<\/p>\n\n\n\n
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. Quantitative Aptitude is one section that involves a lot of calculation and practice; failing which, you will not be able to clear the exam. Many banking exams, like RBI, SEBI, SBI, LIC AAO, IBPS RRB Exams, include tough quantitative aptitude questions. Quant is an integral part of the competitive bank examinations, and one must clear both the prelims and mains stages to crack the exam. As far as prelims and mains are 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.<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n\n\n\n
Banking Exams Quant Essential Concepts & Practice Questions PDF – Download Now<\/strong><\/a><\/p>\n\n\n\n It is important to understand these critical concepts or topics for quant preparation especially for banks. Thus, it\u2019s important for the candidates to prepare the below mentioned topics thoroughly. <\/p>\n\n\n\n Banking Exams Reasoning: Essential Concepts, Practice Questions & Preparation Tips Free PDF<\/a><\/p>\n\n\n\n Basics:<\/p>\n\n\n\n In a partnership, the ratio of profit shares of two different persons \u2018A\u2019 and \u2018B\u2019 is the respective ratio of the product of their investment and time period of investment. Suppose two different persons \u2018A\u2019 and \u2018B\u2019 started a business by investing their capitals in the ratio p:q, respectively And, the ratio of time period of investment of \u2018A\u2019 and \u2018B\u2019 is m:n, respectively, Then, ratio of profit shares of \u2018A\u2019 and \u2018B\u2019, respectively = (p \u00d7 m):(q \u00d7 n)<\/p>\n\n\n\n Practice Questions on Partnership for RBI Assistant Prelims Exam<\/a><\/p>\n\n\n\n Basics:<\/p>\n\n\n\n Time taken to travel a certain distance = Distance travelled \u00f7 average<\/p>\n\n\n\n Speed of travel = Distance travelled \u00f7 Time taken for the distance travelled<\/p>\n\n\n\n Distance travelled = Speed at which the distance was travelled \u00d7 Time taken for the travel<\/p>\n\n\n\n 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.<\/p>\n\n\n\n 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.<\/p>\n\n\n\n Suppose to travel a distance of \u20182d\u2019 km, if a person travelled \u2018d\u2019 km at a speed of \u2018p\u2019 km\/h and remaining distance at a speed of \u2018q\u2019 km\/h, then <\/p>\n\n\n\n average speed of the journey = total distance travelled \u00f7 total time taken<\/p>\n\n\n\n Time taken to travel first \u2018d\u2019 km = (d\/p) hours<\/p>\n\n\n\n Time taken to travel next \u2018d\u2019 km = (d\/q) hours<\/p>\n\n\n\n Problems based on boat and stream is solved mostly using the same concepts as standard time, speed and distance problems.<\/p>\n\n\n\n Basics:<\/strong><\/p>\n\n\n\n The speed of a boat changes with respect to the direction of travel of the boat with respect to the stream.<\/p>\n\n\n\n Let the speed of a boat in still water be \u2018x\u2019 km\/h<\/p>\n\n\n\n Let the speed of the stream in which the boat is sailing be \u2018y\u2019 km\/h<\/p>\n\n\n\n Then, speed of the boat while sailing against the stream i.e. in upstream = (x \u2013 y) km\/h<\/p>\n\n\n\n Speed of the boat while sailing with the stream i.e. in downstream = (x + y) km\/h<\/p>\n\n\n\n Time, Speed & Distance Questions for Bank Exams<\/a><\/p>\n\n\n\n Compound interest:<\/strong><\/p>\n\n\n\n Interest = Amount \u2013 Principal<\/p>\n\n\n\n If Rs. \u2018P\u2019 is invested at a rate of \u2018r%\u2019 p.a. compounded