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Quantitative Aptitude Flash Cards

How to Check Whether Number is Multiple of 3

Sum of digits is multiple of 3.




¨ Last two digits are multiple of 4.¨ The number can be divided by 2 twice.




¨ Number is multiple of 3 and 2.




¨ Sum of digits is multiple of 9.




¨ Sum of digits is multiple of 3, and last two digits multiple of 4.



Odd and Even Numbers¨ Odd + Odd = Even ¨ Even + Even = Even ¨ Odd + Even = Odd ¨ Even × Even = Even¨ Odd × Odd = Odd¨ Odd × Even = Even¨ Any multiplication involving an even number creates

an even product.



Prime Numbers¨ A prime number is a positive integer that has

exactly two different positive divisors: 1 and itself.¨ 1 is not prime.¨ 2 is both the smallest prime and the only even

prime.



Check for Prime Number¨ Pick a number n.¨ Start with the least prime number, 2. See if 2 is a factor of

your number. If it is, your number is not prime.¨ If 2 is not a factor, check to see if the next prime, 3, is a

factor. If it is, your number is not prime.¨ Keep trying the next prime number until you reach one that

is a factor (in which case n is not prime), or you reach a prime number that is equal to or greater than the square root of n.

¨ 5. If you have not found a number less than or equal to the square root of n, you can be sure that your number is prime.



Squaring FractionsWhen positive fractions between 0 and 1 are squared or cubed, they get smaller.¨ (1/4) sqr = (1/16) ¨ (2/3) sqr = (4/9) ¨ (1/2) cube = (1/8)



ApproximationsTry to use approximations during exams where arriving to closest option is important¨ Square root of 2 = 1.4¨ Square root of 3 = 1.7¨ Square root of 5 = 2.25 ~ 9/4¨ 2 to the power 9 = 500¨ 2 to the power 10 = 1000¨ (1.1) to the power 8 = 2



Common Factors¨ Break down both numbers to their prime factors to

see what factors they have in common. Multiply shared prime factors to find all common factors.

¨ CORRECT: What factors greater than 1 do 135 and 225 have in common?

¨ 135 = 3 x 3 x 3 x 5, 225 = 3 x 3 x 5 x 5 ¨ Both share 3 x 3 x 5 in common—find all

combinations of these numbers: 3; 5; 3 x 3 = 9; 3 x 5 = 15; 3 x 3 x 5 = 45



Highest Common Factor (HCF)¨ Start by writing each number as product of its prime

factors.¨ Write so that each new prime factor begins in same

place.¨ Greatest Common Factor (GCF) is found by multiplying

all factors appearing on BOTH lists. ¨ 60 = 2 x 2 x 3 x 5 ¨ 72 = 2 x 2 x 2 x 3 x 3 ¨ HCF = 2 x 2 x 3 = 12





Least Common Multiple (LCM)¨ Start by writing each number as product of its prime

factors.¨ Write so that each new prime factor begins in same

place.¨ Lowest common multiple found by multiplying all factors

in EITHER list. ¨ 60 = 2 x 2 x 3 x 5 ¨ 72 = 2 x 2 x 2 x 3 x 3 ¨ LCM = 2 x 2 x 2 x 3 x 3 x 5 = 360



Application of LCM¨ Find the smallest natural number which is when

divided by x, y and z individually, leaves the same remainder r in each case?

¨ Here important point is that (number x – remainder r) is divisible by all x, y and z.

¨ So smallest such number will be the (L.C.M. of x, yand z) + r.



HCF and LCM of fractions



Averages



Application on Averages¨ Think of averages as balancing.Example: The average of 3, 4, 5, and x is 5. What is x? ¨ 3 is 2 less than 5 ¨ 4 is 1 less than 5 ¨ 5 is the average. ¨ x = 5 + 3 = 8



Mean

For 2 numbers a and b,



Application on MeanNumber Added or Deleted

Use mean to find number that was added or deleted from a group of numbers.¨ Total = mean x (number of terms)¨ Number deleted = (original total) – (new total)¨ Number added = (new total) – (original total)



Sum/Average of Consecutive Numbers Sum of n consecutive natural numbers = n/2*(a+l)Average of n consecutive natural numbers = (a+l)/2¨ The average of a set of evenly spaced consecutive

numbers is the average of the smallest and largest numbers in the set.

