ANALYTICAL SKILLS

                                                                   ANALYTICAL SKILLS

RATIO & PROPORTION

Ratio : A ratio is a comparison of two numbers (quantities ) by division. The ratio of a to b is written as   a : b or 

PROPORTION :  A Proportion is an expression which states that two ratios are equal. For ex 3/12 = ¼ or 3:4 : : 1 : 4

Each quantity in proportion is called term or proportional. If  =  then we say a : b and c : d are said to be in proportion and denoted by a : b = c : d or a : b : : c : d

The 1^st and last terms in a proportion namely a and d are called extremes and second and third terms namely b and c are called means.

 a, b, c, d are in proportion if product of means = product of extremes ( ad = bd )

a is called 1^st proportional

b is called 2^nd proportional

c is called 3^rd proportional

d is called 4^th proportional

Continued proportion :

Three quantities a, b, c, are said to be in continues proportion if a : b = b : c ---à b² = ac

a, b, c are in continued proportion if b² = ac or b =  where b is called mean proportion between a and c i.e between 1^st and 3^rd quantities.

Short cuts

1. If If  and , then

Ex: 1.  A : B  3 : 4, B : C = 5 : 6   find A : B : c

     =  3 x 5 : 4 x 5  : 4 x 6  = 15  : 20  : 24    

2. If a : b = 3 ; 13  b : c = 2 ; 11 find A :B : C  ------      6 : 26  : 143

2.  If  and  and , then

Ex : If  A ; B  = 2 : 3  B : C = 4 :5  C : D = 9 ; 7  find A : B : C :D

 and  and   = 2 X 4 X 9 : 3 X 4 X 9 : 3 X 5 X 9 : 3 X 5 X7 =

72 : 108: 10 : 105

Try Yourself :  A ; B  = 3:5  B : C = 7 :2  C : D = 1 ; 8  find A : B : C :D   =  > 21 : 35 : 10 : 80

PROBLEMS:

1. Find the fourth proportional to 5, 6, 150

            Let fourth proportional be x

            5 : 6 : : 150 : x   or 5 : 6  = 150 : x

            X =     = 180

      Fourth proportional  = 180

2. Find the fourth proportional of 6, 8, 12      =>       Ans:  16

3. Find the mean proportional of 0.2 and 0.8

       Mean proportional between a and c is       b =    =   =   = 0.4

4. Find the mean proportional of  9 and 25     =>  Ans : 15

Note : If b is the mean proportion of a and c i.e a : b = B:c then c is called third proportional to 1^st  and 2^nd  i.e a and b.

5. Find third proportional to 16 and 24

            16  : 24   =  24 : c   => product of means  = product of extremes

                                                24  x  24   =  16  x   c

                                                C  =   =  36

6.  Find the third proportional to 16  and 4     =>   Ans  ; 1

Note: In a ratio a : b the first quantity is called antecedent and second quantity is called consequent

7. In a ration which is equal to 3 : 7 , if the antecedent is 33 then what is the consequent ?

           

Let consequent be x

            3 : 7  = antecedent  : consequent  = 33  :  x

            3  : 7  = 33 : x  =>    =  77

8. Two sums of money are in the ration 2 : 5, if the second sum is Rs. 95 find the 1^st sum ?  Ans: 38

9. The ratio of number of boys and girls in a school is 4 : 3. If there are 480 boys find the number of girls?

            Ans: 360

10. the ration between number of boys and girls is 2 : 5. If there are 350 students in the school find the number of girls in the school ?

Ratio of number of boys and girls  = 2  :  5

Total number of students  = 350

No. of boys  =    =    =  100

           

No. of boys  =    =    =  250

11. A number 351 is divided in the ratio 2 : 7. Find the product of the numbers?   Ans  :  21,  294

12. The marks obtained by suresh in English, mathematics and Science are in the ratio  :     :  . If his total score is 860. Find his marks in English.     Ans : 300

PERCENTAGE

1. Percent means hundred and is denoted by %, x% means 

2. Fraction   in terms of percent is x 100%

            Express  as percentage   =  x100   =  66%

3. Express 0.4 as percentage    = Ans : 40%

4. Express 1.23 as percentage  =  1.23 x 100%  =>  123%

5.  Express 22 as fraction  =>  9/40

6.  Find 10% of 5% of 320 ?

            xx320  =  

7.  125% of 64   =>   80

8. 36 is what % of 144  ?

            Let us suppose x% of 144 = 36

            x144  = 36   =>  x =  = 25

9. 2.5 is 5% of what ?     =>   Ans: 50

10.   If  P1% and P2% of a number N are x1 and x2 respectively. Then

            1. N  = 

          2. ==

Ex: If 15% of a number is 60 then what is 25% of the number ?

            P1 = 15, x1 = 60    P2 = 25    x2 = ?

            =   =>   =   =>    x2 = 4 x 25   =>  100

11. If N₁ exceeds N₂  by P% then N₂ short than N₁   by  %

12. . If N₁ short of  N₂  by P% then N₂ exceeds N₁   by  %

Ex: A number  A exceeds B by 25% by what percent is B short of A ?

       P%  = 25%      =>    %  =>   % =>   20%

Ex: A number X is short of Y by 40% then by what percent is Y excess of X ?