annually for \u2018t\u2019 years, then the amount received after \u2018t\u2019 years will be = P \u00d7 {1 + (r\/100)} t<\/p>\n\n\n\n If the interest is not compounded annually, then<\/p>\n\n\n\n Effective rate of interest = Actual rate \u00f7 Number of times the interest is compounded in a year<\/p>\n\n\n\n Effective time period = Actual time period \u00d7 Number of times the interest is compounded in a year<\/p>\n\n\n\n Simple interest:<\/strong><\/p>\n\n\n\n If Rs. \u2018P\u2019 is invested at simple interest of \u2018r%\u2019 p.a. for \u2018t\u2019 years, then the amount received after \u2018t\u2019 years will be = P + {(P \u00d7 r \u00d7 t)\/100}<\/p>\n\n\n\n And, the simple interest received = {(P \u00d7 r \u00d7 t)\/100}<\/p>\n\n\n\n Relationship between simple interest (SI) and compound interest (CI):<\/p>\n\n\n\n When a principal (P) is invested at the rate of \u2018r%\u2019 p.a., then<\/p>\n\n\n\n Difference between SI and CI after two years = P \u00d7 (r\/100) 2<\/p>\n\n\n\n Difference between SI and CI after three years = P \u00d7 (r\/100) 2 \u00d7 {(r\/100) + 3}<\/p>\n\n\n\n <\/p>\n\n\n <\/p>\n\n\n\n Mensuration deals with measurement of areas and volumes of different figures. These figures include both \u20182D\u2019 (two dimensional) and \u20183D\u2019 (three dimensional) figures (also known as solids).<\/p>\n\n\n\n Note: By definition, \u20182D\u2019 figures exist in a two dimensional world and hence have no volume.<\/p>\n\n\n\n Let\u2019s take a brief look at the formulae for areas of various \u20182D\u2019 figures, and areas and volumes of \u20183D\u2019 figures.<\/p>\n\n\n\n 2D Figures:<\/strong><\/p>\n\n\n\n Polygons: A Polygon is a closed figure made up of at least 3 line segments in a two-dimensional plane.<\/p>\n\n\n\n Triangle: A triangle is a polygon containing 3 sides. The area of a triangle is represented by the figure \u0394.<\/p>\n\n\n\n Formulas associated with triangles:<\/p>\n\n\n\n Area:<\/strong><\/p>\n\n\n\n I. For any triangle in general;<\/strong><\/p>\n\n\n\n For a triangle having sides of length \u2018a\u2019 cm, \u2018b\u2019 cm and \u2018c\u2019 cm;<\/p>\n\n\n\n Perimeter of triangle = Sum of length of three sides = (a + b + c) cm.<\/p>\n\n\n\n Area = \u221a{s \u00d7 (s \u2013 a) \u00d7 (s \u2013 b) \u00d7 (s \u2013 c)}<\/strong><\/p>\n\n\n\n Where \u2018s\u2019 is the semi-perimeter of the triangle. \u2018s\u2019 = {(a + b + c)\/2}<\/p>\n\n\n\n This formula is universally known as heron\u2019s formula.<\/p>\n\n\n\n II. For a triangle whose length of base and length of altitude (Height) to that base is given;<\/p>\n\n\n\n Area = (1\/2) \u00d7 base \u00d7 height<\/p>\n\n\n\n III. Equilateral triangle:<\/strong> If all three sides of a triangle are equal, then it is an equilateral triangle.<\/p>\n\n\n\n For an equilateral triangle with side \u2018a\u2019 cm, Area = (\u221a3\/4) \u00d7 a 2<\/p>\n\n\n\n Using (II), we have; (1\/2) \u00d7 a \u00d7 height = (\u221a3\/4) \u00d7 a 2<\/p>\n\n\n\n Or, height = (\u221a3\/2) \u00d7 a<\/p>\n\n\n\n IV. Isosceles triangle:<\/strong> If any two sides of a triangle are equal, then it is an isosceles triangle.<\/p>\n\n\n\n Area = (a\/4)\u221a(4c 2 \u2013 a 2 ) {where \u2018c\u2019 is the length of each of the two equal sides and \u2018a\u2019 is the length of the third (un-equal) side}<\/p>\n\n\n\n Quadrilaterals:<\/strong> Any polygon having for sides is called a quadrilateral.