formula & tip for cat
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Formulae
&
Tips for CATBrought to you by
The Competition Master 126, Industrial Area-1, Chandigarh - 160002 India
© The Competition Master
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There are several thingsthat you have to keep inmind:
a ) D i f f e r e n t i a t ebetween speed andhurry: Many students thinkthat speed means makingguesses and blunderingon. This is wrong. Do theq u e s t i o n s w i t hconcentration. Don’t worryabout questions you couldnot attempt, because youpossibly cannot do 180questions in two hours.Speed means avoidingu n n e c e s s a r ycalculations.
☛ Hot tip:
Allow yourself 30seconds to glance over thesection before starting.This will help you locate theeasy questions.
b) Keep your cool.There will be some difficult
questions and some easyones. When you hit thedifficult ones, you muststay calm. Skip a fewquestions. Don’t panic,because if you do, youhave almost certainly lostthe battle.
☛ Hot tip:If you become tense,
just close your eyes andtake a few deep breaths.
Return to the paper onceyou are composed again.c) Manage your time:
Keep a watch before you.Write down the time whenyou should move over tothe new section. Don’t be
tempted to stick to theprevious section when thetime is over. Just keepmoving on.
☛ Hot tip:Each section should be
1
FORMULAE
AND TIPS FOR CAT
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attempted in 28 minutes. Allow yourself 2 minutes togo back on questions youcould not solve, or for
making some intelligentguesses.
COMPREHENSION
☛ Tips for speed readingNext time you are
reading a paper, try this.
First stage: Sweep yourhand from left to rightacross the page so yourfinger underlines each line.Follow your finger with youreyes. The difficulty of thematerial determines the
speed, but move your handa little faster than your eyecan follow comfortably.Make your eyes work tokeep up.
Soon you will findyourself reading whole
phrases and ideas. As itgets easier, keepincreasing the speed.
As you improve, insteadof sweeping whole lineswith your finger, just sweepthe middle part. Your eyes
will scan the beginning andthe end of the lineautomatically.
When you’ve mastered
this, your reading speedwill be between 500 and1,000 words per minute.This is the practical limit.
Second Stage: Thenext speed increase comesfrom area reading. Insteadof sweeping lines, usebroader hand motions tomake a series of zigzags ofS’s down the page andread several lines at once. You can even readbackward as your handmoves to the left. You’llcomprehend very little atfirst, but if you keep youreyes following your finger,you’ll find yourselfabsorbing whole chunks ofthe page at once.
Six hot tips forcomprehension:
1. Read the questionsbefore you read thepassage. This will help youlocate the ideas faster.
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2. Force your eyes tosweep entire phrases andideas. Don’t read word forword. Avoid pronouncing
words as you read.3 . D o n ’ t r e r e a dsentences. Don’t skip lines.
4. Let the context defineunfamiliar words. Don’tstop to look for them.
5. Adjust reading speedto the type of material.
6. Look for key ideasand phrases in eachparagraph.
What you should read:Read the editorial page ofone good Englishnewspaper everyday by theabove method. This shouldnot take more than 15minutes everyday. Make ita point to read theeconomic articles andbusiness news. Look updifficult words that youencounter in a dictionary. Also read a magazine oncurrent affairs. Your bestbet is THE COMPETITION
M ASTER, which has ar e g u l a r m a n a g e m e n tsection. Read the debates
regularly as well as theeditorials, features andcurrent affairs andBusiness GK. This will help
you form your opinions andincrease your awareness. Anumber of people whohave cleared CAT have toldus that the magazine waspriceless for theirpreparation.
Additional reading andexercises:
Norman Lewis: How to
Read Better and Faster.
VERBAL ABILITY
It is important to have agood vocabulary, know
good grammar and again,
have the reading habit.
Vocabulary: A good
vocabulary is built up over
time. Reading helps. Check
up important words in adictionary. Or take a good
word-list, such as the one
published in Master Guide
for MBA Entrance, and
learn the usage of the
words. Knowing about
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roots of words helps.
