There are several things
that you have to keep in
mind:
a) Differentiate
between speed and
hurry: Many students think
that speed means making
guesses and blundering
on. This is wrong. Do the
questions with
concentration. Don’t worry
about questions you could
not attempt, because you
possibly cannot do 180
questions in two hours.
Speed means avoiding
unnecessary
calculations.
☛ Hot tip:
Allow yourself 30
seconds to glance over the
section before starting.
This will help you locate the
easy questions.
b) Keep your cool.
There will be some difficult
questions and some easy
ones. When you hit the
difficult ones, you must
stay calm. Skip a few
questions. Don’t panic,
because if you do, you
have almost certainly lost
the battle.
☛ Hot tip:
If you become tense,
just close your eyes and
take a few deep breaths.
Return to the paper once
you are composed again.
c) Manage your time:
Keep a watch before you.
Write down the time when
you should move over to
the new section. Don’t be
tempted to stick to the
previous section when the
time is over. Just keep
moving on.
☛ Hot tip:
Each section should be
1
FORMULAE
AND TIPS FOR CAT
attempted in 28 minutes.
Allow yourself 2 minutes to
go back on questions you
could not solve, or for
making some intelligent
guesses.
COMPREHENSION
☛ Tips for speed reading
Next time you are
reading a paper, try this.
First stage: Sweep your
hand from left to right
across the page so your
finger underlines each line.
Follow your finger with your
eyes. The difficulty of the
material determines the
speed, but move your hand
a little faster than your eye
can follow comfortably.
Make your eyes work to
keep up.
Soon you will find
yourself reading whole
phrases and ideas. As it
gets easier, keep
increasing the speed.
As you improve, instead
of sweeping whole lines
with your finger, just sweep
the middle part. Your eyes
will scan the beginning and
the end of the line
automatically.
When you’ve mastered
this, your reading speed
will be between 500 and
1,000 words per minute.
This is the practical limit.
Second Stage: The
next speed increase comes
from area reading. Instead
of sweeping lines, use
broader hand motions to
make a series of zigzags of
S’s down the page and
read several lines at once.
You can even read
backward as your hand
moves to the left. You’ll
comprehend very little at
first, but if you keep your
eyes following your finger,
you’ll find yourself
absorbing whole chunks of
the page at once.
Six hot tips for
comprehension:
1. Read the questions
before you read the
passage. This will help you
locate the ideas faster.
2
2. Force your eyes to
sweep entire phrases and
ideas. Don’t read word for
word. Avoid pronouncing
words as you read.
3. Don’t reread
sentences. Don’t skip lines.
4. Let the context define
unfamiliar words. Don’t
stop to look for them.
5. Adjust reading speed
to the type of material.
6. Look for key ideas
and phrases in each
paragraph.
What you should read:
Read the editorial page of
one good English
newspaper everyday by the
above method. This should
not take more than 15
minutes everyday. Make it
a point to read the
economic articles and
business news. Look up
difficult words that you
encounter in a dictionary.
Also read a magazine on
current affairs. Your best
bet is THE COMPETITION
MASTER, which has a
regular management
section. Read the debates
regularly as well as the
editorials, features and
current affairs and
Business GK. This will help
you form your opinions and
increase your awareness. A
number of people who
have cleared CAT have told
us that the magazine was
priceless for their
preparation.
Additional reading and
exercises:
Norman Lewis: How to
Read Better and Faster.
VERBAL ABILITY
It is important to have a
good 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 a
dictionary. 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
3
roots of words helps.
☛ Hot tip:
The best way to learn a
word list is through
vocabulary cards. Making
them is an investment,
because they will be helpful
over a period of time.
Analogies, Odd man
out: This means finding
relations among words.
What is the relation
between SALT : SALTY?
We can say that salty
things contain salt. The
best choice would be
COW : BOVINE, because
that best suits the relation-ship. A good vocabulary
helps in such questions.
☛ Hot tip:
Do as many analogies
as you can, to know the
kind of relations that can be
asked. The Test of
Reasoning in THE
COMPETITION MASTER con-tains analogies regularly. If
you do it every month, you
will have a good
understanding of
analogies.
Arranging sentences,
blanks: These are tricky,
especially if they contain
large sentences. The
selections are usually from
current newspapers, so
extra reading increases
speed in such questions.
☛ Hot tip:
Usually, the answer can
be obtained by getting at
the central idea and
thinking what it should start
or end with, or by working
from the choices. Do not
waste time arranging the
entire paragraph.
