10 seconds part 2

5/19/2018 10 seconds part 2

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10 seconds for aptitude


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Written to sharpen young minds who are thinkers of the future


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Note about the author:


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With great gratitude, I express my thanks to

U.V.C.E, B.H.S F.G.C, J.E.M.S, B.K.M.H

That provided me the platform to rise and gave me confidence. I

thank my teachers who have helped me build a personality. To

Rao sir who helped me in times of need. Thanks to my friendsAjitHegde, Anirudh Bhat, Deepak B, Naveen H C, Taranath Aithal,Karthik

C V, Chetan L, Shridhar NC, Pramod Aithal, Medha.S.A, Priyanka

Sarafand many others along with my first batch of dear students

who were the first ones to recognize the potential in me and gaveme hope.


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Preface

10 seconds is a book meant to help students understand the concept. I believe that understanding is

more important than the formula used. The best methods have been introduced to make problem

solving simple and easy. The best approach to read this book would be in the sequence given in the

book. It works out well both for an engineering student learning it overnight for placements and for

a student who wishes to learn the topics in detail. An average student can solve a moderate question

within 10 seconds if enough practice is done with the methods given. Topics that are conventionally

boring and tedious such as calendars are explained with interesting methods found by me. Thesemethods have been found whenever a student told me that conventional method was tedious or

while teaching in the class. Memory systems like the link and peg method are inbuilt and youll not

find it tough to remember.

The reason that this book is written is because of my students who felt the need for it.

-

Sudharshan K.P.


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Contents

1) Clocks

2)

Blood relations3) Short cut techniques

4) Time and work

5) Pipes and cisterns

6) numbers

7) Calendars

8) Business mathematics

9)

Permutations and combinations

10) Probability

11) Speed. Time and distance

12) Averages and allegation

13) Data analysis

14) Syllogism

15) Analytical reasoning


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Clocks

This topic is explained by first giving you an abstract view to solve the problems. Later explanation of how

these methods have been derived is given.

Abstract view:

The Master formula that works on most problems on clocks is as follows.

Hr*30 ~ min*11

/2=

When the time is Hr:min, is the angle between hands of the clock.

In general we are asked to find at what time hands of the clock overlap, or time at which angle between

hands of clock is 90oor when the minute spacing is given.

Solving problems with the abstract view:

Find angle between hands of the clock at a) 4:32 b)5:36 c)6:54 d)7:37

solution: a) 4 : 32

*30 *11

/2


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120 176

120~176 =56othat is the acute angle between hands of the clock.

b) 5 : 36

*30 *11

/2

150 198

The difference = 48o.

c) 6 : 54

*30 *11

/2

180 297

The difference = 117o.

d) ) 7 : 37

*30 *11

/2

210 203.5

The difference = 6.5o.

to multiply by 11/2, first take half the

number. Here it is 32/2=16. Next

answer = 16+16*10 = 16+160 = 176.


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To find time when angle between hands of the clock is given:

-} the time now is between 4 and 5 O Clock. Find time if the hands of clock are at right angles.

Solution: between 4 and 5 the time is 4:m where m is some number of minutes.

Hr*30 ~ min*11

/2=

there are 2 values of time for which angle between hands of the clock is 90o. they are given by 2 formulae

given below

Hr*30 - min*11

/2=90.1.

min*11/2 - Hr*30 =90.2.

1 : 4*30 - m*11

/2= 90o.

min = 55/11.

2 : min*11

/2 - Hr*30 =90.

min = 382/11.

Time is either4:055/11 or 4:38

2/11.


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Understanding the clock

A clock has a round dial with 60 divisions indicating minutes. Each space is called a minute space. After 12

hrs the hour hand returns to its original position whereas the minute hand returns after 60 minutes(1 hour.

S0, we do know that minute hand travels faster.

The Lead:-

Consider the case when time is 12 noon. After 1 hou, the hour hand would be right backat 12 travelling 60

m.s. whereas the hour hand would have travelled 5 m.s. we can say that minute hand has travelled 55 m\s

more than the hour hand. While the time elapsed is 60 minute

55 m.s is lead in 60 minutes

When the lead becomes equal to length of the track, the faster object meets the slower one

from behind. The length of circular track is 60 m.s.

55 m.s in 60 minutes

60 minutes in ? minutes

?*55 = 60*60

? =60*60/55=720/11=655/11


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now we found thathands of the clock overlap after every 655/11 minutes.After 12 noon, they meet

for the first time at 1:055/11

hands of the clock overlap 11 times in 12 hours.

Angle between hands of the clock.