☛ Hot tip:The best way to learn a
word list is throughvocabulary cards. Makingthem is an investment,because they will be helpfulover a period of time.
Analogies, Odd manout: This means findingrelations among words.What is the relationbetween SALT : SALTY?We can say that saltythings contain salt. Thebest choice would beCOW : BOVINE, becausethat best suits the relation-ship. A good vocabularyhelps in such questions.
☛ Hot tip:Do as many analogies
as you can, to know thekind of relations that can beasked. The Test ofReasoning in THE
COMPETITION M ASTER con-tains analogies regularly. Ifyou do it every month, youwill have a goodunderstanding of
analogies.Arranging sentences,
blanks: These are tricky,especially if they contain
large sentences. Theselections are usually fromcurrent newspapers, soextra reading increasesspeed in such questions.
☛ Hot tip:Usually, the answer can
be obtained by getting atthe central idea andthinking what it should startor end with, or by workingfrom the choices. Do notwaste time arranging theentire paragraph.
QUANTITATIVE
ABILITY
The importance ofknowing your tables,decimals, fractions and
formulae cannot be over-e m p h a s i s e d . M a n yquestions can be solved bylooking at the choices.Develop this ability andyour speed will surelyincrease. We give below
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some things which can beremembered easily.
1. Numbers: Remem-ber the power of 10 in the
following: millions (6),lakhs (5), billions (9) andcrores (7). Some questionsmay confuse on the units.
2. Rational and
irrational numbers:
Numbers which can be
expressed in the form
where p and q are
integers and
Examples of rational
numbers are
Irrational numbers arethose which whenexpressed in decimal
would be in non-terminating and non-repeating form. Examplesof irrational numbers are: 2,3, 5, 7, and so on.
3. Division:
Dividend = (Divisor ×
Quotient) + Remainder.4. Tests of divisibility:
A number can be checkedfor divisibility by the
following methods:By 2: If the last digit ofthe given number is evenor zero.
By 3: If the sum of thedigits of the number aredivisible by 3.
By 4: If the sum of thelast two digits of thenumber is divisible by 4.
By 5: If the last digit ofthe number is either zero or5.
By 6: If the number isdivisible by 2 as well as 3.
By 8: If the sum of thelast three digits of thenumber is divisible by 8.
By 9: If the sum of thedigits is divisible by 9.
By 10: If the last digit iszero.
By 11: If the difference
between the sum of the
odd digits and the even
digits in a number is either
zero or divisible by 11.
5
p
q,
q 0≠ .
1 3 8, , , 0, 3, 150 etc,
2 5 5
−
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5. Short cuts for
multiplying: Large multipli-
cation should be avoided.
Instead, look for shortcuts
to do the sums:(a) To multiply by 99,
999, 9999 ... : Place as
many zeroes after the
number and subtract the
number.
(b) To multiply by 5n:
Put n zeroes to the right ofthe number and divide it by
2n.
6. HCF and LCM: The
HCF of two or more
numbers is the greatest
number that divides each
one of them exactly.
The LCM of two or more
numbers is the product of
the highest powers of all
the prime factors that occur
in the numbers.
Product of twonumbers= HCF × LCM.HCF of Fractions
LCM of Fractions
7. Simplification: Tosimplify an expression,always use the orderspecified in BODMAS:Brackets, Of, Division,Multiplication, Addition andSubtraction.
8. Square roots: Learnthe square roots upto 16and squares upto 32. Makememory cards to helpremember these figures.
☛ Hot tips:
1. Square roots can be
approximated by using
, where a is the
nearest root of the number
and r is the remainder.
Hence 85 = 92 + , which
gives the square root of 85
as 9.22 approximately. One
can approximate square
roots easily by this method.2. To square a number,
try putting it in the form(a + b)2.
6
.LCM of numerators
HCF of denominators=
HCF of numeratorsLCM of denominators
=
418
r2a2a
+
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Thus 1152 = (100 + 15)2
= 1002 + 152 + 2(100 x 15),which can be easilycomputed.