¨ Example: What is the average of all integers from 13 to 77?

¨ Ans: Avg of numbers = (13 + 77)/2 = 90/2 = 45



Properties of Surds





Laws of Indices



Laws of Indices



Percentage Change

For two successive changes of a% and b%



Percent Increase vs. Percent of¨ Be careful when calculating percent increase vs.

percent of.¨ % increase = (amount of change) / (original

amount)¨ % of = portion / whole



PercentageTo make a percentage, multiply by 100%: ¨ 1/400 = (1/4) % = 0.25%To drop a percent, divide by 100%: ¨ (1/2) % * 1/100 = 1/20000

Question: 15 is 3/5 percent of what number? ¨ 3/5 percent = 3/500 ¨ 15 = (3/500) x number ¨ number = 15x500/3 = 2500



Useful Percentages to remember

¨ 1/3 = 33.33% 1/11 = 9.09%¨ 1/6 = 16.66% 1/13 = 7.69%¨ 1/7 = 14.28% 1/17 = 5.89%¨ 1/8 = 12.5% 1/19 = 5.26%¨ 1/9 = 11.11% 1/23 = 4.35%



Useful Tricks to calculate PercentagesDivide the denominator into fractions or use subtraction method to arrive at smaller number¨ 1/12 = ½ * (1/6)% = 8.33%¨ 1/15 = 1/3 * (1/5)% = 1/3 * 20% = 6.66%¨ 5/6 = 1 – (1/6) = 100 - (1/6) % = 83.3%¨ 1/27 = 1/3* (1/9)% = 1/3*(11.11)% = 3.70%¨ 1/33 = 1/3 * (1/9)% = 1/3*(9.09)% = 3.03%



Simple and Compound Interest

Final Amount = Principal + Interest



Compound Interest¨ Usually you don’t need to calculate compound

interest. Try finding simple interest and looking for the answer that is a little bit larger.

¨ If there are more than 1 options close to each other, then calculate simple interest again on final amount.



Interest Problem¨ Rs.10,000 is invested at 10% interest rate,

compounded semi-annually, what is the balance after 1 year?

¨ Solution 1: 10,000 + (10,000)(0.05) = 10,500Then, 10,500 + (10,500)(0.05) = Rs.11,025¨ Solution 2 : 10,000 (1 + 0.10/2) sqr = Rs.11,025 ¨ For Compound Interest: Divide interest by # of times

compounded in 1 year to find interest for the compound period.



Population Formula

Here, P = Original population, P’ = population after n years, r% = rate of annual change



Growth RateAbsolute Growth = Final Value – Initial Value

Here, S. A. G. R. = Simple Annual Growth Rate A. G. R. = Average Annual Growth Rate C. A. G. R. = Compound Annual Growth Rate







Buy x and get y Free

If articles worth Rs. x are bought and articles worth Rs. y are given free along with x articles, then the discount is equal to yand discount percentage is given by:



RatiosIf a : b = c : d, then a : b = c : d = (a + c) : (b + d)



Proportions

a - bb



More Proportions



Successive Replacement

Where, x = Original quantity, y = Quantity that is replaced and n = Number of times the replacement process is carried out.



Allegation RuleThe ratio of the weights of the two items mixed will be inversely proportional to the deviation of attributes of these two items, from the average attribute of the resultant mixture.



Mixture Problem¨ Example: How many liters of a solution that is 15%

salt must be added to 5 liters of a solution that is 8% salt so that the resulting mixture is 10% salt?