The angle about a point is 360o. The minute hand covers the angle in 60 minutes so, in a minute it

covers 360/60=6o.the hour hand covers 360

oin 12 hours i.e, 360/12=30

oan hour. Since both of them move

in parallel, if 30 minutes have elapsed,so the minute hand travels 30*6=180oand the hour hand travels 30

degree per hour*30/60hours = 15o. counting from 112 noon, the time that we have considered is 12:30.

Angle between hands of the clock is 180-15=165o.

Angle between hands of the clock when time is Hr:Min = difference between angle covered by

minute hand and angle covered by hour hand.

Angle covered by min hand = min*6o.

Angle covered by hour hand = hr*30+(min/60)*30.

Net angle between hands = hr*30+(min/60)*30 ~ min*6

= hr*30+min/2 ~ min*6

= hr*30 ~ min*11

/2.


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The time is between 3 O clock and 4, given angle between hands of the clock is 90o, find the exact

time.

Solution: between 3 and 4 the time always is 3:? (? Refers to some no of mins)

To find ?,

= 90 = hr*30 ~ min*11

/2.

90 = 3*30 ~ min*11

/2

90 = 90 ~ min*11

/2

90 = 90 - min*11/2 and90 = min*11/2 -90 are 2 forms that give us 2 different values of time.

min = 00 min = 328/11are 2 values that we arrive at.

Time can be one of these 3:00 or 3: 328/11

Incorrect clocks:

An ideal clock should be such that hands of the clock overlap after every 655/11 minutes.

If the clock is such that its hands overlap earlier than 655/11minutes then it gains time. If it takes longer

then it loses time.


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If a clock that was faster becomes slower than actual time it would have shown the correct time

somewhere in between.

-} a clock is 3 minutes slow at 8 am today and it was 2 minutes fast at 10 pm yesterday. When did it

show the correct time?

Solution: time elapsed = 10 hours(from 10 pm to 8 am)

Net variation in clock = 3+2 = 5 minutes

When the clock slows down by 2 minutes after 10 pm it shows correct time

Clock slows down by 5 minutes in 10 hours.

It slows down by 2 minutes in 10/5*2 = 4 hours. 4 hours from 10 pm the time will be 2 am this morning.

-} a clock overlaps after every 65 minutes. How many minutes would it gain or loose in 24 hours ?

Solution: a clock overlaps 22 times in 24 hours

Time gained a overlap is5/11minutes

Total time gained = 5/11*22 = 10 minutes. The clock will be 10 minutes fastafter 24 hours.


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The time in a clock is 20 minute past 2. What is the angle between the minute and hour hands of the

clock?

Given that 27th February 2003 is a Thursday. What day of the week was 27th February 1603?

integral number value?

k and then gains 2% time during the next one week. If the clock

was set right at 12 noon on a Sunday, what will be the time shown by the clock exactly 14 days from thetime it was set right?

At how many points between 10 O'clock and 11 O'clock are the minute hand and hour hand of a clock at

an angle of 30 degrees to each other?

What is the angle between the minute hand and the hour hand when the time is 1540 hours?


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Blood relations:

Create a rough picture of the relationship based on symbols given below standardize them and use to solveall the problems

+ Male

- FemaleHusband and wife

Generation gap

Siblings

-A +B

- C E+

F D+

The above chart is an example of a family chart written with the symbols.

A is female and grandmother of D(2 generation gaps)


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Both F and D are the grandchildren of A and B, they are the children of C and E

The relationship of F to others cannot be clearly said because; male or female is not specified. F may be the

brother of D or sister.

Only when a solution is 100% right youll mark it right.

A is the mother in law of E.

Mothers brother is the maternal uncle

Fathers brother is called paternal uncle.

Sisters son is called nephew.

The questions (1-8) are based on the following statements.

a) Seeta,Rajinder and Surinder are children of Mr. and Mrs.Maudgil

b) Renu,Raja and Sunil are children of Mr. and Mrs.Bhaskar

c) Sunil and Seeta are married and Ashok and Sanjay are their children

d) Geeta and Rakesh are children of Mr. and Mrs.Jain

e) Geeta is married to surinder and has three children named Rita,Sonu

and Raju.