8. Percentages: Learnthe fraction equivalents.
Many questions can besolved faster if we knowthese figures.
To find growth percent-age or percentage change,always use:
9. Averages: Averagesare found by adding up thevalues and dividing by thenumber of values.
10. Ratio and propor-
tion: Can be written as a : b
or . If a : b = c : d, then
bc = ad.11. Partnership: The
share of profits dividedbetween two partners is:
(Amount of money investedby A × No. of monthsinvested by A) : (Amount ofmoney invested by B × No.
of months invested by B).12. Shares: It isimportant to know thefollowing terms:
Face value: The price atwhich shares are issued. Always a round figure.
Market value: The priceat which shares are traded.Will fluctuate and willseldom be a round figure.
If market value = facevalue, the shares aretraded at par.
If market value > facevalue, the sharecommands a premium.
If market value < face
value, the share is at a
discount.
Return: The interest
earned by the shares afterone year. Always calcu-
lated on face value. Yield: Return calculated
on what is actuallyinvested. Calculated bydividing return by market
7
1 1 1= 25%, = 33%,
4 3 51
= 20%, = 16% etc.6
New Quantity – Old quantity×100
Old Quantity
ab
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value.Brokerage: When you
buy a share, the buyer hasto pay the brokerage,
which is added to themarket price. When sharesare sold, the seller gets theprice after the brokerdeducts his brokeragefrom the market priceobtained.
13. Profit and Loss:Profit = SP - CP.Loss = CP - SP.Gain or loss per cent
Marked price is what ismarked in the shop. It isneither CP nor SP.
14. Interest:
It is advisable never touse this formula but workfrom simple interest, bycalculating interest oninterest.
15. Time and distance:Remember the formula,
Distance = Speed ×Time
To convert km/hr intom/s, multiply by .
To convert m/s to km/hr,
multiply by
To calculate average
speed, use the formula: Average speed
where x and y are the
speeds.To calculate when two
bodies will cross eachother, use the formula:
Speed is added whenbodies are going in
opposite directions andsubtracted when bodiesare going in the samedirection, to find the relativespeed.
16. Boats and streams: A boat rowing in still water
8
Gain or Loss= × 100
CP
P × R × TSI =
100In compound interest,
R n A = P(1+ ) .100
518
185
2 xy(x + y)
=
DistanceTime = .
Relative Speed
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at the rate of x km/hr will beaffected if it goes into astream which is flowing. Ifthe rate of the stream is y
km/hr, the rate of the boatwhen it goes downstreamwill increase and will be(x + y) km/hr. However, ifthe boat goes against thecurrent, its speed willdecrease and will be givenby (x - y) km/hr.
Rate in still water isgiven by:
{(rate with the current)
+ (rate against the
current)}
Rate of current is givenby:
{(rate with the current)
–(rate against the current)}.
17. Time and work: a
person can do a piece of
work in x days, the work
done by him in 1 day will be
Conversely, if 1 day’s
work of a person is , then
he can finish the work in x
days.If A is faster than B and
is twice as good in his
work, the ratio of the workdone by A and B will be2:1.
However, the ratio oftime taken by A and B willbe in the ratio of 1:2.
18. Areas and
volumes:Rectangle A = L × Ba) Area = Length x
Breadthb) Diagonal2 = Length2
+ Breadth2
Square Area = Side2
Diagonal2
Four walls of a room Area= 2(Length+Breadth) ×
HeightTriangle with sides a,
b, c
Area
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1
2
1
2
1.
