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Math, DI & LR

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NUMBERS & ALGEBRA Logbook

The Number Glossary

1. Factor: A positive integer ‘f’ is said to be thefactor of a given positive integer ‘n’ if f divides‘n’ without leaving a remainder e.g. 1, 2, 3, 4,6 and 12 are the factors of 12.

2. Rational Number: Real numbers which can be

expressed in the form of qp

, where p and q are

integers and q ≠ 0. Rational numbers haverepeating or terminating decimals.

3. Irrational Numbers: Real numbers which havenon-terminating and non-repeating decimals areirrational numbers.

4. Prime Numbers: A prime number is a positive integerwhich has only two distinct factors, itself and unity.

5. Composite Numbers: A composite number is anumber which has more than two factors.

6. Factorial: For a natural number ‘n’, its factorialis defined as: n! = 1 × 2 × 3 × ... ... ... × n.(Note: 0!=1).

7. Absolute value OR Modulus of a Number: For a realnumber ‘a’, modulus is defined |a| = a when a ≥0 |a| = –a when a < 0.

Tests of Divisibility

1. A non-negative integer is divisible by 2, 4, 8and so on, if and only if the number formedby the last digit, last two digits, last three digitsand so on, respectively, is divisible by 2, 4, 8and so on respectively.

2. A non-negative integer is divisible by 3 and 9if the sum of all the digits of the integer isdivisible by 3 and 9 respectively.

3. A non-negative integer is divisible by 11, if andonly if the difference between the sum of thedigits at the odd places and the sum of the digitsat the even places is divisible by ‘11’ or is ‘0’.

4. A non-negative integer is divisible by a compositenumber, if and only if it is divisible by two ormore of its relatively prime factors e.g. a numberis divisible by 12 if it is divisible by both 3and 4, similarly a number is divisible by 30if it is divisible by 2, 3, and 5.

Factors and Multiples

1. HCF/GCD: HCF of two or more numbers is thegreatest number which divides each of them exactly.

2. LCM: LCM is the least number which is exactlydivisible by each of the given numbers.

3. Product of two numbers = HCF × LCM

4. HCF of Fractions= rsdenominato the of LCM

numerators theofHCF,

when all the fractions are expressed in theirlowest forms.

5. LCM of Fractions= rsdenominato the of HCF

numerators theofLCM,

when all the fractions are expressed in theirlowest forms.

6. (a + b)(a – b) = a2 – b2

7. (a + b)2 = a2 + 2ab + b2

8. (a – b)2 = a2 – 2ab + b2

9. (a + b)3 = a3 + 3ab(a + b) + b3

10. (a – b)3 = a3 – 3ab(a – b) – b3

11. a3 + b3 = (a + b)(a2 – ab + b2)12. a3 – b3 = (a – b)(a2 + ab + b2)13. a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2)14. a3 + b3 + c3 = 3abc, if a + b + c = 0

1. am × an = am+n

2. am ÷ an = am – n

3. am × bm = (a × b)m

4. (am)n = amn

5. m

ba

= m

m

b

a

6. p1

a = p a

7. a0 = 1; a1 = a

8. a–p = pa

1; ap = pa

1−

9. qp

a = q pa =

10. am = an and a ≠ –1,0, 1 then m = n

11. am = bm, m ≠ 0i) a = +b, m is evenii) a = b, m is odd

Roots & Indices and Number Properties

12. A perfect square should always end with 0,1, 4, 5, 6 or 9.

13. A perfect square ending with zero should endwith even number of zeros.

14. A perfect square ending with 5 must always end with 25.15. A perfect square ending with 6, must have its

last but one digit odd.16. A perfect square ending with 1, 4, or 9 must

have its last but one digit even.17. The digit sum of perfect squares should be 1,

4, 7 or 9.

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18. Cyclicity Rule:The units digit of successive powers of ‘n’follow a pattern as given in the table below:

The fifth power of any number has the sameunits place digit as the number itself

19. To find the powers of a prime number ‘p’contained in n!, where n is a natural number.Highest power of prime number ‘p’ in n!= [n/p] + [n/p2 ] + [n/p3 ] + [n/p4 ] + ...Where the value of the term in the [ ] is thegreatest integral value less than or equal tothe terms n/p, n/p 2, n/p3, n/p 4, ...

