Quantitative aptitude Crash course June 2017 CA. … Crash course June 2017 Quant.… · Standard steps in Simplifications • Fractions • Algebra formulas • Cross Product rule

Quantitative aptitude

Crash courseJune 2017

CA. Chinmaya Hegde

Few tips for exam approach• Statistics take less time to solve

• Use shortcuts in mathematics for some problemsSubstitution method instead of solvingOther shortcuts

• Statistics has more of conceptual questions and no shortcuts available for practical question

• Finish economics quickly so as to allocate more for quant

• Don’t attempt if you are not sure of any options

• Take a guess if you are sure about 50 : 50

How to start

How???

Identify chapter

Substitution of options

Check conditions in Questions, by substituting each options

SolveIdentify formula/conceptSimplify

Standard steps in Simplifications• Fractions

• Algebra formulas

• Cross Product rule

• Simplifying square roots

Standard steps in Simplifications• Fractions

• Algebra formulas

• Cross Product rule

• Simplifying square roots

Standard procedures - Fractions

ퟏ −ퟐퟕ

ퟓퟕ

ퟏ −ퟑퟏퟕ

ퟏퟒퟏퟕ

ퟏ −ퟖퟑퟖퟓ

ퟐퟖퟓ

ퟏ −ퟑퟓ

ퟐퟓ

ퟏ −ퟒퟏퟏ

ퟕퟏퟏ

ퟐ +ퟑퟓ

ퟐ ∗ ퟓ + ퟑퟓ

ퟏퟎ + ퟑퟓ

ퟒ +ퟖퟗ

ퟒ ∗ ퟗ + ퟖퟗ

ퟑퟔ + ퟖퟗ

ퟑퟓ+

ퟕퟗ

ퟑ ∗ ퟗ+ ퟓ ∗ ퟕퟓ ∗ ퟗ

ퟐퟕ + ퟑퟓퟒퟓ

Standard procedures – Algebra Formulas

(풂 + 풃)ퟐ

풂ퟐ + 풃ퟐ + ퟐ풂풃

(풂 − 풃)ퟐ

풂ퟐ + 풃ퟐ − ퟐ풂풃

(풙 − ퟓ)ퟐ

풙ퟐ + ퟓퟐ − ퟐ ∗ ퟓ풙

(ퟐ풌 − ퟏ)ퟐ

ퟒ풌ퟐ + ퟏ − ퟒ풌

풂ퟐ-풃ퟐ

(풂 + 풃)(풂 − 풃)

풙ퟐ-ퟖퟏ

(풙 + ퟖ)(풙 − ퟖ)

ퟒ풙ퟐ-ퟗ

(ퟐ풙 + ퟑ)(ퟐ풙 − ퟑ)

풂 + 풃 풂 − 풃

풂ퟐ-풃ퟐ

풙 + ퟑ풚 풙 − ퟑ풚

풙ퟐ-ퟗ풚ퟐ

ퟕ − ퟒ풌 ퟕ + ퟒ풌

ퟒퟗ-ퟏퟔ풌ퟐ

(풂 + 풃)ퟑ

풂ퟑ + 풃 + ퟑ풂풃(풂 + 풃)

Standard procedures – Cross Product rule

10 1530 ?

30 * 1510

17 14? 3

17 * 314

5 ?8 3

5*38

Square root simplification

Square both sides of equations

풙 = ퟒOn squaring 풙 = ퟒퟐ

x = 16

풙 + ퟏ = ퟖOn squaring 풙 + ퟏ = ퟖퟐ

IndexSl No List of Topics Page Number No of

questions1 Ratio and Proportions, Indices and logarithms2 Equations3 Inequalities4 Interest and Annuities5 Basics of permutations and combinations6 Progressions7 Sets, relations and functions8 Limits And Continuity-Intuitive Approach9 Basics Of Differential And Integral Calculus10 Statistical description of data11 Measures of central tendency and dispersion12 Correlation and Regression13 Probability and expected value of mathematical

expectation14 Theoretical distributions15 Sampling theory16 Index Numbers

Total

May Aug Nov Feb Jun Dec Jun Dec Jun Dec Jun Dec Jun Dec Jun Dec Jun Dec Jun Dec

07 07 07 08 08 08 09 09 10 10 11 11 12 12 13 13 14 14 15 15 Med

Section A Mathematics

Ratios,&Indices 4 5 5 5 4 2 6 3 4 4 2 3 3 3 3 5 7 3 4 4 4Equations 2 3 2 2 2 3 3 5 1 4 3 4 5 5 3 4 3 2 5 4 3Inequality 1 2 1 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1Int&Annuities 4 4 3 3 4 3 5 2 3 4 3 2 2 2 2 3 3 3 2 3 3Perm&Comb 4 3 3 2 2 3 5 4 3 3 2 3 2 3 3 3 2 3 3 3 3AP,GP 3 4 3 3 3 4 3 2 3 1 4 6 3 3 4 2 3 4 3 3 3Sets,Functions 2 1 2 1 1 3 3 3 4 3 3 3 3 3 3 3 4 1 3 3Limits 3 2 2 2 2 3 2 3 3 2 2 1 1 1 2 2 1 3 2 2 2Calculus 5 4 5 4 4 2 4 5 3 3 5 4 4 4 4 2 2 2 4 3 4

Section B Statistics

Stats-Description 4 3 4 4 4 4 3 4 3 4 4 3 3 1 1 3 4 4 5 4 4Centrl Tend&Disp 3 4 5 5 4 5 5 4 3 4 4 4 4 7 5 5 4 5 4 6 4Correla 3 2 3 3 3 4 3 4 4 5 3 4 6 4 4 4 4 3 2 2 3Probab 4 3 4 4 3 3 3 3 4 5 6 4 5 4 5 4 3 3 4 4 4Theoritical Dis 3 4 3 4 4 4 1 3 4 4 3 3 3 4 5 1 3 5 3 3 3Sampling 3 3 3 4 4 3 3 3 4 2 2 2 3 3 2 5 3 2 4 4 3Index Nos 2 3 2 3 4 3 4 2 4 1 3 3 2 2 3 3 4 2 3 1 3TOTAL 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50

Chapter 10 :Statistical description Of data

Statistical description of data

Basics Classification of data

Presentation Frequency distribution

Definition Limitations

SingularQuantitative

PluralQuantitative&Qualitative

Deals only with the aggregates

Mostly concerned with quantitative data

Based on measurement

Based on Source of collection

Attributes Qualitative data

Variables-Quantitative

Discrete Continuous

Absolute Range

Primary

Secondary

Interview

Telephone Interview

Mailed questionnaire

Observation

Govt

Quasi govt

Private organisations

Textual

Tabular

Diagrammatic

Pie Bar diagram

Line diagram

Frequency

Class limits(with gap)

Class boundary(Without gap)

Class width

Frequency density = f/w

Relative frequency = f/N

Class mark – mid value

Graphical presentation of frequency distribution

Histogram

Frequency polygon

Ogive curves - cumulative frequency graphs

Parts of table

Examination CandidatesAppeared

Candidatespassed

Pass Percentage

Dec 2012 111961 30305 27.07June 2012 149348 56091 37.56Dec 2011 115984 40975 35.33

Source : WWW.ICAI.ORG

Data on Pass % in CA CPTNo. 1

Table No.

Title

Row caption

Column CaptionsStub entries

Body

Foot notes

Box head

1• Information from the internet is•• (a) primary (b) secondary (c) tertiary (d) none of these

• (b)

2• The most appropriate diagram to express monthly expenditure is•• (a) histogram (b) pie diagram (c) frequency (d) line graph•• (b)

3• The chart that use Logarithm of the variable is known as: •• (a) Line chart (b) ratio chart (c) Multiple line chart (d) Component line chart

• (b)•

4• Find the number of observation between 250 and 300 from the following data: •• Value > 200 > 250 > 300 >.350 • No. of observations 56 38 15 0 •• (a) 56 (b) 23 (c) 15 (d)

38-15=23(b)

5• The number of observation between 150 and 200 based on the following data is:•• Value >100 > 150 > 200 > 250• No Observ 70 63 28 05•

• (a) 46 (b) 35 (c) 28 (d) 2363-2835B)

6• No. of accidents 0 1 2 3 4 5 6 7• Frequency 12 9 11 13 8 9 6 3•• In how many cases 4 or more accidents occur?•• (a) 32 (b) 41 (c) 26 (d) 18

8+9+6+326c)

7• The curve obtained by joining the points, whose x-coordinates are the upper limits of the class-

intervals and y coordinates are the corresponding cumulative frequencies is called:•• (a) Ogive (b) Histogram • (c) Frequency Polygon (d) Frequency Curve

• (a)

8• Histogram is used for the presentation of the following type of series:•• (a) Time series (b) Continuous frequency series• (c) Discrete series (d) Individual series•• (b)

9• The perpendicular line drawn from the intersection of two ogives which touches at _________

point in X-axis:•• (a) Median (b) Mode (c) Third quartile (d) First quartile

• (a)

10• The most appropriate diagram to represent 5 year plan outlay of India in different economic

sectors is •• (a) Pie diagram (b) Histogram (c) Line diagram (d) Frequency polygon

• (a)

11• For construction of Histogram the class intervals of frequency distribution is•• (a) Equal (b) Unequal (c) Either Equal or Unequal (d) None

• (c)

12• 100 persons are divided into number of male/female and employed/un-employed it refers to•• (a) Cardinal Data (b) Ordinal Data(c) Spatial Data (d) Temporal Data• Cardinal = numbers• (a)

13• If the fluctuations in the observed values are very small as compared to the size of the items, it is

presented by •• (a) Z-Chart (b) Ogive chart (c) False-Base Line (d) Control Chart• (c)

14• The following data related to the marks of group of students•• Marks No. of students• More than 70% 7• More than 60% 18• More than 50% 40• More than 40% 60• More than 30% 75• More than 20% 100•• How many students have got marks less than 50% ?•• (a) 60 (b) 82 (c) 40 (d) 53• (a)

15• To draw Histogram the frequency distribution should be•• (a) Inclusive type (b) Exclusive type (c) Inclusive and Exclusive type (d) None of the

above

(b)

Chapter 11A :Measures of central Tendency

Central tendency – Concentration of values

Means

AM

GM-Ratio

HM- Speed

Ungrouped∑ x

n

Grouped∑f x

n

Combined`x12 = `x1n1 +` x2n2

n1+n2

Weighted= ∑wx

∑w

∑( X-`X ) = 0

∑( X-`X )2 isminimum

2aba+b

Median

Ungrouped Grouped

(n + 1)th2

ObservationIn a array

L + (N/2 - m ) C f

Mode

Ungrouped Grouped

Most frequent

= L + f1 - f0 *c2f1-f0-f2

Other positional measures

Quartiles

Deciles

Octiles

Percentiles

i (n+1) th item4

i (n+1) th item10

i (n+1) th item8

i (n+1) th item100

Chapter 11A :Measures of Dispersion

Dispersion – Scatterness, Variation

Absolute measureHas units of measurement

Relative measureIndependent of unit of measurement

Base on selected items

Based on all items Base on selected items

Based on all items

Range H - L

Co efficient of Range = H – L *100

H + L Quartile rangeQ3 – Q1

Semi Quartile range

Q.D

Q3 – Q12

CO-efficient of Q.D

Q3 – Q1 *100Q3 + Q1

Mean deviation ∑| x – Average

n

Average = Mean/Median/Mode

CO-efficient of MD

MD *100 .Average

Standard deviation CO-efficient of variation

S.D * 100Mean

More the Cov, less the consistency Variance = S.D2

All these measures always positive

Change of origin and scale

All the observations are added or subtracted

All the observations are multiplied or divided

Central tendency

Dispersion

Changes Remains same

Central tendency

Dispersion

Changes Changes

Ignore constant in equations

1• G is G.M. between a and b then +

•• (a) G2 (b) 3G2 (c) 1/G2 (d) 2/G2

Let a = 1, b = 4, then GM = 2

1퐺 − 푎 +

1퐺 − 푏 =

1ퟐ − ퟏ +

1ퟐ − ퟒ

=ퟏퟑ −

ퟏퟏퟐ

=ퟒ − ퟏퟏퟐ =

ퟏퟒ =

ퟏ푮ퟐ

(c)

2• A lady traveling with a speed of 20Kmph, and returns quickly with average speed 24Kmph. Find

her return speed •• (a) 30 kmph (b) 25 kmph (c) 38 kmph (d) 22 kmph

HM for 2 variables = 2ab/(a+b)Try option a 2*20*30/(20+30) = 24 Right optionTry option b 2*20*25/(20+25) ≠ 24 wrong option

(a)

3• Mode can be obtained from•• (a) frequency polygon (b) histogram (c) ogive (d) all of the above

(b)

4• Variance of (3,5,8,4) is ____•• (a) 4.5 (b) 3.5 (c) 5.5 (d) 6

휎 =∑푥푛 −

∑푥푛

휎 =∑푥푛 −

∑푥푛

휎 =3 + 5 + 8 + 4

4 −3 + 5 + 8 + 4

4

휎 =114

4 −204

휎 = 3.5(b)

5• If V(x) = 23 Find variance of 2x+10: •• (a) 104 (b) 110 (c) 92 (d) 85

V(x) = 23

SD(x) = √23 = 4.7958

SD(2x+10) = 2(SD of ) + 10=2(4.7958)

= 9.59 Variance(2x+10) =9.592 =92(c)

6• There were 50 students in a class. 10 failed whose average marks were 2.5. The total marks of

class were 281. Find the average marks of students who passed? •• (a) 6.4 (b) 25 (c) 256 (d) 86

Total Marks 281failed passed 40 students10*2.5 = 25 281 – 25 = 256

average 256/40 = 6.4(a)

7• The mean salary for a group of 40 female workers is 5,200 per month and that for a group of 60

male workers is 6,800 per month. What is the mean of the combined salary?• (a) 6160 (b)6280 (c)6890 (d) 6920

Using combined AM40 * 5200 + 60 * 6800

40+606160

(a)

8• The G.M. of 4, 6 and 8 is:•• (a) 4.77 (b) 5.32 (c) 6.14 (d) 5.77•• GM = (4*6*8)1/3

• GM = (192)1/3

• Using calculator• 192• √ 15 times• -1• * 1/3• +1• X = x= 15 times• 5.77• (d)

9• G.M of a, b, c, d is 3 then G.M of 1/a, 1/b, 1/c, 1/d is •• (a) 1/3 (b)3 (c) 1/81 (d) 81