<\/p>\n\n\n\n I. Area of a normal quadrilateral with vertices \u2018A\u2019, \u2018B\u2019, \u2018C\u2019 and \u2018D\u2019 (As<\/p>\n\n\n\n shown in the figure)<\/p>\n\n\n\n Area of quadrilateral ABCD = (1\/2) \u00d7 Length of BD \u00d7 (AF + DE), where AF and DE are perpendicular to BC.<\/p>\n\n\n\n II. Trapezium: <\/strong>A trapezium is a quadrilateral having exactly two parallel sides.<\/p>\n\n\n\n Area of trapezium = (1\/2) \u00d7 sum of parallel sides \u00d7 Height of the trapezium.<\/p>\n\n\n\n III. Parallelogram:<\/strong> A parallelogram is a quadrilateral whose opposite sides are parallel as well as equal to each other.<\/p>\n\n\n\n Area = length of base \u00d7 height (perpendicular distance)<\/p>\n\n\n\n IV. Square<\/strong>: A square is a parallelogram whose all four sides are of equal length and angles are of 90 o each.<\/p>\n\n\n\n Area = (a) 2 {where \u2018a\u2019 is the length of each side}<\/p>\n\n\n\n V. Rhombus:<\/strong> A Rhombus is a parallelogram whose all four sides are equal.<\/p>\n\n\n\n Area = (1\/2) \u00d7 product of length of diagonals<\/p>\n\n\n\n VI. Rectangle:<\/strong> A rectangle is a parallelogram that has all four angles of 90 o<\/p>\n\n\n\n For a rectangle with length \u2018L\u2019 cm and Breadth \u2018B\u2019 cm,<\/p>\n\n\n\n Area = L \u00d7 B<\/p>\n\n\n\n And, Perimeter = 2(L + B)<\/p>\n\n\n\n Circle:<\/strong><\/p>\n\n\n\n For a circle with radius \u2018r\u2019 units<\/p>\n\n\n\n Area = \u03c0r 2<\/p>\n\n\n\n Circumference = 2\u03c0r<\/p>\n\n\n\n Area of sector = (\u03b8 o \/360 o ) \u00d7 \u03c0r 2 {Where \u03b8 is the angle subtended inside the sector}<\/p>\n\n\n\n Length of arc = (\u03b8 o \/360 o ) \u00d7 2\u03c0r<\/p>\n\n\n\n Area and Volumes of solids:<\/strong><\/p>\n\n\n\n Cube: A cube is a solid shape with six square faces.<\/p>\n\n\n\n Total surface area of cube = 6a 2 {Where \u2018a\u2019 is the length of edge of the cube}<\/p>\n\n\n\n Volume of cube = a 3<\/p>\n\n\n\n Cuboid<\/strong>: A cuboid is bound by 6 rectangular faces. The opposite faces<\/p>\n\n\n\n of a cuboid are equal rectangles lying in parallel planes.<\/p>\n\n\n\n Volume of a cuboid with length \u2018x\u2019 cm, breadth \u2018y\u2019 cm and height \u2018h\u2019 cm = (x \u00d7 y \u00d7 h)<\/p>\n\n\n\n Total surface area of cuboid = 2(xy + yh + xh)<\/p>\n\n\n\n Lateral surface area of a cuboid = 2xh + 2yh = 2h(x + y)<\/p>\n\n\n\n Cone:<\/strong> A cone is a solid which has a circle at its base and a slanting lateral surface that converges at the apex.<\/p>\n\n\n\n Curved surface area of the cone = \u03c0 \u00d7 r \u00d7 s {Where \u2018r\u2019 is radius and \u2018s\u2019 is slant height}<\/p>\n\n\n\n Slant height of a cone = \u221a(r 2 + h 2 ), where \u2018h\u2019 is height of the cone.<\/p>\n\n\n\n Total surface area of cone = Curved surface area + area of base = \u03c0rs + \u03c0r 2 = \u03c0r(r + s)<\/p>\n\n\n\n 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.<\/p>\n\n\n\n Curved surface area of frustum = \u03c0 \u00d7 s(R + r) {Where \u2018s\u2019 is slant height ,\u2018R\u2019 is radius of the base and \u2018r\u2019 is the radius of the top.}<\/p>\n\n\n\n Total surface area of the frustum = \u03c0(R 2 + r 2 + Rs + rs)<\/p>\n\n\n\n Volume of the frustum = (1\/2) \u00d7 \u03c0 \u00d7 h \u00d7 {R 2 + r 2 + Rr} {Where \u2018h\u2019 is the height of the cone}<\/p>\n\n\n\n s 2 = (R 2 \u2013 r 2 ) + h 2<\/p>\n\n\n\n Cylinder: <\/strong>Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.