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PERCENTAGES, PROFIT&LOSS, PARTNERSHIP: MEANING: The word percentage indicates the contribution of a quantity to every 100. (ie) suppose we say I scored 60% then I may have scored 60/100, 120/200 ,180/300…….. etc. Thus the total marks and marks obtained may not be the same in all the cases but the percentage is the same. % is just an indicator. PERCENTAGE INCREASE/DECREASE: This is also an indicator which shows by how much % the value changes when compared with a value already available. This is mainly used for comparison purposes. % increase = (final value – initial value ) *100 Initial value % decrease= (initial value– final value) *100 Initial value As the formula suggests when the initial value> final value then it is %decrease when the initial value< final value then it is %increase Let initial value be X The % increase is given as p% then the new value is given by X(1+p/100) The % decrease is given as p% then the new value is given by 1
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Quantitative Aptitude and Logical Reasoning

Oct 15, 2014

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Mithun Singh
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Page 1: Quantitative Aptitude and Logical Reasoning

PERCENTAGES, PROFIT&LOSS, PARTNERSHIP:

MEANING: The word percentage indicates the contribution of a quantity to every 100.(ie) suppose we say I scored 60% then I may have scored 60/100, 120/200 ,180/300…….. etc. Thus the total marks and marks obtained may not be the same in all the cases but the percentage is the same.% is just an indicator.

PERCENTAGE INCREASE/DECREASE:This is also an indicator which shows by how much % the value changes when compared with a value already available. This is mainly used for comparison purposes.% increase = (final value – initial value) *100 Initial value% decrease= (initial value– final value) *100 Initial valueAs the formula suggests when the initial value> final value then it is %decreasewhen the initial value< final value then it is %increase

Let initial value be XThe % increase is given as p% then the new value is given byX(1+p/100)The % decrease is given as p% then the new value is given byX(1-p/100)

When there are successive increase in the % by p%,q%,r% then the effective % increase is given as 100* {(100+p)(100+q)(100+r)} - 1 100*100*100

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If the value of an item goes up/down by x% the % reduction /increment to be now made to bring it back to the original level is

100x %(100±x)

If A is x% more/less than B, then B is 100x %100±x

If the price of an item goes up/down by x% then the quantity consumed should be reduced/increased by100x % so that the total expenditure is the same.100±x

PROFIT AND LOSS

There are different prices that one needs to know.COST PRICE:The cost price is the price at which one buys from a shop.SELLING PRICE:Selling price is the price at which one sells an article.

The discount that the shopkeeper gives his customer is on the marked price(list price) of the product.

We say a profit is incurred by shopkeeper when his selling price is greater than the cost price. In the reverse situation it is a loss.

The profit or the loss % is calculated on the cost price of the item.Profit % = 100 * (S.P – C.P) C.P

Loss % = 100* C.P-S.P C.P

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When 2 articles are sold at the same price such that there is a profit of p% on one article and p% loss on the article then the net resultant to the person is a loss % ofLoss % = (common profit or loss percentage) 2 100

Selling price = Marked price – DiscountThus Discount % = 100* Marked price – Discount Marked Price

When successive discounts are given say p%,q%,r% then the selling price is S.P = M.P (100-p)(100-q)(100-r) 100*100*100

PARTNERSHIP

There are basically 2 types of partners in a business. One is the active partner who participates actively in the business and the other the sleeping partner who invests his capital and stays away from the day to day running of the business.

AMOUNT AND TIME PERIOD SAMEWhen people invest the same amount of money for the same period then the profit is shared in 1:1:1 ratio (ie) equally among the partners.

AMOUNT SAME AND TIME PERIOD DIFFERENTWhen the money invested is the same but the period for which it is invested is different, then the profit is shared in the ratio of the time period for which the amount is invested.TIME PERIOD SAME AND AMOUNT DIFFERENTWhen the time period of the amounts invested is the same but the amounts are different then the profit is shared in the ratio of the amounts invested respectively.

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AMOUNT AND TIME PERIOD DIFFERENTWhen both the amount invested and the time period for the amount invested is different, then the profit is shared in the ratio of the product of the time period and the amount.

PROFIT SHARING BY ACTIVE PARTNER:Another important fact to be considered is when there exists a working partner he may be paid for actively putting efforts into daily business. Thus, after paying the active partner the remaining profit will be shared among all the partners in the way as told above based on the amount and time period of investment.