Rules of Inequalities

1. If a > b and c is any numbera + c > b + ca – c > b – c

2. If a > b and c > 0ac > bc

ac >

bc

3. If a > b and c < 0, then ac < bc

4. If a – c > b, then a > b + c or –c > b – a.5. If x2 < a, where a is a positive number

⇒ a− < x < a ( a is the positve square

root of a)

6. If x2 > a, where a is a positive number

⇒ x > a or x < – a ( a is the positve

square root of a)

7. |x| < a ⇒ –a < x < a

8. |x| > a ⇒ x > a or x < –a

9. Quadratic inequalities:(x – α )(x – β ) > 0 if x < α or x > β ( α < β )

< 0 if α < x < βQuadratic Equations

1. For the quadratic polynomial ax2 + bx + c,let α and β be the roots.

Then ax2 + bx + c = a (x2 + ab

x + ac

)

= a (x – α ) (x – β )

= a [x2 – ( α + β ) x + αβ ]Hence, we have:

i) α + β = –ab ; ii) αβ = a

c

2. The roots α , β are given by:

α , β = a2

b ∆±−; where ∆ = b2 – 4ac is

the discriminant.

3. Properties of the discriminant ∆ :i) If ∆ ≥ 0, then α , β are real.

ii) If ∆ = 0, then α = β .

iii) If ∆ < 0, then α , β are complex conjugates.

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Percentages, Profit & Loss,Simple & Compound Interest

1. %increase = valueinitial

valueinitialvaluefinal − × 100

2. %decrease = valueinitial

valuefinalvalueinitial − × 100

3. If ‘n’ is increased by a%, and then by b%,and then c%, then the new value n1 will be

n1 =

+

1001

a

+

1001

b

+

1001

c

4. %profit = icePrCost

icePrCosticePrSelling − × 100

5. %loss = icePrCost

icePrSellingicePrCost − × 100

6. %discount = icePrMarked

icePrSellingicePrMarked − × 100

7. Simple Interest = PNR100

P ⇒ PrincipalN ⇒ Number of yearsR ⇒ Rate of interest per Annum (%)

The sum of the principal and interest is the AmountA = P + S.I.8. Compound Interest

A = PNR1

100 +

; where A = amount, P = Prin-

cipal, R = rate of interest per Annum (%), N =number of year.C.I. = A – P

Averages, Mixture & Alligation

1. Arithematic mean/average

= elementsofnumberTotalelementstheallofSum

2. If two quantities e1 and e2 are mixed, suchthat their average is ‘a’, then where e2 > e1,then these two quantities should be mixed

in the ratio 1

2

2

1

eaae

qq

−−

= .

3. If a vessel contains ‘a’ litres of wine and if‘b’ litres are withdrawn and replaced withwater, and this operation is continued ‘n’ times,

then VolumeTotal

operationsnafterleftWine =

n

aba

−

ARITHMETIC Logbook

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Ratio, Proportions & Variations; Partnerships

1. Ratio is the comparison between similar typesof quantities; it is an abstract quantity anddoes not have any units.

2. If ba

= dc

, then a, b, c, d are said to be in

proportion.

3. If ba

= dc

, then

a) a × d = c × b

b) ab

= cd

, invertendo

c) ca

= db

, alternendo

d) bba +

=d

dc + , componendo

e) b

ba − =

ddc −

, dividendo

f) baba

−+

= dcdc

−+

, componendo & dividendo

4. If ba

= dc

= fe

= hg

= K

then )atormindenotheallofsum(hfdb

)numeratorstheallofsum(geca

++++++

= K

5. If a∝ b, provided c is constant and a ∝ c,provided b is constant,then a ∝ b × c, if all three of them are varying.

6. If A and B are in a business for the same time,thenProfit distribution ∝ Investment (Time isconstant).

7. If A and B are in a business with the sameinvestment, thenProfit distr ibution ∝ Time of investment(Investment is constant).

8. Profit Distribution ∝ Investment × Time.

Time, Speed and Distances; Work,Pipes and Cisterns

1. Distance = Speed × Time

2. If a distance is covered at ‘a’ kmph and thereturn journey is made at ‘b’ kmph, then the

speed of the entire journey is ba

ab2+

kmph.

3. If two bodies move in the opposite direction,their relative speed = sum of their speeds.

4. If two bodies move in the same direction, theirrelative speed = difference of their speeds.

5. If the speed of a boat in still water is ‘b’,and the speed of the stream is ‘s’, thena. Upstream speed (u) = b – s

b. Downstream speed (d) = b + s

c. b = 2

du +d. s =

2ud −

6. If A can do a piece of work in ‘a’ days andB can do the same work in ‘b’ days then A

and B together can do the work in ba

ab+

days.

7. If the A and B can do a work in a and b daysrespectively, then their efficiency will be inthe ratio b: a, and their wages will be in theratio of their efficiencies, if they are workingtogether.

8. If pipe A is ‘x’ times bigger than pipe B, then

A will take th1

x of the time taken by pipe B,

to fill a cistern.