• G.M of a, b, c, d is 3• Let a, b c, d 3 each• GM of 1/a,1/b, 1/c,1/d = 1/3 each• GM =1/3

• (a)

10• In a class of 11 students, 3 students failed in a test. 8 students who passed secured 10, 11, 20,

15, 12, 14, 26 and 24 marks respectively. What will be the median marks of the students:•• (a) 12 (b) 15 (c) 13 (d) 13.5

• ArrayF1,F2,F3,10,11,12,14,15,20,24,26

Median = (n+1)/2 = (11+1)/2= 6th value= 12

(a)

13• G.M. is a better measure than others when,•• (a) Ratios and percentages area given (b) Interval of scale is given• (c) Both (a) and (b) (d) Either (a) or (b)

• (a)

14• Inter Quartile Range is ____ of Quartile Deviation.•• Half (b) Double (c) Triple (d) Equal•• (b)

15• The equation of a line is 5x + 2y = 17. Mean deviation of y about mean is 5. Calculate mean

deviation of x about mean.•• (a) -2 (b) 2 (c) -4 (d) None of these

5x + 2y = 175(MD of X) + 2(MD of y) = 05(MD of x) + 2(5) = 0MD of x = -10/5 = - 2MD of x = 2

(b)

16• If all the observations are increased by 10, then•• (a) SD would be increased by 10• (b) Mean deviation would be increased by 10• (c) Quartile deviation would be increased by 10• (d) All these three remain unchanged•

• (d)

17• Coefficient of variation is a/an•• (a) Absolute measure (b) Relative measure• (c) Both a and b (d) None of these

• (b)•

18• Which of the following measure(s) satisfies (satisfy) a linear relationship between two variables?•• (a) Mean (b) Median • (c) Mode (d) All of these•• (d)

19• Which measures of dispersion is not affected by the presence of extreme observations?•• (a) Range (b) Mean deviation (c) Standard deviation (d) Quartile

deviation

• (d)

20• The formula for range of middle 50% items of a series is•• (a)Q3-Q1 (b)Q3-Q2 (c)Q2-Q1 (d)(Q3-Q1)/2•• (a)

21• The difference between maximum and minimum value of the data is known as :•• (a) Range (b) Size (c) Width (d) Class

• (a)

22• Which of the following measures of central tendency cannot be shown by graphical method?•• (a) Mean (b) Median (c) Mode (d) Quartiles•• (a)

23• The SD of X is known to be 10 then the SD of 50+5X is:•• (a) 50 (b) 100 (c) 10 (d) 500

• SD of x = 10• SD of 50+5X = 50 + 5(SD of x)

= 5 * 10= 50

(a)

24• The pair of averages whose value can be determined graphically?•• (a) Mean & Median (b) Mode & Mean (c) Mode & Median (d) None of the above

• (c)

25• Which of the following measures of dispersion is used for finding consistency between the

series?•• (a) Q.D (b) S.D (c) Coefficient of variation (d) None of these

• (c)

Chapter 12 :Correlation and regression

Correlation and regression

CorrelationExtent of relationship

RegressionAverage relationship

MeaningMeasurement

Direction

Degree

Positive

Negative

-1

+1

Perfect negative

-0.75

Highnegative

-0.5 Moderate negative

-0.25

Lownegative

Perfect postive

Scatter diagram

Linear & Non-linear

Karl pearson

SySxyxCov

*),(r

nyyxx

))(-(

y)Cov(x,

2222 y)(y.x)(x

yxxyr

nn

n

Spearmen rank

)1(

...)1212

(61r 2

23

213

12

nn

mmmmd

Concurrent deviation

nnc

2r

Regression line y on x

Y = a + byxx

sxsyrbyx .

22 )( xxnyxxyn

byx

xbya yx

Regression line y on x

≠

Regression line x on y

properties

r =0, perpendicular

r = + 1lines coincide

yxxyxy bbr

All 3 will have same signs

Lines intersect at`x,`y

Independent of change of origin, scale and UoM(but not sign)

Independent of change of origin only. Dependent on scale

0 No correlation

To find x on y or y on xbyx*bxy < 1

If equation is x on y , then y : xIf equation is y on x, then x : y

2x + 5y – 9 = 0 3x + y – 5 = 0

Assume x on y Assume y on x

bxy is 5/2 byx is 3/1

ퟓퟐ*ퟑퟏ

= ퟕ.ퟓAssumption

is wrong2x + 5y – 9 = 0 is y on x3x + y – 5 = 0 is x on y

7x+2y+15 =0 2x+5y+10 = 0

Assume x on y

bxy is 2/7

Assume y on x

byx is 2/5

ퟐퟕ*ퟐퟓ

= ퟒퟑퟓ

Assumption is right

1• ___________ gives the mathematical relationship of the variables.•• (a) Correlation (b) Regression (c) both (d) None of these.

• (b)

2• Correlation coefficient between X and Y will be negative when:•• (a) X and Y are decreasing• (b) X is increasing but Y is decreasing• (c) X and Y are increasing • (d) None of these

• (b)•

3• The coefficient of correlation r between x and y when:• Cov (x,y) = -16.5, Var (x) = 2.89, Var (y) = 100 is:•• (a) -0.97 (b) 0.97 (c) 0.89 (d) -0.89

푟 =퐶표푣휎 휎

푟 =−16.5

1.7 ∗ ퟏퟎ

푟 = −0.97(a)

• 4• Two regression lines are x+2y-5=0, 2x+3y-8=0 find regression line of y

on x•• (a) x+2y-5=0 (b) 2x+3y-8=0 (c) any of these (d) none of these

x+2y-5=0, 2x+3y-8=0x on y y on x2 : 1 2 : 3

r =21 ∗

23 = 1.15

Assumption is NOT correct

X+2y – 5 = 0 is y on x(a)

5• The two regression lines are:• 16x – 20y + 132 =0 and 80x -30y – 428 = 0, the value of correlation coefficient is• (a) 0.6 (b) -0.6 (c) 0.54 (d) 0.45

16x – 20y + 132 =0 80x -30y – 428 = 016 : 20 30 : 80

r =1620 ∗

3080 = 0.54

(c)

6• Two regression equations are x+y=6 and x+2y=10, then correlation coefficient between x and y

is •• (a) -1/2 (b) ½ (c)-1/√2 (d) 1/√2

x+y = 6 x+2y = 101 : 1 2 : 1

ퟏퟏ ∗

ퟐퟏ = ퟐ

Correct r =-1/ ퟐ(c)

7• Two variables X and Y are related as 4x + 3y = 7 then correlation between x and y is ______•• (a) Perfect positive (b) Perfect negative (c) Zero (d) None of these• (b)

8• Correlation coefficient between x and y is 1, then correlation coefficient between x-2 and (-y/2)

+1 is•• (a) 1 (b) -1 (c) -1/2 (d) ½

9• If the sum of the product of deviations of x and y series from their means is zero, then the

coefficient of correlation will be•• (a) 1 (b) -1 (c) 0 (d) None of these

(c)

10• If 2 variables are uncorrelated, their regression lines are:•• (a) Parallel (b) Perpendicular (c) Coincident (d) Inclined at 45 degrees

• (b)

11• The covariance between two variables is•• (a) Strictly positive (b) Strictly negative (c) Always 0 (d) Either positive or negative or

zero

• (d)

12• What are the limits of the correlation coefficient?•• (a) No limit (b) –1 and 1 (c) 0 and 1, including the limits (d) –1 and 1, including the limits

• (d)

13If x, y denotes the arithmetic means, σx, σy denotes the standard deviations. bxy, byx denote the regressioncoefficients of the variables ‘x’ and ‘y’ respectively, then the point of intersection of regression line x and y &y on x is _____•• (a) (`x,` y) (b) (σx, σy) (c) (bxy, byx) (d) (σx

2, σy2)

• (a)•

14• If the value of correlation coefficient is positive, then the points in a scatter diagram tend to

cluster•• (a) From lower left corner to upper right corner (b) From lower left corner to lower right

corner• (c) From lower right corner to upper left corner (d) From lower right corner to upper right

corner

• (a)

15• When we are not concerned with the magnitude of the two variables under discussion, we

consider•• (a) Rank correlation coefficient (b) Product moment correlation coefficient• (c) Coefficient of concurrent deviation (d) (a) or (b) but not (c).

• (c)

16• For 10 pairs of observations, No. of concurrent deviations was found to be 4. What is the value

of the coefficient of concurrent deviation?•• (a) √0.2 (b) √– 0.2 (c) 1/3 (d) –1/3

√ [ 2c – n ] = √ [ 2*4 – 9 ] = √ [ 8 – 9] = √ [ - 1]n 9 9 9

-1/3

• (d)

17• 푥̅ =16 휎 = 4.8, 푦 = 0.2 휎 = 9.6 r = 0.6 then regression coefficient of x on y•• (a) 0.03 (b) 0.3 (c) 0.2 (d) 0.05

푏 =푟휎 휎

푏 =0.6(4.8)

(9.6) = 0.3

(b)

18• bxy & byx are zero then r =•• (a) 1 (b) 0 (c) -1 (d) none

• (b)

Chapter 13 :Probability

Probability and expected value

Probability

FormulasType of problems

n(favorable events)n(Total Events)

Odds in favour m:nm

m+n

Odds against m:nn

m+n

P(A U B) = P(A) + P(B) – P(A∩B)

P(A U B) = 1 – P(A’ ∩ B`)

P(A)+P(A’) = 1

Conditional probabilityP(S/M) = P(S∩M)

P(M)

Independent events P(A∩B) = P(A)*P(B)

Coin = 2n

Die = 6n

Balls problemsApply combination if selection > 1

Cards

26 – red -13 diamond, 13 heart26 – black -13 club, 13 spaded

Mathematical expectation

Mean = μ = E(x) =Σ pi x

Σ pi = 1 always

Variance = V(x) = σ2

= E(x2) – E(x)2

E(x + y) = E(x) + E(y) E(xy) = E(x) E(y) E(k. x) = k.E(x)

1• If P (A/B) = P (A) then A & B are•• (a) mutually exclusive (b) dependent (c) independent (d) composite•• (c)

2• If odds in favor of A is 5:7 and odds against B is 9:6 of solving a problem then probability that a

problem can be solved is ____•• (a) 147/180 (b)122/180 (c) 117/180 (d) none

• P(at least one solves) = 1 – P(Both don’t solve)• P(AUB) = 1 – P(A`∩ B`)

= 1 – (7/12)*(9/15)= 1 – 0.35= 0 .65

147/180 = 0.816122/180 = 0.67117/180 = 0.65(c)

3• If P(A ∩ B) = 0, then the two events A and B are•• (a) Mutually exclusive (b) Exhaustive (c) Equally likely (d)

Independent

• (a)

4• A letter is taken out at random from the word RANGE and another is taken out from the word

PAGE. The Probability that they are the same letter is:•• (a) 1/20 (b) 3/20 (c) 3/5 (d) ¾

RANGE PAGEA 1/5 and A 1/4 = 1/20

Or G 1/5 and G 1/4 = 1/20Or E 1/5 and E 1/4 = 1/20

3/20(b)

5• If P(A) = 5/9, then the odds against the event A is•• (a) 5 : 9 (b) 5 : 4 (c) 4 : 5 (d) 5 : 14

• 5 favourable• 9 total• 4 unfavourable

Odds againstunfavourable : favourable4 : 5

(c)

6• A player tosses 3 coins. He wins Rs. 5 if 3 heads appear, Rs. 3 if two heads appear, Rs. 1 if one

head appear and a loss of Rs. 15 if no head appear. Find his expected gain in Rs.•• (a) 0.5 (b) 0.25 (c) 0.2 (d) None of these•

No of Heads3210

Amount(Rs)531-15

P(x)1/83/83/81/8

x*p(x)5/89/83/8-15/82/80.25(c)

7• There are two boxes containing 5 white and 6 blue balls and 3 white and 7 blue balls

respectively. If one of the boxes is selected at random and a ball is drawn from it, then the probability that the ball is blue is

•• (a) 115/227 (b) 83/250 (c) 137/220 (d) 127/250

first box is selected or Second box is selectedand drawn is blue and drawn is blue1/2 + 1/2* 6/11 *7/106/22 + 7/20

60+77220137/220(c)

8• A random variable X has the following probability distribution:• X : -2 3 1• P(X=x) : 1/3 ½ 1/6 Find E(X2) and E(2x+5)•• 6 and 7 respectively (b) 5 and 7 respectively (c) 7 and 5 respectively (d) 7 and 6 respectively

X-231

p2/63/61/6

x2

491

p*x2

8/627/61/636/6=6E(x2)(a)

2x+51117

2x+5*p2/633/67/642/6=7E(2x+5)

9• The probability distribution of a random variable is as follows:• x : 1 2 4 6 8• P : k 2k 3k 3k k• The variance of x is• (a) 2.1 (b) 4.41 (c) 2.32 (d) 2.47

x12468

pK2K3K3KK1

k + 2k + 3k +3k +k = 110k = 1k = 0.1

p0.10.20.30.30.11

px0.10.41.21.80.84.3E(x)

x2

14163664

x2P0.10.84.810.86.422.9E(x2)

V(x) = E(x2) – E(x)2

= 22.9 – 4.32

= 22.9 – 18.49V(x) = 4.41(b)

10• E (13x +9) = ___•• (a) 13x (b) 13E(x) (c) 13E(x) + 9 (d) 9

• (c)

11• If two random variables x and y are related as y = –3x + 4 and standard deviation of x is 2, then

the standard deviation of y is•• (a) – 6 (b) 6 (c) 18 (d) 3.50

• y = –3x + 4• SD of y = -3(SD of x) + 4

= - 3*2 = - 6= 6

(b)

12• Probability of the sample space is•• (a) 0 (b) ½ (c) 1 (d) None of these

• (c)

13• In a class 40 % students read Mathematics, 25 % Biology and 15 % both Mathematics and

Biology. One student is select at random. The probability that he reads Mathematics if it is known that he reads Biology is

•• (a) 2/5 (b) 3/5 (c) 4/5 (d) None of these•• Reads biology, out of 25• Read mathematics also , 15• Probability 15/25• 3/5• (b)

14• An urn contains 3white, 5 black and another urn contain 4 white, 2 black. One ball is drawn from

each of the bag find the probability of getting both white balls•• (a) 1/4 (b) 1/2 (c) 213/420 (d) 215/415

3 * 48 6

124814

(a)

15• If P (A) = 0.45, P (B) = 0.35, P (A and B) = 0.25 then P(A /B) =•• (a) 1.4 (b) 1.8 (c) 0.714 (d) 0.556•P(A/B) = P(A and B)

P(B)= 0.25/0.35= 0.714(c)

16• Two coins are tossed simultaneously then the probability of getting exactly one head is•• (a) 3/4 (b) 2/3 (c) 1/4 (d) ½

(d)

Chapter 14 :Theoretical distributions

Theoretical distributions

Binomial Poisson Normal

Fixed number of trials, represented as n.