x
1
x
1
2=
( )
= s(s -a)(s -b)(s -c)
1where s a b c
2= + +
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where s = (a+b+c)
Triangle with base b
and height h
Area = × b × h
Equilateral triangle
with side x
Area = (x2)
Parallelogram Area = Base × HeightRhombus with
diagonals d1 and d2
Area = (d 1 × d2)
Trapezium
Area = (sum of
parallel sides) × heightQuadrilateral with
diagonal d
(d)(sum of perpendi-
culars on d from opposite
vertices)Circle with radius rCircumference = 2r Area = r2
Area of sector = r2/360Volumes are given by
the following table:
SolidsCube with side xVolume = x3
Surface area = 6x2
Longest diagonal = 3xCuboid with length l,breadth b and height hunits
Volume = l x b x hSurface area =
2(lb + bh + lh)Longest diagonal =
l2 + b2 + h2
Cylinder with radius rand height h
10
1
2
1
2
3
4
1
2
1
2
2Volume = r h
Curved surface area = 2 rh
Total surface area
2= 2 rh + 2 rSphere with radius r
4 3Volume = r3
2Surface area = 4 r
Cone with radius r and1 2height h Volume = r h3
2 2Slant height l = r +h
Curved
π
π
π π
π
π
π
surface area = rl
19.
In AP, to find the nth term
and sum of the series, use
the following:
π
AP, GP:
1
2
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21. Permutations andcombinations:
Circular permutations:The number of circular
permutations of n differentobjects is (n–1)!
For example, if 20people are invited to aparty, to find out how manyways can they and the hostbe seated at a circular
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nth term = a + (n -1)d,
where a is the first term
and d is the common
difference.
nSum = {2a + (n -1)d}, or2n
Sum = (a + l)2
where l is the last term.
n-1In GP, nth term = a.r ,
where r is the common ratio.
Sum of n termsn(r .1)
= a if r >1,r-1
n2(1-r )and if r<1.
1-r
Sum to infinity, if r <1,
of a GP is given by :
aS = .
1-r
: If two
numbers a and b are given,
their geometric mean is
given by ab .
The reciprocals of an AP
form a harmonic
progression.
1 1 1Thus, , ,
3 6 9
Geometric mean
Harmonic Progression:
1, , ...12
is an example of HP, try to
get the series into AP, do
the calculations and then
change into fractions again.
20. If an event
can happen in y ways and
the number of ways that a
particular event can
occur is x, then the
probability of t
Probability :
he event
xhappening is .y
If x is the probability of an
event happening, the
probability that it will not
happen is (1 x).−
n
r
n!P =
(n-r)!
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table, since there are 21people to be seatedaround a circular table. Sothe number of ways that
they can be seated aroundit is 20!.Combinat ions: The
number of combinations ofn distinct objects taken r ata time, is given by:
22. Calendar: Everyyear which is divisible by 4is a leap year. Every fourthcentury is a leap year butno other century is a leap
year. Thus 400, 800, 1200and so on are leap yearsbut 700, 900, 1100 are notleap years, even thoughthey are divisible by 4.
To solve calendar sums,the number of odd days will
give the answer as to theday of the week that aparticular day should be.The week starts withSunday (0 odd day) andgoes till Saturday (6 odddays).
23. Clocks: A clock hasa dial divided into 60minutes. Each minute willthus subtend an angle of 6°
at the centre, since totalcircle is 360°. Each five-minute interval subtends a30° angle.
A minute hand moves 6°
every minute. The hour
hand moves a distance of 5
minutes or 30° in one hour.
In one minute, he
moves
In one hour, the minutehand moves 60 minutes,
while the hour hand moves5 minutes. The minutehand thus gains 55 minutesover the hour hand.
24. Logs: The followingrules are important:
1. Log of 1 to any base
is 0.(x
0
= 1 for any x).2. Log of any number to
the same base is 1.(Loga a = 1, since a
1
=a).
3. Log of any number is
12
n
r
n!C =
r!(n-r)!
30 1degrees, or °.
60 2
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the sum of the logs of itsfactors.
(log mn = log m + logn).
4. Log of a fraction is thelog of numerator minus log
of denominator. (log =
log m - log n).
5. Log ax
= x log a.6. Loga b x logb a = 125. Calculus: Rules for
derivatives:1. The derivative of a
constant is zero. If y = 10,= 0, since it does notchange with respect to x.
3. The derivative of aterm equals the derivativeof each term addedtogether.
4. When y = u.v where uand v are functions of x,
then
GEOMETRY
1. Parallel Lines: Thefollowing rules are useful:
( i ) C o r r e s p o n d i n gangles are equal.