9. If pipe A can fill a cistern in ‘a’ mins andB can empty it in ‘b’ mins, then together they

can fill the cistern in ab

ab−

mins.

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MODERN MATH Logbook

SET THEORY

1. Set A is said to be a subset of Set B if each andevery element of Set A is also contained in SetB. Set A is said to be a proper subset of Set Bif Set B has at least one element that is notcontained in Set A.

2. The Universal set is defined as the set of allpossible objects under consideration. Everyother set is then a subset of the universal set.

3. Union of two sets is represented as A ∪ B andconsists of elements that are present in eitherSet A or Set B or both.

4. Intersection of two sets is represented asA ∩ B and consists of elements that arepresent in both Set A and Set B.

5. Venn Diagram: A venn diagram is used tovisually represent the relationship betweenvarious sets.

What do each of the areas in the figurerepresent?only A - aonly B - bonly C - cA and B, not C - dA and C, not B - eB and C, not A - fA, B and C - gNone of A, B or C - h

6. Some important properties:n(A ∪ B) = n(A) + n(B) – n(A ∩ B)n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B)

–n(A ∩ C) – n(B ∩ C) – n(A ∩ B ∩ C)

LINEAR ARRANGEMENT1. Linear arrangement (or permutation) of ‘n’

distinct items among themselves = n!

2. Linear arrangement ( or permutation) of ‘r’

out of ‘n’ distinct items = nPr = )!rn(

!n−

3. Linear arrangement of ‘n’ items out of which

‘p’ are alike. ‘q’ are alike, ‘r’ are alike = !r!q!p!n

4. Circular arrangement of ‘n’ distinct items

= (n – 1)!

5. Circular arrangement of ‘n’ distinct beads in a

necklace = 2

)!1n( −

6. Selection of ‘r’ items out of ‘n’ distinct items

= nCr =

)!1n(!r!n−

7. Selection of atleast one item from ‘n’distinct items = nC

1 + nC

2 + nC

3 + nC

4 + ... + nC

n

= 2n – 1

PRINCIPLE OF COUNTING:

If there are two jobs which can be performedindependently in m and n ways, then

i) either of the two jobs can be performed in m + nways (OR)

ii) both of the jobs can be performed in m × n ways(AND) e.g. if there are 15 boys and 12 girls in aclass then

i) a boy OR a girl representative can be chosen in15 + 12 = 27 ways.

ii) a boy AND a girl representative can be chosen in15 × 12 = 180 ways.

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PROBABILITY

Probability deals with the chance or likelihoodof a particular event happening or not happening.If an event can happen in ‘a’ ways and fail in‘b’ ways and each is equally likely to happen,then the probability or chance of its happening

is +a

a b i.e., P(a) =

+a

a b and the probability of

failure is P(b) = +b

a b. (P(a) < 1 and P(b) < 1,

P(a) + P(b) = 1, P(a) ≠ 0, P(b) ≠ 0)

Instead of saying the probability of the happening

of an event is +a

a b, the probability of its not happening

is − +

a1

a b =

+b

a b. It is sometimes stated that

the odds are ‘a’ to ‘b’ in favour of the event or ‘b’

to ‘a’ against the event. Two events A and B occur

with probabilities p(A) and p(B). Then, the probability

that A and B occur is P(A + B) = P(A) + P(B) – P(AB);

where P(AB) is the probability of AB occuring together.

P(AB) = P(A) × P(B), if the two events are independent

of each other.

LOGARITHMS

If an = b, then =log b naConventionally, log b represents log10b

1] =blog 1 0

2] =alog a 1

3] alog b = b

1log a

OR 1alogblog ba =×

4] × = +b b blog (m n) log m log n

5] = −

b b b

mlog log m log n

n

6] =nb blog (m) nlog (m)

7] = = ×ab a b

a

log mlog m log m log a

log b

8] =βα

bb

αlog (a ) log a

β

9] =x xlog a log b if and only if a = b

10] logyx > 0 if both x, y > 1 or both x, y < 1 < 0 if x < 1 < y or y < 1 < x

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GEOMETRY Logbook

GEOMETRY

For a triangle ( ∆ ABC) with sides a, b, c :

1. Sum of interior angles = 180°.

2. Side opposite to the greatest angle is longest,and side opposite to the smallest angle issmallest.

3. Sum of any two sides of the triangle is alwaysgreater than the third side.

4. Difference between any two sides of the triangleis always lesser than the third side.

5. Measure of exterior angle = Sum of remoteinterior angles.

6. Area = 21

× (base) × (height)

= )cs)(bs)(as(s −−− , where s = 2

cba ++

= r × s, r = radius of the incircle

= R4

abc, R = radius of the circumcircle

7. If BE is the angle bisector then, BCAB

= ECAE

.