Each trial has two possible outcomes, a “success” and a “failure”.

p ( x ) = nCx px qn-x

n : number of trials,P : Probability of success , q = 1-pX = 0,1,2,….n

Mean = np

Variance = npq

Distribution is symmetrical if p = 0.5

Large number of trials n ∞ and p 0

P(x) = e - µ µx

x!

for x = 0,1,2, ….n

Mean = Variance = µ

µ = n.p (from binomial distribution)

e = 2.7138

Normal curve is bell shaped has one peakMean = Median = Mode

Curve is symmetric in nature

The total area of the normal curve. Is one

The area between – ∞ to µ = the area between µ to ∞ = 0.5

QD =

MD =

Q 1 = Mean – 0.6745 SDQ3 = Mean + 0.6745 SDQ2 = Mean = Median = Mode

Calculating probability

Step 1 convert to Standard normal variate, (x – Mean)/S.D

Step 2 Identify area under curveφ (2) = 0.9772. It implies that area under the curve from -∞ to 2 is 0.9772

68 + 95 + 99.73 Rule

1• Mean of a binomial distribution is 6 and variance is 2. find P•• (a) 2/3 (b) 1/3 (c) 1 (d) none

np = 6npq = 2Dividing the equationsnpq = 2np 6q = 1/3p = 1 – 1/3 = 2/3(a)

2• In the normal distribution P (μ -3 σ < x < μ + 3 σ ) is equal to•• (a) 0.9973 (b) 0.9546 (c) 0.9899 (d) 0.9788•• 68+95+99.73 rule

• (a)

3• Wages paid to workers follows –•• (a) Binominal distribution (b) Poisson distribution (c) Normal (d) Chi-Square.

• (c)

4• In a Binomial Distribution, if mean is k-times the variance, then the value of ‘k’ will be __•• (a)p (b) 1/p (c) 1-p (d) 1/1-p

Mean = k*variancenp = k*npqk = 1/qk = 1/1-p(d)

5• Standard Deviation of binominal distribution is-•• (a) npq (b) (npq)2 (c)Square root of npq (d) n2p2q2

• (c)

6• In the Binomial distribution the parameters are n and p, then X assumes value:•• (a) Between 0 and n (b) Between o and n both inclusive (c) Between 0 and 1 (d) Between o and

∞

• (b)

7• In ______________ distribution, Mean = Variance:•• (a) Binomial (b) Poisson (c) Normal (d) None of these

• (b)

8• Under normal curve: μ +3s covers ________ of the area of items•• (a) 100% (b) 99% (c) 99.73% (d) 99.37%

• (c)

9• If 5% of the families in Kolkata do not use gas as a fuel, what will be the probability of selecting

10 families in random samples of 100 families who do not use gas as fuel? [Given: e-

5 = 0.0067]•• (a)0.038 (b) 0.028 (c) 0.048 (d) 0.018•• Poisson distribution• P(x) = e- µ µx

x!• P(10) = e- 5 510

10!= 0.0067*9765,625

36,28,800= 0.018(d)

10• If the mean of a poisson variable x is 1, what is P (x = at least one)?•• (a)0.456 (b) 0.821 (c) 0.632 (d) 0.254•

P(x) = e- µ µx

x! P(x≥1)= 1 – P(x=0) 1- e- 1 10

0!1 – 1

e1 – 1

2.71831 – 0.36790.632(c)

11• In a sample of 800 students, the mean weight and standard deviation of weight are found to be

50 kg and 20 kg respectively. On the assumption of normality, what is the number of students weighing between 46 Kg and 62 Kg? [Given area of the standard normal curve between z = 0 to z = 0.20 = 0.0793 and area between z = 0 to z = 0.60 = 0.2257.]

• (a) 250 (b) 244 (c) 240 (d) 260Z = X – 50

20P(46 < X < 62)P( 46 – 50 < X – 50 < 62 – 50)

20 20 20P(-0.2 < Z < 0.6)

0-∞ ∞-0.2 0.6

0.0793 0.2257

0.0793+0.2257=0.305*800Students= 244(b)

12• If the inflexion points of a Normal Distribution are 6 and 14. Find its Standard Deviation?•• (a) 4 (b) 6 (c) 10 (d) 12•• Inflexion points are Mean – SD and Mean + SD

Mean – SD = 6Mean + SD = 14

Subtract - - --2SD = -8SD = 4(a)

13Let the distribution function of a random variable X be F(x) = P(X x) then F(5)-F(2)

•

• (a) P(2 x<5) (b) P(2<x 5) (c) P(2 x 5) (d) P(2<x<5)

• F(5) – F(2)

P(X5 { 0,1,2,3,4,5}

P(X2 { 0,1,2}

P(X5 P(X2 { 3,4,5}

P(2<x 5)

(b)

14• 5000 students were appeared in an examination. The mean of marks was 39.5 with standard

deviation 12.5 marks. Assuming the distribution to be normal, find the number of students recorded more than 60% marks. [Given when Z = 1.64 area of normal curve = 0.4494]

•• (a) 1000 (b) 505 (c) 252 (d) 2227

Z = X – 39.512.5

P(X > 60)P( X – 39.5 > 60 – 39.5)

12.5 12.5P(Z > 1.64)

0-∞ ∞1.64

0.4494

?

0.5

0.5-0.4494= 0.0506

0.5-0.4494= 0.0506*5000= 252(c)

15• In a certain poisson frequency distribution, the probability corresponding to two successes is half

of the probability corresponding to three successes. The mean of the distribution is•• (a) 6 (b) 12 (c) 3 (d) 2.95• P(x) = e- µ µx

x!P(2) = ½ P(3)e-µ µ2 = 1 e-µ µ3

2! 2 3!3! = µ3

µ2

µ = 6

(a)

16• Which of the following is false in case of normal distribution.•• (a) It is multi model (b) mean = median = mode (c) It is symmetric (d) Total area is 1•• (a)

17• In a normal distribution Q.D is 6 , then S.D is•• (a) 4 (b) 9 (c) 7 (d) 6•• QD = 0.6745SD• 6 = 0.6745SD• SD = 6/0.6745• SD = 9• (b)

18• If a random variable x follows poission distribution such that E(x2) = 30, then the variance of the

distribution is•• (a) 7 (b) 5 (c) 30 (d) 20

• Variance = E(x2) – [E(x)]2

• v(x) = 30 – [v(x)]2

• (A) 7 = 30 – 72 , 7 ≠ 30 – 49 LHS ≠ RHS • (B) 5 = 30 – 52 , 5 ≠ 30 – 25 LHS = RHS

• (b)

19• For Normal Distribution•• (a) 1st and 2nd Quartiles are equidistant from median • (b) 2nd and 3rd Quartiles are equidistant from median• (c) 1st and 3rd Quartiles are equidistant from median • (d) None of these

• (c)

20• If parameters of a binomial distribution are n and p then this distribution tends to a poission

distribution when•• (a) n→∞,p→0 (b) p→0,np = λ • (c) n→∞,np = λ (d) n →∞, p →0,np = λ where λ is a finite constant

• (d)

Chapter 15 :Sampling theory

Sampling theory

Objective to understand population

Standard errorS.D of sampling distribution Estimation

Types of sampling

StatisticalMeasures

Samplesize

Random (or Probability) Sampling

Purposive or Judgment Sampling

Stratified Sampling

Systematic Sampling

For Population- Parameter

For sampling- Statistic

With replacement = Nn

Without replacement = Ncn

For Mean

For Proportion

With repetition∂/√n,

Without repetition∂ √ (N- n )√n √ (N – 1)

With repetition√(PQ/n )

Without repetition√ PQ√ (N- n )

n √ (N – 1)

point Interval

For MeanMean + (Standard error * table vales)

For ProportionPropn + (Standard error * table vales)

95% confidence level -table value is 1.96

-99% confidence level - table value is 2.58

1• The selection procedure of sample having no involvement of probability is known as•• (a) Purpose sampling (b) judgment sampling (c) subjective sampling (d) all of the above

• (d)

2• The hypothesis which is tested for possible rejection under the assumption that it is true is

known as ___•• (a) biased hypothesis (b) alternative hypothesis (c) null hypothesis (d) none of

these•

• (c)

3• The statistical measure computed from the sample observation alone have been termed as ____•• (a) estimate (b) parameter (c) both (d) none•• (a)• Also called as Statistic

4• It is a measure of precision achieved by sampling. •• (a) Standard error (b) Sampling distribution (c) Sampling Fluctuation (d) Expectation•• (a)

5• If a random sample of size 2 with replacement is taken from a population containing the units 1,

3 and 5 then the sample space would be:•• (a) (1,3) (1,5) (3,5)• (b) (1,1) (3,3) (5,5)• (c) (1,3) (1,5) (3,5), (3,1) (5,1) (5,3), (1,1) (3,3) (5,5)• (d) (1,3) (3,1) (1,5) (5,1) (3,5) (1,1) (3,3) (5,5)•• (c)

6• Which sampling provides separate estimates for population means for different segments and

also an overall estimate?•• (a) Multistage sampling (b) Stratified sampling• (c) Simple random sampling (d) Systematic sampling• (b)

7• If the Expected value of an estimator is equal to the value of the parameter. Then the estimator

is•• (a) Biased (b) Unbiased (c) Both (a) & (b) (d) Neither (a) nor (b)

• (b)

8• Method used to test the human blood is called in statistical terminology ______•• (a) Census investigation (b) Blood investigation (c) Sample investigation (d) None

9• A parameter may be defined as a characteristic of a population based on ____•• (a) Sample units (b) All the units (c) Few units (d) Any of the above

• (b)•• I

10• If it is known that the 95% LCL and UCL to population mean are 48.04 and 51.96 respectively. If

the sample size is 100, what will be the value of population SD?•• (a) 8 (b) 10 (c) 12 (d) 12.5

At 95% UCL =Mean +1.96SE = 51.96LCL = Mean -1.96SE = 48.04

Subtract - + -3.92SE = 3.92

SE = 1SE = SD/√n1 = SD/ √100SD = 10(b)

11• Which of the following is non-probability sampling?•• (a) Systematic sampling (b) Quota sampling (c) Cluster sampling (d) Stratified

sampling•

• (b)

12• Standard deviation of a sampling distribution is known as:•• (a) Standard Error (b) Sampling Error (c) Probable Error (d) Mean Deviation•• (a)

1399% confidence limits of population mean are•• (a) `x ±1.96SE (b) `x ±2.58SE (c) `x ±3 SE (d) None•

• (b)

14• A sampling technique providing separate estimates for population means for different segments

and also an overall estimate is _____•• (a) Stratified sampling (b) Random sampling (c) Systematic sampling (d) None of these•• (a)

Chapter 16 :Index numbers

Index numbers

Meaning MeasurementBase year is always 100

TestingApplication

specialized averag

UnweightedWeighted

Simple aggregate

Simple relative

Laspeyres Index

Paasche’s Index

Fisher's Index

Dorbish-Bowley’s

Unit test

Time reversal test

Factor reversal set

Circular test

Chain index = LR of CY * Chain of PY

100

LR =Current year price *100Previous year price

Deflating time series Real Value = Current year price

Index number

Shifting And Splicing Of Index Numbers

Shifted price = Original price index*100new base year index

1• P10 is the index for time:•• (a) 0 on 1 (b) 1 on 0 (c) 1 on 1 (d) 0 on 0•• (a)

2• Laspayer’s = 90, Paasche’s = 160 then Fisher’s Index number is =•• (a) 120 (b) 340 (c) 360 (d) 400

• 퐹 = 퐿 ∗ 푃• F= 90 ∗ 160• =120• (a)

3• Consumer Price index number for the year 1957 was 313 with 1940 as the base year. The

Average Monthly wages in 1957 of the workers in to factory be Rs. 160/- their real wages is: •• (a) Rs. 48.40 (b) Rs. 51.12 (c) Rs. 40.30 (d) None of these•

Year 1957 1940Index 313 100Wages 160 ?

100*160/31351.12(b)

4• _____________ play a very important role in the construction of index numbers:•• (a) Weights (b) Classes (c) Estimate (d) None of these

• (a)

5• Factor reversal test is:•

• (a) ∑∑ (b)∑∑ (c)∑∑ ∗ ∑∑ (d)None of these

•• (a)

6• If with a rise of 10% in prices the wages are increased by 20% the real wage increases by:•• (a) 10% (b) More than 10% (c) 20% (d) Less than 10%

Index number 100 110Wages 100 120Real wages 120*100

110109.09

Real increase 9.09%(d)

7

pn q0 =1180,p0 q0 =1170,pn qn =1064, p0 qn =1100 then Fisher ideal index number is

•• (a) 96.73 (b) 98.795 (c ) 98.77 (d) 100.86

퐹 =∑푃 푄∑푃 푄

∗∑푃 푄∑푃 푄

∗ 100

퐹 =11801170

∗10641100

∗ 100

= 98.77(c)

8• When the prices are decreased by 30% then the index number is now•• (a) 50 (b) 60 (c) 70 (d) 30

• (c)

9• Circular test is satisfied by which index number?•• (a) Laspayere’s (b) Paasahe’s (c) Fisher’s (d) None of the above

• (d)

10• Fisher’s Index number is _____ of Laspayere’s and Paasehe’s Index numbers•• (a) A.M (b) G.M (c) H.M (d) None

• (b)

11• Which of the following statement is true?•• (a) Paasche’s index number is based on base year quantity• (b) Fisher’s index satisfies the circular test• (c) Arithmetic mean is the most appropriate average for constructing the index number• (d) Splicing means constructing one continuous series from two different indices on the basis of

common base• (d)

12• Monthly salary of on employee was Rs.10,000 in the year 2000 and it was increase to Rs.20,000

in the year 2013 while the consumer price index number is 240 in year 2013 with the base year 2000, what should be his salary in comparison of consumer price index in the year 2013?

