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"Quantitative Ability (QA) "

CAT 2016 seems to be approaching faster than the speed of time and Quantitative Ability (QA) section of CAT gives the impression to be the favorite of the exam setters. It seems as if over 50 per cent of paper setting efforts is devoted to Quant sections.
Quantitative Ability (QA)

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Quantitative Ability




CAT 2016 seems to be approaching faster than the speed of time and Quantitative Ability (QA) section of IIM CAT Exam gives the impression to be the favorite of the exam setters. It seems as if over 50 per cent of paper setting efforts is devoted to Quant sections. Perhaps, that is the reason why solving techniques and the ultimate solutions to this logical, tricky and seemingly lengthy form of aptitude problems have been consistently changing. Hence, it's a must for us to identify and strengthen the strategies in the quantitative that can help us maximize our score in this section as much as possible.

Questions with Solutions

What is Quant (QA) all

There is no any fix syllabus for this section but with the expert analysis of previous CAT papers, following topics are of utmost importance -

1.   Number System:

·        Number Base System

·        BODMAS

·        HCF/LCM

·        AP/GP

·        AM/GM/HM

·        Central Tendency (Mean/Median/Mode)

·        Average

·        Ratio and Proportion

·        Percentage

2.   Algebra:

·        Equation

·        In-Equation

·        Quadratic Equation

3.   Time, Speed and Distance

4.   Work and Time

5.   Profit and Loss

6.   Installment Payment

7.   Clock

8.   Mensuration

9.   Allegation and Mixture

10. Work, Pipes and Cisterns

11. Geometry:

·        Line

·        Angle

·        Triangle

·        Spheres

·        Rectangle

·        Cube, Cone, etc

12.   Set Theory:

·        Venn diagram

·        Probability

·        Permutation and Combination

NOTE: This list is not inclusive but only indicative

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Quantitative Aptitude for CAT can be broadly divided under three main heads:

I. Geometry, Coordinate Geometry and Mensuration:

These topics are grouped together since they deal with the portion of QA that can be visualized. Of the three, maximum weightage is given to geometry, although every CAT paper will have 3-4 questions on mensuration, as well as a couple of questions on coordinate geometry, totaling about 25-30% of questions in the QA section. Topics that need to be covered in geometry are basic theorems involving triangles, circles and parallel lines. A common type of question that is often asked in CAT is to find the value of certain angles or length of certain sides. Therefore, make sure that you cover topics such as congruency and similarity of triangles.

The only things that you need to do in coordinate geometry are straight lines and circles. Given the equation of a circle, you should be able to comment on the centre and radius of the circle and draw it on a piece of graph paper, and nothing more. Similarly, you should know what the slope and y-intercept of a given straight line equation is, and be able to draw the line on a piece of graph paper.

For mensuration, flip through a school level textbook for basic formulae on areas, surface areas and volumes of triangles, circles, cylinders, cones, cuboids and spheres. Mensuration problems are calculation intensive, and require lots of practice.  

II.  Algebra and Number Theory:

Algebra and number theory provide the major chunk of questions in any CAT QA section - 55-60%. Topics that you need to look at are Permutations and Combinations, Probability (very basic, including die and card problems and perhaps Bayes' theorem), Functions, Progressions (A.P, G.P. H.P. and A.G.P), Logarithms, Equations (Quadratic and Linear/Simultaneous) and most importantly, Number Theory.  

Number Theory problems are usually very simple. They require certain tricks that you can pick up from any good textbook. Number theory contributes 3-4 questions to every CAT, and so it is a very important topic. You should be comfortable writing numbers in their algebraic form (e.g. a three digit number having digits xyz can be represented as 100x + 10y + z). You should also learn about divisibility tests.  

III.  Arithmetic and Miscellaneous:

15-20% of questions in any CAT paper fall under this head. Major topics that you need to cover are Set Theory (especially Venn diagrams) and problems on Time, Speed and Distance, both of which are always asked. Both of these topics are covered as part of the school syllabus, but may need some brushing up on. Sometimes, questions on topics such as Linear Programming are also asked.  