(ii) Alternate angles areequal.
(iii) Interior angles onthe same side aresupplementary.
2. Triangle: A triangle isa three sided figure. It has
13
mn
dy2. If y = xn, then
dx
n-1= n.x .
I 2f y = x +2x-1,
dy d 2then = (x )dx dx
d d+ (2x) - (1)
dx dx=2x+2.
dy= v.du + u.dv.
dxu
5. If y = where u and vv
are functions of x, then
dy d.du-u.dv= .
2dx v
dyx x6. (a) If y = e , then = e .dx
dyx(b) If y = a , thendx
x= a log a.
(c) If y = log x,edy 1
then = .dx x
(d) If y = log x,a
dy 1then = = log a.
dx x
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the following properties:(i) The sum of all the
angles is 180°.(ii) The exterior angle is
equal to the sum of theinterior opposite angles.There are 6 exterior
angles of the triangle.(iii) An interior and
exterior angle issupplementary.
(iv) The sum of any twosides is always greater thanthe third side.
(v) The difference of any
two sides is always lessthan the third side.
(vi) The side opposite
the greater angle will be thegreatest side.
(vii) A triangle has atleast 2 acute angles.
A median of a triangle isthe line from a vertex to the
midpoint of the opposite
side. The centroid is thepoint at which the mediansof the triangle meet. The
centroid divides the
medians in the ratio 2:1.The median bisects the
area of the triangle.
Theorem of Appolonius:Sum of the squares of twosides of a triangle =2(median)
2
+ 2(half the
third side)
2
.The orthocentre is thepoint where the threealtitudes of the trianglemeet.
The circumcentre is thepoint where theperpendicular bisectors ofall the sides meet. A circledrawn with thecircumcentre as the centre,can circumscribe thetriangle.
The incentre is the point
where the three bisectors
of a triangle meet. The
inradius of the circle is the
perpendicular distance
from the incentre to any of
the sides of the triangle.
The incentre divides the
bisector of any angle in theratio of (b+c) : a.
Angle bisectortheorem: The bisector ofany angle of a triangledivides the opposite side inthe ratio of the two adjacent
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sides.Area of a triangle:
There are 2 ways to find thearea of a triangle:
where a,b,c are the sides ofthe triangle and
The isosceles triangle:Is a triangle in which twosides are equal and twoangles are also equal.
Equilateral triangle: Is atriangle in which all sidesare equal and all angles arealso equal (60°).
Right angled triangle:
The Theorem of
Pythagoras is repeatedly
used, which states that the
square of the hypotenuse
equals the sum of the
squares of the other twosides.
The median to thehypotenuse bisects thehypotenuse, which is alsothe circumradius of thetriangle.
Pythagorean triplets:The following are someexamples of Pythagoreantriplets:
3, 4, 55, 12, 137, 24, 25
8, 15, 179, 40, 4111, 60, 6112, 35, 3716, 63, 6520, 21, 29.Congruency: Two trian-
gles are congruent if:1. Two sides and the
included angle of onetriangle are respectivelyequal to the two sides andthe included angle of thesecond triangle (SAS).
15
1(a) Area = (base) (height)2
(b) s(s-a)(s-b)(s-c)
a+b+cs= .
2
3Height = side.
2
3 2 Area = side .4
1Inradius = (height)3
2Circumradius = (height).
3
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2. Three sides of the firstare respectively equal tothe three sides of thesecond triangle (SSS).
3. Two angles and aside of the first arerespectively equal to thetwo angles and one side ofthe other triangle (AAS).
4. The hypotenuse andone side of a right angledtriangle are respectivelyequal to the hypotenuseand one side of anotherright angled triangle (RHS).
Similarity: Two trianglesare similar if:
1. Three angles of onetriangle are respectivelyequal to three angles of thesecond (AAA).
2. Two angles of one
triangle are respectively
equal to two angles of the
second (AA).