Examples:Case 1: All of them investing for 12 months each with a capital of 10000 each.The ratio is 1:1:1

Case 2: A -- Rs 4000 for 12 months B --- Rs.4000 for 6 months C ---- Rs.4000 for 5 monthsThe ratio of sharing profit is 12:6:5

Case 3: A --- Rs.8000 for 12 months B --- Rs.4000 for 12 months C --- Rs.5000 for 12 monthsThe ratio of sharing profit is 8:4:5

Case 4: A---- Rs. 8000 for 4 months B ----Rs. 9000 for 5 months C --- Rs.5000 for 6 monthsThe ratio of profit sharing is 8*4 :9*5 : 5*6

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NUMBERS:

The broad classification of numbersReal numbersComplex numbers (general form a±ib)

Real Numbers:The real numbers are classified as rational and irrationalRational:A number that can be expressed in the form p/q where p and q are integers and q≠0 is called a rational number.Eg: ¾ , 2/11 ,-8/23

Recurring decimals also come under the rational numbers.Eg: .3333…. is nothing but 1/3 .6666…… is nothing but 2/3 Hence recurring decimals come under rational numbers.

Irrational numbers:Any non-terminating and non-recurring decimal is called irrational numbers.Eg:√2, √3

Integers:Integers includePositive integersNegative integersZero

Prime numbers:These are numbers which have 1 and itself as its factors.Eg: 17, 13 , 11, etcAny prime number greater than 3 can be represented of the form (6k-1) or (6k+1)

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Composite Numbers:The number that has factors other than 1 and itself is called composite.Eg: 8, 10, 20Note: 1 is neither prime nor even. 2 is the only even prime number.

Relative prime:Two numbers are said to be relative primes then they have only 1 as the common factor.Eg: 15, 14Neither of them are individually prime but the only common factor between them is 1 thus they are relative primes.Multiples:If one number is divisible by another the first number is called as the multiple of the second.Eg: 15 is a multiple of 3.

Factors:If one number another divides without leaving a remainder the first number is called as the factor of the second.Eg: 3 is a factor of 15.

Even and odd numbers:Any number divisible by 2 is called as even number.Any number not divisible by 2 is called as odd number.

The sum of even number of odd numbers is always even.Eg: 3+5+7+9= 24The sum of odd number of odd numbers is always odd.Eg: 3+5+7=13

The product of any number of numbers with at least one even number is always even.

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Perfect number:A number is said to be a perfect number if the sum of all of its factors excluding itself is equal to the number itself.Eg: 28 – (factors are 1,2 4,7,14) ( when we add the factors we get 28 the number itself.)

BODMAS:Simplification of an expression has to be done according to the precedence of operations.B—brackets The order under brackets is

(i) Simple bracket (ii) Flower bracket(iii) Square brackets

O- ofD- divisionM—MultiplicationA—AdditionS—Addition

Divisibility rules:Divisibility by 2:Any number having even number as the last digit is divisible by 2Divisibility by 3:Any number whose sum of digits is divisible by 3 is divisible by 3Divisibility by 4:Any number whose last 2 digits of a number is divisible by 4 is divisible by 4Divisibility by 5:Any number whose last digit is 5 is di divisible by 5Divisibility by 6:If a number is divisible by 2 and 3 then it is divisible by 6.Divisibility by 8:A number whose last 3 digits are divisible by 8 is divisible by 8.Divisibility by 9:A number whose sum of digits is divisible by 9 is divisible by 9.

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Divisibility of 11:The difference between the sum of the digits of the alternate digits is 0 or multiples of 11 then the number is divisible by 11.Eg: 121 ---(1+1)-2 =0 89394811 ---|(8+3+4+1)-(9+9+8+1)|=11

Conversion of recurring decimals into fraction:0.6 means 0.666666…..Let x=0.666666… ----110x=6.6666666….. ---2Thus 9x= 6 (by eqn (2)- eqn(1))x=6/9 =2/3

0.81= .818181….. ----1X=.818181……. ---2100x=81.8181 Thus 99x=81 (by eqn (2)- eqn(1))X=81/99=9/11

0.156=x10x=1+.56.56=y100y=56.565656…100y-y = 5699y=56Y=56/99thus 10x=1+56/99=155/99X=155/990

Number of factors of a number:N=ap×bq×cr

Where a, b, c are prime numbers.(Ie) we need to express the number in terms of the product of different prime numbers.Thus the total number of factors is (p+1)(q+1)(r+1)Eg: 24=23×31

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Thus the total number of factors are (3+1)(1+1)=8They are 1,2,3,4,6,8,12,24.

Number of ways of writing a number as product of 2 co primes.Let N= ap×bq×cr

The number of ways of writing it as a product of 2 co-primes is 2n-1.