¨ Solution: 0.15n + 0.08(5) = 0.1(n + 5) ¨ 15n + 40 = 10n + 50 ¨ 5n = 10 ¨ n = 2 liters



Balancing Method for Mixtures/Dilutions¨ Questions are based on mixing 2 solutions with

different ratios and coming to desired solution(percent/price difference between weaker solution and desired solution ) x ( amount of weaker solution )=( percent/price difference between stronger solution and desired solution ) x (amount of stronger solution)



Example on Balancing Method for Mixtures/Dilutions

¨ How many liters of a solution that is 10% alcohol by volume must be added to 2 liters of a solution that is 50% alcohol by volume to create a solution that is 15% alcohol by volume?

¨ n(15 – 10) = 2(50 – 15) ¨ 5n = 2(35) => n = 70/5 => 14 liters¨ So 14 Ltrs of 10% solution must be added.



Successive Replacement

Where, x = Original quantity, y = Quantity that is replaced and n = Number of times the replacement process is carried out.



Time, Speed and Distance

Important Conversion Factors



Average Speed

Here, d1, d2, d3.. are the distances travelled for t1, t2, t3.. time respectively in the journey



Application of Average Speed

If the distance is constant, then average speed is given by harmonic mean of two speeds:

If the time is constant, then average speed is given by arithmetic mean of two speeds:



Relative Speed¨ It’s the speed of one object with reference to other

object (either moving or stationary)Relative speed for 2 objects travelling in same direction:¨ S (relative) = S1 – S2Relative speed for 2 objects travelling in opposite directions:¨ S (relative) = S1 + S2



Application of Relative Speed

For Trains:

Where, L1 and L2 are the lengths of TrainsS1 and S2 are speed of the trains



Application of Relative Speed

For Boats and Streams:

When going in the stream’s direction, speeds get addedWhen going opposite of the stream, speeds get subtracted



Time and Work

¨ Consider work done in one hour.¨ Inverse of the time it takes everyone working

together = Sum of the inverses of the times it would take each person working individually.

¨ Example: You have worker A and worker B doing a job:

¨ 1/A + 1/B = 1/T



ProbabilityProbability of an event or outcome, given certain conditions equals:

(No. of favorable outcomes) (No. of possible outcomes)



Odds in Favour & Odds AgainstIn Some questions you are asked to calculate Odd in favour and odds against you:



Combined EventsFor events E and F:¨ not E = P(not E) = 1 – P(E)¨ E or F = P(E or F) = P(E) + P(F) – P(E and F)¨ E and F = P(E and F) = P(E)P(F)



Dependent Events¨ Two events are said to be dependent events if the

outcome of one event affects the outcome of the other event.

¨ Here Outcome of 2nd event will be dependent on 1st

event.



1st rule of Probability¨ Basic rule: The probability of event A occurring is

the number of outcomes that result in A divided by the total number of possible outcomes.



2nd rule of Probability¨ Complementary Events: The probability of an event

occurring plus the probability of the event not occurring = 1.

¨ P(E) = 1 – P(not E)



3rd rule of Probability¨ Conditional Probability: The probability of event A

AND event B occurring is the probability of event A times the probability of event B, given that A has already occurred.

¨ P(A and B) = P(A) × P(B|A)



4th rule of Probability¨ The probability of event A OR event B occurring is

the probability of event A occurring plus the probability of event B occurring minus the probability of both events occurring.

¨ P(A or B) = P(A) + P(B) – P(A and B)



Probability of Multiple EventsRules:¨ A and B < A or B¨ A or B > Individual probabilities of A, B¨ P(A and B) = P(A) x P(B) • “fewer options”¨ P(A or B) = P(A) + P(B) • “more options”



Indistinguishable Events¨ To find the number of distinct permutations of a set

of items with indistinguishable (“repeat”) items, divide the factorial of the items in the set by the product of the factorials of the number of indistinguishable elements.

¨ Example: How many ways can the letters in TRUST be arranged?