          Y is excess of X by  %   =>  % =>  %  =>  66%

Ex: If A’s salary is 20% below B’s salary then by how much percent is B’s salary above A ?  Ans: 25%

Ex: If A is short of B by 30% then B is excess of A by what percent ?  

13.  If a number N is increased by P% then the number after increase is N X %

14. If a number N is decreased by P5 then the number after decrease is  N x %

Ex: In 2011 the total enrolment in a school was 1500. Next year it increased by 10% then what is the enrolment in 2012?

          P%  = 10%   N = 1500

          Enrolment in 2012   = N x      =1500 X   =>  1500  x   =  1650

Ex. the daily wage is increased by 15% and a person now gets Rs.203 perday. What his daily wage before increase?    Ans;  20

15. If a number N is increase by P% and then decreased by P% or 1^st decreased by p% and then increased by p% then original number N decreases% in both the cases.

Ex: When the price of a pressure cooker was increased by 15% sale of pressure cookers decreased by 15%. What was the net effect on the sales ?

          P% = 15% ,  p  =  15  

          Overal effect on sales  = %   => % =>  % =>  2.25% decrease

Ex:  If man’s wages are increased by 10% and then decreased by 10% then, what is the total change in percent ?        Ans:   1% decrease.

16.  The two split parts of N such that on part is p% of the other are   X N  and   X N

Ex:  Split 42 into two parts such that one is 2% of the other.

          100 X  =  100 X   =>     and

 X N  =>  x 42   =>   

17. If the price of an article is increased by P% then % decrease in consumption. So as to have no extra expenditure.

Ex :  If the price of coffee is increased by 3% by how much % must a housewife reduce her consumpetion of coffee, so as to have no extra expenditure?

            % decrease in consumption  = %  =>  % =>  %  =>  2

Ex;  The price of milk increases by 20%. If a housewife wants to spend the same amount of money then how much % she must reduce her consumption ?

PROFIT AND LOSS

I. If an article is purchased at Rs C  and its sold at Rs S.

If S > C then gain = S –C

If S< C then Loss = C – S

          % of gain  =

            % Loss =

II. If an article purchased at Rs. C is sold at Rs   and Rs.  , gain or loss  incurred on s and gain or loss  incurred on s then       1.

                                    2. C   = 

III. If two different articles are sold at same price such that

          a) One article incurs a gain loss of % and other article incurs a gain or loss of %  then overall gain % or  loss %     

          b) One article incurs a gain of x% and other article incurs a loss of x% then overall effect is loss and

                   Loss %  =

IV. If selling price of  N₁  articles is equal to cost price of  N₂  articles then % loss or gain % 

                                              

          If it is –ve then there is Loss, if it is +ve then there is gain

PROBLEMS :

1. If C.P  = 120  S.P = 90 then what is Loss% ?

          Loss  = C.P – S.P  = 120-90   = 30

          Loss%  = =  = 25%

2. if C.P = 80  S.P = 100 then what is Loss% ?       Ans : 25%

3. If C.P = 125    S.P = 96   then what is Loss% ?    

4. By selling an article for Rs.450 a man loses 25%. At what price he will sell in order to gain 25%?

            = =

                    =     =  750

5. A person by selling on article for Rs. 450 loses 20% . In order to make a profit of 20%, What price he must sell out ?          Ans : 675

6. By selling 100 mangoes a fruit seller gains selling price of 20 mangoes. Find his gain%?

          No of articles on S.P  = =100

            No of articles on C.P = = 20

            Effect is loss %   = = = 80%

7. The cost price of 25 articles is equal to the selling price of 20 articles find gain% ?  Ans: 25%

8. If selling price of 5 pens is equal to the cost price of 4 pens, find the gain% or loss% ?  Ans: 20% Loss

9. A man sold two steel chairs for Rs. 500 each on one he gain 20% and on other he loses 12%. How much does he gain or loss in the whole transaction?

          Overall gain% or loss%  =   (  x1=20, x2=12)

             =   = -1.5% loss

          Since the output is –ve overall effect = Loss and %Loss = 1.5

10. A man sold two wathes for Rs1000 each on one he gains 25% and on the other 20% loss. Find how much % does he gain or lose in the whole transaction ?    Ans. 2

13. Two articles were sold at Rs.480 each on one a gain of 20% is made and on the other a loss of 20%. How much % loss or gain% is made in the whole transaction ?

            Since  = s1 = s2    and  p1 = p2 = 20

            Overall effect is loss and % of loss  =   = 4%

14. A man sold two watches for Rs.1000 each one he gains 12% and the other 12% loss, find how much percent he gain or loss in the whole transaction ?     Ans : 1.44% Loss

         

         