•• (a) 2,000 (b) 16,000 (c) 24,000 (d) None

• Year 2000 2013• Salary 10000 20000• IN 100 240

240*10000/10024000

(c)

13• What is the formula for calculating the deflated index :•• (a) Current Value/Index of current year *100 • (b) Current Value/Index of last year *100• (c) Current Value/Index of current year • (d) Current Value/Index of last year • (a)

Chapter 1A : Ratios and proportions

Ratios and proportions

Ratios Proportions

Meaning Types

Different form of expression of a fraction

Ratio a: bFraction a/bDecimalPercent

Ratio is 2 : 3Number is 2x &3x

If a : b and c:d are two numbers

• Compound ratio: ac : bd.

• Duplicate ratio: a2 : b2

• Triplicate ratio : a3 : b3

• Sub-Duplicate ratio: a1/2 : b1/2

• Sub-Triplicate ratio : a1/3 : b1/3

• Inverse ratio : b : a

Meaning Types

Equality of ratios

a : b = c : d

Cross product rulead = bc

• Fourth proportionala : b = c : x

• Third proportionala : b = b : x

• Mean proportionala : x = x : b

1A person has assets worth Rs.1,48,200. He wish to divide it amongst his wife, son and daughter in the ratio 3:2:1 respectively. From this assets the share of his son will be

(a) Rs.74,100 (b) Rs.37,050 (c) Rs.49,400 (d) Rs.24,700

Wife Son Daughter Total3 2 1 6

? 148200148200*2/6

49400(c)

2• The ratio of numbers is 1:2:3 and sum of their squares is 504 then the numbers are•• (a) 6,12,18 (b) 3,6,9 (c) 4,8,12 (d) 5,10,15Ratio

1 : 2 : 3 1 : 2 : 3 1 : 2 : 3 1 : 2 : 3Sum of squares

62 + 122 + 182 32 + 62 + 92 42+82+122 52+102+152

504 126 224 350(a)

3• If one type of rice of cost Rs. 13.84 is mixed with another type of rice of cost Rs. 15.54. the

mixture is sold at Rs. 17.6 with a profit of 14.6% on selling price then in which proportion the two types of rice mixed?

• (a) 3 : 7 (b) 5 : 7 (c) 7 : 9 (d) 9 : 1• Solution

Cost + Profit = Sales85.4 14.6 100? 17.685.4*17.610015.0304Average cost should be 15.03

Type 1cost13.84

Type 2cost15.54

Average cost15.03

15.03-13.84=1.19

15.54-15.030.51

Ratio of 0.51 : 1.19 0.51/1.19 = 0.428573/7 = 0.428573 : 7(a)

4• The first, second and third month salaries of a person are in the ratio 2:4:5. The difference

between the product of the salaries of first 2 months & last 2 months is 4,80,00,000. Find the salary of the second month

•• (a) Rs.4,000 (b) Rs.6,000 (c) Rs.12,000 (d) Rs. 8,000

• Let salaries be 2x,4x,5x(5x*4x) -(2x)*(4x) = 480,00,00020x2 – 8x2 = 480,00,00012x2 = 480,00,000x2 = 40,00,000x = 2000

Second month salary, 4x4*20008000(d)

5• In a mixture 60 litres, the ratio of milk and water is 2:1. If the this ratio is to be 1:2, then the

quantity of water to be further added is:•• (a) 20 (b) 30 (c) 40 (d) 60

Old Milk Water Total

Ratio 2 1 3Litre ? ? 60

60*2/3 60*1/340 20

New

Ratio 1 2Litre 40 ?

40*2/180 To be added = 80 – 20 = 60

(d)

6• Anand earns Rs. 80 in 7 hours and Promod Rs. 90 in 12 hours. The ratio of their earnings is•• (a) 32: 21 (b) 23: 12 (c) 8: 9 (d) None of these•• Ratio of earning per hour

80/7 : 90/1211.43 : 7.5

11.43/7.51.524

Try all options32 : 21 23 : 12 8 : 932/21 23/12 8/91.524 1.916 0.88(a)

7• The ratio compounded of 4 : 9, the duplicate ratio of 3 : 4, the triplicate ratio of 2 : 3 and 9 : 7 is•• (a) 2 : 7 (b) 7 : 2 (c) 2 : 21 (d) None of these•

4 : 9Duplicate of 3 : 4 9:16Triplicate of 2 : 3 8 : 27

9 : 7compounded 2592 : 27216

0.095Try all options

2 : 7 7 : 2 2 : 210.285 3.5 0.095

(c)

8• If P is 25% less than Q and R is 20% higher than Q the Ratio of R and P•• (a) 5:8 (b) 8:5 (c) 5:3 (d) 3:5

P Q75 100

Q R100 120

R : P120 : 758 : 5(b)

9• If A : B = 3 : 2 and B : C = 3 : 5, then A:B:C is•• (a) 9 : 6 : 10 (b) 6 : 9 : 10 (c) 10 : 9 : 6 (d) None of these

A : B B : C 3 : 2 3 : 5*3 *3 *2 *29 : 6 6 : 10

Continuous ratio9 : 6 : 10

(a)

10• Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should

these be mixed to get an alloy 15 times as heavy as water?•• (a) 3:2 (b) 2:3 (c) 4:3 (d) 3:4

Gold Copper19 9

Mix1515- 9 19 - 156 4

Ratio 6 : 43 : 2

(a)

11• The ratio of third proportional to 12 and 30 and the mean proportional between 9 and 25 is •• (a) 3:2 (b)2:3 (c) 5:1 (d) 1:5••

Third proportional to 12 and 3012 : 30 = 30 : xx = 30*30/12x = 75

Mean Proportional to 9 and 259 : x = x : 25x2 = 25*9x = 15

75 : 155 : 1(c)

Chapter 1B : Indices and logs

Indices and Logs

Indices Logs

Exponential form

an=a x a x a…….to n factors

base is ‘a andpower is ‘n’.

Log form

Loga x = n

base is ‘a andpower is ‘n’.

Conversion

Exponential to log Log to exponential

an =y

= na

Log y

Log x = na

an= y

Laws

Indices Logarithms

am X an =a m+n loga mn = loga m + loga n

am / an =a m-n loga m/n = loga m - loga n

(am)n = amn loga mn = n loga m

(a.b)n = an .bn (a.b)n = n log a +n log b

a –m = 1am

Log b a = 1 .Log b a

a1 = a Log a a = 1

a0 = 1 Log a 1 = 0

Special concepts in log• Change Of Base

Log b m = log a mlog a b

• Log base10 : Normal Logarithme : Natural logarithm(ln)

• Log with base as 10Characteristic : One less than the number of characters before decimal.Mantissa : Obtained from log table

– Note : `4.15 is different from -4.15

12• If px = q, qy = r, rz = p6 then the value of xyz is:•• (a) 0 (b) 1 (c) 3 (d) 6

px = q

rz = p6

Raising power to 1/z both sidesr = p6/z

qy = r

(px)y = p6/z

xy = 6/zXyz = 6(d)

13• If 25 150 = (25x)50, then x = •• (a) 35 (b) 54 (c) 52 (d) 5

• 25 150 = (25x)50

• 2550*25*100 = (25)50(x)50

• 25 100 = (x)50

• (25 2)50 = (x)50

x = 252

X = 54

(b)

14• If x = 3 + 2√2, then value of (√x - 1 ) is • √x • (a) 2 (b) 3 (c) 1 (d) 0

x = 3 + 2√2X = 5.828√x = 2.41421/ √x = 0.4142

(√x - 1 ) = 2.4142 – 0.4142√x

= 2(a)

15

• Value of x, if 푥 ∗ 푥 = 푥• (a) 3 (b) 4 (c) 2 (d) 6

푥 ∗ 푥 = 푥

푥 = 푥

푥 = 푥

= 푥

x = 4(b)

16• If x = 31/3+3-1/3, then 3x3 -9x is•• (a) 15 (b) 10 (c) 12 (d) None of these

32 + 110(b)

17• If x = 51/3+5-1/3, then 5x3 -15x is•• (a) 25 (b) 26 (c) 27 (d) None of these•52 + 126(b)

18• log2 8 is equal to•• (a) 2 (b) 8 (c) 3 (d) None of these•

• log2 8 = x• 2x = 8• 2x = 23

• X = 3• (c)

19• If log x = m + n, log y = m – n the log (10x/y2):•• (a) 1-m + 3n (b) m-1+3n (c) m+3n+1 (d) None of these

log (10x/y2):log 10 + log x - log y2

log 10 + Log x – 2 log y1 + m+n – 2(m-n)1+ m + n – 2m + 2n1 – m + 3n

(a)

20

• 퐼푓 푙표푔 2 = 푙표푔 2, 푡ℎ푒푛푥 =•• (a) 16 (b) 32 (c) 8 (d) 4

푙표푔 2 = 푙표푔 2

푙표푔 2푙표푔 푥

= 푙표푔 2

푙표푔 2푙표푔 푥

= 1

푙표푔 212 푙표푔 푥

= 1

푙표푔 212

= 1

4푙표푔 2 = 1

푙표푔 2 = 1

푥 = 2

푥 = 16(푎)

21• If log x + log y = log (x+y), y can be expressed as• (a) x–1 • (b) x • (c) x/x–1 • (d) none of these• Solution

log x + log y = log (x+y)

log x y = log (x+y)

xy = x+y

xy –y = xy(x –1) = x

y = x .(x –1)

22• 1 + 1 + 1 = • Log ab abc Log bc abc Logca abc•• (a) 1 (b) 0 (c) -1 (d) 2

• Log abc ab+ Log abc bc + Logabc ca

• Log abc ab*bc*ca

• Log abc (abc)2

• 2Log abc abc• 2• (d)

23• On solving the equation log3 [log2 (log3 t )]=1, the value of t is •• (a) 8 (b) 18 (c) 81 (d) 6561

• log3 [log2 (log3 t )]=1

log2 (log3 t ) = 31

(log3 t ) =2t =3

t = 6561(d)

24• The value of a log b/c. b log c/a c log a/b

•• (a) 0 (b) 1 (c) –1 (d) None of these

• a log b/c. b log c/a c log a/b

• a log b-log c. b log c-log a c log a-log b

a log b. b log c c log a

a log c. b log a c log b

In log , x log y = y log x

a log b. b log c c log a

c log a. a log b b log c

1(b)

Chapter 2 : Equations

Equations

Linear equations Quadratic equations Co-ordinate geometry

Application problems

Simultaneous equations

Substitute options in question for x and y

Solving x

Factorization

OR

Use formula

푥 =−푏 ± 푏 − 4푎푐

2푎

ax2+bx+c=0

Properties

Sum of roots = -b/a

Product of roots = c/a

x2-(sum)x + product =0

Nature of rootsb2-4ac

=0 >0 <0

Equal Real

Perfect square

Rational

Not Perfect square

irrational

Imaginary

Irrational roots occur in pairsIf 2+√3 is one root, 2-√3 will be the another root

Distance formula(푥 푥 ) + (푦 푦 ) units

Straight lineY = mx + c

m is slope C is y intercept

a point on SLSubstitute x and y

If LHS = RHSThen point is on SL

Slope = Y2 – Y1

X2 – X1

Parallel linem1 =m2is ax + by + k=0

Perpendicularm1 *m2 = -1is bx– ay +k =0

Point of intersection Solve simultaneous equation

Intercept form of equationx + y = 1a b

Y = 3 x + 53 is slope

3x - 4 y + 7 = 0Slope is 3/4

Line 2x + 2 - y = 0Point 3,8 is on linePoint 1,3 is not on line

1• If |x − 2 |+ |x – 3| = 7 then x =•• (a) 6 (b) -1 (c) 6 & -1 (d) None

• Substitute all options• (a) 6

|6 − 2 |+ |6 – 3| = 7 4 + 3 = 7

LHS = RHS• (b) -1

|-1 − 2 |+ |-1 – 3| = 7 3 + 4 = 7

LHS = RHS

(c)

2• Three persons Mr. Roy, Mr. Paul and Mr. Singh together have Rs. 51. Mr. Paul has Rs. 4 less

than Mr. Roy and Mr. Singh has got Rs. 5 less than Mr. Roy. What is the money each has?• (a) (Rs. 20, Rs. 16, Rs. 15) (b) (Rs. 15, Rs. 20, Rs. 16) • (c) (Rs. 25, Rs. 11, Rs. 15) (d) None of these.•• Mr. Roy, Mr. Paul and Mr. Singh

4 less 4 More

(a)

3• If x+5y=33 and (x+y)/(x-y) = 13/3 then x and y is •• (a) (4,8) (b) (8,5) (c) (4,6) (d) (16,4)• x+5y=33• Substitute by all options• (a) x = 4, y = 8 4 + 5(8) = 33

4 + 40 ≠ 33, LHS ≠ RHS • (b) x = 8, y = 5 8 + 5(5) = 33

33 = 33, LHS = RHS

• (c) x = 4, y = 6 4 + 5(6) = 334 + 30 ≠ 33, LHS ≠ RHS

• (d) x = 16, y = 4 16 + 5(4) = 3316 + 20 ≠ 33, LHS ≠ RHS

(b)

4• A railway half ticket costs half the full fare and the reservation charge is the same on half ticket

as on full ticket. One reserved first class ticket from Chennai to Trivandrum costs Rs. 216 and one full and one half reserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the reservation charge?