Where n is the number of distinct prime factors of the given number NEg: 24=23×3 here the distinct prime factors are 2 and 3 so the number of ways 22-1=2

Least Common Multiple (LCM) and Highest Common Factor (HCF)

Product of two numbers = LCM * HCF

HCF of Fractions = HCF of Numerators LCM of Denominators

LCM of Fractions = LCM of Numerators HCF of Denominators HCF by long division:Divide the larger number by the smaller then divide the divisor by the remainder; divide the divisor of this division by the next remainder and so on until the reminder is 0. The last divisor is the HCF of the two numbers taken . Eg: Find 180 and 22

22)180(8 176 4)22(5 20

2)4(2 4

0

Thus the HCF of 180 and 22 is 2 (Last divisor)

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Finding the last digit of any power

The last digits of the powers of any number follow a cycle pattern. They repeat after certain number of steps. Eg; Last digit of 21=2 Last digit of 22=4 Last digit of 23=8 Last digit of 24=6 Last digit of 25=2 Thus we find that the last digit of powers of 2 repeat itself after every 4 times . For eg find the last digit of 288 88 has 22 4’s thus the last digit is 6 find the last digit of 294 94 /4 = 23 remainder =2 . Thus the last digit is 4 Check for the remainder when the power is not exactly divisible by the number after which the last digit of the power of number repeats itself .

Remainder theoremThis is to find the remainder when a number is divided by another .

Eg: Divide 254 by 7 By remainder theorem method R(f(x)/(x-a)) is given as f(a).Let us consider 23 as x because the divisor 7 can be expressed in term of 23 as 23 -1Thus Nr: 254 = (23)18= x18

So the division becomes x18/ (x3-1)By remainder theorem method here it is (1)54 =1 is the remainder.

To find the largest power of a number in N!Find the largest power of 7 in 256! 7 256 7 36 7 5 Thus considering only the quotient when the number is successively divided and ignoring the remainder we get 36+5=41.

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The largest power of 7 in 256! is 41.This successive division method is applicable only when the divisor is a prime number.Else the divisor is split into its prime factors and the prime factor which gives the least value by the above said method is alone considered and the least value is taken as the largest power of the number in N!.Eg: Find the largest power of 10 in 20!10 is split into its prime factors as 2,5Thus 2 202 102 52 22 1We get the value as 10+5+2+1 =18

5 205 4Here we get 4 We should always consider the least value given by the prime factors.Thus the highest power of 5 in 20! is 4.

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GEOMETRY AND MENSURATION:TRIANGLES:

The sum of the 3 angles in a triangle is 180 degrees.

The exterior angle is equal to the sum of the interior opposite angles.

A line that is perpendicular to a side is called as the altitude.

A line that is perpendicular and bisects the side it is called as the perpendicular

bisector.

A line drawn from a vertex that bisects the side opposite to the vertex is called

median.

A line which bisects the angle is called the angle bisector.

In any triangle the sum of 2 sides is always greater than the third side.

Acute angle triangle

The square of the greatest side is less than the sum of the squares of the other two

sides in an acute angle triangle.

Obtuse angle triangle

The square of the greatest side is more than the sum of the squares of the other

two sides in an obtuse angle triangle.

Right angle triangle

The square of the greatest side is equal to the sum of the squares of the other two

sides in a right angle triangle

Centroid is point where all the medians intersect.

The centroid divides the median in the ratio of 2:1 where the greater length is

towards the vertex.

Area of the triangle is ½ ×base×heightArea = (s(s-a)(s-b)(s-c)) where s-semiperimeter = (a+b+c)/2

The area of the equilateral triangle is √3 a2/4 a- Side of the triangle.

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The height of the equilateral triangle is √3 a/2

An isosceles triangle is one in which two sides are equal or two angles are equal.The median to the base of the isosceles triangle is also the perpendicular bisector, altitude, median.Area of isosceles triangle:b/4 * √(4a2-b2) a-common side b-base

QUADRILATERAL:In a cyclic quadrilateral the sum of the opposite angles is equal to 180 degrees.The exterior angle is equal to the interior opposite angle.Area of a quadrilateral:Area = 1/2×one diagonal × sum of the offsets drawn to that diagonal from the other 2 vertices.Offset: it is the perpendicular drawn to the diagonal from the vertex.

RHOMBUS:Side of the rhombus: {(1/4) (sum of squares of diagonals)}1/2

RECTANGLE:Length of the diagonal is (a2+b2)1/2

Area: length* breadth

POLYGON:The sum of interior angles in a polygon is (2n-4)90°A polygon with n sides has n(n-3) diagonals. 2Area= ½ * perimeter* perpendicular distance from the centre of the polygon to any side.

PARALLELOGRAM:Area: base * heightAlso it can be given as area= product of two sides * sine of the included angle.

TRAPEZIUM:Area: ½ * sum of parallel sides * distance between them.