<\/p>\n\n\n\n Curved surface area of a cylinder = 2\u03c0rh {Where \u2018r\u2019 is the radius of the base and \u2018h\u2019 is the height of the cylinder}<\/p>\n\n\n\n Total surface area of the cylinder = 2\u03c0rh + 2\u03c0r 2 = 2\u03c0r(r + h)<\/p>\n\n\n\n Volume of cylinder = \u03c0r 2 h<\/p>\n\n\n\n Sphere:<\/strong> 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).<\/p>\n\n\n\n Volume of sphere = (4\/3) \u00d7 \u03c0r 3<\/p>\n\n\n\n Surface area of a sphere = 4\u03c0r 2<\/p>\n\n\n\n Hemisphere:<\/strong> Where a sphere is divided into two equal and identical parts, the new formed parts are called hemispheres.<\/p>\n\n\n\n Volume of hemisphere = (2\/3)\u03c0r 3<\/p>\n\n\n\n Total surface area of hemisphere = 3\u03c0r 2<\/p>\n\n\n\n Curved surface area of hemisphere = 2\u03c0r 2<\/p>\n\n\n\n SBI Clerk Prelims Free Mock Test<\/a><\/p>\n\n\n\n Factorial:<\/strong><\/p>\n\n\n\n Let \u2018n\u2019 be a natural number, then n! = n \u00d7 (n \u2013 1) \u00d7 (n \u2013 2) \u00d7 \u2026 \u00d7 1<\/p>\n\n\n\n Permutation:<\/strong><\/p>\n\n\n\n Permutation is the arrangement of different number of things where we take into consideration the order or arrangement.<\/p>\n\n\n\n Combination:<\/strong><\/p>\n\n\n\n The number of different groups or selections that can be formed by taking some or all items at a time, is called combination.<\/p>\n\n\n\n SBI Clerk Prelims Free Mock Test<\/a><\/p>\n\n\n\n 1. Ratio: The ratio of two quantities \u2018a\u2019 and \u2018b\u2019, in the same units, is the fraction (a\/b) and we write it as a:b.<\/p>\n\n\n\n 2. Proportion: The equality of two ratios is called proportion.<\/p>\n\n\n\n If a:b = c:d, we write, a:b::c:d and we say that \u2018a\u2019, \u2018b\u2019, \u2018c\u2019, \u2018d\u2019 are in proportion. Here \u2018a\u2019 and \u2018d\u2019 are called extremes, while \u2018b\u2019 and \u2018c\u2019 are called mean terms. Also, Product of means = Product of extremes.<\/p>\n\n\n\n Thus, a:b::c:d = (b \u00d7 c) = (a \u00d7 d)<\/p>\n\n\n\n 3. Third Proportional: If a:b = b:c, then \u2018c\u2019 is called the third proportional to \u2018a\u2019 and \u2018b\u2019.<\/p>\n\n\n\n 4. Fourth Proportional: If a:b = c:d, then \u2018d\u2019 is called the fourth proportional to \u2018a\u2019, \u2018b\u2019 and \u2018c\u2019.<\/p>\n\n\n\n 5. Mean Proportional: Mean proportional between \u2018a\u2019 and \u2018b\u2019 is \u221aab.<\/p>\n\n\n\n 6. If a > b and \u2018x\u2019 is a positive quantity, then (a\/b) > {(a + x)\/(b + x)} and (a\/b) < {(a \u2013 x)\/(b \u2013 x)}<\/p>\n\n\n\n 7. If a < b and \u2018x\u2019 is a positive quantity, then (a\/b) < {(a + x)\/(b + x)} and (a\/b) > {(a \u2013 x)\/(b \u2013 x)}<\/p>\n\n\n\n 8. If (a\/b) = (c\/d) = (e\/f) = k, then:<\/p>\n\n\n\n i. {(a + c + e)\/(b + d + f)} = k<\/p>\n\n\n\n ii. {(pa + qc + re)\/(pb + qd + rf)} = k, where all \u2018p\u2019, \u2018q\u2019 and \u2018r\u2019 are either positive or negative real numbers. <\/p>\n\n\n\n SBI Clerk Prelims Free Mock Test<\/a><\/p>\n\n\n\n Time and work questions deals with:<\/p>\n\n\n\n i) Time taken by a person (or) a group of persons to do a work.<\/p>\n\n\n\n ii) Total work taken to complete a task<\/p>\n\n\n\n iii) Ratio of work efficiencies of 2 or more persons doing a work<\/p>\n\n\n\n iv) Quantity\/proportion of work completed by a person or a group of persons in a certain time interval.