on Partnership for RBI Assistant Prelims Exam

Time, Speed & Distance

Basics:

Time taken to travel a certain distance = Distance travelled ÷ average

speed of travel

Speed of travel = Distance travelled ÷ Time taken for the distance

travelled

Distance travelled = Speed at which the distance was travelled × Time

taken for the travel

Trains:

Trains are objects with significant length. So while calculating measures

such as speed, distance travelled or time taken by a train, we also take

into consideration the length of the train.

For example, the time taken by the train to cross a pole is equal to the

time taken by the just one point of the train to cross a distance that is

equal to the length of the train.

Average speed:

Suppose to travel a distance of ‘2d’ km, if a person travelled ‘d’ km at a

speed of ‘p’ km/h and remaining distance at a speed of ‘q’ km/h, then

average speed of the journey

= total distance travelled ÷ total time taken

Time taken to travel first ‘d’ km = (d/p) hours

Time taken to travel next ‘d’ km = (d/q) hours

Boat and stream:

Problems based on boat and stream is solved mostly using the same

concepts as standard time, speed and distance problems.

Basics:

The speed of a boat changes with respect to the direction of travel of the

boat with respect to the stream.

Let the speed of a boat in still water be ‘x’ km/h

Let the speed of the stream in which the boat is sailing be ‘y’ km/h

Then, speed of the boat while sailing against the stream i.e. in upstream

= (x – y) km/h

Speed of the boat while sailing with the stream i.e. in downstream = (x

+ y) km/h

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Time, Speed & Distance Questions for Bank Exams

Simple Interest Compound Interest

Compound interest:

Interest = Amount – Principal

If Rs. ‘P’ is invested at a rate of ‘r%’ p.a. compounded annually for ‘t’

years, then the amount received after ‘t’ years will be = P × {1 +

(r/100)} t

If the interest is not compounded annually, then

Effective rate of interest = Actual rate ÷ Number of times the interest is

compounded in a year

Effective time period = Actual time period × Number of times the

interest is compounded in a year

Simple interest:

If Rs. ‘P’ is invested at simple interest of ‘r%’ p.a. for ‘t’ years, then the

amount received after ‘t’ years will be = P + {(P × r × t)/100}

And, the simple interest received = {(P × r × t)/100}

Relationship between simple interest (SI) and compound interest (CI):

When a principal (P) is invested at the rate of ‘r%’ p.a., then

Difference between SI and CI after two years = P × (r/100) 2

Difference between SI and CI after three years = P × (r/100) 2 × {(r/100)

+ 3}

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Mensuration

Mensuration deals with measurement of areas and volumes of different

figures. These figures include both ‘2D’ (two dimensional) and ‘3D’

(three dimensional) figures (also known as solids).

Note: By definition, ‘2D’ figures exist in a two dimensional world and

hence have no volume.

Let’s take a brief look at the formulae for areas of various ‘2D’ figures,

and areas and volumes of ‘3D’ figures.

2D Figures:

Polygons: A Polygon is a closed figure made up of at least 3 line

segments in a two-dimensional plane.

Triangle: A triangle is a polygon containing 3 sides. The area of a

triangle is represented by the figure Δ.

Formulas associated with triangles:

Area:

I. For any triangle in general;

For a triangle having sides of length ‘a’ cm, ‘b’ cm and ‘c’ cm;

Perimeter of triangle = Sum of length of three sides = (a + b + c) cm.

Area = √{s × (s – a) × (s – b) × (s – c)}

Where ‘s’ is the semi-perimeter of the triangle. ‘s’ = {(a + b + c)/2}

This formula is universally known as heron’s formula.

II. For a triangle whose length of base and length of altitude (Height) to

that base is given;

Area = (1/2) × base × height

III. Equilateral triangle: If all three sides of a triangle are equal, then it is

an equilateral triangle.

For an equilateral triangle with side ‘a’ cm, Area = (√3/4) × a 2

Using (II), we have; (1/2) × a × height = (√3/4) × a 2

Or, height = (√3/2) × a

IV. Isosceles triangle: If any two sides of a triangle are equal, then it is

an isosceles triangle.