3. Two sides of onetriangle are proportional to
two sides of the other and
the included angles are
equal (SAS).In a right angled
triangle, the altitude to the
hypotenuse separates thetriangle into two triangleswhich are similar to eachother and to the original
triangle.Midpoint theorem: Theline joining the midpoints ofany two sides of a triangleis parallel to the third sideand equal to half of it.
Basic proportionality
theorem: A line parallel toone side of a triangle
divides the other two sides
proportionally. In the figure,
DE is parallel to BC. Then,
AD/BD = AE/EC.
3. Polygons: A polygon
is any closed plane figure. A triangle is a polygon with
3 sides, a quadrilateral with
4 sides, a pentagon with 5
sides and a hexagon with 6
sides. A polygon with
infinite sides is a circle.
A regular polygon is one
which has all sides and
angles equal.
In a polygon, the sum of
all the interior angles is
(2n – 4) right angles.
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Straight lines joining the
midpoints of the adjacent
sides of any quadrilateral
forms a parallelogram.
4. Circles: Somequalities of circles are given
below.
1. A tangent touches a
circle at only one point. A
chord is any line joining
any two points on thecircle. When the chord
passes through the centre,
it becomes the diameter.2. A tangent is
perpendicular to the radius.3. A perpendicular from
the centre of the circle tothe mid-point of a chord isperpendicular to the chord.Equal chords are
equidistant from the centre.The reverse is also true.
4. There is only one
circle that can pass
through three non-collinear
points.
5. Tangents drawn from
an external point are equal.6. The angle subtended
by an arc of a circle at the
centre is double the angle
subtended by it at any
point on the remaining part
of the circle.
7. Angles in the samesegment are equal.
8. The angle in a semi
circle is a right angle.9. In a cyclic
quadrilateral, the sum ofthe opposite angles is
1800. If one side of cyclicquadrilateral is produced,then the exterior angle isequal to the interioropposite angle. Thequadrilateral formed byangle bisectors of a cyclical
17
1 Area = (perimeter)
2
(perpendicular from centre
to any side).
Quadrilaterals : In aquarilateral, the sum of all
four angles is 360°.
1 Area = (diagonal)
2
(sum of perpendiculars
on it from oppositevertices)
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quadrilateral is also cyclic.10. Equal arcs make
equal chords.11. When two circles
touch, their centres and thepoint of contact arecollinear. If they touchexternally, the distancebetween their centres isequal to the sum of radiiand if the cicles touchinternally, the distancebetween the centres equalsthe difference of the radii.
12. If from the point ofcontact of a tangent, achord is drawn then theangle which the chordmakes with the tangent isequal to the angle formedby the chord in thealternate segment.
TRIGONOMETRY In a right angled
triangle, three ratios must
be learnt:
1. Sin A= Opposite/ Hypotenuse
2. Cos A
= Adjacent/Hypotenuse
3. Tan A
= Opposite/Adjacent
Some important ratios
are given in the followingtable:
18
2
2
13. Area of the circle is r .
Area of sector with angle
= r × .360
θθ π
Angle Ratio
Sin Cos Tan
0 0 1 0
1 3 1
30 2 2 3
1 145 1
2 2
3 160 3
2 2
90 1 0 unde -
fined
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ALGEBRA
Quadratic equations:The equation ax
2
+ bx + c= 0 where a, b, c are real
numbers and , is aquadratic equation.Quadratic equations canbe solved by factorising.Two solutions are obtained,which are also called rootsof the equation.
If the equation ax
2
+ bx+ c = 0 cannot befactorised, the roots areobtained by the formula
a) If b2
– 4ac is positive,the roots and are both realand unequal.
b) If b2
– 4ac is a perfectsquare, then the roots arerational and unequal.
c) If b2
–4ac is zero, thenthe roots are real andequal.
d) If b2
–4ac is negative,
the roots are complex andunequal.
The value of b2
–4ac iscalled the discriminant.
Formulae: It is useful toremember the followingformulae:
1. (x+y)
2
= x
2
+ 2xy +y
2
.
2. (x-y)2
= x2
–2xy + y2
.