<\/p>\n\n\n\n Basics:<\/p>\n\n\n\n Suppose to complete a certain task, it takes \u2018B\u2019 units of work<\/p>\n\n\n\n And, person \u2018P\u2019 takes \u2018A\u2019 days to complete the entire task<\/p>\n\n\n\n Then, efficiency of person \u2018P\u2019 = (B\/A) units\/day<\/p>\n\n\n\n So, efficiency = Total work \u00f7 time taken to complete the work<\/p>\n\n\n\n Total work = efficiency of given person(s) \u00d7 time taken to complete the work by him\/them<\/p>\n\n\n\n II:<\/p>\n\n\n\n If person \u2018A\u2019 takes \u2018x\u2019 days to complete a work <\/p>\n\n\n\n Then, work done by \u2018A\u2019 in 1 day = (1\/x) units\/day <\/p>\n\n\n\n And person \u2018B\u2019 takes \u2018y\u2019 days to complete the same work<\/p>\n\n\n\n Then, work done by \u2018B\u2019 in 1 day = (1\/y) units\/day<\/p>\n\n\n\n Then, work done by person \u2018A\u2019 and \u2018B\u2019 together in 1 day =<\/p>\n\n\n\n III:<\/p>\n\n\n\n If the ratio of work efficiencies of persons \u2018A\u2019 and \u2018B\u2019 is x:y, respectively, then the ratio of time taken by \u2018A\u2019 alone and \u2018B\u2019 alone to complete the same work is y:x, respectively.<\/p>\n\n\n\n IV:<\/p>\n\n\n\n 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.<\/p>\n\n\n\n V:<\/p>\n\n\n\n If \u2018x\u2019 men working \u2018y\u2019 hours per day for \u2018z\u2019 days can complete \u2018u\u2019 units of work and \u2018p\u2019 women working \u2018q\u2019 hours per day can complete \u2018v\u2019 units of the same work in \u2018r\u2019 days and efficiency of each man and woman is \u2018m\u2019 units per day and \u2018w\u2019 units per day, then<\/p>\n\n\n\n {(x \u00d7 y \u00d7 z \u00d7 m)\/u} = {(p \u00d7 q \u00d7 r \u00d7n)\/v}<\/p>\n\n\n\n VI:<\/p>\n\n\n\n Suppose men and women are two different types of workers with different efficiencies.<\/p>\n\n\n\n If \u2018A1<\/sub>\u2019 men and \u2018B1<\/sub>\u2019 women take \u2018D1<\/sub>\u2019 days to do a certain work, and \u2018A2<\/sub>\u2019 men and \u2018B2<\/sub>\u2019 women take \u2018D2<\/sub>\u2019 days to do the same work, then the time taken by \u2018A3<\/sub>\u2019 men and \u2018B3<\/sub>\u2019 women to do the same work is:<\/p>\n\n\n\n SBI Clerk Prelims Free Mock Test<\/a><\/p>\n\n\n\n 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.<\/p>\n\n\n\n Basics:<\/p>\n\n\n\n Time taken by a pipe to fill a tank = Capacity of the tank \u00f7 quantity of water inlet by the pipe per unit time<\/p>\n\n\n\n Time taken by a pipe to empty a tank = Capacity of the tank \u00f7 quantity of water outlet by the pipe unit time<\/p>\n\n\n\n Quantity of water inlet by a pipe per unit time = Capacity of the tank \u00f7 time taken by the pipe to fill the tank<\/p>\n\n\n\n Quantity of water outlet by a pipe per unit time = Capacity of the tank \u00f7 time taken by the pipe to empty the tank<\/p>\n\n\n\n Combined efficiency of inlet and outlet pipes<\/p>\n\n\n\n Let inlet pipe \u2018A\u2019 take \u2018x\u2019 minutes to fill a tank<\/p>\n\n\n\n Then, efficiency of pipe \u2018A\u2019 = (1\/x) units\/minute<\/p>\n\n\n\n Let outlet pipe \u2018B\u2019 take \u2018y\u2019 minutes to empty the same tank<\/p>\n\n\n\n Then, efficiency of pipe \u2018B\u2019 = -(1\/y) units\/minute {Negative sign shows that \u2018B\u2019 is an outlet pipe}<\/p>\n\n\n\n So, combined efficiency of pipes \u2018A\u2019 and \u2018B\u2019<\/p>\n\n\n\n = (1\/x) \u2013 (1\/y) = [(y \u2013 x)\/xy] units\/minute<\/p>\n\n\n\n Cross sectional area and rate of flow:<\/p>\n\n\n\n Suppose the cross-sectional area of the opening of a pipe is given to be \u2018x2<\/sup>\u2019 units<\/p>\n\n\n\n And, the rate of flow of water from the pipe is given to be \u2018y\u2019 units\/second<\/p>\n\n\n\n Then, quantity of water inlet by the pipe per second = x2<\/sup> \u00d7 y = x2<\/sup>y cubic units<\/p>\n\n\n\n SBI Clerk Prelims Free Mock Test<\/a><\/p>\n\n\n\n Questions based on probability can be asked in the following ways<\/p>\n\n\n\n i) To find the probability of an event<\/p>\n\n\n\n ii) To find the probability of a combination of different events<\/p>\n\n\n\n iii) To find the difference in the probability of occurrence of two or more different events.