1. How is Rajinder related to Raju?

a)brother b)uncle c)brother in law d)cousin


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e)maternal uncle

2.How is Rajinder related to Ashok?

a)brother in law b)father in law c)cousin

d)uncle

e)maternal uncle

3.How is Rakesh related to surinder?

a)brother b)cousin c)uncle d)maternal uncle

e)brother in law

4.How is rakesh related to raju?a)brother b)cousin c)uncle d)maternal uncle

e)brother in law

5.what is sanjay's surname?

a)bhaskar b)jain c)maudgil d)surinder

e)none of these

6.Renu is sanjay's

a)sister b)sister in law c)cousin d)niece

e)aunt

7.Raju's surname is


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a)Jain b)bhaskar c)maudgil d)surinder

e)none of these

8.Sunil and Rakesh are related asa)brothers b)cousins c)uncle and cousin d)brother in law

e)none of these

Passage (questions 9 to 12):

Amit is the son of Rahul.Sarika,Rahul's sister has a son sonu and a daughter Rita.

Raja is the maternal uncle of sonu.

9. How is Amit related to sonu?

a)Nephew b)Cousin(brother) c)uncle d)brother

e)none of these

10. How is rita related to raja?

a)sister b)daughter c)niece d)aunt

e)none of these

11.How many nephews does raja have?

a)1 b)2 c)3 d)4 e)none

12.What is the relationship of Raja with Rita?

a)uncle b)brother c)maternal uncle d)nephew


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e)cant be determined

Directions:Following questions pertains to Ques 13 - 15

There are six persons S1,S2,S3,S4,S5 and S6

S3 is the sister of S6

S2 is the brother of S5's husband

S4 is the father of S1 and grandfather of S6.

There are 2 fathers, one mother and 3 brothers in the family

13. Who is S5's husband?

a)S2 b)S3 c)S1 d)S4 e)S614.Who is the mother?

a)S1 b)S2 c)S3 d)S5 e)cannot be determined

15.How many male members are there?

a)1 b)2 c)3 d)4 d)cannot be determined

Passage (Questions 16-20)

Mr and Mrs sharma have two children Asha and Shashi. Shashi married Radha,

daughter of Mrs Mahajan. Suresh , son of Mrs Mahajan married Rita. Sonu and

Rocky are born to Suresh and Rita. Uma and Sudha are the daughters of Shashi

and Radha.


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16. What is Sudha's relation to Asha?

a) Sister b)niece c)Aunt d)Daughter

e)none of these

17.How is Sonu related to Mr Mahajan?

a)son in law b)sib c)grandson d)none of these

e)cannot be determined

18.How is Asha related to Radha?

a)mother in law b)aunt c)sister in law d)niece

e)none of the above

19.What is the surname of sonu?

a)Mahajan b)sharma c)shashi d)cannot be detemined

e)none20.How is suresh related to sudha?

a)brother b)maternal uncle c)uncle d)cousin

e)cannot be determined

Following information pertains to ques 21-25:

-In a family of seven three generations are living together

-The family consists two married couples having two children each

-Gopal is lucky t have two grandchildren

-There are two housewives and both are beautiful

-Gopal who is Manoj's father , is a lawyer and earns the most.


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-jyotsna is the sister of a lecturer and herself is a nurse

-Anuradha is married to a lecturer who is Nidhi's son

-Jyothika is the grand daughter of one of the housewives and is a

classical dancer

21. What is Manoj's profession?

a)student b)lecturer c)lawyer d)cannot be determined

e)none of these

22.How many male members are there in the family?

a)2 b)3 c)4 d)cannot be determined e)none of these

23.Which of the following statements is not true?

a)The nurse is sister in law of the housewivesb)Gopal has two grand children

c)Nidhi has a son and a daughter

d)Gopal has two children

e)Anuradha has a son and a daughter

24.Who are the children of Nidhi?

a)jyotsna and manoj b)anuradha and jyotsna c)anuradha and manoj

d)cannot be determined e)none

25.who among the following is one of the married couples?

a)Gopal-jyotika b)Nidhi-Gopal c)Manoj-Jyotika d)cannot be determined

e)none

26.Anil introduces Rohit as the son of the only brother of his fathers wife.How


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is Rohit related to Anil?

a)cousin b)uncle c)brother d)father

27.Pointing out to a photograph , a man tells his friend,"she is the daughter of

the only son of my fathers wife.How is the girl in the photograph related to

the man?

a)niece b)daughter c)mother d)none of these

28.X introduces Y saying ,"He is the husband of the granddaughter of the father

of my father".How is Y related to X?

a)brother b)brother in law c)daughter d)datainadequate

Questions 29-31: I)In a family of six persons A,B,C,D,E and F there are two married

couplesII)D is grandmother of A and mother of B

III)C is wife of B and mother of F

IV)F is the grand daughter of E

now answer the following questions based on the above conditions.