NUMBERS

Divisibility by…	Rule	Example	Explanation
2	Unit’s digit of the number should be zero or divisible by 2.	4, 2, 102, etc.
3	Sum of the digits in the number should be divisible by 3.	1782	1+7+8+2 = 18 which is divisible by 3 hence 1782 also divisible by 3.
4	Number formed by the last two digits should be divisible by 4 or are both zero.	4784, 300, etc.	4784 à Since 84 is divisible by 4, 4784 is also divisible by 4.
5	Unit’s digit of the number should be 0 or 5.	120, 625, etc.
6	Should satisfy divisibility rules of 2 and 3.	4518
7	The unit digit of the given number is doubled and then it is subtracted from the number obtained after omitting the unit digit. If the result is divisible by 7, then the given number is also divisible by 7.	448	448 à 44 – 8(2) = 44 – 16 = 28 which is divisible by 7 and hence 448 is also divisible by 7.
8	Number formed by the last three digits should be divisible by 8. or zero’s	1576	1576 à 576 is divisible by 8 and hence 1576 is also divisible by 8.
9	Sum of the digits in the number should be divisible by 9.	1395	1395 à 1+3+9+5 = 18 is divisible by 9 and hence 1395 is also divisible by 9.
10	Number should end in zero.	1000
11	Sum of digits at odd places – Sum of digits at even places should be 0 or divisible by 11.	38797	38797 à Sum of digits at odd places = 3+7+7 = 17 Sum of digits at odd places = 8+9 =17 and 17 – 17 =0, hence 38797 is divisible by 11.
12	Should satisfy divisibility rules of 3 and 4.	156	156 is divisible by 2 and 3 hence 156 is also divisible by 12.
14	Should satisfy divisibility rules of 2 and 7.	322	322 is divisible by 2 and 7 hence 322 is also divisible by 14.
25	Last two digits in the number should be 0 or divisible by 25.	175	175 à 75 is divisible by 25 and hence 175 is also divisible by 25.
125	Last three digits in the number should be 0 or divisible by 125.	2250	2250 à 250 is divisible by 125 and hence 2250 is also divisible by 125.

1.    The average of first n natural numbers is .

2.    The average of squares of natural numbers till n is .

3.    The average of cubes of natural numbers till n is .

4.    The average of odd numbers from 1 to n is

5.    The average of even numbers from 1 to n is

6.    If n is odd: The average of n consecutive numbers, consecutive even numbers or consecutive odd numbers is always the middle number.

7.    If n is even: The average of n consecutive numbers, consecutive even numbers or consecutive odd numbers is always the average of the middle two numbers.

8.    The average of first n consecutive even numbers is (n + 1).

9.    The average of first n consecutive odd numbers is n.

10.  The average of squares of first n consecutive even numbers is.

11.  The average of squares of consecutive even numbers till n is.

12.  The average of squares of consecutive odd numbers till n is .

Note : The place value of any digit is itself, but place value is obtained as below

             Place value of units digit  = unit digit  x 1

             Place value of tens digit = tens digit x 1

             Place value of hundred’s digit  =( hundred’s digit) X 100

PROBLEMS:

 1. Find the difference between the place value and face value of 6 in the numeral 556973

                      Face value of 6 = 6

                      Place value of 6 = 6 X 1000  = 6000

                      Difference   =  6000 – 6   = 5994

2. Find the difference between the place values of two sevens in the numeral 69758472

3. Find 287 X 287  + 269 X 269  - 2 X 287 X 269   = ?

                      a²   + b²

LCM & HCF

LCM : Least Common Multiple

HCF/GDC : Highest Common Factor / Greatest Common Divisor

LCM of two or more numbers is the product of the highest powers of all the prime factors that occur in these numbers.

HCF of two or more numbers is the product of highest (maximum) number of common factors in the given numbers.

Find the L.C.M. of 12, 15, 18 and 20.

                                                               So, the L.C.M. = 2 x 2 x 3 x 5 x 3 = 180

Find the H.C.F. of 50 and 70

Sol:  50 = 2 x 5 x 5

   70 = 2 x 5 x 7

Common factors are 2 and 5.

So, H.C.F. = 2 x 5 = 10

Ex: Find LCM and HCF of 36, 48, 64, and 72   

EX: Find LCM and HCF of Rs.2, Rs.2.40Paise, Rs.3.20paise

             (Hint : Rs.2 = 200p, Rs.2.40 = 240p, Rs.3.20 = 320p)

             Solve LCM and HCF of  200, 240, 320   =>  ?

Ex: Find LCM and HCF of 6hrs 4mints, 2hrs 42mints

Ex: Find  HCF of 2metrs 28cms,  3m24cms and 4m 86cms (hint : 1m = 100cms)    Ans  ; 6met 

Method II:   

1.     L.C.M. and H.C.F. of fractions

    

2.     Product of two numbers = L.C.M. of two numbers x H.C.F. of two numbers.

3.     To find the greatest number that will exactly divide a, b and c, simply find the H.C.F. of a, b and c.

4.     To find the greatest number that will divide x, y and z leaving remainders a, b and c respectively, find the H.C.F. of (x - a), ( y-b) and (z - c).

5.     To find the least number which is exactly divisible by a, b and c, simply find the L.C.M. of a, b and c.

6.     To find the least number when divided by a, b and c leaving remainders x, y and z respectively, find the (L.C.M. of a, b and c) – k, where k = (a – x) = (b – y) = (c – z).

PROBLEMS :

1. Find the greatest number that will exactly divide 200 and 320 ?

   The required number = H.C.F. of 200 and 320 = 40.

2. E.g.  1. Find the H.C.F. of 3332, 3724.

Sol:

      

                                                         

So, the H.C.F. of 3332, 3724 is 196.    

3. Find LCM of 28, 35 , 56, 84 using division method?   Ans : 840

4. Find the greatest number that will divide 148 , 246, 623 leaving remainders 4, 6, 11 respectively.

          Requried number  = HCF OF (x - a), ( y-b) (z - c).  = ( 148-4, 246-6,  632-11) = 144, 240 , 612

          HCF of 144 , 240, 612

          HCF of 144, 240   = 48

          HCF of 48, 612   = 12

          Ans : 12

5. find the greatest number that will divide 1375, 4935 leaving in each case same remainder 3 ?