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SQUARE:Area= side2

=1/2 * (diagonal)2

CIRCLE:Area=A=π r2

Circumference:2πr

Length of arc= 2πr * Ѳ 360Area of sector:π r2* Ѳ 360ELLIPSE:Area : πab where a and b are the semi major and semi minor axis.Perimeter= : π (a+b )

AREA OF SOLIDS:

PRISM:Lateral surface area: perimeter of base * heightTotal surface area: LSA+2*area of baseVolume: area of base * height

CUBOID:LSA: 2(l+b)hTSA: 2(l+b)h+2lbVolume:lbhLongest diagonal of the cuboid: √(l2+b2+h2) l- length,b-breadth,h-height

CUBE:volume=a3

LSA=6a2

TSA=4a2

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a-side

CYLINDER:LSA: 2πrhTSA: 2πrh+2π r2

Volume: π r2h

r-radius, h-height

CONE:CSA: πrlTSA: πrl+π r2

Volume = 1/3* π r2h

L= slant height: √( r2+ h2)

SPHERE:Surface area of sphere: 4π r2

Volume: 4/3 * πr3

HEMISPHERE:Surface area = 2πr2

Total surface area : 3πr2

Volume: 2/3 πr3

TIME AND WORK

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 The 3 important factors are1)time required2) men available3) work to be completed When time is constant:Men and work vary in direct proportion (ie) when work to be done increases number of men required to complete it within time also increases. When work is constant:Men and days are inversely proportional (ie) when number of men available is high then the number of days required is less. When number of men is constant:Work and days are directly proportional (ie) when work is increased the number of days required also increases. CONCEPT OF MANDAYS: When the work done is the same the product of men and days is constant.(ie) when a piece of work can be done in 12 days by 8 men then the same work can be completed by 6 men in :Total work = 12 * 8 =96Thus 6* days =96Days= 16 If M1 men can do W1 work in D1 days working H1 hours/day and M2 men can do work W2 in D2 days working H2 hours /day thenM1*D1*H1/W1   =    M2*D2*H2/W2 If A can do a piece of work in p days and B can do a piece of work in q days then the no. of days required to complete the work together isp*q/(p+q) 

SHARE OF WORK:

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The money paid for a job is distributed among the members in the ratio of their efficiencies.

Efficiency is inversely proportional to the no. of days required.Suppose A, B, C can do the job in 3, 4, 6 days respectively then the amount paid for the work is shared in the ratio of their efficiencies

(ie) 1/3:1/4:1/6(ie) 4/12 : 3/12: 2/12= 4:3:2 PIPES AND CISTERNS:

The only difference here is that there can be pipes that are emptying in that case negative work is done hence we indicate this by a ‘minus sign’.

TIME AND DISTANCE

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Speed=distance/time Conversion (km/hr to m/s)Multiply by 5/18 (ie 1000/3600) Conversion(m/s to km/hr)Multiply with 18/5 Average Speed: Total distance/total time If a body travels from point A to B with a speed p km/hr and point B to A with speed q km/hr then the average speed of the body is 2pq/ (p+q)

(Or) If a body travels a certain distance at speed p and the remaining distance at speed q and the distances covered are in the ratio m:n then the avg speed for the entire journey is(m+n)pq/(mq+np) When the distance d is travelled at a speed u km/hr  the person reachesp hours late and when travelled at a speed v km/hr the person reachesq hours before.Then the distance = vu (p+q)/(v-u) Relative speed:The relative speed when two objects travel in the opposite direction is the sum of their speeds.The relative speed when two objects travel in the same direction is the difference of their speeds. When 2 persons are separated by a distance‘d’ and they travel towards each other with speeds ‘a’ km/hr and ‘b’ km/hr then the time taken to meet is given asT=distance/ relative speedT=d/ (a+b). 

Circular tracks:

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Considering that the race is not for one round When 2 persons with speeds ‘a’ km/hr and ‘b’ km/hr start running in the same direction the time taken to meet for the first time isLength of the track=LT=L/ (a-b) When 2 persons with speeds ‘a’ km/hr and ‘b’ km/hr start running in the opposite direction the time taken to meet for the first time isLength of the track=LT=L/ (a+b) When 2 persons with speeds ‘a’ km/hr and ‘b’ km/hr start running in the same / opposite direction the time taken to meet for the first time at the starting point isLength of the track=LT=LCM (L/a, L/b) (in both cases-à same and opposite) The same logic applies for any number of people running (for example if 2 people ram ,shyam run in the same direction with speeds ‘a’ km/hr, ‘c’ km/hr  and akash runs in the opposite direction with speed ‘b’ km/hr the time taken to meet for the first time(all 3 of them)  is LCM (L/ (a-b), L/ (b-c)) The same logic applies for any number of people running (for example if 2 people ram ,shyam run in the same direction with speeds ‘a’ km/hr, ‘c’ km/hr  and akash runs in the opposite direction with speed ‘b’ km/hr the time taken to meet for the first time at the starting point isLCM (L/a, L/b,L/c) Boats and streams: Upstream speed (u) = speed of boat in still water - speed of water current Downstream speed (v)= speed of boat in still water + speed of water current Speed of boat in still water= (u+v)/2 Speed of water current = (v-u)/2 Clocks:

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The angle between any two consecutive hours is 30 degrees. The minute hand moves 6 degrees every minute and the hour hand moves by ½ degrees every minute. Thus the relative speed between the hour and the minute hand is5.5 degrees/min The time (in mins) between any 2 coincidences of the minute and hour hand is65 5/11 mins There are 22 coincidences in a day when the angle between the minute and the hour hand is 0 degrees. When the time (in mins) between any 2 coincidences of the minute and hour hand is less than 65 5/11   mins then the clock is fast . When the time (in mins) between any 2 coincidences of the minute and hour hand is more than 65 5/11   mins then the clock is slow. PROBLEMS ON TRAINS: When two trains of lengths L1 and L2 travel in the opposite directions with speeds ‘a’ km/hr and ‘b’ km/hr then the time taken to meet isT=d/ (a+b)d-distance btw them The time taken to cross each other isT= (L1+L2)/ (a+b) When two trains of lengths L1 and L2 travel in the same directions with speeds ‘a’ km/hr and ‘b’ km/hr then time taken to meet isT=d/ (a+b)The time taken to cross each other isT= (L1+L2)/ (a-b)

AVERAGES AND MIXTURES:

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Average: sum of all the items in the group Number of items in the group

If the value of each of the item is increased by the same value p,then the average of the group or items will also increase by p.

If the value of each of the item is decreased by the same value p,then the average of the group or items will also decrease by p.

If the value of each of the item is divided by the same value p,then the average of the group or items will also be divided by p.

If the value of each of the item is multiplied by the same value p,then the average of the group or items will also be multiplied by p.

Weighted Average:When there are 2 groups then the average can be calculated atleast when the ratio of members in the groups are given.Suppose the number of members in group 1 is 100 in group 2 is 50 and the average is 40 and 20 respectively then weighted avg is 100* 40 + 50*20 40+20(Or) even if the ratio is given as 4:2 then the average can be calculated.100*4+50*2 6

If q1 is the quantity of one particular item of quality p1, q2 be the quantity of the second item of quality p2are mixed then the weighted average value of the quality of the mixture is P=p1*q1+p2*q2 q1+q2

The above formula can be extended upto any number of quantities.

SIMPLE INTEREST & COMPOUND INTEREST:

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SIPMLE INTERESTA sum of money when invested in simple interest means that the interest is calculated on the same principal every year (ie) the principal deposited initially.

Let us take the principal invested as 100Rs.The interest rate to be 5%Then the S.I at the end of the year can be given as S.I= p*n*r 100The simple interest remains the same for any number of years.Amount= principal + S.IAmount= p+ p*n*r 100COMPOUND INTEREST:A sum of money when invested in compound interest means that the interest is calculated on the amount that results after every year.Thus the above statement can be restated as the interest for the first year is calculated on the principal initially invested. Then for the next year the interest already calculated and interest on this interest is calculated. Thus the process continues.

Eg:Rs.1000 is invested initially at the rate of 5% compound interest.1 st yr interest is the same as simple interest given by 1000*5/100=50 Rs.For the 2 nd yr it is 50+50*5/100 = 50+2.5=52.5 (I st yr interest + interest on 1 st yr interest)Thus the collective interest for two years is 50+52.5=102.5Rs

We also have interest being compounded every 6 months or 3 months etc.In such cases we need to divide the rate percent for a year by number of times it is compounded in a year.

Suppose the principal is Rs.5000 at rate percent 12% per annum and is told interest is compounded half yearly then after a year the interest will beRate for 6 months= 12%/2 = 6%Thus interest for 6 months= 5000*6/100=Rs.300

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For the next 6 months it is 300+300*6/100 = Rs. 318Thus after a year the interest is Rs.618 (300+318) Hence we find that when interest is compounded half yearly it is more than that of interest compounded annually.

When compounded annually:C.I = 5000*12/100= Rs.600

Hence if we take the case of interest getting compounded half yearly it is equal to a effective 12.36% under interest being compounded annually in the above example.

Also amount under C.I when compounded annually can be given as A=P+C.I = P(1+(r/100))n

Where P = principal,r-rate %, n- no. of years.