Miscellaneous problems are those problems which do not fall under any head. They are rarely asked, and even when they do appear in a CAT paper they do not number more than one or two. They are purely tests of mathematical aptitude, and you cannot learn how to solve them. The advice, for dealing with these problems, is to try back-substitution of answer choices, or to avoid these problems altogether.  

Quick ways of solving Quant problems -

Short-Cut methods:  

The CAT-Crackers ought to keep certain methods in mind while tackling questions related to Quantitative Aptitude.

Direct method: You may also call it "The Conventional Method" of solving questions. We can solve the question based on the information given by applying the formula. But we no longer get questions in CAT requiring direct procedure.  

Substitution method: This method is useful in solving problems related to algebra. A few values in the question given can be swapped with the other values to get the answer.  

'Elimination' method: This is considered to be the most effective one as far as CAT questions are concerned. Options give some inkling of the answer and through expertise gained by regular practice you can easily tackle the questions.  

'Observation' method: In this method, a repeat pattern can be observed and answer can be formalized accordingly.

Simultaneous method: Two methods can go concurrently. First half of the questions can be done using direct method and when the picture becomes clear, elimination procedure can be used.  

Understanding accelerators & speed breakers

CAT is more about smartness than about intelligence. In CAT nobody expects you to answer all the questions. There should be an elimination round for the questions as well. You should know what to select and what to reject. You have to be choosy while attempting your paper. There is no fun in getting 4 marks for a question and losing one mark for the next. So, unless you are very sure of the solution to the question, don't attempt it.  


If you want to be the smartest, give priority to "accuracy". The act of choosing a question, in a way, is related to accuracy as you will select and reject a question based on the accuracy level you can achieve in solving the given question. 47 - 57 per cent of the paper done with 80 per cent accuracy will be much better than 80 per cent of the paper done with 47 - 57 per cent accuracy.  

Improving speed of calculation

Do put in some efforts to improve your calculation speed. For attempting a question, we can divide time into two parts, to grasp the question and to calculate it further. Fast calculation will definitely save your time which you can then allocate to other questions.  

Handling traps

Many a times there are language traps laid by the paper setter. Even simple questions confuse you with their tricky language which is a deliberate attempt to affect your understanding.  

Modified CAT pattern

As we all know CAT likes to give us surprises. Over the years the pattern has shifted from pure calculation based questions to pure logic based ones. So, handle the questions accordingly and ensure a proper balance between attempt and accuracy.  


The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10n on the nth day of 2007 (n = 1, 2, ..., 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n on the nth day of 2007 (n = 1, 2, ..., 365). On which date in 2007 will the prices of these two varieties of tea be equal?

1) May 21  

2)  April 11  

3)  May 20  

4)  April 10  

5)  June 30  


The greatest temptation while solving this question is to equate the prices of both teas. This should never be done because the price of Darjeeling tea is constant after 100 days. So, on the 100th day, its price will be 110 and it is constant throughout the rest of the year. The price of Ooty tea will become 110 on the 140th day. That is on May 20. So Option (3) is the right answer.  


Mock tests  

You may have mastered the concepts theoretically, but you will not gain confidence unless you take various "Mock Tests". Take as many mock tests as you can. Remember, Practice makes a person perfect. Understanding things theoretically is far different from doing them practically. So, put your theories in practical framework and check out if your strategy is working effectively. Optimal use of time to attempt optimum number of questions is desirable as well as fruitful.  

Important Points to Remember -  

The importance of knowing your tables, decimals, fractions and formulae cannot be over emphasized. Many questions can be solved by looking at the choices. Develop this ability and your speed will surely increase. We give below some important points which can be remembered easily.

1. Numbers: Remember 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 p/q where p and q are integers and q ≠ 0. Examples of rational numbers are 1/8, 3/5, -8/5, 0,3,149, etc.

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 is 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. Short cuts for multiplying: Large multiplication 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 numerators
                              HCF of denominators       

LCM of Fractions = HCF of numerators
                              LCM of denominators        

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.  