Area = (a/4)√(4c 2 – a 2 ) {where ‘c’ is the length of each of the two equal

sides and ‘a’ is the length of the third (un-equal) side}

Quadrilaterals: Any polygon having for sides is called a quadrilateral.

I. Area of a normal quadrilateral with vertices ‘A’, ‘B’, ‘C’ and ‘D’ (As

shown in the figure)

Area of quadrilateral ABCD = (1/2) × Length of BD × (AF + DE),

where AF and DE are perpendicular to BC.

II. Trapezium: A trapezium is a quadrilateral having exactly two

parallel sides.

Area of trapezium = (1/2) × sum of parallel sides × Height of the

trapezium.

III. Parallelogram: A parallelogram is a quadrilateral whose opposite

sides are parallel as well as equal to each other.

Area = length of base × height (perpendicular distance)

IV. Square: A square is a parallelogram whose all four sides are of

equal length and angles are of 90 o each.

Area = (a) 2 {where ‘a’ is the length of each side}

V. Rhombus: A Rhombus is a parallelogram whose all four sides are

equal.

Area = (1/2) × product of length of diagonals

VI. Rectangle: A rectangle is a parallelogram that has all four angles of

90 o

For a rectangle with length ‘L’ cm and Breadth ‘B’ cm,

Area = L × B

And, Perimeter = 2(L + B)

Circle:

For a circle with radius ‘r’ units

Area = πr 2

Circumference = 2πr

Area of sector = (θ o /360 o ) × πr 2 {Where θ is the angle subtended inside

the sector}

Length of arc = (θ o /360 o ) × 2πr

Area and Volumes of solids:

Cube: A cube is a solid shape with six square faces.

Total surface area of cube = 6a 2 {Where ‘a’ is the length of edge of the

cube}

Volume of cube = a 3

Cuboid: A cuboid is bound by 6 rectangular faces. The opposite faces

of a cuboid are equal rectangles lying in parallel planes.

Volume of a cuboid with length ‘x’ cm, breadth ‘y’ cm and height ‘h’

cm = (x × y × h)

Total surface area of cuboid = 2(xy + yh + xh)

Lateral surface area of a cuboid = 2xh + 2yh = 2h(x + y)

Cone: A cone is a solid which has a circle at its base and a slanting

lateral surface that converges at the apex.

Curved surface area of the cone = π × r × s {Where ‘r’ is radius and ‘s’

is slant height}

Slant height of a cone = √(r 2 + h 2 ), where ‘h’ is height of the cone.

Total surface area of cone = Curved surface area + area of base = πrs +

πr 2 = πr(r + s)

Frustum of a cone: If a cone is cut into two parts by a plane parallel to

the base, the portion with the upper tip of the cone still remains a cone;

however the portion that contains the base is called the frustum of the

cone.

Curved surface area of frustum = π × s(R + r) {Where ‘s’ is slant height

,‘R’ is radius of the base and ‘r’ is the radius of the top.}

Total surface area of the frustum = π(R 2 + r 2 + Rs + rs)

Volume of the frustum = (1/2) × π × h × {R 2 + r 2 + Rr} {Where ‘h’ is

the height of the cone}

s 2 = (R 2 – r 2 ) + h 2

Cylinder: Cylinder is a three-dimensional solid that holds two parallel

bases joined by a curved surface, at a fixed distance.

Curved surface area of a cylinder = 2πrh {Where ‘r’ is the radius of the

base and ‘h’ is the height of the cylinder}

Total surface area of the cylinder = 2πrh + 2πr 2 = 2πr(r + h)

Volume of cylinder = πr 2 h

Sphere: Sphere is a three dimensional solid, that has all its surface

points at equal distances from the centre. The distance between the

centre and a surface point of the sphere is called its radius (r).

Volume of sphere = (4/3) × πr 3

Surface area of a sphere = 4πr 2

Hemisphere: Where a sphere is divided into two equal and identical

parts, the new formed parts are called hemispheres.