When interest is compounded k times a year the amount is given as A=P (1+(r/(100*k))k*n

n- no. of yearsk= no. of times interest is compounded in a year.P=principal

DIFFERENCE BETWEEN THE COMPOUND INTEREST AND SIMPLE INTEREST FOR TWO YEARS.

Difference between C.I and S.I for 2 yrs is P(r/100)2

PERMUTATION AND COMBINATION:

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PERMUTATION:The word permutation deals with the way of arranging things.In case of permutations the arrangement of items change the meaning or the ultimate aim of the experiment.For example: when the no. of possible words that can be formed using the alphabets a,b,I,s,p,c is asked without repeating the alphabets , the position of the alphabet matters as “bicasp” is different from “capsib”

Any experiment where the outcome is different when the arrangement is changed permutation plays a role.

COMBINATIONS:Even when the arrangement of items is changed the outcome doesn’t change then combination plays a role.For example:When there is a bag of 5balls--- 1 white 1-blue 1 red 1-yellow,1-greenAnd the experiment is you need to pick 3 balls:It is the same if you pick1st- red,2nd- yellow and 3rd- white (or) 1st –yellow 2nd—red,3rd- white Because you have ultimately picked up 3 balls whose colors are red, white and yellow.But the picking of red, yellow and green balls is different from the above said combination.That is the outcome should differ by at least one color from the previous one.Hence the combinations that are possible are Red, blue, yellowRed, blue, green,Red, blue, whiteRed, yellow, greenRed, yellow, whiteYellow, green, blueYellow, green, whiteBlue, white, yellowBlue, white, green

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Blue, green, yellowHence 10 combinations are possible.FORMULA---- PERMUTATION:The formula for Permutation is nPr = n! (n-r)! Where n- total no.of itemsr- no. of items selected at a time according to the experiment.This is the general formula and is applicable only when theItems are used only onceAll the items are distinct and dissimilar.Eg: there are 4 letters a,b,c,dWe need to form different words from it. Thus The first blank can be filled in four ways as you have 4 options either a or b or c or d.After filling one of the blanks we are left with 3 letters for the second blank.Thus we are left with 2 options for third blank and only one option for the fourth blank.Hence the no. of ways the words can be arranged is 4*3*2*1=24.

Number of arrangements of n items of which p are of one type, q are of one type, and the others are distinct: n! p!*q!This can be extended to any number of groups having identical elements.

Eg: there are 4 balls of red color and 3 of white color 2 of green we need to find out in how many ways balls can be arranged.

Thus no. of arrangements are: 9! 4!*3!*2!

Until now we have seen about linear arrangement of items but if items need to be arranged in a circular fashion then the total no. of ways the n items can be arranged is (n-1)!

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FORMULA-----COMBINATION:The formula for COMBINATION = n! (n-r)!Where n- total no.of itemsr- no. of items selected at a time according to the experiment.

The total number of combinations possible for an experiment is given as nC1+nC2+nC3+……….. nCn=2n-1

Here also the n items must be distinct.

Dividing the given items into groups:Dividing (p+q) items into 2 groups one of p members and other of q members.The number of ways the total no.of items can be split into groups of p items and q items is (p+q)! p!*q!This formula can be extended to the splitting of n items into various groups of desired no. of items for each group.

PROBABILITY:

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There are two types of experiments: Deterministic Random

Deterministic Experiment: An experiment where the outcome is certain or unique is called deterministic experiment.

Random:An experiment where the result is not unique and there are several possible outcomes then such an experiment is called Random experiment.

Biased and Un-biased experiments:When we throw a die we cannot say whether one of the numbers will turn up more likely or frequently than other numbers. Thus it is an un-biased die. But if suppose assume that there is number 2 on three faces of the die then number 2 is likely to come more no. of times than other numbers.This is a biased die.

What is an event?It is defined as the collection of occurrences.Now let us consider an example for an event:Getting an even number when we throw a die.This event consists of 3 occurrences namely getting a 2 (or) getting a 4 (or) getting a 6 (as 2, 4, 6 are all even numbers)

Probability of an event may be defined as:The number occurrences in favour of the event Total no. of occurrences possible

Suppose we take ‘m’ as the favourable occurrences and total no. of occurrences are ‘n’ then P(E)= probability of success: m/nP(E) = probability of failure : (n-m)/nWhen the probability of an event is P(E)=1 then it is termed as a certain eventWhen the probability of an event is P(E)=0 then it is termed as an impossible event.

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COLLECTIVELY EXHAUSTIVE EVENTSWhen a certain no. of events together cover all possible occurrences of an experiment they are termed as collectively exhaustive events.Eg: getting a number>=3, getting a number<=3.