•• (a) Rs. 105 and Rs. 6 (b) Rs.216 and Rs. 12 (c) Rs. 210 and Rs. 12 (d)Rs. 210 and Rs. 6

• One reserved ticket costs Rs. 216• Try all options• 105 + 6 ≠ 216• 216 + 12 ≠216• 210+12 ≠216• 210 + 6 = 216• (d)

36x + 12 = 304

36x + 12 = 12036x = 108x = 3

5• In a school number of students in each section is 36. If 12 new students are added, then the

number of sections are increased by 4, and the number of students in each section becomes 30. The original number of sections at first is:

•• (a) 6 (b) 10 (c) 14 (d) 18Trying all options

Number of sections6101418

Students in each section 36Total if 12 new students6*36 + 12 = 22810*36 + 12 = 37214*36 + 12 = 51618*36 + 12 = 660

New section increased by410141822

Students per section228/10 = 22.8372/14 = 26.57516/18 =28.67660/22 = 30(d)

6• A number consisting of two digits is four times the sum of its digits and if 27 be added to it the

digits are reversed. The number is •• (a) 63 (b) 35 (c) 36 (d) 60• 27 added numbers reversed• Try all options• 63 + 27 = 90• 35 + 27 = 62• 36 + 27 = 63• 60 + 27 = 87• (c)

7• Mrs. B. invested Rs.30,000; part at 5%, and part at 8%. The total interest on the investment was

Rs.2,100. How much did she invest at each rate?•• (a) Rs 10000, 20000 (b) Rs 15,000, 20,000 (c) Rs 15000, 25000 (d) Rs 10,000, 30,000

• Investment 30,000• Try all options

• 10,000+20,000 =30,000• 15000 + 20000 ≠ 30000• 15000+ 25000 ≠ 30000 • 10000 + 30000 ≠ 30000• (a)

8• 3x–4y+70z = 0, 2x+3y–10z = 0, x+2y+3z = 13•• (a) (1, 3, 7) (b) (1, 7, 3) (c) (2, 4, 3) (d) (–10, 10, 1)•• 3x–4y+70z = 0• Try all options• 3(1)–4(3)+70(7) = 481≠ 0• 3(1)–4(7)+70(3) = 185≠ 0• 3(2)–4(4)+70(3) =200 ≠ 0• 3(-10)–4(10)+70(1) = 0• (d)

9• One root of x2 – 3x+ k = 0 is 2 then k is•• (a) -10 (b) 0 (c) 2 (d) 10• x2 – 3x+ k = 0 • X = 2• 22 – 3(2)+ k = 0 • 4 – 6 + k = 0• K = 2• (c)

10• If , be the roots of a quadratic equation if + = -2, = -3 Find quadratic equation: •• (a) x2 + 2x – 7 = 0 (b) x2 + 2x – 3 = 0 (c) x2 - 2x – 3 = 0 (d) x2 - 2x + 7 = 0

• X2 – Sum of roots (x) + Product of roots = 0• X2 – (-2)x + (-3) = 0• X2 + 2x – 3 = 0 • (b)

11• If the roots of the equation 2x2 + 8x – m3 = 0 are equal then value of m is•• (a) – 3 (b) – 1 (c) 1 (d) – 2

• 2x2 + 8x – m3 = 0• a =2, b=8, c= – m3

• Roots equal, b2 -4ac = 0• (8)2-4(2)(– m3) = 0• 64 + 8m3 = 0• 8m3 = -64• m3 =-8• m = -2• (d)

12• If α, β are roots of x2 + 7x +11 = 0 then the equation whose roots as (α + β)2 &(α − β)2 is _•• (a) x2 − 54x + 245 = 0 (b) x2 −14x + 49 = 0 (c) x2 − 24x +144 = 0 (d) x2 − 50x + 49 = 0•• x2 + 7x +11 = 0• a=1, b = 7, c = 11• Sum of roots = -b/a

α+ β = -7/1 = -7

• Product of roots = c/aα β = 11/1 = 11

roots (α + β)2 (α − β)2

(α + β)2 = (-7)2 = 49(α − β)2 = (α + β)2 – 4(α β)

= 49 – 4*11= 5

Sum of roots (α + β)2+ (α − β)2

49 + 5 = 54

X2 – Sum of roots (x) + Product of roots = 0

X2 – 54(x) + 245= 0(a)

Product of (α + β)2 (α − β)2

49*5245

13• α,β are the roots of the equation 2x2 + 3x + 7 = 0 . Then the value of αβ−1 + βα−1 is•• (a) 2 (b) 3/7 (c) 7/2 (d) -19/14• 2x2 + 3x +7 = 0• a=2, b = 3, c = 7• Sum of roots = -b/a

α+ β = -3/2 = -1.5

• Product of roots = c/aα β = 7/2 = 3.5

αβ−1 + βα−1

α + ββ α

α2 + β2

βα

(α + β)2-2αβαβ

(-1.5)2-2(3.5)3.5=-1.357= 19/14(d)

14• Roots of the cubic equation x3 – 7x + 6 = 0 are ____________:• (a) 1,2,3 (b) 1,-2,3 (c) 1,2,-3 (d) 1,-2,-3•• x3 – 7x + 6 = 0• Substitution method(a) 1 2 3

13 – 7(1) + 6 23 – 7(2) + 6 33 – 7(3) + 60 0 12 not correct option

(b) 3 not correct option(c) -3

-33 – 7(-3) + 60

(c)

15• The roots of equation y3 + y2 – y – 1 = 0•• (a) 1,1,-1 (b) -1,-1,1 (c) 1,1,1 (d) None

y3 + y2 – y – 1 = 0Y2 (y+1) – (y+1) = 0(y+1) (y2-1) = 0(y+1) (y+1) (y-1) = 0y = -1,-1,1(b)

15• If 3y = 2+3x, then slope of the line is•• (a) 3 (b)1 (c) 2 (d) None of these

• General form y = mx +c• 3y = 2+3x• y = (2/3)+(3/3)x• y = (2/3)+1x• m = 1• (b)

17• The equation of line joining the point (3, 5) to the point of intersection of the lines 4x + y –1 = 0

and 7x – 3y – 35 = 0 is• (a) 2x – y = 1 (b) 3x + 2y = 19 (c) 12x – y – 31 = 0 (d) None of these.Point of intersection of the lines 4x + y –1 = 0 and 7x – 3y – 35 = 0Solve simultaneous equation 4x+ y – 1 = 0 *3 12x +3y – 3 = 0

7x – 3y -35 = 0 *1 7x – 3y – 35 = 0add 19x – 38 = 0

x = 2x = 2 in 4x+y-1 = 0

4(2) + y – 1 = 0y = 1 – 8y = -7

Two points (3,5) (2.-7)Slope y2 – y1 = -7 – 5 = - 12 = 12 (c)

x2 – x1 2 – 3 -1

18• Equations 3x-4y+5=0, 9x-12y+5=0 •• (a) Intersect (b) Don’t intersect (c) Cant say (d) None of these•• 3x-4y+5=0, 9x-12y+5=0

• Slope ¾ slope 9/12• Slopes are equal• Lines are parallel• Lines don’t intersect• (b)••

19• If the equations kx + 2y = 5 , 3x + y = 1 has no solution then the value of k is•• (a) 5 (b) 2/3 (c) 6 (d) 3/2•• No solution means parallel lines• Slopes are equal

kx + 2y = 5 3x + y = 1slope k :2 slope 3 : 1

k = 32 1k = 6

(c)

20A manufacturer produces 80 T.V. sets at a cost Rs. 220000 and 125 T.V. sets at a cost of Rs. 287500. Assuming the cost curve to be linear find the equation of the line and then use it to estimate the cost of 95 sets.

• (a) 242500 (b) 1500 (c) 67500 (d) insufficient data

TV Sets 80 125Cost 220,000 287,500

45

67500

95

15

?67500*15/45=22500220,000 + 22500242500(a)

Chapter 3 : Inequalities

Inequalities

Solving inequalities Representing on graph

To check substitute for x and y

If LHS < RHS orIf LHS > LHS as case may be

Solution is satisfied

Single line Multiple line

Test Point method

Origin (0,0)

If inequality is satisfied then area towards origin

If inequality is not satisfied then area away origin

Line passes through origin (1,0)

If inequality is satisfied then area below line

If inequality is not satisfied then area above line

Usually,> is area away n from origin< is area towards origin

Examples

3x + 2y < 12 3x + 2y > 12

1• The solution of the inequality 8x+6 < 12x +14 is•• (a) (-2,2) (b) (-2,0) (c) (2, ∞) (d) (-2, ∞ )

8x +6< 12x +148x – 12x < 14 – 6-4x < 8-x < 2x> -2(d)

2• A firm is engaged in producing two products A and B. Each unit of product A requires 2 kg of raw material

and 4 labour hours for processing, where as each unit of B requires 3 kg of raw materials and 3 labour hours for the same type. Every week, the firm has an availability of 60 kg of raw material and 96 labour hours. This can be expressed as

•• (a) 2X1 + 3X2 < 60 (b) 2X1 + 3X2 > 60 (c) 2X1 + 3X2 > 60 (d) 2X1 + 3X2 < 60

4X1 + 3X2 < 96 4X1 + 3X2 > 96 4X1 + 3X2 < 96 4X1 + 3X2 > 96• (a)

3• As per companies act, a legal association can have at most 20 members (without being a

company). This can be expressed as•• (a) X ≥ 20 (b) X ≤ 20 (c) X = 20 (d) None of these

• (b)

4• The equation X = 15 when represented in a graph•• (a) is parallel to Y axis (b) is Perpendicular to X axis • (c) Both (a) & (b) (d) is parallel to X axis

• (c)

X = 15

5• A seller makes an offer of selling certain articles that can be described by the equation x=25-2y

where x is price per unit and y denotes the no. of units. The cost price of the article is Rs.10 per unit. The maximum quantity that can be offered in a single deal to avoid loss is

•• (a) 6 (b) 7 (c) 8 (d) 9•

Cost 10SP = 25 – 2y ≥ 10

25 – 10 ≥ 2y15 ≥ 2y7.5 ≥ yy ≤ 7.5(b)

6• By lines x + y = 6, 2x – y = 2, the common region shown is the diagram refers to•

• (a) x + y ≥ 6, 2x – y ≤ 2, x ≥ 0, y ≥ 0 • (b) x + y ≤ 6, 2x – y ≤ 2, x ≥ 0, y ≥ 0 • (c) x + y ≤ 6, 2x – y ≥ 2, x ≥ 0, y ≥ 0 • (d) None of these• (b)

Chapter 4 : Interest and annuities

Interest and annuities

Interest Annuities-Series of payments

SimpleCompound

Interest remains same every year

Interest on principal and interest till previous period

F = P (1+n.i)

Interest amount

I =P.n.i OR I = F -P

F = P (1+i)n

I = F - P

Compounded more than once a year

Effective interest rate

Alter the formula, F =

(1+ i )n – 1,n = number of conversion period in a year

Half yearly

Monthly Quarterly

P (1 + i )2n

2P (1+ i )12n

12P (1+ i )4n

4

Annuity regularEnd of each period

Annuity ImmediateBeginning of each period

F =A [ (1+i) n - 1 ]i

P= A [ (1+i) n - 1 i (1+i)n

F =A [ (1+i) n - 1 ] (1+i) i

P= A [ (1+i) n - 1 (1+i) i (1+i)n

Apply P incase of loan problems and purchase problems

Calc shortcut using GT/MRC

Special problems

Difference between SI and CI

CI – SI =

FCI- FSI

Reverse calculation for P

• Assume P as 100

• Find FCI – FSI

• Use cross multiplication method to calculate p

Doubling of money problems

Under CIn * i =72

Under SIn * i =100

Multiple interest rates

Under CIF = P (1 + i1)n1 (1+i2)n2 (1+i3)n3

Under SIF = P[1+(i1n1)+(i2n2) (i2n2)]

Depreciation

SLM - simple

WDV -compound

For both methodsP is original costn is number of yearsF is WDV after nth year

If n is useful life then F is scrap value

LEASE or BUY

1. PV of Lease2. Purchase price3. Opt whichever is less

Use calcshortcut for quick PV calculation

Remember

If type of interest not specified

Then compound interest

If it involves

Period payments, periodic investments, yearly savings

Annuity

Loan, Hire purchase, lease or borrow

Present value

Savings after retirement, sinking fund, amount accumulated, redemption of debentures

Future value

1• Investment required to yield an annual income of Rs.420 at the rate of 7% p.a by Simple interest

is•• (a) 6000 (b) 6420 (c) 5580 (d) 5000

• 420/7%• 6000• (a)

2• The S.I on a sum of money is 4/9 of the principal and the no. of years is equal to the rate of

interest per annum. Find the rate of interest per annum•• (a) 5% (b) 6% (c) 20/3% (d) 22/7%

SI = P T R100

4/9P = P*i*i100

i2 = 400/9i = 20/3(c)

3• A sum of money doubles itself in 10 years (Simple interest). The number of years it would triple

itself is•• (a) 25 years. (b) 15 years. (c) 20 years (d) None of these•

Today After 10 Years another 10 years100 200 300

Interest 100 100

(c)

4• No. of years a sum 4 times itself at 12% pa at simple interest:•• (a) 20 (b) 21 (c) 25 (d) 30

P = 100F = 400SI = F - P = 400 – 100

= 300SI = P*n*I300 = 100*n*0.12n = 300______

100*0.12= 25

(c)

5• In what time will a sum of money double itself at 6.25% p.a simple interest•• (a) 5 Yrs (b) 8 Yrs (c) 12yrs (d) 16Yrs•P = 100F = 200SI = F - P = 200 – 100

= 100SI = P*n*I100 = 100*n*0.0625n = 100______

100*0.0625= 16

(d)

6• A sum of 44,000 is divided into 3 parts such that the corresponding interest earned after 2 years,

3 years and 6 years may be equal at the rate of simple interest are 6% p.a., 8% p.a., & 6% p.a. respectively. Then the smallest part of the sum will be:

•• (a) Rs. 4,000 (b) Rs. 8,000 (c) Rs. 10,000 (d) Rs. 12,000

interest on x = interest on y = interest on zx*2*6% y*3*8% z*6%*60.12x = 0.24y = 0.36z =

Three parts assume to be x,y,zx+y+z = 44000

0.12x = 0.24yx = 0.24y/0.12x = 2y

0.24y = 0.36z0.24y = z0.36Z = 0.6667y

2y + y+ 0.6667y = 440003.6667y = 44000Y = 12000

Smallest part Z = 0.6667y= 0.6667*12000=8000(b)

7• A man invests half his capital at the rate of 10% per annum, 1/3 at 9%and the rest at 12%per

annum. The average rate of interest per semi-annum, will he gets, is•• (a) 9% (b) 4.5 % (c) 5.5% (d) 5%

capital Interest period Product1/2 10% 1/2 2.5%1/3 9% 1/2 1.5%

Rest 1 – 1 - 1 = 6 – 3 – 2 1/6 12% 1/2 1%2 3 6 5%

(d)

8• Find compound interest on a sum of Rs.2500 invested for 4 years at 12% rate of interest

compounded yearly•• (a) 1420 (b)1434 (c) 450 (d) None of these

2500 + 12% + 12% + 12% + 12%- 25001434(b)