QUANTITATIVE
ABILITY
The importance of
knowing your tables,
decimals, fractions and
formulae cannot be over-emphasised. Many
questions can be solved by
looking at the choices.
Develop this ability and
your speed will surely
increase. We give below
4
some things which can be
remembered easily.
1. Numbers: Remem-ber the power of 10 in the
following: millions (6),
lakhs (5), billions (9) and
crores (7). Some questions
may 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 are
those which when
expressed in decimal
would be in non-terminating and non-repeating form. Examples
of irrational numbers are: 2,
3, 5, 7, and so on.
3. Division:
Dividend = (Divisor ×
Quotient) + Remainder.
4. Tests of divisibility:
A number can be checked
for divisibility by the
following methods:
By 2: If the last digit of
the given number is even
or zero.
By 3: If the sum of the
digits of the number are
divisible by 3.
By 4: If the sum of the
last two digits of the
number is divisible by 4.
By 5: If the last digit of
the number is either zero or
5.
By 6: If the number is
divisible by 2 as well as 3.
By 8: If the sum of the
last three digits of the
number is divisible by 8.
By 9: If the sum of the
digits is divisible by 9.
By 10: If the last digit is
zero.
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
,
q0≠ .
13 8
, , , 0, 3, 150 etc,
25 5
−
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 of
the 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 two
numbers
= HCF × LCM.
HCF of Fractions
LCM of Fractions
7. Simplification: To
simplify an expression,
always use the order
specified in BODMAS:
Brackets, Of, Division,
Multiplication, Addition and
Subtraction.
8. Square roots: Learn
the square roots upto 16
and squares upto 32. Make
memory cards to help
remember 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 numerators
LCM of denominators=
4
18
r2a
2a+
Thus 1152
= (100 + 15)2
= 1002
+ 152
+ 2(100 x 15),
which can be easily
computed.
8. Percentages: Learn
the fraction equivalents.
Many questions can be
solved faster if we know
these figures.
To find growth percent-age or percentage change,
always use:
9. Averages: Averages
are found by adding up the
values and dividing by the
number 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 divided
between two partners is:
(Amount of money invested
by A × No. of months
invested by A) : (Amount of
money invested by B × No.
of months invested by B).
12. Shares: It is
important to know the
following terms:
Face value: The price at
which shares are issued.
Always a round figure.
Market value: The price
at which shares are traded.
Will fluctuate and will
seldom be a round figure.
If market value = face
value, the shares are
traded at par.
If market value > face
value, the share
commands a premium.
If market value < face
value, the share is at a
discount.
Return: The interest
earned by the shares after
one year. Always calcu-lated on face value.
Yield: Return calculated
on what is actually
invested. Calculated by
dividing return by market
7
11 1= 25%, = 33%,
43 5
1= 20%, = 16% etc.
6
New Quantity – Old quantity
× 100
Old Quantity
a
b
value.
Brokerage: When you
buy a share, the buyer has
to pay the brokerage,
which is added to the
market price. When shares
are sold, the seller gets the
price after the broker
deducts his brokerage
from the market price
obtained.
13. Profit and Loss:
Profit = SP - CP.
Loss = CP - SP.
Gain or loss per cent
Marked price is what is
marked in the shop. It is
neither CP nor SP.
14. Interest:
It is advisable never to
use this formula but work
from simple interest, by
calculating interest on
interest.
15. Time and distance:
Remember the formula,
Distance = Speed ×
Time
To convert km/hr into
m/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 each
other, use the formula:
Speed is added when
bodies are going in
opposite directions and
subtracted when bodies
are going in the same
direction, to find the relative
speed.
16. Boats and streams:
A boat rowing in still water
8
Gain or Loss
= × 100
CP
P×R×TSI =
100
In compound interest,
R nA=P(1+ ) .
100
5
18
18
5
2xy
(x + y)=
DistanceTime = .
Relative Speed
at the rate of x km/hr will be
affected if it goes into a
stream which is flowing. If
the rate of the stream is y
km/hr, the rate of the boat
when it goes downstream
will increase and will be
(x + y) km/hr. However, if
the boat goes against the
current, its speed will
decrease and will be given
by (x - y) km/hr.
Rate in still water is
given by:
{(rate with the current)
+ (rate against the
current)}
Rate of current is given
by:
{(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 work
done by A and B will be
2:1.
However, the ratio of
time taken by A and B will
be in the ratio of 1:2.