          Req number  = 1375 – 3, 4935 – 3   = 1372, 4932   

          HCF of 1372, 4932   is 4

6. Four bells first begin to toll together and then at intervals of 6, 7, 8, 9 seconds. Find how many times the bells toll together in two hours and at what intervals they toll together ?

          Interval  = LCM of 6, 7, 8, 9   = 504seconds

All the bells toll together after each interval of 504 seconds.

No of times  They all toll together in two hours  =   = 14times

7. Find the least number which when divided by 27, 35, 45, 49 leaves the remainder 6 in each case ?

          LCM of 27, 35, 45, 49    = 6615

          Req Number  = 6615 +6   = 6621

8. Find the least number which when divided by 36, 48, 64 leaves the remainders 25, 37, 53 respectively ?

          LCM of  36-25, 48 – 37, 64 – 53  =>  K = 11

          LCM of 36, 48, 64   = 576

          Req Number  : 576-11  = 565

9. find the greatest possible length of a scale that can be used to measure exacte the length of cloth 3cm, 5m , 10m, 12m 90cm ?   (Hint : Find HCF of 300cm, 500cm  1000cm, 1290cm)

10.  Find LCM & HCF of   , ,

   =   =   =140

   =  =

Note:  Product of numbers = LCM  x  HCF  

11. If HCF, LCM of two numbers are 16, 240. If one number is 48 find the other?

            16 X 240   = 48 X x

            X  = = 80         Second Number   = 80

PROBABILITY

          Today we will discuss and understand the concept of probability, for this let us once again consider the simple case.

Random Experiment

An experiment whose outcome cannot be predicted with certainty is called a random experiment. In other words, if

an experiment is performed many times under similar conditions and the outcome of each time is not the same, then

this experiment is called a random experiment.

Example: a) Tossing of a fair coin

b) Throwing of an unbiased die

c) Drawing of a card from a well shuffled pack of 52 playing cards

Let us consider an example in which a person is given 10 bullets to hit the given target. He fires the shots one by one, he finds only 5 bullets have hit the target

          Now he again fire the another set of 10 bullets to the target again he finds that only five bullets have hit the target so his target hitting accuracy or chance of hitting the target is 50%, as he can hit five out of ten times.

          Now again if he tries to hit the target with 10 bullets we can very well say that most probably he will have five hits at the target out of ten so this probably word stated above in mathematics is known as Probability.

          Here the event was hitting the target. He was given total ten chances and he managed to hit only five shots accurately at the target i.e his accuracy in percentage was  or simply probability of his hitting the target is  i.e

                So, for example if this person is given 20 bullets, we can say that most probably he will fire 10 shots accurately or if he is given 100 bullets, most probably he will fire 50 shots accurately. But practically, one may suppose that it may not happen always that this person can hit with 50% accuracy. This we are talking in terms of chances only. It may be possible that he may hit 4 out 10 or 6 or 7 out of 10 times. But again the most probably seeing his record one can say his accuracy is approximately 50%.

                Now if we have followed the above event  or case or an experiments then we can define and try to understand the probability mathematically and also some of basic terms related to this topic.

A) Probability (of any given event A ) = P(A), is ratio of chances favorable to given event (A) to total chances of happening of given event (A). It is denoted by P(A) read as Probability of (A)

                P(A)  =    Chances in favor of (A) / total number of equally likely cases

                Remember that to find the probability of any ;event there must be some definite and unbiased outcomes (results). Also there must not be unknown external factors affecting the outcomes or result or given event or events. Otherwise we may not arrive at correct answer mathematically as well as in real application or probability.

B) Experiment : it is process which can be repeated and in any trial we can expect fair unbiased and definite outcome or result, though we may not be knowing the value outcome or what that outcomes would be unless. It is calculated or the value outcome or what that outcomes would be unless. It is calculated or shown.

                For e.g. if coin is tossed, we know that either head or tail, will come but what actually has come, can be known after seeing the tossed coin. So there was certain and fair outcome, no matter it could be either head or tail Therefore tossing of a coin was an Experiment.

C) Sample space : It is the set or collection of total fair possible outcomes of any given experiment. Usually the set of outcomes is written with in bracket {   }. For e.g. if a coin is tossed there can be only two outcomes i.e head or tail.

                  Sample space   =   { H T } or {T H }

            The elements of sample space are called sample points i.e. (H T )

D) Simple Events : When there is single possible outcome of an experiment it is a simple event for e.g., if we toss a coin then getting a tail is simple or in case of tossing of two coins getting only two tails or only two heads is simple event.

E) Compound event : When there are possibilities of more than one outcome for an event then it is a compound event. For e.g. in a toss of two coins the probability of getting at least one head is compound as we have two possibilities for this (1) one head, one tail, 92) two heads i.e it consists of two simple events.

F) Exhaustive events : When there are fair, unbiased and certain possible outcomes, events are exhaustive. For e.g., in tossing of a coin, outcome will either be head or tail. Also there can be no other possible outcome

G) Mutually exclusive events : The events are said to be Mutually exclusive events fif only one and one can occur, not each of them can occur simultaneously for e.g., in a toss of coin either head will come or tail will come i.e., both head and tail cannot come up simultaneously

H) Equally likely events: Events are equally likely if the probability of each possible outcome is equal. For e.g., in a toss of a coin, the probability of getting a head or a tail is same or equal.