9• On a certain sum rate of interest @ 10% p.a., S.I=Rs. 90 Term = 2 year, Find Compound interest

for the same:•• (a) 544.5 (b) 94.5 (c) 450 (d) 18

SI = P.n.i90 = P*2*0.1P = 90___

2*0.1P = 450

CI = 450 + 10% + 10% - 450= 94.5

10• Effective rate of 6% of compounded half yearly•• (a) 6.06% (b) 6.09% (c) 6.08% (d) 6.07%

100 + 3% + 3% - 1006.09(b)

11The cost of machine 1,25,000/- useful life is estimated 20 years and the rate of Depreciation of its cost is 10% p.a. its scrap value is ____

(a) 15102 (b) 15400 (c) 15300 (d) 15200

F = P(1-r)n

= 125000*(1 – 0.1)20

= 125000*0.920

= 15197= 15200

12• If P invested 10% simple interest & Q invest 5% compounded annually. After two years P & Q

interests are same then P=(a) (41/80)Q (b) (41/40)Q (c) (41/100)Q (d) None

SI = p*n*i CI = P(1+i)n - P= P*2*0.1 = Q(1+0.05)2 - Q

SI = 0.2P CI = Q1.1025 – QCI = 0.1025Q

SI = CI0.2P = 0.1025QP = 0.1025Q/0.2P = 0.5125QP =( 41/80) Q(a)

13• If the sum of money when compounded annually become 1140 in 2 years and 1710 in 3 years at

rate of interest•• (a) 30% (b) 40% (c) 50% (d) 60%

F2 = 1140 F3 = 1710Interest = 1710-1140

= 570Rate of interest = 570/1140

= 0.5or 50%(c)

14• The Partners A & B together lent Rs. 3903 at 4% p.a interest compounded annually. After a span

of 7 years, A gets the same amount as B gets after 9 years. The share of A in the sum of Rs.3903/- would have been

•• (a) Rs.1875 (b) Rs.2280 (c) Rs.2028 (d) Rs.2820

Trying all options

a)b)c)d)

A1875228020282820

B = 3903-A2028162318751083

A interest =A(1.04)7

2467300026693711

B interest =A(1.04)9

2886231026691541(c)

15• The difference between CI and SI on a certain sum of money for 2 years at 4% per annum is

Rs.1. The sum is•• (a) 625 (b) 630 (c) 640 (d) 635

Fci –Fsi = CI – SIP(1+0.04)2 – P(1+0.04*2) = 1P[ 1.042 – 1.08] = 1P[ 0.0016 ] = 1P = 1/0.0016P = 625(a)

Shortcut 1 / 4% / 4% = 625applicable only for two years

16• The difference between and C.I & S.I at 7% p.a for 2 years is Rs. 29.4. then principal is•• (a) Rs.5,000 (b) Rs.5,500 (c) Rs.6,000 (d) Rs.6,500

29.4 / 7% / 7%= 6000(c)

17• The present value of an annuity of Rs.1,000 made annually for 5 years at the rate of interest

14% compound annually is • (a) Rs.5610 (b) Rs.6610 (c) Rs.3433 (d) Rs.5160

• Present value of an annuity = A [(1+i)n - 1]i (1+i)n

= 1000 [(1+0.14)5 - 1]0.14 (1+0.14)5

= 7142.86 [ 1.145 – 1]1.145

= 7142.86*(1.9254 – 1)1.9254

= 3433(c)

Shortcut using GT1/1+ i= = ..no of years

A * GT = P

Shortcut using GT1/1.14= = = = =

1000 * GT = 3433

Shortcut using MRC1/1+ i=M+ =M+ ..no of years

A * MRC = P

Shortcut using MRC1/1.14 = M+ = M+= M+= M+= M1000 * MRC = 3433

18• The Future of an annuity of Rs.1,000 made annually for 5 years at the rate of interest 14%

compound annually is • (a) Rs.5610 (b) Rs.6610 (c) Rs.3433 (d) Rs.5160Future value of an annuity = A [(1+i)n - 1]

i= 1000 [(1+0.14)5 - 1]

0.14= 7142.86 [ 1.145 – 1]

= 7142.86*(1.9254 – 1)

= 6610(b)

Shortcut using GT1/1+ i= = ..no of years

A * GT = + i for number of years

Shortcut using GT1/1.14= = = = =

1000 * GT = 3433+14%+14%+14%+14%+14% 6610

Shortcut using MRC1/1+ i=M+ =M+ ..no of years

A * MRC = P+I for number of years

Shortcut using MRC1/1.14 = M+ = M+= M+= M+= M1000 * MRC = 3433 + 14%+14%+14%+14%+14%6610

19• Suppose your mom decides to gift you Rs. 10,000 every year starting from today for the next

sixteen years. You deposit this amount in a bank as and when you receive and get 8.5% per annum interest rate compounded annually. What is the present value of this money: Given that P (15, 0.085) = 8.304236

•• (a) 83042 (b) 90100 (c) 93042 (d) 10100

PV of annuity immediatePV = Annuity * annuity factor [ i%, n]

= 10,000 * AF[ 8.5%, 16 ]= 10,000 * [AF(8.5%, 15) + 1]= 10000 * [ 8.304236 + 1]= 93042

(c)

20• X limited is planning to purchase a machinery costing Rs 50000 with useful life of 5 years .

Company has two optionsOption A : Borrow Rs.50000 at interest rate of 10% and buy machinery and repay in 5 yearsOption B : Obtain the same on lease for 5 years with Rs 12000 per annum as lease rental

• The company should opt for • (a) Borrowing (b) Leasing (c) Can’t say (d) Either borrow or lease

Option ABorrow today 50,000

Option BLease 12000 p.a for 5 yearsEquivalent borrowing today = PV of annuity1/1.1 = = = = =GT * 1200045489Leasing is cheaper(b)

Chapter 5 : Permutations and combinations

P&C

Permutations

Meaning

Selection

and

arrangement

Problems on

Digits

Alphabets, vowels

Arrangement of people

Combinations

Meaning

Selection

Problems on

Selection of team

Committees

Straight lines, triangles

Permutations and combinations

Counting principle

Doing m and n = “m*n ”

Doing m or n= “m + n ”

Factorialn! = n(n-1)(n-2)…

8! = 8*7*6*5…

PermutationsSelection & arrangement

CombinationsONLY selection

Normal

Special

nPr = n!n-r!

8p3 = 8*7*610p2 = 10*9

Box method

Circular(n-1)!

Like elementsn!

r1! r2! r3!....

Always togetherEntity/Unit method

Never togetherGap method

With repitionnr

nCr = n!n-r!r!

10C2 = 10*9/(2*1)8C3 = 8*9*7(3*2*1)

nCr = nCn-r

nCr + nCr+1 = n+1cr

nC0 = 1

nCn = 1

Selecting one or more thingsnC1 + nC2 + nC3+…..nCn

2n-1

nPr = r! * nCr

Unit method

9 people 4 are together

1 Unit 5 Units 6 Units

6 P 6

4 together

4P 4Among single unit

Both in

6P6 * 4P4

Gap method

6 boys and 3 girls Boys stand in gap6 P 6

7 P 3

Both6P6 * 7P3

1• If there are 6 trees in a row, then no. of ways 5 students can be arrange between 6 trees to take

a photograph •• (a) 120 (b) 720 (c) 1440 (d) none

5P55*4*3*2*1120

(a)

2• How many different words can be formed with the letters of the word “LIBERTY”•• (a) 4050 (b) 5040 (c) 5400 (d) 4500•L I B E R T Y7P77*6*5*4*3*2*15040(b)

3• If 12 school teams are participating in a quiz contest, then the number of ways the first, second

and third positions may be won is•• (a) 1230 (b) 1320 c) 3210 (d) None of these

12P312*11*101320(b)

4• A question paper consist 10 questions, 6 in math and 4 in stats. Find out number of ways to

solve question paper if at least one question is to be attempted from each section.•• (a) 1024 (b) 950 (c) 945 (d) 1022

6 Math questions 4 stat questionsone or more one or more

26 – 1 * 24 – 164 – 1 * 16 -1

63 * 15945

(c)

5• In how many ways the word “ARTICLE” can be arranged in a row so that vowels occupy even

places?•• (a) 132 (b) 144 (c) 72 (d) 160•

A R T I C L Eodd even odd even odd even odd

Remaining letters R,T,C,L =4Remaining places 7 – 3 = 4Number of ways of arranging Remaining places = 4P4

Vowels : A, I, E = 3Even places = 3Number of ways of vowels occupying even places = 3P3

Total arrangements3P3*4P46 * 24144(b)

6• A man has 3 sons and 6 schools within his reach. How many ways can his sons go to school, if

no two of them are in same school.•• (a)6P2 (b) 6P3 (c) 63 (d) 36

1st Son * 2nd Son * 3rd Son• 6 schools * 5 schools * 4 schools• 6P3

• (b)

7• A student has 3 books on computer, 3 books on Economics, 5 on Commerce. If these books are

to be arranged subject wise then these can be placed on a shelf in the ____________ number of ways:

•• (a) 25,290 (b) 25,920 (c) 4,230 (d) 4,320

3 computer = 1 unit3 economics = 1 unit5 commerce = 1 unit

3 unitsunit arrangement 3!Internal arrangement 3! And 3! and 5!Total arrangements = 3!*3!*3!*5!

= 6*6*6*120= 25920

(b)

8• A family of 4 brothers and three sisters is to be arranged for a photograph in one row. In how

many ways can they be seated if no two sisters sit together?•• (a) 1440 (b) 2840 (c) 124 (d) 320• no two sisters together• Brother sit first with a gap between them. Sisters fill gaps

Brother arrangement = 4P4 = 24

Sister arrangement = 5P3 = 60

Total arrangement = 60*24= 1440

(d)

9• 5 persons are sitting in a round table in such way that Tallest Person is always on the right side

of the shortest person. The number of such arrangements is•• (a) 6 (b) 8 (c) 24 (d) None of these

Tallest and shortest together = 1 unitremaining 5 – 2 = 3 units

total 4 unitsunit arrangement (4-1)! =3! =6 waysInternal arrangement 1 ways( because order is fixed)

total arrangement 6*16(a)

10• The number of ways of arranging 6 boys and 4 girls in a row so that all 4 girls are together is:•• (a) 6!, 4! (b) 2(7!. 4!) (c) 7!. 4! (d) 2(6!.4!)•

4 girls together = 1 unit6 boys = 6 units

7 unitsunit arrangements 7!internal arrangement 4!total arrangements 7!*4!

(c)

11• In how many ways can six digit telephone numbers be formed by using 10 distinct digits?•• (a) 106 (b) 610 (c) 10C6 (d) 10P6

• (d)

12• 15 C3 +15C13 is equal to•• (a) 16C3 (b) 30C16 (c) 15C16 (d) 15c15

15C3 +15 C13 using nCr =nCn-r15C3 + 15C2 using nCr + nCr-1 = n+1 C r 15+1 C316C3(a)

13• If 15 C3r = 15Cr+3, then r = •• (a) 2 (b)3 (c)4 (d) 5

• 15C3r = 15Cr+3

• 15 C3r = 15Cr+3 using nCr = nCn-r• 15C15-3r= 15Cr+3

• 15 – 3r = r + 3• 15 –3 = r + 3r• 12 = 4r• r =12/4• r = 3• (b)

14• The no. of ways that the team of 11 players to be selected from 15 players in which one

particular player is excluded •• (a) 364 (b) 728 (c) 1001 (d) 1234•

Available = 15-1 = 14to be selected = 1114C1114C314 * 13 * 123 * 2 *1

364(a)

15• The number of ways in which a person can chose one or more of the four electrical appliances :

T.V, Refrigerator, Washing Machine and a cooler is•• (a) 15 (b) 25 (c) 24 (d) None of these•• Choosing one or more = 2n - 1• 24 – 1• 15• (a)

16• Every two persons shake hands with each other in a party and the total number of handshakes is

66. The number of guests in the party is•• (a) 11 (b) 12 (c) 13 (d) 14

nC2 = 66 n = ?Try all options

11C2 = 5512C2 = 66

(b)

17• In how many ways can a family consist of 3 children have different birthdays in a leap year•• (a) 366 x 365 x 364 (b) 366C3 (c) 365C3 (d) 365C3 – 3

• (a)

18• There are 12 points in a plane of which 5 are collinear. The number of triangles is•• (a) 200 (b) 211 (c) 210 (d) None of these

12C3 – 5C312*11*10 – 5 * 4 * 33 * 2 *1 3* 2 * 1220 - 10210(c)

Chapter 6 :Progressions

Progressions

Types of progressions Applying formulas

Special cases

AP GP

Difference between two consecutive terms are equal

Ratiobetween two consecutive terms are equal

4 terms a,b,c,d

b-a = c-b = d-c

4 terms a,b,c,d

b = c = da b c

To find position

To find a positional value

To find sum of terms upto a position valuen

TnSn

AP GP

Tn = a+(n-1)d

Tn = a*r(n-1)

APGP

Sn= n (a+l)

2

Sn = n[2a+(n-1)d

2

Sn = a [rn-1]

r - 1

S∞= a

1-r

Natural numbers

∑n = n(n+1)/2

∑n2 = n(n+1)(2n+1)

6

=∑n3 = [n(n+1)]2

2

Inserting means

Intermediate terms

Inserting 4 A.M. impliesThere are 6 terms in AP

Tn = Sn - Sn-1

1• An AP has 13 terms whose sum is 143. The third term is 5, then first term is•• (a) 4 (b) 7 (c) 9 (d) 2

S13 = 143 T3 = 5 a = ?

Sn = n [ 2a+(n-1)d] 2

143 = 13 [2a+(13-1)d]2

22 = 2a + 12da+6d = 11

Tn = a+(n-1)d5 = a+ (3-1)d5 = a+2d

a+2d = 5

a+6d = 11 *2a+2d = 5 *6

2a+12d = 226a+12d = 30

Subtract - - --4a = -8

a = 2(d)

2• In an A.P. if the sum of 4th & 12th terms is ‘8’ then sum of first 15 terms is _______•• (a) 60 (b) 120 (c) 110 (d) 150

T4 + T12 = 8a+3d + a+11d = 82a+ 14d = 8

Sn = n [ 2a+(n-1)d] 2

S15= 15 [2a+(15-1)d]2= 15 [ 2a+ 14d]

2= 15 * 8

2= 60

3• The two arithmetic means between –6 and 14 are• (a) 2/3,1/3 (b) 2/3, 7 1/3 (c) -2/3, -7 1/3 (d) None of these

T1 T2 T3 T4-6 Arithmetic means 14

a = -6d = Difference between nth term / difference between number of termsd = 14 –(-6) = 20

4 – 1 3 Series

-6 -6+ 20 2 + 20 143 3 3

- 6 2/3 22/3(b)

4• There are ‘n’ AMs between 7 & 71 and 5th AM is 27 then ‘n’ = ______• (a) 15 (b) 16 (c) 17 (d) 18

a = 75th AM = T6 = 27d = difference between Tn

difference between n= 27 – 7

6 – 1= 20

5= 4

n=

Last term – first term + 1d

71 – 7 + 14

16 + 117 terms-215 AM

5• A contractor who fails to complete a building in a certain specified time is compelled to forfeit Rs

200 for the first day of extra time required and thereafter forfeited amount is increased by Rs 25 for every day. If he loses 9,450, for how many days did he over-run the contract time?

