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.
Banking Exams Quant Essential Concepts & Practice Questions PDF – Download Now
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.
SBI Clerk Prelims Free Mock Test
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
SBI Clerk Prelims Free Mock Test
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}
SBI Clerk Prelims Free Mock Test
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
SBI Clerk Prelims Free Mock Test
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.
SBI Clerk Prelims Free Mock Test
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.
SBI Clerk Prelims Free Mock Test
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:
SBI Clerk Prelims Free Mock Test
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
SBI Clerk Prelims Free Mock Test
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).
SBI Clerk Prelims Free Mock Test
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)}
SBI Clerk Prelims Free Mock Test
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’ ……… ‘xn’ are assigned weights ‘w1’, ‘w2’ …………., ‘wn’, respectively, then
Weighted average = {(w1 x1 + w2x2 + …………… + wnxn)/(w1 + w2 + ……….. + wn)}
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.
SBI Clerk Prelims Free Mock Test
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
Banking Exams Quant Essential Concepts & Practice Questions PDF – Download Now
This brings us to the end of the complete Banking Exams Quant Essential Concepts, Practice Questions & Preparation Tips Free PDF. Download this free PDF now and take your SBI Clerk prelims exam preparation to another level.
Number Series Quant Practice Questions for both Prelims & Mains with Detailed Solutions
Explore the NIACL AO Mains Exam Analysis 2024 including good attempts, difficulty level, new exam…
Solve the IDBI ESO Previous Year Question Papers provided in this blog. Candidates must solve…
Cover all the topics via RRB ALP Study Plan 2024 for CBT 1 exam by…
Hurry up and follow this amazing IDBI ESO 2024 Study Plan for 15 Days. This…
Grab the best RRB JE Study Plan 2024 For 1 month complimented by the simplest…
Here we are providing the ECGC PO Expected Cut Off 2024. Candidates can check Category…