<\/p>\n\n\n\n Basics:<\/strong><\/p>\n\n\n\n Probability is the measure that tells us how likely is an event to happen.<\/p>\n\n\n\n Let \u2018P\u2019 be the value of probability of occurrence of an event.<\/p>\n\n\n\n Then, 0 \u2264 P \u2264 1<\/p>\n\n\n\n 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 \u20181\u2019.<\/p>\n\n\n\n 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 \u20180\u2019.<\/p>\n\n\n\n Probability of two related events:<\/strong><\/p>\n\n\n\n Consider the event of writing an examination. The given event has only two possible outcomes: passing the examination and failing the examination.<\/p>\n\n\n\n Let \u2018P\u2019 be the probability of passing the examination and let \u2018Q\u2019 be the probability of failing the examination. Then (P + Q) = 1.<\/p>\n\n\n\n Following the same principle, if \u2018P\u2019 be the probability of a person passing the exam, then the probability of the person failing the exam is (1 \u2013 P).<\/p>\n\n\n\n Classical definition of probability = Number of favorable events in the outcome \u00f7 total number of events in the outcome<\/p>\n\n\n\n The set of all possible events is called the sample space<\/strong>.<\/p>\n\n\n\n The set of events which is of interest to us is called the event.<\/p>\n\n\n\n Then, probability of a given event = n(E) \u00f7 n(S)<\/p>\n\n\n\n Bias \u2013 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 probabilities of getting a head and a tail are not equal, then the coin is said to be a biased coin.<\/p>\n\n\n\n Let \u2018E 1 \u2019 and \u2018E 2 \u2019 be two different events where the probability of occurrence of \u2018E 1 \u2019 and \u2018E 2 \u2019 is \u2018p\u2019 and \u2018q\u2019, respectively<\/p>\n\n\n\n Then, the probability of occurrence of both \u2018E 1 \u2019 and \u2018E 2 \u2019 = (p \u00d7 q).<\/p>\n\n\n\n Some basic terms, associated with profit and loss:<\/p>\n\n\n\n Cost price (CP) = The price for which a commodity\/article is bought is called cost price.<\/p>\n\n\n\n Selling price (SP) = The price for which a commodity\/article is sold is called selling price.<\/p>\n\n\n\n Profit\/Gain = The difference (SP \u2013 CP) between selling price and cost price is called the profit.<\/p>\n\n\n\n Loss = The difference (CP \u2013 SP) between cost price and selling price is called loss.<\/p>\n\n\n\n Marked price (MP) = The price at which a seller has labelled an article is called marked price.<\/p>\n\n\n\n Discount = The reduction made on the marked price of the article is called discount.<\/p>\n\n\n\n Some important formulae associated with profit and loss.