18. Areas and
volumes:
Rectangle A = L × B
a) Area = Length x
Breadth
b) Diagonal2
= Length2
+ Breadth2
Square
Area = Side2
Diagonal2
Four walls of a room
Area
= 2(Length+Breadth) ×
Height
Triangle with sides a,
b, c
Area
9
1
2
1
2
1
.
x
1
x
1
2
=
()
= s(s - a)(s - b)(s - c)
1
where s a b c
2
=++
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 × Height
Rhombus with
diagonals d1 and d2
Area = (d1 × d2)
Trapezium
Area = (sum of
parallel sides) × height
Quadrilateral with
diagonal d
(d)(sum of perpendi-culars on d from opposite
vertices)
Circle with radius r
Circumference = 2r
Area = r2
Area of sector = r2
/360
Volumes are given by
the following table:
Solids
Cube with side x
Volume = x3
Surface area = 6x2
Longest diagonal = 3x
Cuboid with length l,
breadth b and height h
units
Volume = l x b x h
Surface area =
2(lb + bh + lh)
Longest diagonal =
l2
+ b2
+ h2
Cylinder with radius r
and 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 r h + 2 r
Sphere with radius r
4 3Volume = r
3
2Surface area = 4 r
Cone with radius r and
1 2height h Volume = r h
3
22Slant height l = r + h
Curved
π
π
ππ
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
21. Permutations and
combinations:
Circular permutations:
The number of circular
permutations of n different
objects is (n–1)!
For example, if 20
people are invited to a
party, to find out how many
ways can they and the host
be seated at a circular
11
n t h t e r m = a + ( n - 1 ) d ,
where a is the first term
and d is the common
difference.
nS u m = { 2 a + ( n - 1 ) d } , o r
2
nS u m = ( 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 terms
n(r .1)= a i f 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
: I f t w o
numbers a and b are given,
their geometric mean is
given by ab .
The reciprocals of an AP
form a harmonic
progression.
111Thus, , ,
369
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
canhappeniny waysand
the number of ways that a
particular event can
occur is x, then the
probability of t
Probability :
he event
x
happening 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)!
table, since there are 21
people to be seated
around a circular table. So
the number of ways that
they can be seated around
it is 20!.
Combinations: The
number of combinations of
n distinct objects taken r at
a time, is given by:
22. Calendar: Every
year which is divisible by 4
is a leap year. Every fourth
century is a leap year but
no other century is a leap
year. Thus 400, 800, 1200
and so on are leap years
but 700, 900, 1100 are not
leap years, even though
they are divisible by 4.
To solve calendar sums,
the number of odd days will
give the answer as to the
day of the week that a
particular day should be.
The week starts with
Sunday (0 odd day) and
goes till Saturday (6 odd
days).
23. Clocks: A clock has
a dial divided into 60
minutes. Each minute will
thus subtend an angle of 6°
at the centre, since total
circle is 360°. Each five-minute interval subtends a
30° 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 minute
hand moves 60 minutes,
while the hour hand moves
5 minutes. The minute
hand thus gains 55 minutes
over the hour hand.
24. Logs: The following
rules 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
the sum of the logs of its
factors.
(log mn = log m + log
n).
4. Log of a fraction is the
log of numerator minus log
of denominator. (log =
log m - log n).
5. Log a
x
= x log a.
6. Loga b x logb a = 1
25. Calculus: Rules for
derivatives:
1. The derivative of a
constant is zero. If y = 10,
= 0, since it does not
change with respect to x.
3. The derivative of a
term equals the derivative
of each term added
together.
4. When y = u.v where u
and v are functions of x,
then
GEOMETRY
1. Parallel Lines: The
following rules are useful:
(i) Corresponding
angles are equal.
(ii) Alternate angles are
equal.
(iii) Interior angles on
the same side are
supplementary.
2. Triangle: A triangle is
a three sided figure. It has
13
m
n
dy2. If y = xn, then
dx
n-1= n . x .
I 2f y = x +2x -1,
dy d 2then = (x )
dx dx
dd+(2x)-(1)
dx dx
=2x+2.
dy = v . d u + u . d v .
dx
u5. If y = where u and v
v
are functions of x, then
dy d.du-u.dv=.2dx v
dyxx6 . ( a ) I f y = e , t h e n = e .
dx
dyx(b) If y = a , then
dx
x= a l o g a .