29.What is C to A?

a)mother b)grandmother c)daughter d)granddaughter

30.How many male members are there in the family?

a)cannot be determined b)2 c)3 d)0

31.Who are the two couples?

a)BC and DE b)AC and DB c)cannot be determined d)none of these


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32.What will be the daughter of the woman who is the mother of the husband of my

mother to me?

a)mother b)aunt c)grandmother d)niece

33.A woman sees the photograph of a man and saus."this mans sister is my mother in

law". HOw is the womans husband related to the man in the photograph?

a)uncle b)brother c)father d)none of these

34.How is your mothers sisters brothers wifes child related to you?

a)sister b)brother c)cousin d)none of these

Directions(Questions 35-39):

There are six persons A,B,C,D,E and F.C is the sister of F.B is the brother

of E's husband.D is the father of A and grandfather of F.There are two fathers , threebrothers and a mother in the group.

35.Who is the mother?

a)A b)B c)D d)E

36.Who is E's husband?

a)B b)C c)A d)F

37.How many male members are there in the group?

a)one b)two c)three d)four

38.How is F related to E?

a)uncle b)husband c)son d)daughter

39.Which of the following is a group of brothers?


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a)ABF b)ABD c)BFC d)BDF

40.A party consists of grandmother,father,mother,four sons and their wives and one

son and two daughters to each of the sons.How many females are there is all?

a)14 b)26 c)18 d)none of these


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Short cut techniques

** a student by himself can design shortcuts, that would be the best way to pace up. Those that are

given here are just to tell you that you can have shortcuts.**

Specialty of certain numbers:

11: multiplication by 11is fast and easy

2 3 4 5 2 3 6 2 4 5 2 7 2 1 2 3 2 3 * 11

Answer: add

2 5 7 9 7 5 9 8 6 9 7 9 9 3 3 5 5 5 3

Find a) 232456*11

b) 5342.56*11

15: multiplication by 15

Multiply the number by 10 add result to half of itself

23402*15 = 234020+1/2(234020) = 351030.

9: division by 9


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Write the first number. Add this to second number if sum does not exceed 9 write it in place of result if it

exceeds 9 add 1to the previous place and write the remainder of sum divided by 9

3 2 8 / 9

3 513

/9 = 364/9

vedic method for multiplication:

multiply the numbers in units place write units place of result in the answer. 10s place is carried forward.

select 2 places next. for 3 digit by 3 digit multiplication selection of places is 1st one each, 1st two, all three, last two, last one

5 3 4 5 3 4 5 3 4 5 3 4 5 3 4

*2 3 2 2 3 2 *2 3 2 *2 3 2 *2 3 2

5*2 3*2 + 4*3 5*2+3*3+4*2 3*2+4*3 4*2

= 10 = 18 = 27 =18 = 8

answer: 1 2 0 8 8 8.

Other techniques are

base method for multiplication, common multiple methods(in this book), multiplication with fingers,

graphical methods of calculation.

But, the ones that are self designed are the best.


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TIME AND WORK

The concept of time and work has been a very necessary part of thinking. It is in the estimation of time

taken for a job since time unknown. From when Taj Mahal was built till todays tunnels, railway lines and

every imaginable improvement needs this estimation.

This unit will take you from the simplest examples to some of the complex problems that are faced.

-} Vikram ate 2 loaves of bread in some time. In the

same time Ram could eat 3. If they together ate a

pound in 12 min, in what time can ram alone eat

it.(pound has 10 loaves)

Did you find the answer in 10 seconds?? If not check

this out.

Soln: their capacity to eat is in the ratio

2:3::Vikram:Ram the no of loaves each one ate will also be in the same ratio so, Vikram ate 4 while Ram

ate 6. This happened in 12 minutes. Ram eats 6 loaves in 12 mins so he can eat a loaf in 2 mins and 10

loaves in 20 mins.

Conclusion: when time given is

same, work done by different

individuals is proportional to

their capacity to do work.


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Based on this conclusion we have different methods of analyzing problems. I shall introduce you to some

of them and well then follow the best.

-} Ram can do a job in 10 days. Lakshman can do it in 15 days. In how many days will they together

complete the job?

1 Per unit method:

The unit here refers to days, hours, mins depending upon the situation.

In the problem given above days is the unit.

If Ram completes 1 job in 10 days, he will do 1/10th

of it in a day. Similarly, Lakshman will do 1/15th

of it.

Working together, 1/10+1/15= 1/6th

of the work is done in a day.

If 1/6th

work is done in a day, in 6 days1 work is done.

Drawback: dealing with fractions.