8. A line segment joining the midpoints of anytwo sides is parallel to the third side andhas half the length of the thirdside.

9. If DE||BC, then DBAD

= ECAE

(BPT)

10. If AD is the median .i.e. BD =DC, then AB2 + AC2 = 2(AD2 + DC2)

11. If ∠ ABC = 90°, then AC2 = AB2 + BC2

12. In a 30°- 60°- 90° triangle, the sides are in

the ratio 1 : 3 : 2

13. In a 45°- 45°- 90° isosceles triangle, the sides

are in the ratio 1 : 2

Note: The above two properties are also used tosolve Trignometry based questions in CMATrelating to Angle of Elevation and Angle ofDepression.

14. If ∆ ABC is an equilateral triangle:

a. Each angle = 60°,

b. Height = 23

× side,

c. Area = 43

× (side)2 ,

d. Inradius(r) = 3

height ,

e. Circumradius (R) = 32

× height .

15. ∆ ABC is similar to ∆ PQR if ∠ A = ∠ P, ∠ B

= ∠ Q, ∠ C = ∠ R and is represented as ∆ ABC

≈ ∆ PQR.

a.PQAB

= QRBC

= PRAC

i.e ratio of their sides

remains constant, say, k.

b. PQRofPerimeter

ABCofPerimeter

∆∆

= PQAB

= k

c. PQRofArea

ABCofArea

∆∆

= 2

PQAB

= k2

d. All other linear geometric dimensions willbe in the same ratio as is the side.

For a circle with center "O" and radius "r":

16. Chords equidistant from the center are equal.

17. Perpendicular from the center bisects thechord, conversely, the line segment joiningthe center and the midpoint of the cord isperpendicular to it.

A

D E

B C

A

°

E

B C°

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18. Equal chords subtends equal angle at thecenter.

19. In the adjoining Figure, Chord AB divides thecircle into two parts:

PQ

R

BxA

O

y

a. minor arc - AXB

b. major arc - AYC

The area bounded by OAXB is a sector.

20. Measure of arc AXB = ∠ AOB = θ

21. Length (arc AXB) = 360

θ × 2 π r

22. Area (sector OAXB) is 360

θ × π r2

23. Angles subtended by the arc at all the pointson the alternates segments are equal. ThusAPB = AQB = ARB

24. Angle subtended by the arc at the center istwice than that subtended at the alternatesegment. Thus ∠ APB = ∠ AQB = ∠ ARB

25. Angle inscribed in a semicircle or thatsubtended by the diameter is a right angle.

TANGENT PROPERTIES

PA and PB are tangents to the given circle, OA isthe radius.

1. PA = PB.

2. OA ⊥ PA..

Secant, Tangent and Chord Properties

3. If two secants viz. AB and CD, intersect at P, thenAP × BP = CP × DP.

Note: This equation will hold even if the secantsmeet outside the circle.

4. If a tangent (OC) and a secant (AB) meet externallyat O, then OC2 = OA × OB.

(Tangent – Secant theorem)

5. The angle made by chord (AB) with the tangent atA (AD) is equal to the angle subtended by it on theopposite arc.

m ∠ BAD = m ∠ ACB.

(Tangent Chord Property)

POLYGONS AND THEIR PROPERTIES

For any regular polygon:

(A polygon which has all its sides and angles equal)

1. Sum of internal angles = 180°(n – 2).

2. Measure of an internal angle = n

)2n(180 −°.

(where n is the number of sides)

Properties of some special polygons:

Parallelogram:

3. Opposite sides are parallel and congruent.

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CMAT

SOLID FIGURES

Euler’s formula:

Number of faces + Number of vertices =Number of edges + 2

1. Cube: If each edge of a cube is ‘a’ units,then

• Volume of cube = a3

Also, edge of thecube

=3 cube theof Volume

• Total surface area of the cube = 6a2

• Lateral surface area = 4a2

• Longest diagonal of the cube = a 3

2. Cuboid/Rectangular Parallelopiped:

Let length, breadth and height of a cuboidbe l , b and h, respectively, then

• Volume of cuboid =l × b × h

• Total surface area= 2( l b + bh + h l )

• Lateral surface area = 2h( l + b)

4. Opposite angles are congruent.

5. Diagonals bisect each other.

6. Area of parallelogram = Base × height .

Rhombus:

7. Opposite sides are parallel and all sides areequal.

8. Opposite angles are congruent.

9. Diagonals bisect each other at 90°.

10. Area = 21

× Product of diagonals.