Important Tips:  

1. Square roots can be approximated by using a2 + r/2a where, 'a' is the nearest root of the number and r is the remainder. Hence 85 = 92+4/18, that 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. Thus 1152 = (100 + 15)2 = 1002 + 152 + 2(100 x 15), which can be easily computed.

8. Percentages: Learn the fraction equivalents.

1/4 = 25%, 1/3 = 33%, 1/5 = 20%, 1/6 = 16%, etc.

Many questions can be solved faster if we know these figures.

To find growth percentage or percentage change, always use:

New Quantity - Old quantity X 100

Old Quantity

9. Averages: Averages are found by adding up the values and dividing by the number of values.

10. Ratio and proportion: Can be written as a : b or a/b. 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 calculated on face value.

Yield: Return calculated on what is actually invested. Calculated by dividing return by market 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 = Gain or Loss X 100

 Marked price is what is marked in the shop. It is neither CP nor SP.

14. Interest:

In Simple Interest (SI) = P * R * T

In compound interest, Amount = P (1 + R/100)n

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 5/18

To convert m/s to km/hr, multiply by 18/5

To calculate average speed, use the formula:

Average speed = 2xy/(x+y)

where x and y are the speeds. To calculate when two bodies will cross each other, use the formula:

Time = Distance/Relative Speed  

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 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: 1/2 {(rate with the current) + (rate against the current)}

Rate of current is given by: 1/2 {(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 1/x.

Conversely, if 1 day's work of a person is 1/x, 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


Area = Side2 = 1/2 Diagonal2  

Four walls of a room

Area = 2(Length + Breadth) × Height

Triangle with sides a, b, c

Area = s(s - a)(s -b)(s - c)

Where, s = 1/2 (a + b + c)

Triangle with base b and height h

Area = 1/2 × b × h

Equilateral triangle with side x

Area = 3/4 (x2)


Area = Base × Height

Rhombus with diagonals d1 and d2

Area = 1/2 (d1 × d2)


Area = 1/2 (sum of parallel sides) × height

Quadrilateral with diagonal d

1/2 (d)(sum of perpendiculars 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:


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

Volume = p r2h

Curved surface area = 2p rh

Total surface area = 2p rh + 2p r2

Sphere with radius r

Volume = 4/3 p r3

Surface area = 4pr2

Cone with radius r and height h

Volume = 1/3 pr2h

Slant height l = Ö r2+h2

Curved surface area = prl

19. AP, GP:

In AP to find the nth term and sum of the series, use the following:

nth term = a + (n -1)d, where a is the first term and d is the common difference.

Sum = n/2 {2a + (n-1)d}, or

Sum = n/2 (a + l)

where l is the last term.

In GP, nth term =a.rn-1 where r is the common ratio.


Geometric mean: If two numbers a and b are given, their geometric mean is given by Öab .

Harmonic Progression: The reciprocals of an AP form a harmonic progression. Thus, 1/3, 1/6, 1/9, 1/12,.. is an example of HP.

20. Probability : 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 the event happening is x/y.

If x is the probability of an event happening, the probability that it will not happen is (1−x).

21. Permutations and combinations: nPr = n!/(n-1)!

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 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 30/60 degrees or ½ degrees

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.

2. Log of any number to the same base is 1.

3. Log of any number Is the sum of the logs of its factors.

4. Log of a fraction is the log of numerator minus log of denominator.

5. Log ax = 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.

2. If y = xn, then dy/dx = n.x n-1

3. The derivative of a term equals the derivative of each term added together.

If y = x2+2x -1, then dy = d (x2) + d (2x) - d (1) = 2x+2

  dx dx dx


4. When y = u.v where u and v are functions of x, then

dy/dx = v.du + u.dv.


5. If y = u/v where u and v are functions of x, then dy/dx = d.du-u.dv/ v2


 6. (a) If y = ex, then dy/dy = e x

(b) If y = ax, then dy/dx =ax log a

(c) If y = logex, then dy/dx = 1/x.

 (d) If y = logax, then dy/dx = 1/x= log a.


1. Parallel Lines: The following rules are useful:

(i) C o r r e s p o n d i n g 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 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 sides.