MUTUALLY EXHAUSTIVE EVENTS:When the occurrence of one event precludes the possibility of the occurrences of any other event such events are called mutually exclusive events.Eg: getting an even number, getting an odd number.Here there are common occurrences like we saw in the above case where getting a 3 was common to both events.

In generalP(A U B) = P(A) + P(B) – P(A п B)In case of mutually exclusive events P(A п B)In case of independent events P(AпB) = P(A) *P(B)

PROGRESSIONS:

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Progressions are of 3 types: Arithmetic Harmonic Geometric

Arithmetic Progression:In case of arithmetic progression we find that the next term in a series is either increased or decreased by a constant value.For eg: 2,6,10,14…………..The nth term in the series can be identified using the formula Tn=a+(n-1)dWhere a= first termD=common differenceTn = nth termThe sum of n terms is Sn=n(2a+(n-1)d) 2When there are three numbers in A.P say a, b, cThen the arithmetic mean is given by: b= a+c 2The general representation of numbers in A.P can be expressed as a-2d,a-d,a,a+d,a+2d……..The common difference can either be negative or positive.

Geometric progression:In this series we find that the next term in the series is scaled by a common ratio.(Ie) multiplied by a common ratio that can be greater or lesser than 1.The general representation may be given as a/r2 , a/r , a , ar, ar2, ar3 ……………

The nth term of a G.P can be expressed as Tn= ar(n-1)

a-first term

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r-common ratiosum of n terms can be given asSn= a(1-r n ) or a(r n -1) 1-r r-1The geometric mean of a series can be given asG.M= a1*a2*a3*a4*a5*…………..*an

When the G.P series continues till ∞ then the sum of the series is given bySn= a 1-rSome important results:∑n= n(n+1) 2∑n2= n(n+1)(2n+1) 6∑n3= (n(n+1)) 2 4

RATIO AND PROPORTIONS:

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The ratio of numbers is obtained when we cancel the common factor between the 2 numbers.(Eg):We have 2 numbers 24 and 18 the highest common factor is 6Hence the ratio of the numbers is given as 4:3Thus when a ratio is given the original numbers may be taken as 4x and 3x.

Let us consider the ratio a:bIf a > b then it is called a ratio of greater inequality.If a < b then it is called a ratio of lesser inequality.If a=b then it is called a ratio of equality.Also If a< b then (a+x):(b+x) > a:bIf a > b then (a+x):(b+x) < a:bIf a=b then (a+x):(b+x) = a:b

PROPORTIONS:

When two ratios are equal, then the four quantities involved in the ratios are set to be proportional (i.e) If a/b = c/d , then a, b, c, d are proportional. Also if a:b=c:d then b: a =d:c --------- 1 a: c = b:d --------- 2 (a+b): b= (c+d): d --------- 3 (a-b): b= (c-d): d --------- 4

Dividing equation 3 by equation 4 we get (a+b):(a-b)=(c+d): (c-d)

The two types of proportions are direct and indirect proportions.

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Direct variation:

When a quantity increases as a result of which the other quantity increases it is called direct variation and can be expressed as a α b .Thus a=kb, where k is a constant.

Inverse variation:

When a quantity increases as a result of which the other quantity decreases it is called inverse variation and can be expressed as a α 1/b. Thus a = k/b ,where k is a constant .

LOGICAL REASONING:

DEDUCTIONS:

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The statements given in the question is called as the premises.

The premise consists of a subject and a predicate.

A premise can be of the two forms:

Universal and

Particular

The qualifier that identifies if a premise is universal or not is “ALL”

The word ‘some’ identifies a particular premise.

The premise can also be classified as positive or negative

SUBJECT PREDICATEUniversal affirmative XUniversal negative Particular affirmative X XParticular negative X

Rules for deduction:

Every deduction should contain three and only three distinct terms. The middle term must be distributed atleast once in the premises. If one premise is negative then the conclusion must be negative. If one premise is particular then the conclusion must be particular. If both the premises are negative no conclusion can be drawn. If both the premises are particular no conclusion can be drawn. No term can be distributed in the conclusion if it is not distributed in the

premises.

Eg: all dogs are cats.

All refers to universal affirmative hence the subject dogs is distributed (indicated by a tick mark in the table for universal affirmative.)

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The predicate cats is not distributed (as indicated by a cross mark under universal affirmative for predicate)

Consider 2 statements:

All dogs are cats.

All cats are pigs.

The common term cats must be distributed atleast once. In this case it is true.

The deduction should thus be

All dogs are pigs and not all pigs are dogs because the term which is not distributed in the premises cannot be distributed in the deduction.

If we write all pigs are dogs we make pigs distributed in the deduction which is not distributed in the premises. So only solution is

All dogs are pigs

The common term should not come in the deduction.