Square:

11. All sides are congruent and opposite sidesare parallel. All angles are right angles.

12. Diagonals are congruent and bisect each other

at 90°. l (diagonal) = 2 × (side).

13. Area = (side)2.

Kite:

14. Two pairs of adjacent sides are congurent.

15. Diagonals intersect each other at 90° andlonger diagonal bisects shorter diagonal.

16. Area = 21

× product of diagonals.

Isosceles Trapezium:

17. One pair of opposite sides is parallel.

18. Non-parallel sides are congruent.

19. Area = 21

× sum of parallel sides × height.

Regular Hexagon:

20. Area = 2

33 (side)2.

21. Six equilateral triangles are formed byjoining the opposite vertices of the hexagon.

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• Longest diagonal of cuboid =

222 hb ++l

3. (Right) Circular Cylinder: Let ‘r’ be the radiusof the circular base and ‘h’be the height of a right cir-cular cylinder, then

• Area of each circular face= π r2

• Lateral surface area =Circum-ference ofbase × height = 2 π rh

• Total surface area = Curved surfacearea + Area of two circular bases = 2 π rh+ 2 π r2 = 2 π r(h + r)

• Volume of cylinder = (Area of base) ×height = π r2h

4. Right Circular Cone: If h is the height, lis the slant height and r is the radius ofthe base of the right circu-lar cone, then

• Volume of cone = 31

π r2h

• Lateral surface area =π r l

• Total surface area = Curved surface area+ Area of circular base

= π r l + π r2 = π r( l + r)

• l 2 = h2 + r2 i.e., l = 22 rh +

5. Sphere: If r is the radiusof a sphere, then

• Volume of sphere =

34

π r3

• Surface area of sphere= 4 π r2

6. Hemisphere: If r isthe radius of a hemi-sphere, then

• Volume of hemisphere = 32

π r3

• Lateral/Curved surface area of hemi-

sphere = 2 π r2

• Total surface area of hemisphere = 2 π r2

+ π r2 = 3 π r2

7. Spherical Shell: Let R and r be the outerand the inner radius of a spherical shell,then

• Volume of spherical shell = 34

π (R3 – r3)

8. Solid Ring: Let R and r be the outer andthe inner radius of a ring, then

• Volume of a solid ring = 4

2π(R – r)2(R + r)

• Lateral/Curved surface area of solid ring = π 2(R2 – r2)

CO-ORDINATE GEOMETRY

Distance between two points

The distance between two points (x1, y1) and (x2,

y2) = 221

221 )yy()xx( −+− .

Section FormulaIf P is a point dividing the joint of two pointsA(x1, y1) and B(x2, y2) internally in the ratiom : n (i.e., PA : PB = m : n), then the co-ordinates(x, y) of P are given by:

++

++

=nmnymy

,nmnxmx

)y,x(P 1212

If P(x, y) divides the joint of A(x1, y1) and B(x2,y2) externally in the ratio m : n(i.e., PA : PB = m : n), then the co-ordinates (x,y) of P are given by:

−−

−−

=nmmymy

,nmnxmx

)y,x(P 1212

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CMAT

I(x, y) =

++++

++++

cbacybyay,

cbacxbxax 321321

where a, b and c are the lengths of the threesides opposite to ∠ BAC, ∠ ABC and ∠ ACB.

Area of a triangle

If A(x1, y1), B(x2, y2) and C(x3, y3) are thethree vertices of a triangle, then

Area of the triangle = 111

yyyxxx

21

321

321

= )]yy(x)yy(x)yy(x[21

213312321 −+−−−

If area = 0, then the three points are col-linear.

Parallel and Perpendicular lines

i) Two lines whose slopes are m1 and m2 areparallel to each other, if and only if m1= m2, or both m1 and m2 does not exist.

i i ) Two lines whose slopes are m1 and m2 areperpendicular to each other, if and onlyif either m1 × m2 = –1 or if m1 = 0 and m2does not exist.

Thus, if the slope of a line is m, then the

slope of a line perpendicular to it is m1

− .

Midpoint of a SegmentIf P is the midpoint of the segment joining thepoints A(x1, y1) and B(x2, y2), then the co-ordinates(x, y) of P are given by:

P(x, y) =

++

2yy,

2xx 2121

Centroid and Incentre of a triangle

If A(x, y), B(x2, y2) and C(x3, y3) are the verticesof a triangle, then

i) The co-ordinates of the centroid G(x, y) of

∆ ABC are:

G(x, y) =

++++

3yyy,

3xxx 321321

i i )

The co-ordinates of the incentre I(x, y) of ∆ ABCwith vertices A(x1, y1), B(x2, y2) and C(x3, y3)are:

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®CMAT

TRIGNOMETRY

1. Angle of Elevation:

O

B

Lineof

sight

AHorizontalθ

O

B

Lineof

sight

AHorizontal

Angle of elevationθ

If a person at a lower level looks up atan object at a higher level, the line of sightmakes an angle with the horizontal linewhich is called the angle of elevation.