¨ Answer: 5!/2! = 120/2 = 60



Multiple Event ProbabilityTo determine multiple-event probability where each individual event must occur in a certain way:¨ Figure out the probability for each individual event.¨ Multiply the individual probabilities together.



Trial Problems¨ Look at the probability of NOT OCCURRING.¨ P(Event Not Occurring) = 1 – P(Event Occurring)¨ Example: What is the probability that No 6 is shown

up when 2 dice are thrown¨ Solution: P = 1 – (5/6)*(5/6) = 1 – 25/36

= 11/36



Probability and Geometry¨ If a point is chosen at random within a space with

an area, volume, or length of Y and a space with a respective area, volume, or length of X lies within Y.

¨ Then the probability of choosing a random point within Y is the (area, volume, or length of X) divided by the (area, volume, or length of Y).



Combinations: Order doesn’t matter¨ Combination of r items from the n items equals:

nCr = n!/r!*(n – r)!

¨ n = number of items¨ r = number of items chosen



Permutations: Order matters¨ Number of permutations of r objects from a set of n

objects equals:nPr = n!/r!

¨ n = number of items¨ r = number of items chosen



P & C : Important Properties

Number of ways of distributing n identical things among r persons when each person may get any number of things =





Circular Permutations

¨ The number of ways to arrange n distinct objects along a fixed circle is: (n – 1)!

¨ Logic is simple here ¨ n objects can be arranged in a straight line in n!

ways and there be n ways when all n objects are arranged in same order (when seen circular)

¨ So P = n!/n = (n – 1)!



Binomial Theorem

If n is a natural number that is greater than or equal to 2, then according to the binomial theorem:



Laws of Logarithms



Laws of Logarithms



Roots of Quadratic EquationThe two roots of the quadratic equation, ax2 + bx + c = 0 are given by:







Arithmetic Progression

Where, a = First term of the APd = common differencen = number of terms



Geometric Progression

Where, a = First term of the GPr = common ration = number of terms in GP



Harmonic Progression

Where, a = First term of the GPd = common differencen = number of terms in GP



Sum of Important Series

Sum of First n natural numbers

Sum of the squares of the first n natural numbers

Sum of the cubes of the first n natural numbers



Set Theory



Set Theory Problem



Pythagoras Theorem For Right Triangle ABC For acute triangle ABC



Pythagoras Theorem For obtuse triangle ABC



Area of a TriangleWhen lengths of the sides are given



Area of a TriangleWhen lengths of the base and altitude are given

When lengths of two sides and the included angle are given



Area of a TriangleFor Equilateral Triangle For Isosceles Triangle



Apollonius Theorem

If AD is the median, then:

AB2 + AC2 = 2(AD2 + BD2)



Angle Bisector Theorem

If AD is the angle bisector for angle A, then:

AB



Exterior Angle Theorem



Similar Triangle Areas



Frequently Used Triangles



Area of a Quadrilateral

For Cyclic Quadrilateral



Area of a QuadrilateralIf lengths of one diagonal and two offsets are given



Area of a QuadrilateralIf lengths of two diagonals and the included angle are given



Area of a QuadrilateralFor Trapezium



Area of a QuadrilateralFor Parallelogram



Area of a QuadrilateralFor Rhombus For Square



Area of a QuadrilateralFor Rectangle



PolygonsNumber of Diagonals

The sum of all the interior angles

The sum of all the exterior angles



Circle

Circumference

Area



CircleLength of Arc

Perimeter of Sector

Area of Sector



Trigonometric RatiosFor a right triangle, if P is the length of perpendicular, B is the length of base, H is the length of hypotenuse and Theta is the angle between base and hypotenuse



Distance between Points

Distance between two points A (x1, y1) and B (x2, y2) is given by



Cube

L.S.A. = 4a2

T.S.A. = 6a2

Volume (V) = a3



Cuboid

L.S.A. = 2(lh + bh)

T.S.A. = 2(lh + bh + lb)

Volume (V) = lbh



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