<\/p>\n\n\n\n 1. Profit = SP \u2013 CP<\/p>\n\n\n\n 2. Loss = CP \u2013 SP<\/p>\n\n\n\n 3. Profit percentage = {(Profit)\/Cost price} \u00d7 100<\/p>\n\n\n\n 4. Loss percentage = {(Loss)\/Cost price} \u00d7 100<\/p>\n\n\n\n 5. Discount allowed = MP \u2013 SP<\/p>\n\n\n\n 6. Discount percentage = (discount allowed\/marked price) \u00d7 100<\/p>\n\n\n\n Special case:<\/p>\n\n\n\n If an article is sold at cost price, but using a false weight, then the overall profit percentage;<\/p>\n\n\n\n = {(100 + Gain%)\/100} = {(True Scale or Weight) \u00f7 (False Scale or Weight)}<\/p>\n\n\n\n SBI Clerk Prelims Free Mock Test<\/a><\/p>\n\n\n\n Simple Average:<\/p>\n\n\n\n Average of a set = {(Sum of all observations in the set) \u00f7 (Number of observations in the set)}.<\/p>\n\n\n\n Weighted Average:<\/p>\n\n\n\n If the values, \u2018x1<\/sub>\u2019, \u2018x2<\/sub>\u2019 \u2026\u2026\u2026 \u2018xn<\/sub>\u2019 <\/sub>are assigned weights \u2018w1<\/sub>\u2019, \u2018w2<\/sub>\u2019 \u2026\u2026\u2026\u2026., \u2018wn<\/sub>\u2019, respectively, then<\/p>\n\n\n\n Weighted average = {(w1<\/sub> x1<\/sub> + w2<\/sub>x2<\/sub> + \u2026\u2026\u2026\u2026\u2026 + wn<\/sub>xn<\/sub>)\/(w1<\/sub> + w2<\/sub> + \u2026\u2026\u2026.. + wn<\/sub>)}<\/p>\n\n\n\n Income = Expenditure + Savings<\/p>\n\n\n\n If income of \u2018B\u2019 is \u2018x%\u2019 more than that of \u2018A\u2019, then income of \u2018B\u2019 = Income of \u2018A\u2019 \u00d7 {1 + (x\/100)}<\/p>\n\n\n\n If income of \u2018B\u2019 is \u2018x%\u2019 less than that of \u2018A\u2019, then income of \u2018B\u2019 = Income of A \u00d7 {1 \u2013 (x\/100)}<\/p>\n\n\n\n If the price of a commodity increases by \u2018x%\u2019, then percentage reduction in consumption, so as not to increase the expenditure is: [{x\/(100 + x)} \u00d7 100]%<\/p>\n\n\n\n Problems on ages are usually based on concepts of ratio and proportions, percentages or averages.<\/p>\n\n\n\n Basics:<\/p>\n\n\n\n Let the present age of person \u2018A\u2019 be \u2018x\u2019 years<\/p>\n\n\n\n Let the present age of person \u2018B\u2019 be \u2018y\u2019 years<\/p>\n\n\n\n Then, age of \u2018A\u2019, 3 years ago from now = (x \u2013 3) years<\/p>\n\n\n\n Age of \u2018B\u2019, 3 years ago from now = (y \u2013 3) years<\/p>\n\n\n\n If it is said that, 3 years ago from now, the ratio of ages of \u2018A\u2019 and \u2018B\u2019 was p:q, respectively,<\/p>\n\n\n\n We have, (x \u2013 3):(y \u2013 3) = p:q<\/p>\n\n\n\n So, qx \u2013 3q = pq \u2013 3p<\/p>\n\n\n\n Average of ages:<\/strong><\/p>\n\n\n\n If it is said that the average of present ages of \u2018A\u2019 and \u2018B\u2019 is \u2018p\u2019 years<\/p>\n\n\n\n Then, sum of present age of \u2018A\u2019 and \u2018B\u2019 = p \u00d7 2 = \u20182p\u2019 years<\/p>\n\n\n\n Let the present age of \u2018A\u2019 be \u2018x\u2019 years<\/p>\n\n\n\n Then, present age of \u2018B\u2019 = (2p \u2013 x) years.<\/p>\n\n\n\n SBI Clerk Prelims Free Mock Test<\/a><\/p>\n\n\n\n How to Approach Simplification and Approximation questions?<\/strong><\/p>\n\n\n\n 1. BODMAS Rule<\/p>\n\n\n\n The first thing to keep in mind while solving simplification\/approximation questions is the proper application of \u2018BODMAS\u2019 rule.<\/p>\n\n\n\n \u2018BODMAS\u2019 is an acronym that is used to denote the order of operations to be followed while calculating the simplified value of a given expression. Mathematical expressions having multiple operations need to be solved from left to right using this order only i.e. BODMAS<\/p>\n\n\n\n Consider this example:<\/p>\n\n\n\n Find the value of the given expression:<\/p>\n\n\n\n 21.5 of 12 \u00f7 23 \u2013 184 + 15 \u00d7 12 \u2013 (19.25 \u00d7 16)<\/p>\n\n\n\n Solution:<\/p>\n\n\n\n Step 1: First, we solve the expression within the brackets.