2. Angle of Depression:

If a person standing at a higher level observesan object at a lower level, the line of sightmakes an angle with the horizontal line whichis called the angle of depression.

Note:

i] Numerically, the angle of elevation is equalto the angle of depression.

i i ] The angle of elevation and the angle of de-pression are measured with the horizontalline.

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CMAT

DATA INTERPRETATION Logbook

Comparison of fractions

1. If a < b or ba

< 1 then ba

< kbka

++

k being a positive integer.

2. If a > b or ba

> 1 then ba

> kbka

++

k being a positive integer.

3. If ad > cb then ba

> dc

a, b, c, d being positive integers.

General Terms Involved in DI.

1. Percentage change =

ValueInitial

ValueInitial–valuefinal × 100

2. Growth = final value – Initial Value

3. Growth Rate (in percentage) = ValueInitial

GrowthTotal × 100%

4. Average Annual Growth Rate = yearsofNumberTotal

RateGrowthTotal

5.1001–

ValueInitial

Valuefinal

RateGrowthAnnualAverageCumulativeOR

RateGrowthAnnualAverageCompoundedn1

×

=

We need to know the decimal values.

21

= 0.531

= 0.3332

= 0.6641

= 0.2543

= 0.7551

= 0.2

52

= 0.453

= 0.654

= 0.861

= 0.16665

= 0.83371

= 0.142857

72

= 0.28571473

= 0.42857174

= 0.57142875

= 0.71428576

= 0.85714281

= 0.125

83

= 0.37585

= 0.62587

= 0.87591

= 0.1192

= 0.2294

= 0.44

95

= 0.5597

= 0.7798

= 0.88111

= 0.0909112

= 0.1818113

= 0.2727

114

= 0.3636115

= 0.4545116

= 0.5454117

= 0.6363118

= 0.7272119

= 0.8181

1110

= 0.9090

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(5) "A is neither greater than nor equal to B" means"A is less than B" (A < B)

(6) Comparison should always be done in a UniqueMode i.e., either greater than or lesser than.i.e., if A is not less than B, B is less than Cand D is equal to A, then D = A > B < C isconventionally incorrect.

D = A > B or B < A = D

C > B or B < C is conventionally correct.

(7) Two quantities which are not the part of aninequality cannot be compared among them-selves e.g., In the above example C cannot becompared with A and D.

Alphanumeric SeriesThe logics which are generally used to framesuch type of questions can be categorized as

(1) Difference series:– Difference between the successive elementswill lead to the logic behind these types of series.(a) Constant difference series eg. 3, 5, 7, 9,11, 13(b) Increasing difference series eg. 3, 5, 8, 12,17, 23(c) Decreasing difference series eg. 12, 8, 5,3, 2, 2(d) Increasing decreasing difference series

eg. 17, 19, 23, 26, 31, 35

(2) Product Series– Ratio between the successive elements willlead to the logic behind these type of series.(a) Constant ratio series eg. 3, 6, 12, 24, 48,

96(b) Increasing ratio series eg. 3, 3, 6, 18, 72,

360(c) Decreasing ratio series eg. 72, 18, 6, 3, 3,

0(d) Increasing and decreasing ratio series.

eg. 3, 3, 9, 18, 72, 216, 1080

(3) Square/Cube seriesThese types of series can be characterized by

Seating Arrangement(1) In seating arrangement, generally the data will

be given with respect to the subjects or thepeople involved in the puzzle. Hence the leftand right should be considered with respectto the subjects or the people involved in.

(2) The candidate/observer is always assumed tobe facing North Direction.

(3) In case of linear Seating Arrangement if noth-ing is mentioned the subjects are consideredto be facing North Direction. If mentionedotherwise the left of the subject will becomethe right of the observer and vice-versa. Cir-cular seating arrangement will not have thisproblem.The language of the question should be carefullyunderstood.

eg., (i) In the photograph whois to the left of Harbhajan?Ans. Pointing.(ii) Who was standing to theleft of Harbhajan Singh when

the photograph was taken? Ans. Symonds.

(4) In case of linear seating arrangement thestatement, "A is seating to the left of B" doesnot mean "A is seating to the immediate leftof B". But in case of circular seatingarrangement it does mean so.

(5) The term diagonally opposite is generally usedto denote two people who are seated as faras possible.