Area of a triangle: There are 2 ways to find the area of a triangle:

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°).

  Height = Ö3/2 side

Area = Ö3/2 side

Inradius = 1/3 (height)

Circumradius = 2/3 (height)

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 triangles 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).

 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.

 Area = 1/2 (perimeter)

(perpendicular from centre to any side).

 Quadrilaterals : In a quarilateral, the sum of all four angles is 360°.

Area = 1/2(diagonal)

(sum of perpendiculars on it from opposite vertices)

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 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.

13. Area of the circle is r2. Area of sector with angle θ = pr2 × θ/360


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:

























Quadratic equations: The equation ax2 + 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 ax2 + bx + c = 0 cannot be factorised, the roots are obtained by the formula:

a) If b2 - 4ac is positive, the roots and are both real and unequal.

b) If b2- 4ac is a perfect square, then the roots are rational and unequal.

c) If b2-4ac is zero, then the roots are real and equal.

d) If b2-4ac is negative, the roots are complex and unequal. The value of b2-4ac is called the discriminant.

If a and b are the roots of a quadratic equation ax2 + bx + c = 0, then

a + b = -b/a and ab = c/a

Formulae: It is useful to remember the following formulae:

1. (x+y)2 = x2 + 2xy + y2.

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).

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: The following formulae are useful:

1. am × an = am+n

2. am/an = am-n

3. (am)n = amn.

4. (ab)n = an bn and (a/b)n = an/bn

5. a0 = 1.

6. na = a1/n.

7. n a.b = na.nb.


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).

Illustrations: (Based on previous years CAT question paper)

1) A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

A. 120 metres B. 180 metres

C. 324 metres D. 150 metres

Answer: Option D


Speed= (60 * 5/18) m/s = (50/3) m/s

Length of the train = (Speed x Time) = (50/3 * 9) m = 150m

2) A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

A. 12 days B. 15 days

C. 16 days D. 18 days

Answer: Option B


A's 2 day's work = (1/20* 2) = 1/10

(A + B + C)'s 1 day's work = (1/20 + 1/30 + 1/60) = 6/60 = 1/10

Work done in 3 days = (1/10 +1/10) = 1/5

Now, 1/5 work is done in 3 days.

Whole work will be done in (3 x 5) = 15 days.

3) The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:

A. 15   B. 16

C.  18     D. 25

Answer: Option B


Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 - x).

Therefore, (20 - x /x * 100 = 25)

or, 2000 - 100x = 25x

125x = 2000

x = 16.

4) A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was:

A. 14 years     B. 19 years

C. 33 years     D. 38 years

Answer: Option A


Let the son's present age be x years. Then, (38 - x) = x

2x = 38.

x = 19.

Son's age 5 years back (19 - 5) = 14 years.

5) The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?


A. 0     B. 1

C. 10      D. 19

Answer: Option D


Average of 20 numbers = 0.

Sum of 20 numbers (0 x 20) = 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).

6) In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

A. 810   B. 1440

C. 2880    D. 50400

E. 5760

Answer: Option D


In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters = 7!/2! = 2520

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 Ways

Therefore, Required number of ways = (2520 x 20) = 50400.

7) The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000       B. 9400

C. 9600       D. 9800

Answer: Option C


Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.

8) The least perfect square, which is divisible by each of 21, 36 and 66 is:


A. 213444        B. 214344

C. 214434        D. 231444

Answer: Option A


L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 22 x 32 x 72 x 112 = 213444

9) Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?


A. 648        B. 1800

C. 2700       D. 10800

Answer: Option B


Let the required number of bottles be x.

More machines, More bottles (Direct Proportion)

More minutes, More bottles (Direct Proportion)

Machines 6:10 :: 270 : x

Time (in minutes) 1 : 4 :: 270 : x

Therefore, 6 x 1 x x = 10 x 4 x 270

x = (10 x 4 x 270)/ (6)

x = 1800.