<\/p>\n\n\n\n So, 21.5 of 12 \u00f7 23 \u2013 184 + 15 \u00d7 12 \u2013 (19.25 \u00d7 16) = 21.5 of 12 \u00f7 23 \u2013 184 + 15 \u00d7 12 \u2013 308<\/p>\n\n\n\n Step 2: Next, we calculate the power if any, and terms which are linked by the word \u201cof\u201d<\/p>\n\n\n\n So, 21.5 of 12 \u00f7 23 \u2013 184 + 15 \u00d7 12 \u2013 308 = 258 \u00f7 8 \u2013 184 + 15 \u00d7 12 \u2013 308<\/p>\n\n\n\n Step 3: We calculate the division operations<\/p>\n\n\n\n 258 \u00f7 8 \u2013 184 + 15 \u00d7 12 \u2013 308 = 32.25 \u2013 184 + 15 \u00d7 12 \u2013 308<\/p>\n\n\n\n Step 4: We calculate the multiplication operations<\/p>\n\n\n\n 32.25 \u2013 184 + 15 \u00d7 12 \u2013 308 = 32.25 \u2013 184 + 180 \u2013 308<\/p>\n\n\n\n Step 5: We calculate the addition operations<\/p>\n\n\n\n 32.25 \u2013 184 + 180 \u2013 308 = 212.25 \u2013 184 \u2013 308<\/p>\n\n\n\n Step 6: At last, we calculate the subtraction operations<\/p>\n\n\n\n 212.25 \u2013 184 \u2013 308 = -279.75<\/p>\n\n\n\n Banking Exams Quant Essential Concepts & Practice Questions PDF – Download Now<\/strong><\/a><\/p>\n\n\n\n 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 Bank exam preparation to another level.<\/p>\n\n\n\nImportant Quant Concepts & Their Explanations<\/h2>\n\n\n\n
\n
Partnership<\/strong><\/h2>\n\n\n\n
Time, Speed & Distance<\/strong><\/h2>\n\n\n\n
Trains<\/h2>\n\n\n\n
Average speed<\/strong><\/h2>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-us.googleusercontent.com\/U5tTM_R8GWvU-nQauHMrRzUV3loswuW0yPps7xUSxxKfs2n3VCOXN8FcQvWACk7DgunZ4UCSYkT5OJgLgla4XwU8h_aiMNOKGGboT1BgOQJVoDNAbzdNC9aN39jIZX5zVHR2HXXMqniAVowixBI9HTQ\")
Boat and stream<\/strong><\/h2>\n\n\n\n
Simple Interest Compound Interest<\/strong><\/h2>\n\n\n\n
![\"banking](\"https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2025\/08\/Banking-Plus-Course-1-1.png\")
<\/a><\/figure><\/div>\n\n\nMensuration<\/strong><\/h2>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-us.googleusercontent.com\/0mEZNWDJFe_SRHkbVBCcSzDLwwPVC6tWg-FVIeuTEVqA_rHksj5n_ATkFQaajVLFolkqaHq065HLkP_cw3_bWQcPvdJcyBfhl4kutH0NRZQqw0_x_aCguF9cROJ56_UfIWTm4VBmwsTd65JkT0Z9iFo\")
Permutation & Combination<\/strong><\/h2>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-us.googleusercontent.com\/yrDsOJ4VorAzftWd9SRDg5-cy_9W18nu7QMuhUN_Cw9UIsR2BMPtybonthXqHL1qXNxW7JE_6QoNZb6bTZSxWvYhNR-IGOGb3p-_Jiy9DSzjssHmdG6qAp2aoS9gnTVdNYAJakxsYf7wWhzgEpCfu4E\")
Ratio & Proportions<\/strong><\/h2>\n\n\n\n
Time & Work<\/strong><\/h2>\n\n\n\n
![\"\"\/](\"https:\/\/lh7-us.googleusercontent.com\/nFVR3iUDcL9Ssk93LtCV86h4ccLsSDefZIqzdFQZk2fwrrNvuwdjjJxindTFRHIrend6qTFYaVVV0UWvtEFrspRQrX1AGWuDfzJ0ZUO6uAA5OF3y3ERhf9v-XnqyqAtkMyEDW1_TMaiKovsALvshyHQ\")
![\"\"\/](\"https:\/\/lh7-us.googleusercontent.com\/anj5qTIqo8lhKFODvqPb_C2KYC6l7ft0Buelcp47b5fN5Qf1_BQ66-ZVZZfDT6GHSKAxQcfQRLbouUyXXR3DrOBJVhne-dMfw5KNe9jbqBN26v4vChF307Fg00UbuGGP0Z-eg8t1BNZLnEbQOfmHkOQ\")
Pipes and cisterns<\/strong><\/h2>\n\n\n\n
Probability<\/strong><\/h2>\n\n\n\n
Profit & Loss<\/strong><\/h2>\n\n\n\n
Average<\/strong><\/h2>\n\n\n\n
Income & Expenditure<\/strong><\/h2>\n\n\n\n
Ages<\/strong><\/h2>\n\n\n\n
Simplification & Approximation<\/strong><\/h2>\n\n\n\n
It stands for
B \u2013 Brackets
O \u2013 Order of power or roots (it also stands for the expression \u201cof\u201d)
D \u2013 Division
M \u2013 Multiplication
A \u2013 Addition
S \u2013 Subtraction<\/p>\n\n\n\n
![\"banking](\"https:\/\/www.practicemock.com\/blog\/wp-content\/uploads\/2025\/08\/New-Banking-Packages-Banner-1.png\")
<\/a><\/figure><\/div>\n\n\n