Volume of hemisphere = (2/3)πr 3

Total surface area of hemisphere = 3πr 2

Curved surface area of hemisphere = 2πr 2

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Permutation & Combination

Factorial:

Let ‘n’ be a natural number, then n! = n × (n – 1) × (n – 2) × … × 1

Permutation:

Permutation is the arrangement of different number of things where we

take into consideration the order or arrangement.

Combination:

The number of different groups or selections that can be formed by taking some or all items at a time, is called combination.

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Ratio & Proportions

1. Ratio: The ratio of two quantities ‘a’ and ‘b’, in the same units, is the

fraction (a/b) and we write it as a:b.

2. Proportion: The equality of two ratios is called proportion.

If a:b = c:d, we write, a:b::c:d and we say that ‘a’, ‘b’, ‘c’, ‘d’ are in

proportion. Here ‘a’ and ‘d’ are called extremes, while ‘b’ and ‘c’ are

called mean terms. Also, Product of means = Product of extremes.

Thus, a:b::c:d = (b × c) = (a × d)

3. Third Proportional: If a:b = b:c, then ‘c’ is called the third

proportional to ‘a’ and ‘b’.

4. Fourth Proportional: If a:b = c:d, then ‘d’ is called the fourth

proportional to ‘a’, ‘b’ and ‘c’.

5. Mean Proportional: Mean proportional between ‘a’ and ‘b’ is √ab.

6. If a > b and ‘x’ is a positive quantity, then (a/b) > {(a + x)/(b + x)}

and (a/b) < {(a – x)/(b – x)}

7. If a < b and ‘x’ is a positive quantity, then (a/b) < {(a + x)/(b + x)}

and (a/b) > {(a – x)/(b – x)}

8. If (a/b) = (c/d) = (e/f) = k, then:

i. {(a + c + e)/(b + d + f)} = k

ii. {(pa + qc + re)/(pb + qd + rf)} = k, where all ‘p’, ‘q’ and ‘r’ are either positive or negative real numbers. 

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Time & Work

Time and work questions deals with:

i) Time taken by a person (or) a group of persons to do a work.

ii) Total work taken to complete a task

iii) Ratio of work efficiencies of 2 or more persons doing a work

iv) Quantity/proportion of work completed by a person or a group of persons in a certain time interval.

Basics:

Suppose to complete a certain task, it takes ‘B’ units of work

And, person ‘P’ takes ‘A’ days to complete the entire task

Then, efficiency of person ‘P’ = (B/A) units/day

So, efficiency = Total work ÷ time taken to complete the work

Total work = efficiency of given person(s) × time taken to complete the work by him/them

II:

If person ‘A’ takes ‘x’ days to complete a work

Then, work done by ‘A’ in 1 day = (1/x) units/day

And person ‘B’ takes ‘y’ days to complete the same work

Then, work done by ‘B’ in 1 day = (1/y) units/day

Then, work done by person ‘A’ and ‘B’ together in 1 day =

III:

If the ratio of work efficiencies of persons ‘A’ and ‘B’ is x:y, respectively, then the ratio of time taken by ‘A’ alone and ‘B’ alone to complete the same work is y:x, respectively.

IV:

If wages are provided for the work done, then the wage is distributed between the different workers in the ratio of the work done by them. If all workers worked for the same quantity of time, then the wages are distributed in the ratio of the efficiency of the workers.

V:

If ‘x’ men working ‘y’ hours per day for ‘z’ days can complete ‘u’ units of work and ‘p’ women working ‘q’ hours per day can complete ‘v’ units of the same work in ‘r’ days and efficiency of each man and woman is ‘m’ units per day and ‘w’ units per day, then

{(x × y × z × m)/u} = {(p × q × r ×n)/v}

VI:

Suppose men and women are two different types of workers with different efficiencies.