2 Relative Work method:

Name Time Capacity Relative

work (RW)


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Ram 10 days 3 units

30 unitsLakshman 15 days 2 units

Together 6 days 5 units

Take the common multiple of 10 and 15 not necessarily the LCM but a multiple is fine.

Common multiple difference method:

Difference = 5*2=10

10 15 15*2=30

To find capacity , capacity = RW/Time.

When working together, their capacity is summed.

Time = RW/Capacity.

Note: This method is the easiest and most powerful one.


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3. Graphical Method.

10 cm

Any length 6 cm

15 cm

This method is purely of interest for readers and yet to be developed for complex problems.

We have other methods like the percentage method and direct method. But the methods introducedare more than sufficient to solve problems that we come across.


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Sample problems solved : (each problem solved is different so they are to be noticed carefully.)

A and B together complete a work in 12 days, B and C together in 20 days, C and A in 15 days. In how

many days can A, B and C together complete it?

Solution:

A and B 12 5

60B and C 20 3

C and A 15 4

A, B and C 6

Sum of units per day work of A+B + B+C + C+A = 5+3+4 =12 units =2(A+B+C)

A+B+C = 12/2 = 6 units per day.

RW = 60, time(A+B+C) = 60/6 = 10 days.

Group of people with same capacity:

The number of workers required to complete a task when all of them can work at the same rate is

explained here.

Concept: Effort input/Work output = constant.


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Classification:

Input: Number of days.

Number of hours they work per day.

Number of men working.

Output: work done in terms of length of road.

Volume of a structure.

Surface of the interior of the building. (painting)

Area of the floor.

Depending upon the question input and outputs are incorporated while solving the problem

-} 14 men built a tunnel 12 meters long in 45 days working 6 hours a day. How many men are required to

build a tunnel 60 meters long in a month if they work 7 hours a day.

Solution:

Case 1: input = 14 men, 45 days, 6 hours/day

Output = 12 meter long tunnel.


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Case 2: input = ? men, 30 days, 7 hours/day

Output = 60 meter long tunnel.

14*45*6/12 = ?*30*7/60

? = 90.

90 men are required for the job.

Dividend is the amount paid to workers.

When a sum of money has to be distributed among the employees, it is done based on a simpleprinciple that the man who works more earns more.

The concept of ratio is used here. If we have two people working together and both of them have

been working for the entire period. Then, sum of money is divided in the ratio of their capacities to work.

Rahim chopped 2 trees in a day while Azad chopped 3. After working for a week they together

received a sum of 5000 rupees. What amount should Azad get??

We know that in a day 5 trees were cut, Azad contributes to three of them. So he gets 3/5th

of the total

money i.e. 3/5*5000=3000 rupees.


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-}15 women and 12 men together complete a job in 24 days. what is the amount a woman will get when

the total amount paid is 6200 dollars, If 3 days work of a man is done by a women in 4 days.

Solution: we will write no of men in terms of women since we know the ratio of their capacities.

3 men complete a job in a day that will take 4 women.

12 mens work will be done by 16 women (in ratio 3:4).

Total no of workers = 15 + 16 = 31 (in terms of women).

Total Amount paid is 6200 dollars for 31 women

So each women gets 6200/31 = 200 dollars.

[sum of money is divided in the ratio of their capacities to work when time is same.]

Cyclic work

We come across situations where people do not work continuously, in such cases the concept of

cyclic work will help us solve the problem.

Heena and Taj work alternately to complete a job that heena can do in 7 days and taj in 10 days, after

how many days is the job done and who finishes it?


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person Time capacity RW

Heena 7 10 70

Taj 10 7

Difference = 3*7/3=7

7 10 10*7=70 = RW

.

Heena can complete 10 units in a day while Taj can complete 7 units in a day. When 70 units are done the

job is complete.

We shall consider first 2 days (one days work of Heena and next days work of Taj) as one cycle. So, in a

cycle 10+7=17 units of work are done and 2 days are elapsed. Take the multiple of 17 just less than

70(nearest multiple).

17*4=68 , 68 is the multiple nearest to 70. Now 68 units are done in 4 complete cycles.

4 cycles correspond to 8 days and on the 8th

day taj would be at work. Still we have 70-68=2 units of work

left out that Heena completes on the 9th

day because it would be her turn.


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The exact time taken by heena to complete 2 units is 2/10 days = 1/5day. So total time taken is 81/5 days

and heena completes the job.

Ram completes 60% of a task in 15 days and then takes the help of Rahim and Rachel. Rahim is 50%

as efficient as Ram is and Rachel is 50% as efficient as Rahim is. In how many more days will theycomplete the work?