I) Independent events : Two events are said to be independent if the outcome of first is not affecting or related to or having any outcome in common to second and vice-versa for e.g., consider a tossing of coin and throwing a dice, we get either a head or tail in a toss and any number from 1 to 6 in throw of a dice. In both the cases outcome of each event is not having any effect on each other at all i.e., outcome of toss is not affecting the outcome of throw of dice. Therefore the events are independent.

  J)Dependent events : two events are said to be dependent when the outcome is first is related to second or have outcomes in common or vice-versa.

  K) Sum of events : For two events A and B, the sum 0of these two, which is compound means either event A or event B occur i.e., at least one of A or B occurs. It is denoted by A + B (read as A or B)

L) Product of events : for two events A and B the product of these two means that event A and B both will occur. It is denoted by AB (read as A and B )

M) If odds in favor of A is a : b then it means, ratio of chances that event will happen to chances that event will o happen

            P(A)  =    and  P(`A) =   when a+b are total chances.

If odd against A is a:b then, it is ratio of not happening of A to happening of A

 P(A)  =    and  P(`A) =        When a + b are total chances.

N) For nay event A, 0< P(A) <1 nbsp="" u="">i.e., Probability of any event A is always greater than or equal to 0 and it is always less than or equal to 1

O) 1. Total when n coins are tossed once or one coin is tossed n times = 2ⁿ

       2. Total cases when n dice are thrown once or one dice is thrown n times = 6ⁿ                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        

Types of Questions:

1. Simple questions based on concepts of permutation and combinations

2. Based on addition theorem of probability

3. Based on conditional probability

Type I:

1. A dice is thrown what is the probability of getting a six

                Total cases = 6   ( i.e., (1,2,3,4,5,6 as any number from 1 to 6 come in throw of a dice)

                Favorable cases = 1   ( as only one six as a number is there on dice )

                Probability =

2. A coin is tossed, what is probability of getting a head

          When a coin is tossed, total outcomes = 2 i.e., (H,T) favorable outcome for given event = (H) i.e., one out come.           Probability (getting a head)  =

3. Two coins are tossed, what is the probability of getting two tails only.

          Sample space, when two coin are tossed    = (H,H) (T,T) (T,H) (H,T)

          Total outcomes   =    2²     = 4  

            As it clear from the sample space, we have 1 outcome only in which there are exactly two tails.

            Probability (getting two tails) =

4. A bag contains 4 Red, 6 White and 7 blue balls. Two balls are drawn at random, what is the probability that both are red.

          Total balls = 4 + 6 + 7  = 17

          Total cases = Ways of selecting two balls =  17C₂ =

          Let A be event if drawing two red balls.  Also ways of selecting 2 red balls of 4 red, which is also number of favorable outcomes for this question.   = 4C₂ = == 6

          P(A) of drawing 2 red balls = =

5. A card is drawn at random from deck of 52 cards. What is the probability of getting . 

i) an ace    II) a rd card III) a diamond

Since total cards are 52,   total no. of outcomes = 52

i) Favorable outcomes = 4   ( as 4 aces are there in of 52 cards )

            Probability = =

ii) Favorable outcomes = 26  ( as there are 26 black and 26 red cards are in a pack of cards )

          Probability  = =

iii) Favorable outcomes = 13

  (as there are 13 cards each of 4 different suits  13 of diamond, 13 of club, 13 of Heart , 13 of spade )

   Probability = =

7. Two cards are drawn at random from the pack of 52 cards. What is the probability of getting a King or Queen?

        The drawing of king or Queen is an independent event, because there is no card in common between king and Queen or i.e., there is no card on which king and Queen are present or printed simultaneously.

        Therefore by additive rule of probability

        P (King or Queen) = P (King )  + P (Queen)   = +=

8.  3 coins are tossed together. Find the probability of getting  

        i) Exactly one head ii) At least 1 head  iii) At most 1 head

Total number or cases when three coins are tossed together = 2³ = 8.

Now we will solve each of the above questions one by one.

        i) The cases of exactly one head are (HTT) (TTH) (THT)

        Probability of getting at least one head =

ii)  Probability of getting at least one head =  1 – probability of getting no head = 1 -=

iii) The cases of at most one head are (TTT)  (HTT) (TTH) (THT)

            Probability of getting at most one head  =  =

9. A card drawn from the pack of 50 cards numbered 1 to 50. The probability of drawing a number, which is a square.

        Total possible outcomes = 50 (as there are 50 cards)

        Favourable outcomes = Number of cards on which square of numbers from 1 to 50 are there.

        Those can be ( 1, 4 , 9, 16, 25, 36, 49)  i.e. seven favourable outcomes.

TYPE : II   - BASED ON ADDITION THEOREM OF PROBABILITY

1. Notations:

        1) P(AUB)  : Probability of either A or B  (or)    Probability of at least one of A or B

        ii) P(AUBUC) : probability of either A or B or C   (or) Probability of at least one of A or B or C

          iii) P(A ÇB) : Probability of both A and B

                iv) P (A ÇB ÇC) = Probability of simultaneous occurrence of A, B, C

                v) P(A Ç`B)   = Probability of A but not B

            vi) P (`A Ç `B)  = Probability of neither A or B

            Now we will do addition theorems of probability.