Comparisons

(1) "A is neither greater than nor less than B" means"A is equal to B" (A = B).

(2) "A is not less than B" means "A is either greaterthan or equal to B" (A > B)

(3) "A is not greater than B" means "A is eitherless than or equal to B" (A < B)

(4) "A is neither less than nor equal to B" means"A is greater than B" (A > B)

LOGICAL REASONING logbook

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Groups & ConditionalitiesIn the formation of a group, or for an event tooccur or for a particular arrangment, certainconditions are imposed. While forming the group,the conditions have to be taken into account alongwith the directions given in each question.

Types of Conditional Statements

Conditional statements can be classified into fourgroups:

a) If A occurs then B will also occur.A → BThe reverse implication of this statementis, that, if B has not occurred A will alsonot occur. The not statement isdenoted by the symbol ‘~’, i.e., not B isdenoted as ~B.~B → ~AIt is not necessary that B → A.

b) If A occurs then B will not occur.A → ~BThe reverse implication isB → ~AIt is not necessary that ~B → A

c) If A has not occurred then B will occur.~A → B

the presence of peculiar squares or cubes eitherin the series or in the series of the differenceof the elements in the series.eg. 4, 8, 17, 33, 58, 94eg. 4, 9, 1, 6, 2, 5

(4) Miscellaneous(a) Fibbonocci series: Every element is the

sum of two preceding terms.eg. 3, 5, 8, 13, 21, 34

(b) Prime number/Square Cubes of primenumbers

(c) Product of consecutive prime numbers,eg. 6, 15, 35, 77, 143, 221

(5) Combination series(a) Alternate elements forming two different

serieseg. 2, 13, 4, 17, 6, 19, 8

(b) Difference and product series.eg. 2, 4, 12, 45, 206.5, 1140.75

The reverse implication is~B → AIt is not necessary that B → ~A

d) If A has not occurred then B will also notoccur.~A → ~BThe reverse implication isB → AIt is not necessary that ~B → ~A.

Notation1. A belongs to group I.

A ∈ I2. A does not belong to group I.

A ∉ I3. A and B are in the same group.

AB4. A and B are not in the same group.

ABx

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BLOOD RELATIONS

Mother’s or father’s son — BrotherMother’s or father’s daughter — SisterMother’s or father’s brother — UncleMother’s or father’s sister — AuntMother’s or father’s father — GrandfatherMother’s or father’s mother — GrandmotherSon’s wife — Daughter-in-lawDaughter’s husband — Son-in-lawHusband’s or wife’s sister — Sister-in-lawHusband’s or wife’s brother — Brother-in-lawBrother’s son — NephewBrother’s daughter — NieceUncle or aunt’s son or daughter — CousinSister’s husband — Brother-in-lawBrother’s wife — Sister-in-lawGrandson’s or granddaughter’s daughter —Great granddaughter

Notations

1. A is a male

2. A is a female

3. Sex of A is not known

A

4. A and B are married to each other

A = B

5. A and B are siblings

A <–> B

6. A is the child of BB

A

7. A is the uncle / aunt of B

DIRECTIONSLogical reasoning questions based on directionstest your sense of directions and understandingof them.The figure below shows the four main directions(North N, South S, East E and West W) and itscardinal directions (NorthEast NE, NorthWest NW,SouthEast SE and SouthWest SW).

N

NE

SES

SW

W

NW

E

There are 2 types of questions based on Direc-tions, one that tests your sense of directions andthe other that involve calculations.

CALENDARS

1] An ordinary year contains 365 days, i.e., 52weeks and 1 odd day.

2] A leap year contains 366 days, i.e., 52 weeksand 2 odd days.

3] 100 years (a century) contain 76 ordinaryyears and 24 leap years

= (76 × 52) weeks + 76 odd days + (24 ×52) weeks + 48 odd days

= [(76 × 52) + (24 × 52)] weeks + 124 odddays

= [(76 × 52) + (24 × 52) + 17] weeks + 5 odddays

i.e., 100 years contain 5 odd days.

200 years contain 10 and therefore 3 odddays. Similarly, 300 years contain 1 odd day,400 years will have (20 + 1) odd days i.e.,0 odd days. Similarly, the years 800, 1200,

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1600, 2000 each contain no odd days.

4] 1st January, AD was Monday. Therefore, we mustcount days from Sunday, i.e., Sunday for 0 oddday, Monday for 1 odd day, Tuesday for 2 odddays and so on.

5] February in an ordinary year gives no oddday, but in a leap year gives one odd day.

CLOCKS

1] The entire clock can be divided into 60 minsspaces.60 mins space is not a time of 60 mins,but a distance.60 mins space equals 360o turn. So, 1 minutespace equals 6o turn.