10) A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

A. 10        B. 20

C. 21        D. 25

Answer: Option C


Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = (7x -7/12*9)liters = (7x - 21/4) liters

Quantity of B in mixture left = (5x -5/12*9)liters = (5x - 15/4) liters

(7x - 21/4)/ (5x - 15/4)+9 = 7/9

28x - 21/20x + 21 = 7/9

252x - 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 liters of A.

11) A man buys Rs. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is:

A. Rs. 12        B. Rs. 15

C. Rs. 18        D. Rs. 21

Answer: Option B


Dividend on Rs. 20 = Rs. (9/100 * 20) = Rs. 9/5

Rs. 12 is an income on Rs. 100.

Therefore, Rs. 9/5 is an income on Rs. (100/12 * 9/5) = Rs. 15

12) The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 270. The banker's discount is:

A. Rs. 960          B. Rs. 840

C. Rs. 1020         D. Rs. 760

Answer: Option C



 13) If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:

A. 50 km        B. 56 km

C. 70 km        D. 80 km

Answer: Option A


Let the actual distance travelled be x km.

Then, X/10 = X+20/14

14x = 10x + 200

4x = 200

x = 50 km.

14) Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

A. Rs. 6400     B. Rs. 6500

C. Rs. 7200      D. Rs. 7500

E.  None of these

Answer: Option


Let the sum inve

sted in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

Then,( X*14*2/

100) + [(13900 - X) * 11 * 2/100] = 3508

28x - 22x = 350800 - (13900 x 22)

6x = 45000

x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

15) A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A's share is Rs. 855, the total profit is:

A. Rs. 1425          B. Rs. 1500

C. Rs. 1537.50          D. Rs. 1576

Answer: Option B


Let the total profit be Rs. 100.

After paying to charity, A's share = Rs. (95*3/5) = Rs. 57

If A's share is Rs. 57, total profit = Rs. 100.

If A's share Rs. 855, total profit = (100/57 * 855) = 1500

16) What was the day of the week on 17th June, 1998?

A. Monday          B. Tuesday

C. Wednesday          D. Thursday

Answer: Option C


17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June

( 31 + 28 + 31 + 30 + 31 + 17) = 168 days

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3. Given day is Wednesday.

17) A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A. 2.91 m            B. 3 m

C. 5.82 m       D. None of these

Answer: Option B


Area of the park = (60 x 40) m2 = 2400 m2.

Area of the lawn = 2109 m2.

Area of the crossroads = (2400 - 2109) m2 = 291 m2.

Let the width of the road be x meters. Then,

60x + 40x - x2 = 291

x2 - 100x + 291 = 0

(x - 97)(x - 3) = 0

x = 3.

18) What least number must be added to 1056, so that the sum is completely divisible by 23 ?


A. 2      B. 3

C. 18      D. 21

E. None of these

Answer: Option A


23) 1056 (45
Required number = (23 - 21) = 2.
19) What decimal of an hour is a second ?
A  .0025            B  .0256
   C  .00027                D  .000126

Answer: Option C


Required decimal = 1/60*60 = 1/3600 = .00027

20) If 5a = 3125, then the value of 5(a - 3) is:

A. 25       B. 125

C. 625       D. 1625

Answer: Option A


5a = 3125

5a = 55

a = 5.

5(a - 3) = 5(5 - 3) = 52 = 25.

20) A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:

A. 6 hours B.   10 hours

C. 15 hours     D. 30 hours

Answer: Option C


Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

1/x + 1/(x - 5) = 1/(x - 9)

x - 5 + x /x(x - 5) = 1/ (x - 9)

(2x - 5)(x - 9) = x(x - 5)

x2 - 18x + 45 = 0

(x - 15)(x - 3) = 0

x = 15. [neglecting x = 3]

21) Which of the following statements is not correct?

A. log10 10 = 1

B. log (2 + 3) = log (2 x 3)

C. log10 1 = 0

D. log (1 + 2 + 3) = log 1 + log 2 + log 3

Answer: Option B


(a) Since loga a = 1, so log10 10 = 1.

(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3

log (2 + 3) is not equal to log (2 x 3)

(c) Since loga 1 = 0, so log10 1 = 0.

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, (b) is incorrect.

22) From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?