(a) 19 days (b) 21 days (c) 23 days (d) 25 days

a = 200d = 25Sn = 9450n = ?

Sn = n [ 2a+(n-1)d] 2

9450 = n [2(200)+(n-1)25]2

18900 = n[ 400 + 25n – 25]18900 = 400n + 25n2 – 25n18900 = 375n + 25n2

756 = 15 n + n2

15 n + n2 = 756

Try all optionsn = 19n = 21n = 23n = 25

15 (19) + 192 = 64615 (21) + 212 = 756

(b)

6• The sum of n terms of an AP is 3n2 + 5n, which term of AP is 164.•• (a) 25 (b) 27 (c) 29 (d) 31

Sn = 3n2 + 5nS1 = 3(1)2 + 5(1) = 3 + 5 = 8S2= 3(2)2 + 5(2) = 12 + 10 = 22

T1 = 8T2 = 22 – 8 = 14d = 14 – 8 = 6

n = last term – first term + 1d

= 164 - 8 + 16

= 27(b)

7• Three No’s a,b,c are in A.P find a-b+ c•• (a) a (b) –b (c) b (d) c

Let AP be 1,2,3a-b+ c1 – 2 + 32

(c)

8• Find two numbers whose A.M is 10 and G.M. is 8•• (a) [10, 10] (b) [16, 4] (c) [18, 2] (d) [14, 6]

Try all options

(a)(b)(c)(d)

퐺푀 = 10 ∗ 10 = 10퐺푀 = 16 ∗ 4 = 8퐺푀 = 18 ∗ 2 = 6퐺푀 = 14 ∗ 6 = 9.1

(b)

9• Find the sum to infinity of the following series: 1 – 1 +1 – 1 + 1 – 1 +....... ∞•• (a) 1 (b) ∞ (c) 1/2 (d) Does not exist

Sum upto infinity = a/1-r= 1/1-(-1)= 1/2

(c)

10• If x = 1 + 1/3 + (1/3)2 + ….. ∞ and y = 1 + 1/4 + (1/4)2 + ….. ∞ , find xy•• (a) 2 (b) 1 (c) 8/9 (d) ½

x = 1 + 1/3 + (1/3)2 + ….. ∞= 1

1 – 1/3= 1

2/3x = 3/2

y = 1 + 1/4 + (1/4)2 + ….. ∞= 1

1 – 1/4= 1

3/4y = 4/3

X*y = 3/2 * 4/3= 2

(a)

11• Find the product of : (243), (243)1/6, (243)1/36, …….∞•• (a) 1,024 (b) 27 (c) 729 (d) 248

(243), (243)1/6, (243)1/36, …….∞2431 + 1/6 + 1/36 …

1 + 1/6 + 1/36 ….1

1-1/615/6 6/5

2436/5

(35)6/5

36

729(c)

12• 0.7 + 0.77 + 0.777 + ….. to n terms is given by•

• (a) 9n − 1 + 10 (b) 8n − 1 + 10

•

• (c) 4n − 1 + 10 (d) 19n − 1 + 10

• Substitute n = 1, it should give 0.7

• (a) 9(1) − 1 + 10 = 0.7 (b) 8(1) − 1 + 10 = 0.855

•

• (c) 4(1) − 1 + 10 = 0.27 (d) 19(1) − 1 + 10 = 1.56

• (a)

13• Three numbers are in A.P. and their sum is 15. If 8, 6, 4 be added to them respectively, the

numbers are in G.P. The numbers are•• (a) 2, 6, 7 (b) 4, 6, 5 (c) 3, 5, 7 (d) None of these

not in AP not in AP in AP+8 +6 +411,11,11

(c)

14• If the numbers x,y,z are in G.P then the numbers x2+y2, xy+yz, y2+z2 are in ___•• (a) A.P (b) G.P (c) H.P (d) None

Let x = 1, y = 2, z = 4x2+y2, xy+yz, y2+z2

12+22, 1*2+2*4, 22+42

5, 10, 20GP(b)

Chapter 7 :Sets relations and functions

Sets, relations and functions

Sets

Meaning

List of objects{ }

Types

Universal set

Subset 2N

Finite Set:

Null set {} ɸ

Singleton Set:

Equal setequivalent sets

Compliment set

Intersection

Formulas

N(AUB)=N(A) + N(B)– N(A ∩ B)

A – B = A ∩ Bc

( What is there only in Set A }

(A UB)’ = A’ ∩ B’

N (AUBUC ) = N(A) + N(B) + N(C)- N (A∩ B) - N(B∩C)- N (C ∩ A ) + N (A ∩ B ∩ C )

Relations

set of ordered pairs

A B

Domain is set of inputs

Range is set of outputs

co-domain is set of inputs mapped

TypesIf A = A, then it is Reflexive

If A = B and B=A then it is symmetric

If A= B and B= C then A= C then it is transitive

Functions

Meaning1)All elements of inputs are mapped

2)One input can’t have two outputs

Odd functionf(–x) =- f(x),

Even function f(–x) = f(x),

Inverse function1) Assume y = f(x)2) Now express x in terms of y3) Replace y with x 4) We get f-1(x)

Composite functionF(x) , g(x)

Composite = f[g(x)] =fog

x is set of inputs

f(x) is set of outputs

50 65

20Both

30Tea Only

45Coffee only

Tea Coffee

Either Tea or coffee50+65-20

95

1005

1• Find inverse of f(x) = 10x –7•• (a) 1/(10x-7) (b) 1/(10x+7) (c) (x+7)/10 (d) (x-7)/10

10x – 7 = y10x = y + 7x = y + 7

10f(-1)x = x + 7

10

(c)

2• If ƒ(x) = 2 + x , then ƒ-1(x) :

2 - x• (a) 2(X-1) (b) 2(X+1) (c) X+1 (d) X-1

X+1 X-1 X-1 X+1• 2 + x = y

2 – x2 + x = 2y – xyx + xy = 2y – 2

X(1+y) = 2(y-1)X = 2 (y-1)

(y+1)f-1x = 2 (x – 1)

x + 1 • (a)

3• If f (x) = x + 1, g(x)= x2 + 1, fog (-2) =•• (a) 6 (b) 5 (c) -2 (d) none.

f (x) = x + 1 g(x)= x2 + 1f[g(x)] = g(x) + 1fog = x2 + 1 + 1fog(-2) = (-2)2 + 2

= 4 + 2= 6

(a)

4• If f (x-1) = x2-4x+8 then f (x+1) =•• (a) x2+8 (b) x2+7 (c) x2+4 (d) x2-4x+4

f (x-1) = x2-4x+8x2 – 2x -2x + 7 + 1x2 – 2x + 1 – 2x + 7 + 2 - 2(x-1)2 – 2x + 2 + 7 – 2

f(x-1)= (x-1)2 – 2(x – 1) + 5f(x+1)= (x+1)2 – 2(x + 1) + 5

= x2 +2x + 1 – 2x – 2 + 5= x2 + 4

(c)

5• If A B then which of the following is true•• (a) A∩B = B (b) A∪B = B (c) A∩B = Ac (d) A∩B = φ

(b)

B

A

6• In a class of 80 students, 35% play only cricket, 45% only Tennis, How many play Cricket?•• (a) 86 (b) 54 (c) 36 (d) 44

Cricket Tennis

35% 45%100 – 35- 45= 20

35+20 = 55%

*80 = 44(d)

7• A = {2,3}, B={4,5}, C={5,6} then Ax(B∩C)•• (a) {(5,2), (5,3)} (b) {(2,5), (3,5)} (c) {(2,4), (5,3)} (d) {(3,5), (2,6)}

A = { 2, 3}(B∩C) = {5}Ax(B∩C) = {(2,5) (3,5)}

(b)

8

If f(x) = x/(x-1) then

(a) x/y (b)y/x (c) –x/y (d) –y/x•푓 푥 =

푥푥 − 1

푓푥푦 =

푥푦

푥푦 − 1

=

푥푦

푥 − 푦푦

= 푥

푥 − 푦

푓푦푥 =

푦푥

푦푥 − 1

= 푦푥

푦 − 푥푥

= 푦

푦 − 푥

푓 푥푦

푓 푦푥

=

푥푥 − 푦푦

푦 − 푥

= −푥푦

(푐)

9• In a class of 50 students 35 opted for Maths, 37 opted for commerce. The number of such

students who opted for both maths and commerce is•• (a) 13 (b) 15 (c) 22 (d) 28

n(A U B) = n(A) + n(B) – n(A∩B)50 = 35 + 37 - n(A∩B)n(A∩B) = 35 + 37 – 50

= 22(c)

10• If A={1,2,3} and B={4,6,7} then the relation R={(2,4) (3,6)} is•• (a) A function from A to B (b) A function from B to A• (c) Both (a) and (b) (d) Not a function• (d)

11• Of the 200 candidates who were interviewed for a position at call center, 100 had a two wheeler,

70 had a credit card and 140 had a mobile phone. 40 of them had both a two wheeler and a credit card, 30 had both a credit card and mobile phone, 60 had both a two wheeler and a mobile phone and 10 had all the three. How many candidates had none of them?

•• (a) 0 (b) 20 (c) 10 (d) 18

n(A U B U C) = n(A) + n(B)+ n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B ∩C)= 100 + 70 + 140 – 40 – 30 – 60 + 10= 190

n(A` ∩ B` ∩ C`) = n(U) – [n(A U B U C)]= 200 – 190= 10

(c)

12• In a class of 60 students, 40 students like Maths, 36 like Science, and 24 like both the subjects.

Find the number of students who like Science only•• (a) 12 (b) 16 (c) 22 (d) None of these

Maths =40 Science 36

24 36 – 24= 12

(a)

Chapter 8 :Limits and continuity

Break up at first fight

Limits and continuity

Limits

Limit

f(x) = c x -->a

RHL

f(x) = c x -->a+

LHL

f(x) = c x -->a-

Steps in solving limit problems

Factorise to remove 0/0 cases (Substitute value closer to the value given )

Rationalise to remove 0/0 Cases(Substitute value closer to the value given )

Divide by highest power of n to solve ∞/∞ cases If maximum power is same thenco-efficient of the maximum powerIf maximum power is differentthen 0

Lim xn – an = n.a n-1

x--a x – a

Lim (ax - 1 ) = Log e ax--0 x

Lim (1+ 1/x )x = 1x--0

Notex/∞ = 0x/0 = ∞0/x = 0

Continuity

LHS = RHS

Yes

Limit exists

No

Limit don’t exists

LHS = RHS = f(x)

F(X) is continuous

Function existsBut discontinuous

Find the value of f(x) in such way that it becomes continuous

Special points

Find points of discontinuity, implies one has equate denominator to zero

1/∞ means functionf(x) don’t exist

1• A function f(x) is defined as• f(x) = x – 1 for x < 0• = -1/2 for x = 0• = x + 1 for x > 0 then f is•• (a) Continuous at x = 0(b) Discontinuous at x = 0 (c) Un defined at x = 0 (d) None of these

X<0 X = 0 X>0LHL at the point RHLf(x) = x-1 f(x) = -1/2 f(x) = x+1f(0) = -1 f(0) = -1/2 f(0) = 1

LHL is not equal to RHLLimit doesn’t exist and discontinuous (b)

2• A function f(x) is defined as• f(x) = x+2 when x ≤1• = 5-px when x>1 then • find the value of p , for f(x) is continuous at x=1•• (a) 1 (b) 2 (c) -1 (d) -2

LHL = RHLx + 2 = 5 – px at x = 11 + 2 = 5 – PP = 5 – 1 – 2P = 2(b)

3

• lim→

•• (a) ¾ (b) 5/2 (c) 0 (d) 3/2• 푥 = 0.01

• . ( . ).

• = 1.5• 3/2

4

• lim→

= 405.푓푖푛푑푛

•• (a) 4 (b) 5 (c) 3 (d) 1•

• lim→

= 405

Lim xn – an = n*a n-1 = 405x→ a x – aa = 3 n*3n-1 = 405 n = ?

n*3n-1 = 405

4*34-1 = 108

5*35-1 = 405(b)

5• A function f(x) defined as follows f(x) = x + 1 when x ≤ 1• = 3 – px when x > 1• The value of p for which f(x) is continuous at x = 1 is:•• (a) -1 (b) 1 (c) 0 (d) None of these

LHL RHLx+ 1 3-px x = 11+ 1 = 3- pp = 3 – 1 – 1p = 1(b)

6

• lim→

=?

(a) log (3/2) (b) log 3 /log 2 (c) log 6 (d) None of these

• lim→

lim→

(9 −1) − (3 − 1)(4 −1)− (2 −1)

lim→

(9 −1)푥 − (3 − 1)

푥(4 −1)

푥 − (2 −1)푥

=퐿표푔9 − 퐿표푔3퐿표푔4 − log 2 =

퐿표푔93퐿표푔 4

2

=퐿표푔3퐿표푔2

(B)

7

lim→

1 + 2 + 3 + ⋯… 푥푥

=

• (a) 1/3 (b) ∞ (c) -∞ (d) None of these

lim→

1 + 2 + 3 + ⋯… 푥푥 =

lim→

푥 푥 + 1 2푥 + 16푥 =

lim→

2푥3

6푥 = 2/6

(a)

8• If f(x) = (x2 – 25)/(x-5), then f(5) = •• (a) 0 (b) 1 (c) 10 (d) Undefined

• f(x) = (x2 – 25)/(x-5)• f(5) = (52 – 25)/(5-5)

= 0/0undefined(d)

9• Evaluate lim

→

• (a) 1/3 (b) -1/3 (c) 1 (d) None

• lim→

• x = 1.01

• .( . ) ( . )

• ..