           

Theorem : 1  : If A and B be any two events in a sample space S, then the probability of occurrence of at least one of the events A and B is given by  P(AUB) = P(A) + P(B) – P(AÇB)

Theorem : 2 : If A, B and c are any there events in a sample space S, then

                                P(AUBUC) = P(A) + P(B)+ P(C) – P(AÇB) - P(BÇC) - P(CÇA) + P(AÇBÇC)

Ex : 1  If P(A) = ¼,  P(B) = ½  P(AUB) = 5/8  then P(AÇB)  = ?

P(AUB) = P(A) + P(B) – P(AÇB)   =   +-=

Ex: 2 : Twenty tickets are numbered from 1 to 20 and one of them is drawn at random, the probability that number is divisible by 2 or 5 is.

                Total possible outcomes  = 20

                Let A denote the event that number drawn is multiply of 2 and these no. ar4e (2, 4, 6, 8, 10, 12, 14, 16, 18, 20 )

                N(A)  = 10

                B denote the event that number drawn is multiple of 5 and these number are ( 5 10 15 20 )

                n(B) = 4

So we see the from the sample space for multiple of 2 and 5 that two cases are common i.e. (AÇB) = (10 20), n(AÇB) = 2

Therefore the events are not independent.

Required probability  = P(A) + P(B) – P(AÇB)  = +-=

,

                           

1.

               

SIMPLE INTEREST

  When money is loaned, the borrower usually pays a fee to the lender this fee is called “Interest”.  Based on the method of calculation of the amount of interest given on the amount lent or borrowed, the interest can be simple interest or compound interest Simple interest or FLAT RATE interest is the amount of interest paid each year in a fixed percentage of the amount borrowed or lend at the beginning.

 IMPORTANT FACTS AND FORMULAE

1.. Principal: The money borrowed or lent out for a certain period is called the

principal or the sum.

2. Interest: Extra money paid for using other's money is called interest.

3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,

(i)                  S.I. =  (P*R*T )/100

  (ii)        P=(100*S.I)/(R*T) ;R=(100*S.I)/(P*T) and T=(100*S.I)/(P*R)

SOLVED EXAMPLES

Ex.1.Given Principal: ‘P’  interest rate:’R’ =12% =0.12, Repayment time :’T’ = 2 years

PartI: Find the amount of interest paid

              Interest=’I’= PRT

             =1,500 X0.12 X 2 = 360

PartII: Find the Total  amount to be  paid

             Total  amount to be  payable=Principal + Interest

               =1,500 + 360 =1,860

Part III: Calculatethe Weekly payable  amount

                                                                   Total  Repayments

               Weekly payment amount =   --------------------------

                                                                  Loan Period,T, in weeks

                                                                 1860

            Weekly payment amount =   -------------------  = 17.88 per week.

                                                                 2 x 52

Ex. 2. Find the simple interest on Rs. 30,000 at 16 2/3% per annum for 9 months.             

Sol.  P = Rs.30,000,R = 50/3% p.a and T = 9/12 years  = 3/4years.

      \  S.I. = (P*R*T)/100 = Rs.(30,000*(50/3)*(3/4)*(1/100))= Rs.3750

Ex. 3. Find the simple interest on Rs. 6000 at 6 1/4% per annum for the period from

4th Feb., 2005 to 18th April, 2005.

Sol. Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.

         P = Rs.6000 and R = 6 ¼ %p.a = 25/4%p.a        \S.I. = Rs.(6,000*(25/4)*(1/5)*(1/100))= Rs.75.

IMPORTANT : The day on which money is deposited is not counted while the day on which money is withdrawn is counted .

Ex.4. A sum borrowed by Augustine at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum borrowed by Augustine.

Sol.  Let sum be Rs. x then , S.I.=Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50

        \amount = (Rs. x+(27x/50)) = Rs.77x/50

        \ 77x/50 = 2502.50 Û x = 2502.50 * 50    = 1625

                                                             77

         Hence , sum = Rs.1625.

Ex. 5. If a sum of Rs. 800 amounts to Rs. 920 in 3 years at simple interest and if interest rate is increased by 8%, it would amount to how much ?

Sol. S.l. = Rs. (920 - 800) = Rs. 120; p = Rs. 800, T = 3 yrs. _

     . R = ((100 x 120)/(800*3) ) % = 5%.

       New rate = (5 + 3)% = 8%.

       New S.l. = Rs. (800*8*3)/100 = Rs. 192.

:      New amount = Rs.(800+192) = Rs. 992.          

Ex. 6. Prabhakar Rao borrowed some money from Rajaiah at the rate of 6% p.a. for the first two years , at  the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. If he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

Sol. Let the sum borrowed be x. Then,

(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400

Û  (3x/25 + 27x/100 + 14x / 25) = 11400       Û 95x/100 = 11400 Û x = (11400*100)/95 = 12000.

 Hence , sum  borrowed = Rs.12,000.      

Ex.7. If certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Callculate the sum and rate of interests.

Sol.. S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156.

          S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208         

          Principal = Rs. (1008 - 208) = Rs. 800.     

          Now, P = 800, T = 2 and S.l. = 208.

            Rate =(100* 208)/(800*2)% = 13%               

Ex. 8. What will be the  rate percent per annum will make a sum of money double in 16 years.