(c) If y = log x, e
dy 1then = .
dx x
(d) If y = log x,a
dy 1t h e n = = l o g a .
dx x
the following properties:
(i) The sum of all the
angles is 180°.
(ii) The exterior angle is
equal to the sum of the
interior opposite angles.
There are 6 exterior
angles of the triangle.
(iii) An interior and
exterior angle is
supplementary.
(iv) The sum of any two
sides is always greater than
the third side.
(v) The difference of any
two sides is always less
than the third side.
(vi) The side opposite
the greater angle will be the
greatest side.
(vii) A triangle has at
least 2 acute angles.
A median of a triangle is
the line from a vertex to the
midpoint of the opposite
side. The centroid is the
point at which the medians
of 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 two
sides of a triangle =
2(median)
2
+ 2(half the
third side)
2
.
The orthocentre is the
point where the three
altitudes of the triangle
meet.
The circumcentre is the
point where the
perpendicular bisectors of
all the sides meet. A circle
drawn with the
circumcentre as the centre,
can circumscribe the
triangle.
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 the
ratio of (b+c) : a.
Angle bisector
theorem: The bisector of
any angle of a triangle
divides the opposite side in
the ratio of the two adjacent
14
sides.
Area of a triangle:
There are 2 ways to find the
area of a triangle:
where a,b,c are the sides of
the triangle and
The isosceles triangle:
Is a triangle in which two
sides are equal and two
angles are also equal.
Equilateral triangle: Is a
triangle in which all sides
are equal and all angles are
also 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 two
sides.
The median to the
hypotenuse bisects the
hypotenuse, which is also
the circumradius of the
triangle.
Pythagorean triplets:
The following are some
examples of Pythagorean
triplets:
3, 4, 5
5, 12, 13
7, 24, 25
8, 15, 17
9, 40, 41
11, 60, 61
12, 35, 37
16, 63, 65
20, 21, 29.
Congruency: Two trian-gles are congruent if:
1. Two sides and the
included angle of one
triangle are respectively
equal to the two sides and
the included angle of the
second 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 2Area = side .
4
1Inradius = (height)
3
2Circumradius = (height).
3
2. Three sides of the first
are respectively equal to
the three sides of the
second triangle (SSS).
3. Two angles and a
side of the first are
respectively equal to the
two angles and one side of
the other triangle (AAS).
4. The hypotenuse and
one side of a right angled
triangle are respectively
equal to the hypotenuse
and one side of another
right angled triangle (RHS).
Similarity: Two triangles
are similar if:
1. Three angles of one
triangle are respectively
equal to three angles of the
second (AAA).
2. Two angles of one
triangle are respectively
equal to two angles of the
second (AA).
3. Two sides of one
triangle 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 the
triangle into two triangles
which are similar to each
other and to the original
triangle.
Midpoint theorem: The
line joining the midpoints of
any two sides of a triangle
is parallel to the third side
and equal to half of it.
Basic proportionality
theorem: A line parallel to
one 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.
16
Straight lines joining the
midpoints of the adjacent
sides of any quadrilateral
forms a parallelogram.
4. Circles: Some
qualities 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 the
circle. 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 to
the mid-point of a chord is
perpendicular 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 same
segment are equal.
8. The angle in a semi
circle is a right angle.
9. In a cyclic
quadrilateral, the sum of
the opposite angles is
1800. If one side of cyclic
quadrilateral is produced,
then the exterior angle is
equal to the interior
opposite angle. The
quadrilateral formed by
angle bisectors of a cyclical
17
1Area = (perimeter)
2
(perpendicular from centre
to any side).
Quadrilaterals : In a
quarilateral, the sum of all
four angles is 360°.
1Area = (diagonal)
2
(sum of perpendiculars
on it from opposite
vertices)
quadrilateral is also cyclic.
10. Equal arcs make
equal chords.
11. When two circles
touch, their centres and the
point of contact are
collinear. If they touch
externally, the distance
between their centres is
equal to the sum of radii
and if the cicles touch
internally, the distance
between the centres equals
the difference of the radii.
12. If from the point of
contact of a tangent, a
chord is drawn then the
angle which the chord
makes with the tangent is
equal to the angle formed
by the chord in the
alternate 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 following
table:
18
2
2
13. Area of the circle is r .
Area of sector with angle
=r× .
360
θθπ
Angle Ratio
Sin Cos Tan
0 0 1 0
131
30
22 3
11
45 1
22
31
60 3
22
90 1 0 unde -fined
ALGEBRA
Quadratic equations:
The equation ax
2
+ bx + c
= 0 where a, b, c are real
numbers and , is a
quadratic equation.