2] The hour hand goes over 5 mins spaces.The minute hand passes over 60 minsspaces.Thus, in an hour, the minute hand gains(60 – 5) = 55 mins spaces over the hourhand.

3] In one minute, the minute hand moves 6o

and in one minute, the hour hand moves

o12

.

4] In one minute, the minute hand gains 152

o

over the hour hand.Every hour, both hands coincide exactlyonce.Every hour, the two hands are twice at rightangles. In this position they are 15 minsspaces apart.

5] In an hour, the two hands point exactly oncein the opposite direction. In this position,they are 30 mins spaces apart.

6] The hands are said to be in the same straightline, when they are coincident or oppositeto each other.

7] If a clock shows 10:20 when the time is10:00, then the clock is said to be 20 minsfast.

8] If a clock shows 7:45 when the exact timeis 8:00, then the clock is said to be 15 minsslow.

9] If both hands start moving from the same

position, then they coincide in 565

11 mins.

If the hands coincide in time less than 565

11mins, then the clock is fast. If the hands

coincide in time more than 565

11 mins, then

the clock is slow.

10] In a slow clock i.e., a clock that loses time:

Total t ime lost in T hours = (T ×

60)

5x 6511

x

−

mins, where x is the time

in which the hands of slow clock coincide.

11] For a fast clock, i.e., a clock that gains time:

Total t ime gained in T hours = (T ×

60)

565 x11x

−

mins, where x is the time

in which the hands of fast clock coincide.

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VISUAL PUZZLES - BASIC CONCEPTS

A Clockwise Movement:As the hands of a clock move.

e.g., ‹— Begin

The direction in which the ‘+’ moves is clock-wise.

B Anticlockwise Movement:Movement in the reverse direction of clockwisemovement. [ceiling fans move in this fashion].

e.g., ‹— Begin

Taking the same example, the movement of ‘+’is in the reverse direction.

C Rotation:The object rotates by certain degrees in themovement specified. The most commonly usedmovements are: rotations by 45°, 90° and 135°.Remember that these are identifiable with justvisual inspection – a precise measurement isnot required.

e.g.,1] Rotate 90° clockwise =>

to

original position after rotation

2] Rotate 45° anticlockwise =>

to

original position after rotation

3] Rotate 1350 clockwise =>

to

original position after rotation

When movements/rotations are different fromthe ones mentioned above you will have clues– like multiples of 300 in a clock.

D Lateral Inversion:This is basically a mirror reflection or if youlike, think of it as turning over an object onits side while keep it vertical.

e.g.original position laterally inverted

original position laterally inverted

E Vertical Inversion:This is inverting an object vertically - think ofit as a reflection of trees in a pond.

e.g., tooriginal position vertically inverted

or tooriginal position vertically inverted

F Spatial inversion:When you invert an object laterally or verti-cally, one of two things can happen. One, theobject is inverted and is placed in the originalposition; two, the object is inverted on its axisand placed below or to the side of the originalposition as the case may be.

e.g., to

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This is vertical inversion, but the inverted objectis placed in the same position as the originalobject.

But to

Here the object is inverted and moves down.This movement is what we call spatial inversion.There is no fixed type of visual reasoningquestion. Any or all of the types discussed maybe asked. You could find all three forms ofinversion together.

e.g., to

Here the object is vertically and laterallyinverted; it also moves down.By the end of this section, you should not haveany difficulty recognising these movements.

G Move 1 space or move 1/2 space:This is purely our terminology coined for visualreasoning questions.

e.g., or

This is moving one space - from a corner tothe other corner, or from a particular placeon one side to the same place on the adjoiningside. e.g., or

This is half space movement - half of what youwould move in moving one space.

SERIES COMPLETION

Compare each figure to the preceding one andfind out the difference.Numbers of elements in-creasing or decreasing; rotary movements in oneor more elements; elements changing in a definitepattern; common properties in all figures, andso on.

The ideal strategy is these kinds of questionsis elimination. If you can eliminate 2 or 3alternatives using a defined pattern, then yourtask is that much simpler. You do not need toconsider each alternative for all elements.

ANALOGIES

There exists a defined relationship between 2figures and given that you have to identify a pairwhich exhibits/does not exhibit the samerelationship.

ODD FIGURE OUT

You have to identify similarities/patterns in a groupof figures and make the figure which does notfit in with the group. A variation of this type ofquestions is identifying the odd figure in asequence.You have to identify the figure which doesnot follow the set rules of the sequence.The things you would look for remain the same– number of elements, angles, number of straightlines or curves, arrangements and such.