If ‘A₁’ men and ‘B₁’ women take ‘D₁’ days to do a certain work, and ‘A₂’ men and ‘B₂’ women take ‘D₂’ days to do the same work, then the time taken by ‘A₃’ men and ‘B₃’ women to do the same work is:

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Pipes and cisterns:

Pipes and cisterns are a special case of time and work problems. The concepts used to solve pipe and cistern problems are mostly the same concepts used to solve standard time and work problems.

Basics:

Time taken by a pipe to fill a tank = Capacity of the tank ÷ quantity of water inlet by the pipe per unit time

Time taken by a pipe to empty a tank = Capacity of the tank ÷ quantity of water outlet by the pipe unit time

Quantity of water inlet by a pipe per unit time = Capacity of the tank ÷ time taken by the pipe to fill the tank

Quantity of water outlet by a pipe per unit time = Capacity of the tank ÷ time taken by the pipe to empty the tank

Combined efficiency of inlet and outlet pipes

Let inlet pipe ‘A’ take ‘x’ minutes to fill a tank

Then, efficiency of pipe ‘A’ = (1/x) units/minute

Let outlet pipe ‘B’ take ‘y’ minutes to empty the same tank

Then, efficiency of pipe ‘B’ = -(1/y) units/minute {Negative sign shows that ‘B’ is an outlet pipe}

So, combined efficiency of pipes ‘A’ and ‘B’

= (1/x) – (1/y) = [(y – x)/xy] units/minute

Cross sectional area and rate of flow:

Suppose the cross-sectional area of the opening of a pipe is given to be ‘x²’ units

And, the rate of flow of water from the pipe is given to be ‘y’ units/second

Then, quantity of water inlet by the pipe per second = x² × y = x²y cubic units

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Probability

Questions based on probability can be asked in the

following ways

i) To find the probability of an event

ii) To find the probability of a combination of different

events

iii) To find the difference in the probability of occurrence

of two or more different events.

Basics:

Probability is the measure that tells us how likely is an

event to happen.

Let ‘P’ be the value of probability of occurrence of an

event.

Then, 0 ≤ P ≤ 1

If an event is a certain event, i.e. if it is sure that the

event is going to occur, then probability of the given

event is ‘1’.

If an event is an impossible event i.e. if it is sure that the

event cannot occur, then probability of occurrence of the

given event is ‘0’.

Probability of two related events:

Consider the event of writing an examination. The given

event has only two possible outcomes: passing the

examination and failing the examination.

Let ‘P’ be the probability of passing the examination and

let ‘Q’ be the probability of failing the examination. Then

(P + Q) = 1.

Following the same principle, if ‘P’ be the probability of a

person passing the exam, then the probability of the

person failing the exam is (1 – P).

Classical definition of probability = Number of favorable

events in the outcome ÷ total number of events in the

outcome

The set of all possible events is called the sample space.

The set of events which is of interest to us is called the

event.

Then, probability of a given event = n(E) ÷ n(S)

Bias – When the probability of a certain event is different

than the usually expected probability, then the

occurrence of the event is said to be biased. For

example, while tossing a coin, if the probability of getting

a head and a tail are not the equal, then the coin is said

to be a biased coin.

Let ‘E 1 ’ and ‘E 2 ’ be two different events where the

probability of occurrence of ‘E 1 ’ and ‘E 2 ’ is ‘p’ and ‘q’,

respectively

Then, the probability of occurrence of both ‘E 1 ’ and ‘E 2 ’ =

(p × q).

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Profit & Loss

Some basic terms, associated with profit and loss:

Cost price (CP) = The price for which a commodity/article is bought is called cost price.

Selling price (SP) = The price for which a commodity/article is sold is called selling price.

Profit/Gain = The difference (SP – CP) between selling price and cost price is called the profit.

Loss = The difference (CP – SP) between cost price and selling price is called loss.

Marked price (MP) = The price at which a seller has labelled an article is called marked price.

Discount = The reduction made on the marked price of the article is called discount.

Some important formulae associated with profit and loss.