Working together, A and B can do a job in 6 days. B and C can do the same job in 10 days, while Cand A can do it in 7.5 days. How long will it take if all A, B and C work together to complete the job?

Four men and three women can do a job in 6 days. When five men and six women work on the samejob, the work gets completed in 4 days. How long will a woman take to do the job, if she works aloneon it?

Shyam can do a job in 20 days, Ram in 30 days and Singhal in 60 days. If Shyam is helped by Ram

and Singhal every 3rdday, how long will it take for them to complete the job? Pipe A usually fills a tank in 2 hours. On account of a leak at the bottom of the tank, it takes pipe A 30

more minutes to fill the tank. How long will the leak take to empty a full tank if pipe A is shut?

A, B and C can do a work in 5 days, 10 days and 15 days respectively. They started together to do thework but after 2 days A and B left. C did the remaining work (in days).

X alone can do a piece of work in 15 days and Y alone can do it in 10 days. X and Y undertook to do itfor Rs. 720. With the help of Z they finished it in 5 days. How much is paid to Z?

A and B can do a piece of work in 21 and 24 days respectively. They started the work together andafter some days A leaves the work and B completes the remaining work in 9 days. After how manydays did A leave?


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A can complete a project in 20 days and B can complete the same project in 30 days. If A and B startworking on the project together and A quits 10 days before the project is completed, in how many dayswill the project be completed?

Pipes and cisterns

A farmer opens the valve of a tank. He knows in what time it can be emptied. But if the inlet to the

tank is also open, after what time should he return to turn off the valves?

Water from the river enters a reservoir. The gates should be open when reservoir is full to its brim. If

the reservoir is half full now, after what time should the gates open? Will the reservoir overflow even

when gates are open?

Answers to these questions can be found easily. The method used is the relative value method described

while learning time and work.

Consider a simple situation where a tank has an inlet and one outlet. If the inlet alone can fill the tank in 30

minutes and outlet alone can empty it in 40 minutes, in what time will it be full when both inlet and outletare open.


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To solve it, we have to find a common multiple of the time given. LCM of 30 and 40 is 120. This

common multiple is considered to be a relative value of the capacity of the tank. Assume the tanks

capacity is 120 liters. With this assumption, find the rate of water flow in the inlet and outlet. Inlet fills 120

liters in 30 minutes rate of water flow is 120/30 = 4 liters per minute. Outlet empties 120 liters in 40minutes, hence rate of flow is 120/40 = 3 liters per minute. Every minute 4 liters of water enter and 3 liters

flow out hence 1 liter remains per min. water in the tank increases at the rate of 1 liter per minute. To fill

120 liters it takes 120 minutes.

valve Time

(min)

Rate (liters

per min)

Relative

value(liters)

Inlet 30 +4

120Outlet 40 -3

net 120 1

Inlet helps (+), outlet opposes (-), net rate is sum.

-} A hot tap fills a bucket in 30 seconds while a cold tap fills it in 20 seconds. In how many seconds will they

together fill it?

TapTime

secondsRate litersper second

Relative

value liters

Hot inlet 30 +2 60


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Cold inlet 20 +3

Net 12 5

-} A bucket that should fill in 20 minutes actually takes 26 minutes to be full, due to a hole. In how many

minutes can the hole empty a full bucket?

Sol: net time is given, outlets time to be found.

Opening Time Rate Relative

valueInlet 20 13 260

Hole 862/3 -3

Net 26 10

Net is 10 so holes rate should be -3.

-} There are 2 inlets to a tank. One can fill it in 9 minutes and the other in 18 minutes. The operator opens

both and returns at the time when tank should be full by his calculation. He finds that the outlet is open

and he closes it. It takes another 3 minutes to fill the tank. In what time can the outlet alone empty it?

Sol: operators calculation


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40

Time Rate Value

Inlet1 9 2

18Inlet2 18 1

net 6 3

Operator returns after 6 minutes. But, the tank is not full. It takes 3 more minutes when outlet is closed.

From the table above we know that when the only inlets are open rate is 3. For 3 minutes they fill 9 liters.

If 9 out of 18 liters were filled in last three minutes. 9 liters were remaining after first 6 minutes. Inlets

added 18 but only 9 remained so, the outlet should have removed 9 liters in 6 minutes. The outlet can

remove 18 liters in 12 minutes.

Cyclic operation:

The concept is the same as used in time and work but both positive and negative rates are

encountered here.