= −0.33

• -1/3• (b)

10• The points of discontinuity of the function, F(x) = x2+2x+5 are• x2-3x+2•• (a) x = 0, x = 1 (b) x = 1, x = 2 (c) x = 0, x = 2 (d) None of these

• Points of discontinuity denominator = 0• x2-3x+2 = 0• X – 2x – x + 2 = 0• X(x-2) – (x- 2) = 0• (x-2) (x – 1) = 0• X = 2 and 1• (b)

11• The function ƒ(x) = x2-9 is undefined at x = 3. • x-3 • What value must be assigned to ƒ(3), if ƒ(x) is to be continuous at x = 3•• (a) 6 (b) 0 (c) 9 (d) 3

ƒ(x) = x2-9 = (x+3)(x-3)• x-3 (x-3)

= x + 3x = 3

3 + 36

(a)

12• Lim 3x + | x| is equal to• X0 7x-5| x|

• (a)1/6 (b) 1 (c) Does not exist (d) 2

RHL x >03(x) + (x) is equal to7(x) –5(x)4x2x

2

LHL x < 03(-x) + (-x)7(-x) –5(-x)

3(-x) + (x)7(-x) –5(x)

-2x-12x

1/6

LHL ≠ RHLFunction doest exist(c)

Calculus

1• The derivative of x2 log x is:•• (a) 1+2 log x (b) 2 log x (c) x (1+2 log x) (d) None of these•

• 1• The derivative of x2 log x is:• (a) 1+2 log x (b) 2 log x

(c) x (1+2 log x) (d) None of these

Y = x2 log x 푑푦푑푥 =

푑[풙ퟐ풍풐품풙]푑푥

푑푦푑푥 = 풍풐품풙

푑[풙ퟐ]푑푥 + 풙ퟐ

푑[풍풐품풙]푑푥

푑푦푑푥 = 풍풐품풙ퟐ풙+ 풙ퟐ

ퟏ풙

푑푦푑푥 = 풍풐품풙ퟐ풙+ 풙

푑푦푑푥 = 풙(ퟐ풍풐품풙+ ퟏ)

(c)

2If f(x) = x2 +1 then f’(x) is

X2 -1(a) –4x / (x2 – 1)2 (b) 4x / (x2 – 1)2 ( c) x / (x2 – 1)2 (d) None of these

2)()()()()(

)()(

xgxgxfxfxg

xgxf

dxd

푑푦푑푥 =

푑 풙ퟐ + ퟏ풙ퟐ − ퟏ푑푥

푑푦푑푥 =

풙ퟐ − 1 풅(풙ퟐ + ퟏ)풅풙 − (풙ퟐ + ퟏ)

풅 풙ퟐ − ퟏ풅풙

풙ퟐ − 1 ퟐ

푑푦푑푥 =

풙ퟐ − 1 (ퟐ풙 + ퟎ) − (풙ퟐ + ퟏ)(ퟐ풙 − ퟎ)풙ퟐ − 1 ퟐ

푑푦푑푥 =

2풙 − 2푥 − 2풙ퟑ − ퟐ풙풙ퟐ − 1 ퟐ

푑푦푑푥 =

−4푥풙ퟐ − 1 ퟐ

(a)

3Given x = 2t + 5 ; y = t2 – 2, then dy/dx is calculated as:

(a) t (b) 1/t (c) -1/t (d) None of these

푑푦푑풕 = ퟐ풕

푑풙푑풕 = ퟐ ퟏ + ퟎ = ퟐ

푑푦푑풙 =

푑푦/풅풕푑풙/풅풕

푑푦푑풙 =

ퟐ풕ퟐ

푑푦푑풙 = 푡

x = 2t + 5 y = t2 – 2

(a)

4• If x3 – 2x2y2 + 5x + y = 5, then dy/dx at x = 1 and y = 1 is:• (a) 4/3 (b) -5/4 (c) 4/5 (d) -4/3

x3 - 2x2y2 +5x+ y = 5

Differentiate both sides w.r.t x

푑 x3

푑풙 −푑 ퟐ풙ퟐ풚ퟐ

푑풙 +푑 5x푑풙 +

푑 y푑풙 =

푑 5푑풙

ퟑ풙ퟐ − 2 푥풅 푦풅풙 + 푦

풅 푥풅풙 + 5(1) +

푑풚푑풙 = ퟎ

ퟑ풙ퟐ − 2 푥 2푦푑푦푑푥 + 푦 2푥 + 5 +

푑풚푑풙 = ퟎ

x = 1 and y = 1

ퟑ(ퟏ)ퟐ−2 1 2(1)푑푦푑푥 + 1 2(1) + 5 +

푑풚푑풙 = ퟎ

ퟑ − 4푑푦푑푥 − ퟒ + 5 +

푑풚푑풙 = ퟎ

−4푑푦푑푥 +

푑풚푑풙 = −ퟑ+ ퟒ − ퟓ

푑푦푑푥 −ퟒ + ퟏ = −ퟒ

푑푦푑푥 =

−4−3

푑푦푑푥 =

43

(a)

5• The slope of the tangent at the point (2, -2) to the curve x2+xy+y2-4=0 is given by:•• (a)0 (b) 1 (c) -1 (d) None of these

x2 + xy+y2 -4 = 0Differentiate both sides w.r.t x푑 x2

푑풙 +푑 xy푑풙 +

푑 y2

푑풙 −푑 4푑풙 =

푑 0푑풙

ퟐ풙 + 풙풅풚풅풙 + 풚

풅풙풅풙 + 2풚

풅풚풅풙 − 0 = ퟎ

(x=2, y-2)

ퟐ(ퟐ) + (2)풅풚풅풙 + (−2)

풅풙풅풙 + 2(−ퟐ)

풅풚풅풙 − 0 = ퟎ

풅풚풅풙 [−ퟒ] = −ퟒ 풅풚

풅풙 = ퟏ (b)

6• The derivative of 푒 is•• (a) 30(1 –5x)5 (b) (1–5x )5 (c) 6(x–1) e3x^2 -6x+2 d) none of these

푑푒 )푑푥 = (푒 )

푑(3푥 − 6푥 + 2)푑푥

= (푒 )(6푥 − 6(1) + 0)

= (푒 )6 푥 − 1(c)

7• Find the second derivative of y = √(x + 1)•• (a) ½ (x + 1) -1/2 (b) -1/4 (x + 1) -3/2 (c) ¼ (x + 1) -1/2 (d) None of these• 푦 = 푥 + 1

푑푦푑푥 =

푑( 푥 + 1)푑푥

푑푦푑푥 =

12 푥 + 1

푑(푥 + 1)푑푥

푑푦푑푥 =

12 푥 + 1

푑2푦푑푥2 = −

14 푥 + 1 3

2

(b)

8• If log (x / y) = x + y, dy/dx may be found to be•• (a) y (1-x) (b) y (c) (1-x) (d) x_ • x(1+y) x (1+y) x+1

log (x / y) = x + y

log x- log y = x + y

Differentiate both sides w.r.t x

ퟏ풙 −

ퟏ풚풅풚풅풙 =

풅풙풅풙 +

풅풚풅풙

ퟏ풙 − ퟏ =

풅풚풅풙 +

ퟏ풚풅풚풅풙

ퟏ − 풙풙 =

풅풚풅풙 ퟏ +

ퟏ풚

풅풚풅풙

ퟏ+ 풚풚 =

ퟏ − 풙풙

풅풚풅풙 =

풚(ퟏ − 풙)풙(ퟏ + 풚)

(a)

• 9• If y = 1 + x + x2 + x3 + …. + xn+……..∞ , then dy/dx – y is equal to:

2! 3! n!(a)1 (b)-1 (c) 0 (d) None

y = 1 + x + x2 + x3 + …. + xn+……..∞ 2! 3! n!

y = ex

dy/dx = ex

ex – ex

0 (c)

dy/dx – y

10• If ƒ(x) = xC3; then ƒ’(1) = ? •• (a) 1/6 (b) -1/6 (c) 5/6 (d) -5/6xC3 =x(x-1)(x-2)

6=(x2 – x)(x-2)/6= (x3 – x2 -2x2 +2x) / 6= (x3 – 3x2 +2x) / 6

On differentiationf`(x) = 3x2 – 6x + 2

6f`(1) = 3(1)2 – 6(1) + 2

6= -1/6(b)

11• Evaluate ∫ ( x2 - 1 ) dx•• (a)X5 + 2x3 + k (b) x3 – x + k (c) 2x (d) None of these• 5 3 3

• ∫ ( x2 - 1 ) dx• x3 – x + k • 3

• (b)•

12• If ƒ’(x) = 3x2 – 2/x3, ƒ(1) = 0 and ƒ(x) = ____•• x3/3- x-2 -2 (b) x3 + x2 +2 (c) x3 + x-2 – 2 (d) None of these•

• ƒ’(x) = 3x2 – 2/x3

• ƒ(x) = x3 – 2x-3+1 + K -2

f(1) = 13 + (1)-2 + k = 0 k = - 2

• ƒ(x) = x3 +x-2 -2 (c)

13• Evaluate: ∫ x. cx dx•• (a)ex (x + 1) +c (b) ex (x – 1)+c (c) ex + c (d) x – ex + c

• ∫ x ex dx

∫u.v = u∫v dx - ∫ [ du ∫ v dx ] dx dx

st nd st nd st nddI function II function dx I function II function dx I function II function dx dxdx

= 푥 푒 푑푥 −푑(푥)푑푥 푒 푑푥 푑푥

= 푥푒 − 1 ∗ 푒 푑푥

= 푥푒 − 푒

= 푒 푥 − 1 + 퐶

(b)

148푥

푥 + 2 푑푥

푥 + 2 = 푡퐷풊풇풇풆풓풆풏풕풊풂풕풆풘.풓. 풕풙

푑 풙ퟑ + ퟐ푑푥 =

푑푡푑푥

3푥 + 0 =푑푡푑푥

푑푥 =푑푡3푥

=8푥푡

푑푡3푥

=ퟖퟑ 푡 푑푡

=ퟖퟑ풕 ퟑ ퟏ

−ퟑ + ퟏ

=ퟖ풕 ퟐ

−ퟔ

=−ퟒ(푥 + 2) ퟐ

ퟑ + 푪(b)

15• 4• ∫ (2x +5) and the value is• 1• (a) 10 (b) 3 (c) 30 (d) None

4∫ (2x +5) and the value is1

x2 + 5x42 + 5(4) – [12 + 5(1)]16 + 20 – 1 – 530

(c)

16• Evaluate ∫ 1 dx :• (x – 1) (x – 2)• (a) log (x-2) + C• (x-1)

(b) log [(x-2) (x-1)] + C (c) log (x-1) + C(x-2)

(d) None of these

1(푥 − 1)(푥 − 2) =

퐴푥 − 1 +

퐵푥 − 2

1(푥 − 1)(푥 − 2) =

퐴 푥 − 2 + 퐵(푥 − 1)(푥 − 2)(푥 − 1)

1 = 퐴 푥 − 2 + 퐵(푥 − 1)

푝푢푡,푥 = 2 퐵 = 1

푝푢푡,푥 = 1 퐴 = −1

1(푥 − 1)(푥 − 2) =

−1푥 − 1 +

1푥 − 2

= − log 푥 − 1 + log(푥 − 2)

= log푥 − 2푥 − 1 + 푐

(a)

17• 1• The value of ∫ dx is :

0 (1 + x) (2 + x)• (a) log 3/4 (b) log 4/3 (c) log 12 (d) None of these

1(1 + 푥)(2 + 푥) =

퐴1 + 푥 +

퐵2 + 푥

1(1 + 푥)(2 + 푥) =

퐴 푥 + 2 + 퐵(푥 + 1)(푥 + 2)(푥 + 1)

1 = 퐴 푥 + 2 + 퐵(푥 + 1)

푝푢푡, 푥 = −2 퐵 = −1

푝푢푡, 푥 = −1 퐴 = 1

1(푥 + 1)(푥 + 2) =

1푥 + 1 +

−1푥 + 2

= log 푥 + 1 − log(푥 + 2)

= log푥 + 1푥 + 2

X = 1

= log1 + 11 + 2

= log

−

X = 0

log0 + 10 + 2

log

= log2312

= log43

(b)

18• Find slope of tangent of curve 푦 = 푎푡푥 = 2

• (a) 3/16 (b) 5/17 (c) 9/11 (d) None of the above

푦 = 푥 − 1푥 + 2

푑푦푑푥 =

푑 푥 − 1푥 + 2푑푥

푑푦푑푥 =

푥 + 2 푑 푥 − 1푑푥 − (푥 − 1)푑(푥 + 2)

푑푥푥 + 2

푑푦푑푥 =

푥 + 2 (1 − 0)− (푥 − 1)(1 + 0)푥 + 2

X = 2푑푦푑푥 =

2 + 2 − (2 − 1)2 + 2

푑푦푑푥 =

316

(a)

19• ∫ 푑푥 (a) 2 log 3/2 - 1 (b) 2 log 3 + 1 (c) ½ log 3/2 - 1 (d) 2 log 2 – 1 + k

1 − 푥1 + 푥 푑푥 =

2 − 1 − 푥1 + 푥 푑푥

2 − (1 + 푥)1 + 푥 푑푥 =

21 + 푥 푑푥 − 1푑푥

2 log 1 + 푥 − 푥

2 log 1 + 2 − 2 - [2 log 1 + 1 − 1]

2 log 3 − 2 log 2 − 2 + 1

2 [log 3 − log 2] − 1

2 log32 − 1

(a)

20• U= 5t4 + 4t3 + 2t2 + t +4 at t=-1 find du/dt•• (a) -11 (b) 11 (c) -16 (d) 16

U= 5t4 + 4t3 + 2t2 + t +4du/dt = 20t3 + 12t2 +4t + 1 +0

t = -120(-1)3 + 12(-1)2 +4(-1) + 1-20 + 12 – 4 + 1-11

(a)

MODEL PAPER 1 ANSWERS

16 B17 C18 B19 D20 B21 D22 A23 A24 A25 C26 D27 C28 A29 B30 A

1 C2 B3 B4 B5 B6 A7 C8 B9 A10 C11 B12 C13 B14 A15 B

31 C32 A33 B34 B35 B36 A37 A38 D39 B40 C41 A42 B43 C44 B45 A

46 C47 A48 C49 B50 B

[email protected]

Thank You