           Sol.. Let principal = P. Then, S.l. = P and T = 16 yrs.

                    \Rate = (100 x P)/(P*16)% = 6 ¼ % p.a.                            

Ex. 9. The simple interest on a sum of money is 4/9 of the principal .Find the rate per cent and time, if both are numerically equal.                           

         Sol. Let sum = Rs. x. Then, S.I. = Rs. 4x/9     

                    Let rate = R% and time = R years.        

               Then, (x*R*R)/100=4x/9 or R² =400/9 or R = 20/3 = 6 2/3.

               \Rate = 6 2/3 %   and Time = 6 2/3 years = 6 years 8 months.

Ex.10. A person borrowed an amount with simple interest on a certain sum of money for 2 l/2 years at 12% per annum  is Rs. 40 less than the simple interest on the same sum for 3 ½  years at 10% per annum. Find the sum.

        Sol.  Let the sum be Rs. x Then, ((x*10*7)/(100*2)) – ( (x*12*5)/(100*2)) = 40                            

                 Û (7x/20)-(3x/10)=40                 

                     Ûx = (40 * 20) = 800.

                  Hence, the sum is Rs. 800.

Ex. 11. Srinivas invested a sum  put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched him Rs. 360 more. Find the sum.    

        Sol. Let sum = P and original rate = R.

             Then, [ (P*(R+2)*3)/100] – [ (P*R*3)/100] = 360.

        Û 3PR + 6P - 3PR = 36000 Û 6P=36000 Û P=6000 

             Hence, sum = Rs. 6000.

Ex. 12. What annual installment will discharge a debt of Rs. 1092  due in 3 years

at 12% simple interest by a ?      .

    Sol . Let each Instalment be Rs. x

            Then, ( x+ ((x*12*1)/100)) + (x+ ((x*12*2)/100) ) + x = 1092

        Û ((28x/25) + (31x/25) + x) = 1092   Û (28x+31x+25x)=(1092*25)

       Û x= (1092*25)/84 = Rs.325.  

        \ Each instalment = Rs. 325.

Ex. 13. A sum of Rs. 1550 is lent outto Rammurthy to pay  into two parts, one at 8% and another one at

6%. If the total annual income is Rs. 106, find the money lent at each rate.         

      Sol. Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 - x).

      \((x*8*1)/100) + ((1550-x)*6*1)/100=106

     Û8x + 9300 –6x=10600 Û 2x = 1300  Û x = 650.

     \ Money lent at 8% = Rs. 650. Money lent at 6% = Rs. (1550 - 650) = Rs. 900.­

COMPOUND INTEREST                                                            

Compound Interest: When  the borrower and the lender agree to fix up a certain unit of time, say yearly or half-yearly or quarterly to settle the previous amount.              In such cases, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on.

        After a specified period, the difference between the amount and the money borrowed is called the Compound Interest C.I.

IMPORTANT FACTS AND FORMULAE

 Principal = P, Rate = R% per annum, Time = n years.

I. When interest is compound Annually:

                                                Amount = P(1+R/100)ⁿ

II. When interest is compounded Half-yearly:

                                Amount = P[1+(R/2)/100]²ⁿ­             

              III. When interest is compounded Quarterly:  

                Amount = P[ 1+(R/4)/100]⁴ⁿ­

IV. When interest is compounded Annually but time is in fraction, say 3(2/5) years.

Amount = P(1+R/100)³ x (1+(2R/5)/100)

V. When Rates are different for different years, say R 1%, R2%, R3% for 1st, 2nd and 3rd year respectively.

Then, Amount = P(1+R₁/100)(1+R₂/100)(1+R₃/100)

SOLVED EXAMPLES

Ex.1. Calculate the compound interest on the amount borrowed Rs. 7500 at 4% per annum for 2 years, compounded annually.

                Sol.           Amount = Rs [7500*(1+(4/100)²] = Rs (7500 * (26/25) * (26/25)) = Rs. 8112.       

            therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.

Ex. 2. Calculate the  compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.

    Sol.         Time = 2  years  4  months = 2(4/12) years = 2(1/3) years.                      

                    Amount = Rs'. [8000 X (1+­(15/100))² X (1+((1/3)*15)/100)]

                                =Rs. [8000 * (23/20) * (23/20) * (21/20)]

                                = Rs. 11109.                                          .

                   :. C.I. = Rs. (11109 - 8000) = Rs. 3109.

    Ex. 3. Find the compound interest on Rs. 10,000 in 2 years at 4% per annum, the

    interest being compounded half-yearly.                                                           (S.S.C. 2000)

    Sol.

        Principal = Rs. 10000; Rate = 2%  per half-year;  Time = 2 years = 4 half-years.          

       Amount = Rs [10000 * (1+(2/100))⁴] = Rs(10000 * (51/50) * (51/50) * (51/50) * (51/50))             

       = Rs. 10824.32.          :. C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.

   Ex. 4. Find the compound interest on Rs. 16,000 at 20% per annum for 9 months,

    compounded quarterly.

   Sol.           Principal = Rs. 16000; Time = 9 months =3 quarters;

                      Rate = 20% per annum = 5% per quarter.

                  Amount = Rs. [16000 x (1+(5/100))³] = Rs. 18522.  

      CJ. = Rs. (18522 - 16000) = Rs. 2522.

Ex.  5. If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.