3. (x+y)2
– (x–y)2
= 4xy.
4. (x+y)2
+(x–y)2
= 2(x2
+ y2
).
5. (x+y)3
= x3
+ y3
+3xy(x+y).
6. (x–y)3
=
x3
–y3–3xy(x–y).
7. x2
–y2
= (x+y)(x–y).
8. x3
+ y3
=
(x+y)(x2
+y2
–xy).
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2
2
2
–b ± b – 4acx=
2a
If and are the roots so
obtained, then
–b + b – 4acx=
2a
–b – b – 4ac= x=
2a
α β
2
If and are the roots
of a quadratic equation
ax + bx + c = 0, then
-b c+ = and =
a a
α β
α β α β
a 0≠
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9. x3
–y3
= (x–y)(x2
+ y2
+ xy).
10. (x + y + z)2
= [x2
+
y2
+ z2
+2(xy +yz + xz)].
11. x3 +y3 + z3–3xyz =
(x + y + z)(x2
+y2
+ z2
– xy
– yz – zx).
12. If x + y + z = 0,
then x3
+ y3
+ z3
= 3xyz.The converse is also
true.
Surds and indices: Thefollowing formulae areuseful:
1. am
× an
= am+n
.
Sets:n(AUB) = n(A) + n(B).If the sets intersect, thenn(AUB) = n(A) + n(B) –
n(AB).For three sets,
n(AUBUC) = n(A) + n(B) +n(C) – n(AB) – n(BC) –
n(CA) + n(ABC).
DATA
INTERPRETATION
Steps to do DIquestions:
1. Spend half a minute
to look at the table or
graph. Note the years to
which the data refers to
and the units. Sometimes
the figures may be given in
thousands while the
answer may be in millions,resulting in mistakes.
2. Make sure you
understand what the table
says and what it does not.3. The level of approxi-
mation that can be done is
assessed from the choices.If the answers are wide,time should not be wastedin working out exactfigures. If the choice ‘noneof the above’ exists, a closeapproximation may be
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0
ma m–n2. = a .nam n mn3. (a ) = a .
an n n n4. (ab) = a b and ( )b
na= .nb
5. a = 1.
n 1/n6. a = a .n n n7. a.b = a. b.
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required.4. Read the question
carefully. It will give anindication as to which row
and column should beseen.5. There may be one or
two very large questionsrequiring calculations. Attempt these at the last.
6. Revise bar charts, piecharts, statistics andgraphs before attemptingDI questions.
DATA SUFFICIENCY
Steps to solve DSquestions:
1. Read the statementcarefully and understandthe question that must beanswered.
2. Read the first piece ofdata provided, whilecompletely ignoring the
second. If you readeverything at once, it will bedifficult to arrive at ananswer.
3. Can the question beanswered on reading thefirst statement only? If so, it
is sufficient to answer thequestion. The secondstatement must be similarlychecked. If it is not
sufficient, it must bechecked whether theanswer can be provided bycombining it with thesecond.
4. Read the secondstatement while completelyignoring the first. This isimportant, otherwise thedata in the first willinfluence your answer.
5. After both statementshave been consideredindividually, combine themto see whether the answeris obtained by combiningthem. This step is notnecessary if eachstatement is sufficientindependently to answerthe question.
6. Do not waste timetrying to solve a problem;you are only to determinewhether the information issufficient to solve theproblem. The exact answeris not required.
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7. Sometimes choiceshelp. If you are sure aboutthe first statement, theanswer can be A or D. If
you are sure about thesecond one, the answercan be B or D. Looking atthe choices at this stagewill help you tick the rightchoice.
8. When geometricfigures are given, do notassume things going bythe look of the figures. Anangle may look like 90degrees, but the dataprovided may make it just alittle more or a little lessthan a right angle. Similarly,a triangle may lookisosceles, but the dataprovided may beotherwise. Remember, thefigures may not be drawnto scale.
9. Note that datasufficiency problems aretime savers since they donot require longcalculations. So theyshould be attempted first.
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