1. Profit = SP – CP

2. Loss = CP – SP

3. Profit percentage = {(Profit)/Cost price} × 100

4. Loss percentage = {(Loss)/Cost price} × 100

5. Discount allowed = MP – SP

6. Discount percentage = (discount allowed/marked price) × 100

Special case:

If an article is sold at cost price, but using a false weight, then the overall profit percentage;

= {(100 + Gain%)/100} = {(True Scale or Weight) ÷ (False Scale or Weight)}

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Average

Simple Average:

Average of a set = {(Sum of all observations in the set) ÷ (Number of observations in the set)}.

Weighted Average:

If the values, ‘x₁’, ‘x₂’ ……… ‘x_n’ are assigned weights ‘w₁’, ‘w₂’ …………., ‘w_n’, respectively, then

Weighted average = {(w₁ x₁ + w₂x₂ + …………… + w_nx_n)/(w₁ + w₂ + ……….. + w_n)}

Income & Expenditure

Income = Expenditure + Savings

If income of ‘B’ is ‘x%’ more than that of ‘A’, then income of ‘B’ = Income of ‘A’ × {1 + (x/100)}

If income of ‘B’ is ‘x%’ less than that of ‘A’, then income of ‘B’ = Income of A × {1 – (x/100)}

If the price of a commodity increases by ‘x%’, then percentage reduction in consumption, so as not to increase the expenditure is: [{x/(100 + x)} × 100]%

Ages

Problems on ages are usually based on concepts of ratio and proportions, percentages or averages.

Basics:

Let the present age of person ‘A’ be ‘x’ years

Let the present age of person ‘B’ be ‘y’ years

Then, age of ‘A’, 3 years ago from now = (x – 3) years

Age of ‘B’, 3 years ago from now = (y – 3) years

If it is said that, 3 years ago from now, the ratio of ages of ‘A’ and ‘B’ was p:q, respectively,

We have, (x – 3):(y – 3) = p:q

So, qx – 3q = pq – 3p

Average of ages:

If it is said that the average of present ages of ‘A’ and ‘B’ is ‘p’ years

Then, sum of present age of ‘A’ and ‘B’ = p × 2 = ‘2p’ years

Let the present age of ‘A’ be ‘x’ years

Then, present age of ‘B’ = (2p – x) years.

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Simplification & Approximation

How to Approach Simplification and Approximation questions?

1.  BODMAS Rule

The first thing to keep in mind while solving simplification/approximation questions is the proper application of ‘BODMAS’ rule.

‘BODMAS’ is an acronym that is used to denote the order of operations to be followed while calculating the simplified value of a given expression.
It stands for
B – Brackets
O – Order of power or roots (it also stands for the expression “of”)
D – Division
M – Multiplication
A – Addition
S – Subtraction

Mathematical expressions having multiple operations need to be solved from left to right using this order only i.e. BODMAS

Consider this example:

Find the value of the given expression:

21.5 of 12 ÷ 23 – 184 + 15 × 12 – (19.25 × 16)

Solution:

Step 1: First, we solve the expression within the brackets.

So, 21.5 of 12 ÷ 23 – 184 + 15 × 12 – (19.25 × 16) = 21.5 of 12 ÷ 23 – 184 + 15 × 12 – 308

Step 2: Next, we calculate the power if any, and terms which are linked by the word “of”

So, 21.5 of 12 ÷ 23 – 184 + 15 × 12 – 308 = 258 ÷ 8 – 184 + 15 × 12 – 308

Step 3: We calculate the division operations

258 ÷ 8 – 184 + 15 × 12 – 308 = 32.25 – 184 + 15 × 12 – 308

Step 4: We calculate the multiplication operations

32.25 – 184 + 15 × 12 – 308 = 32.25 – 184 + 180 – 308

Step 5: We calculate the addition operations

32.25 – 184 + 180 – 308 = 212.25 – 184 – 308

Step 6: At last, we calculate the subtraction operations

212.25 – 184 – 308 = -279.75

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Number Series Quant Practice Questions for both Prelims & Mains with Detailed Solutions