Answer: Option D


Let S be the sample space.

Then, n(S) = 52C2 =

(52 x 51)

= 1326.

(2 x 1)


Let E = event of getting 2 kings out of 4.

23) Find the odd man out.

8, 27, 64, 100, 125, 216, 343

A. 27       B. 100

C. 125        D. 343

Answer: Option B


The pattern is 23, 33, 43, 53, 63, 73. But, 100 is not a perfect cube.

24) From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:

A. 149 m        B. 156 m

C. 173 m        D. 200 m


Answer: Option C


Let AB be the tower.

Then, angle APB = 30º and AB = 100 m.

AB/AP = tan 30º = 1/ Ö3

AP = (AB x Ö

= 100 Ö3 m

= (100 x 1.73) m

= 173 m.

25) The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:

A. 2 B.        2.1/2

C. 3        D. 4

Answer: Option A


Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000 (1 + 7/100)n = 34347


26) Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

A. 39, 30 B. 41, 32

C. 42, 33 D. 43, 34

Answer: Option C


Let their marks be (x + 9) and x.

Then, x + 9 =56/100(x + 9 + x)

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

27) A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:

A. 145º        B. 150º

C. 155º        D. 160º

Answer: Option C


Angle traced by hour hand in 12 hrs = 360º.

Angle traced by hour hand in 5 hrs 10 min. i.e., 31/6 hrs. (360/12 * 31/6)0 = 1550

28) A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

A. 720        B. 900

C. 1200        D. 1800

Answer: Option C


2(15 + 12) x h = 2(15 x 12)

h = 180/27 m = 20/3 m

Volume = (15*12*20/3)m3 = 1200m3

29) The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?

A. 4         B. 8

C. 16        D. None of these

Answer: Option B


Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.

Let ten's and unit's digits be 2x and x respectively.

Then, (10 x 2x + x) - (10x + 2x) = 36

9x = 36

x = 4.

Required difference = (2x + x) - (2x - x) = 2x = 8.

30) There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:

A. 20               B. 80

C. 100               D. 200

Answer: Option C


Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10 x - y = 20 .... (i)

and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

31) A sum of money is to be distributed among A, B, C, D in the proportion of 5: 2: 4: 3. If C gets Rs. 1000 more than D, what is B's share?


A. Rs. 500               B. Rs. 1500

C. Rs. 2000               D. Rs. 2500

Answer: Option C


Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x - 3x = 1000

x = 1000.

B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

32) A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:

A. 4       B. 5

C. 6       D. 10

Answer: Option B


Let the speed of the stream be x km/hr. Then,

Speed downstream = (15 + x) km/hr,

Speed upstream = (15 - x) km/hr.

30/(15+X) + 30/(15 - X) = 4. 1/2

900/225 - X2 = 9/2

9x2 = 225

x2 = 25

x = 5 km/hr.

33) In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by:

A. 60 m       B. 40 m

C. 20 m       D. 10 m

Answer: Option C


To reach the winning post A will have to cover a distance of (500 - 140)m, i.e., 360 m.

While A covers 3 m, B covers 4 m.

While A covers 360 m, B covers (4/3 * 360)m = 480m

Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.

Therefore, A wins by 20 m.

34) If Rs. 10 be allowed as true discount on a bill of Rs. 110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is:

A. Rs. 20      B. Rs. 21.81

C. Rs. 22      D. Rs. 18.33

Answer: Option D


S.I. on Rs. (110 - 10) for a certain time = Rs. 10.

S.I. on Rs. 100 for double the time = Rs. 20.

T.D. on Rs. 120 = Rs. (120 - 100) = Rs. 20.

T.D. on Rs. 110 = Rs. (20/120 * 110) = Rs. 18.33

Useful tips for Quant Preparation:

·        Practice the new online format adequately;


·        Identify sitters in the exam first;

·        Avoid making random guesses;

·        Put maximum emphasis on accuracy.

The age-old saying 'Practice makes a man perfect' is still applicable here. Working hard on one's fundamentals and focusing on one's accuracy will only go to increase your chances of belling the revamped CAT 2016. Good luck!

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