If there is no common term then no conclusion is possible.

Binary Logic:

In case of binary logic there can be 3 types of people

Truth-teller: who always tells the truth

Alternator: whose statements can be true or false.

Liar: whose statements are always false.

We need to assume the person to be the truth teller or liar or alternator to check if the given statements satisfy the assumption.

For Eg:

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A: I am not the liar. C is the liar.

B: I am the liar. A is the truth teller.

C: I am the alternator, B is the liar.

Find out who is who?

From the 1st statement of B we find that B cannot be the liar or truth teller(because if a liar says I am not a liar he becomes truth teller, if he is truth teller the statement I am a liar cannot be told)

thus he is the alternator.

Since the first statement of B is false the 2nd is true so A is the truth teller and C the liar.

Venndiagrams:

U

Here we find that the entire set is represented as U

Overlapping between the sets may also be possible as shown in the figure.

Thus ,

U= n(A)+n(B)-n(AnB) + n(AUB)|

Suppose we consider there a 100 people

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10 people don’t like either coffee or tea.

40 like coffee and 20 like both coffee and tea.

Then find out the no. of people who like only tea.

100= 40+x-20+10

X=70

The same concept can be applied to 3 sets also.

U= n(A)+n(B)+n(C)-n(AnB)-n(AnC)-n(BnC)+n(AnBnC)+n(AUBUC)|

ORDERING AND SEQUENCING:

Depending upon the conditions imposed we need to sequence the items so that it in turn satisfies the conditions.

For Example:

Seven persons Paul, Queen, rax, sam, tom, unif and vali are sitting in a row. Ram and sam sit next to each other. There must be exactly 4 persons between queen and vali.sam sits to the immediate right of queen.

The above condition is general and applicable to all questions based on the above condition.

1) If paul and Tom are separated exactly by 2 persons, then who sits to the immediate left of vali.The arrangements that are possible are

V R S Q

Or

V R S Q

Since there are 4 people between Q and V the above 2 arrangements are only possible.

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Now, in the given question Paul and Tom are separated by 2 persons which means only the first arrangement is possible.

So the arrangement can now be

T V U P R S Q

P V U T R S Q

Hence in both cases we find unif sitting to the immediate left of vali.

CRITICAL REASONING

Some argument may be stated in general and the argument that supports or weakens it may be asked under this topic.

For years, a considerable number of students on West County High School's track team complained about shin splints (medial tibial syndrome). However, during the most recent season, the number of students who complained about shin splints dropped significantly. School officials assert that this reduction in complaints occurred entirely as a result of the school's decision to build a new running track that provided a softer running surface, which absorbed much of the shock on the knees and shins that occurs when running and causes shin splints.

Which of the following, if true, most severely weakens the school officials' explanation for the decrease in complaints about shin splints?

a) As a result of West County High School's adoption of better medical staff and new medical scanning devices, many students whose complaints would have been diagnosed in years past as an instance of shin splints are now diagnosed with a different condition.

b) West County High School built its track after a number of neighboring schools with similar track teams built new tracks and each school saw the number of complaints about shin splints drop.

c) This past season, members of West County High School's track team received and wore new and highly acclaimed shoes designed to soften the impact of running on the shin and knee.

d) This past season, the total number of students who complained of pain while running rose.

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e) The maker of the new track claims that on average, complaints about shin splints fall 25% when its tracks are implemented.

The answer is c

The reason why option c is correct may be given as:

a) The argument made by West County High School officials is based upon the fact that "the number of students who complained about shin splints dropped significantly." This answer would weaken an argument that dealt with the number of diagnosed instances of shin splints. However, the school administrators make their argument only because of a reduction in the number of "claims."

b) This answer significantly strengthens the argument of the high school administrators by noting that other schools experienced a link between a new track and a decrease in claims about shin splints.

c) This answer calls into question the school's assertion that the new track was "entirely" responsible for the reduction in claims of shin splints. The answer does this through providing an alternative and viable (but not necessarily competing) explanation of the reduction in claims of shin splints.

d) The original argument pertains to complaints about shin splints in particular (not the number of students who "complained of pain while running"). This answer confuses complaints in general with complaints about shin splints in particular.

e) This answer strengthens the school officials' claim instead of weakening it as it provides more evidence that the new track helped decrease complaints of shin splints.

References:

Books:

“ Quantitative Aptitude” by R.S AGGARWAL.

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“Logical and verbal Reasoning ” by R.S AGGARWAL .

“Numerical ability and Mathematical Aptitude” by Dr. A.B .Rao .

Websites:

http://www.indiabix.com/aptitude/questions-and-answers/

http://www.indiabix.com/verbal-ability/questions-and-answers/

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