-} the inlets C and D can fill a tank in 12 minutes and 14 minutes respectively. Another outlet can empty it

in 84 minutes. If they are operated in cyclic manner for 1 minute each, starting from C, which pipe will take

the last turn and time taken? Sol: first well make a table. Find relative capacity, rates this makescalculation simpler.

Pipe Time Rate Value


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C 12 7

84D 14 6

Outlet 84 1

After 1st

minute the tank holds 7 liters

At the end of 2nd

minute it holds 7+6 = 13 liters

Then, after 3rd

minute it holds 13-1 = 12 liters.

12 liters are filled in a 3 minute cycle.

Take the multiple of 12 just less than 84

i.e 6thmultiple = 12*6=72 at the end of 6 complete cycles 72 liters are filled. 6 cycles take 6*3=18 min.

next minute, C operates tank is 72+7 = 79 liters full

later D operates and tank is 79+6 = 85 already full before the minute is over.

Reanalyzing, tank is 79 liters full in 18+1 = 19 min.

Next 1 minute D can fill 6 liters but only 84-79=5 liters are required. So it takes 5/6 minutes more.

Total time taken is 195/6minutes. Dfinally fills it.


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42

Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened

simultaneously, how much time will be taken to fill the tank?

A pipe can fill a tank in 16 hours. Due to a leak in the bottom,it is filled in 24 hours. If the tank is full,how much time will the leak take to empty it ?

A cistern is filled by pipe A in 10 hours and the full cistern can be leaked out by an exhaust pipe B in

12 hours. If both the pipes are opened, in what time the cistern is full?

Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened

simultaneously and it is found that due to leakage in the bottom, 32 minutes extra are taken for the

cistern to be filled up. When the cistern is full in what time will the leak empty it ?

Pipes A and B can fill a tank in 20 hours and 30 hours respectively and pipe C can empty the full tank

in 40 hours. If all the pipes are opened together, how much time will be needed to make the tank full

?

Two pipes A and B can frll a tank in 24 min. and 32 min. respectively. If both the pipes are opened

simultaneously, after how much time B should be closed so that the tank is full in 18 minutes?


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Two pipes A and B can fill a tank in 36 min. and 4S min. respectively. A water pipe C can empty the

tank in 30 min. First A and B are opened. After 7 minutes, C is opened. In how much time, the tank Is

full ?

A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to

empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty

the tank can empty it in 6 hours. If all the pipes are kept open when the tank is full, it will take exactly

6 hours for the tank to empty. How many of these are fill pipes?

Pipe A fills a tank in 30 minutes. Pipe B can fill the same tank 5 times as fast as pipe A. If both the

pipes were kept open when the tank is empty, how much time will it take for the tank to overflow?

Pipe A fills a tank of 700 litres capacity at the rate of 40 litres a minute. Another pipe B fills the same

tank at the rate of 30 litres a minute. A pipe at the bottom of the tank drains the tank at the rate of

20 litres a minute. If pipe A is kept open for a minute and then closed and pipe B is kept open for a

minute and then closed and then pipe C is kept open for a minute and then closed and the cycle

repeated, how long will it take for the empty tank to overflow?

Two pipes A and B can fill a tank in 36 hours and 45 hours respectively.If both the

pipes are opened simultaneously,how much time will be taken to fill the tank?


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44

a)10 hrs b)15 hrs c)20 hrs d)25 hrs

Two pipes can fill a tank in 10 hours and 12 hours respectively while a third pipe

empties the full tank in 20 hours.If all the three pipes operate simultaneously,in

how much time will the tank be filled?a)7 hrs 30 min b)5 hrs c)7 hrs d)none of these

If two pipes function simultaneously,the reserviour will be filled in 12 hours.One

pipe fills the reservoir 10 hours faster than the other.How many hours does it take

the second pipe to fill the reservior?

a)10 b)20 c)30 d)40

A cistern has two taps which fill it in 12 minutes and 15 minutes respectively.There

is also a waste pipe in the cistern.When all the three are opened,the empty cistern

is full in 20 minutes.How long will the waste pipe take to empty the full?

a)5 min b)10 min c)15 min d)20 min

An electric pump can fill a tank in 3 hours.Because of a leak in the tank it took 3 1/2hours to fill the tank.If the tank is full,how much time will the leak take to empty

it?


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a)17 b)18 c)20 d)21

Two pipes can fill a cistern in 14 hours and 16 hours respectively.The pipes are opened

simultaneously and it is found that due to leakage in the bottom it took 32 minutes moreto fill the cistern.When the cistern is full in what time will the leak empty it?

a)110 hrs b)111 hrs c)112 hrs d)none of these

Two pipes A and B can fill a tank in 36 min and 45 min resp.A water pipe C can empty

the tank in 30 min.First A and B are are opened.After 7 min,C is also opened.In how much

time,the tank is full?