Quadratic equations can
be solved by factorising.
Two solutions are obtained,
which are also called roots
of the equation.
If the equation ax
2
+ bx
+ c = 0 cannot be
factorised, the roots are
obtained by the formula
a) If b
2
– 4ac is positive,
the roots and are both real
and unequal.
b) If b
2
– 4ac is a perfect
square, then the roots are
rational and unequal.
c) If b
2
–4ac is zero, then
the roots are real and
equal.
d) If b
2
–4ac is negative,
the roots are complex and
unequal.
The value of b
2
–4ac is
called the discriminant.
Formulae: It is useful to
remember the following
formulae:
1. (x+y)
2
= x
2
+ 2xy +
y
2
.
2. (x-y)
2
= x
2
–2xy + y
2
.
3. (x+y)
2
– (x–y)
2
= 4xy.
4. (x+y)
2
+(x–y)
2
= 2(x
2
+ y
2
).
5. (x+y)
3
= x
3
+ y
3
+3xy(x+y).
6. (x–y)
3
=
x
3
–y3–3xy(x–y).
7. x
2
–y
2
= (x+y)(x–y).
8. x
3
+ y
3
=
(x+y)(x
2
+y
2
–xy).
19
2
2
2
–b ± b – 4ac
x=
2a
If and are the roots so
obtained, then
–b+ b – 4ac
x=
2a
–b – b – 4ac
=x=
2a
αβ
2
If and are the roots
of a quadratic equation
a x + b x + c = 0 , t h e n
-b c
+= and =
aa
αβ
αβ αβ
a0≠
9. x
3
–y
3
= (x–y)(x
2
+ y
2
+ xy).
10. (x + y + z)
2
= [x
2
+
y
2
+ z
2
+2(xy +yz + xz)].
11. x
3
+y
3
+ z
3
–3xyz =
(x + y + z)(x
2
+y
2
+ z
2
– xy
– yz – zx).
12. If x + y + z = 0,
then x
3
+ y
3
+ z
3
= 3xyz.
The converse is also
true.
Surds and indices: The
following formulae are
useful:
1. a
m
× a
n
= a
m+n
.
Sets:
n(AUB) = n(A) + n(B).
If the sets intersect, then
n(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 DI
questions:
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 wasted
in working out exact
figures. If the choice ‘none
of the above’ exists, a close
approximation may be
20
0
ma m–n2. = a .na
mn mn3. (a ) = a .
annn n4. (ab) = a b and ( )
b
na
=.nb
5. a = 1.
n1/n6. a = a .
nnn7. a.b = a. b.
required.
4. Read the question
carefully. It will give an
indication as to which row
and column should be
seen.
5. There may be one or
two very large questions
requiring calculations.
Attempt these at the last.
6. Revise bar charts, pie
charts, statistics and
graphs before attempting
DI questions.
DATA SUFFICIENCY
Steps to solve DS
questions:
1. Read the statement
carefully and understand
the question that must be
answered.
2. Read the first piece of
data provided, while
completely ignoring the
second. If you read
everything at once, it will be
difficult to arrive at an
answer.
3. Can the question be
answered on reading the
first statement only? If so, it
is sufficient to answer the
question. The second
statement must be similarly
checked. If it is not
sufficient, it must be
checked whether the
answer can be provided by
combining it with the
second.
4. Read the second
statement while completely
ignoring the first. This is
important, otherwise the
data in the first will
influence your answer.
5. After both statements
have been considered
individually, combine them
to see whether the answer
is obtained by combining
them. This step is not
necessary if each
statement is sufficient
independently to answer
the question.
6. Do not waste time
trying to solve a problem;
you are only to determine
whether the information is
sufficient to solve the
problem. The exact answer
is not required.
21
7. Sometimes choices
help. If you are sure about
the first statement, the
answer can be A or D. If
you are sure about the
second one, the answer
can be B or D. Looking at
the choices at this stage
will help you tick the right
choice.
8. When geometric
figures are given, do not
assume things going by
the look of the figures. An
angle may look like 90
degrees, but the data
provided may make it just a
little more or a little less
than a right angle. Similarly,
a triangle may look
isosceles, but the data
provided may be
otherwise. Remember, the
figures may not be drawn
to scale.
9. Note that data
sufficiency problems are
time savers since they do
not require long
calculations. So they
should be attempted first.
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