Sol.

Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200.      . .

So principal=RS [100*1200]/3*5=RS 8000

Amount = Rs. 8000 x [1 +5/100]^3 - = Rs. 9261.

.. C.I. = Rs. (9261 - 8000) = Rs. 1261.

Ex. 6. In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?       (S.S.C. 2004)

Sol.

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a.                 Let the time be n years. Then,

[ 1000 (1+ (10/100))ⁿ  ] = 1331 or (11/10)^{n =}  (1331/1000) = (11/10)³

n = 3 years.

Ex.  7.  If Rs. 600 amounts to Rs. 683.20 in two years compounded annually, find the

rate of interest per annum.

                Sol. Principal = Rs. 500; Amount = Rs. 583.20; Time = 2 years.

                    Let the rate be R% per annum.. 'Then,

        [ 500 (1+(R/100)² ] = 583.20 or [ 1+ (R/100)]^{2  =}  5832/5000 = 11664/10000

      [ 1+ (R/100)]² = (108/100)²  or  1 + (R/100) = 108/100  or  R = 8  

                    So, rate = 8% p.a. 

Ex. 8. The difference between the compound interest and simple interest on a

certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.

 Sol. Let the sum be Rs. x. Then,

                    C.I. =  x ( 1 + ( 10 /100 ))² - x  =  21x / 100 , 

              S.I.  = (( x * 10 * 2) / 100) =  x / 5

                    (C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) =  x / 100

                    ( x / 100 )  =  632    x  =  63100.

                    Hence, the sum is Rs.63,100.

Ex.9. The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a. ?                                       (Bank P.O. 2003)

Sol. Let the rate be R% p.a. then,

[ 18000 ( 1 + ( R / 100 )² ) - 18000 ] - ((18000 * R *  2) / 100 ) = 405

18000 [ ( 100 + (R / 100 )²  / 10000)  -  1 - (2R / 100 ) ]  =  405

 18000[( (100 + R ) ² - 10000 - 200R) / 10000 ]  =  405

 9R² / 5  =  405   R²  =((405 * 5 ) / 9) = 225

 R = 15.

 Rate = 15%.

Ex.10.What annual payment will discharge a debt of Rs.7620 due in 3years at

16 2/3% per annum interest?

Sol. Let each installment be Rs.x.

     Then,(P.W. of Rs. x due 1 year hence)+(P>W of Rs.x due 2 years hence)+(P.W of Rs. X due 3  years hence)=7620.

\ x/(1+(50/3*100))+ x/(1+(50/3*100))² + x/(1+(50/3*100))³=7620

Û(6x/7)+(936x/49)+(216x/343)=7620.

Û294x+252x+216x=7620*343.

Û x=(7620*343/762)=3430.

\Amount of each installment=Rs.3430.

PART – II  - REASONING

SECTION  - I   - VERBAL REASONING

UNIT – VIII

a) Number series completion

b) Number Analogy

c) Number Classification

d) Coding & Decoding

a) Number series Completion

          The student after studying the given series will be able to

·      Identify the pattern followed in series

·      Complete the given series with suitable alternative

·      Find the wrong term in the series

Example: Series  : 7, 8, 13, 14, ……

                  2x3+1 = 7

                  3x3-1 = 8

                  4x3+1=13

                  5x3-1=14

                  6x3+1 = 19     Ans: 19

Model 2 : to identify the pattern in  a series

        Ex:  2, 4, 3, 9, 4, 16, 5, 25,       

        The series is in the pattern of n,

Model 3:  To find the wrong number in a series

                  9, 25, 35, 81, 121

          35 is not square, reaming numbers are squares

Exercise:

1. 1, 2, 6, 21, 22, 66, 67,  ?

2. 3, 7, 15, ?, 63, 127

3. 7, 28, 63, 124, 215, 342, 511, find the wrong number.

B) Number Analogy

Objective: At the end of the chapter the student will be able to arrive at the relationship between two quantities and can identify the quantity from the alternative provided.

The relationships may be

a) Between individual & group

b) Animal & their young ones

c) Study & topics  etc. any such relation

Example 1 : Snakes  : Hiss  : : Cat : Mew

Example 2 : Bee : hive : : Bird  : Nest

Example 3 : Carpenter : Saw : : Warrior : Sword3

Working Problems :

1. Pen : Nib : : Pencil :  ?

2. Shirt : trousers : : Saree  : ?

3.Wish : Desire : : Quarrel : ?

C) CLASSIFICATION

        Objective : The student will be able to

·      Assort items of a given group on the basic for common quantity

·      Identify the oddman out

Examples : January, May, July, November

                 Oddman is November as it has 30days & others 31days

Example2 : Bake, Peel, Fry, Boil, Roast

                 Peel is Answer as  all others are different forms of cooking

Exercise :

1. Calendar, Year, Date, Mouth, Day

2. House, Cottage, School, Place

3. Read, Write, Knowledge, Learn, Study

D) Coding & Decoding

Objective : The student will learn how

·      To decipher the rule that codes a particular word or a message

·      To identify the code & decode it

 Example 1 : If HEALTH IS written  GSKZDG  then how will NORTH be written in that code?

          The letter of the given word are wtitten in reverse order and then each letter is moved one step backword to obtain the code.   So the answer would be GSQNM

Example 2 : If ADD is bee  them how will DOG be coded ?     Ans : EPH