Quantitative aptitude Crash course June 2017 CA. … Crash course June 2017 Quant.… · Standard steps in Simplifications • Fractions • Algebra formulas • Cross Product rule
Post on 17-Mar-2018
222 Views
Preview:
Transcript
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
top related