Two pipes A and B can fill a tank in 24 min and 32 min. resp.If both the pipes are

opened simultaneously,after how much time B should be closed so that the tank is full

in 18 min?


Two pipes A and B can fill a tank in 20 and 30 min resp.If both the pipes are used

together , then how long will it take to fill the tank?


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46


Three pipes A,B and C can fill a tank in 6 hours.After working at it together for 2 hrs,

C is closed and A and B can fill the remaining part in 7 hours.The number of hourstaken by C alone to fill the tank is

a)10 b)12 c)14 d)16

A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in

9 hrs.If both the taps are opened simulataneously,then after how much time will the

cistern get filled?

a)4.5 hrs b)5 hrs c)6.5 hrs d)7.2 hrs

Two pipes A and B can fill a cistern in 37 1/2 minutes and 45 minutes resp.Both pipes

are opened.The cistern will be filled in just half an hour,if the pipe B is turned off

after


A tap can fill a tank in 6 hours.After half the tank is filled,three more similar taps

are opened.What is the total time taken to fill the tank completely?


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a)3 hrs 15 min b)3 hrs 45 min c)4 hrs d)4 hrs 15 min

Two pipes can fill a tank in 20 and 24 min resp. and a waste pipe can empty 3 gallons

per minute.All the three pipes working together can fill the tank in 15 min.The capacityof the tank is

a)60 gallons b)100 gallons c)120 gallons d)180 gallons

A water tank is two fifth full.Pipe A can fill a tank in 10 min and pipe B can empty it

in 6 min.If both the pipes are open,how long will it take to empty or fill the tank

completely?

a)6 min to empty b)6 min to fill c)9 min to empty d)9 min to fill

A leak in the bottom of a tank can empty the full tank in 8 hours.An inlet pipe fills

water at the rate of 6 litres a min.When the tank is full,the inlet is opened and due to

the leak,the tank is empty in 12 hrs.How many litres does the cistern hold?

a)7580 b)7960 c)8290 d)8640

Pipe A can fill a tank in 5 hrs,pipe B in 10 hrs and Pipe C in 30 hrs.If all the pipes

are open,in how many hours will the tank be filled?


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48

a)2 b)2.5 c)3 d)3.5

A booster pump can be used for filling as well as for emptying a tank.The capacity of

the tank is 2400 cubic meter.The emptying capacity of the tank is 10 m^3 per min higherthan its filling capacity and the pump needs 8 min lesser to empty the tank than it needs

to fill it.What is the filling capacity of the pump?


Pipes A and B can fill a tank in 5 and 6 hours respectively.Pipe C can empty it in 12 hrs.

If all the three pipes are opened together,then the tank will be filled in

a)1 13/17 hrs b)2 8/11 hrs c)3 9/17 hrs d)4 1/2 hrs

Three taps A,B and C can fill a tank in 12,15 and 20 hours respectively.If A is open all

the time and B and C are open for one hour each alternately,the tank will be full in

a)6 hrs b)6 2/3 hrs c)5 hrs d)7 1/2 hrs

Three pipes A,B and C can fill a tank from empty to full in 30 min,20 min and 10 min resp.When the tank is empty,all the three pipes are opened A,B and C discharge chemical solutions

P,Q and R respectively.What is the proportion of solution R in the liquid in the tank after

3 min


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a)5/11 b)6/11 c)7/11 d)8/11

Two pipes A and B can fill a tank in 6 hrs and 4 hrs resp.If they are opened on alternate

hours and if pipe A is opened first,in how many hours,the tank shall be full?a)4 b)4 1/2 c)5 d)5 1/2

Two pipes A and B can separately fill a cistern in 60 min and 75 min respectively.There is

a third pipe in the bottom of the cistern to empty it.If all the three pipes are simultaneously

opened,then the cistern is full in 50 min.In how much time,the third pipe alone can empty

the cistern?


Two pipes A and B can fill a cistern in 12 min and 15 resp,while a third pipe C can empty

the full tank in 6 min.A and B are kept open for 5 min in the beginning and then C is also

opened.In what time the cistern is emptied?

a)30 min b)33 min c)37 1/2 min d)45 min

A pump can fill a tank with water in 2 hrs.Because of a leak,it took 2 1/3 hours to fill the

tank.The leak can drain all the water of the tank in

a)4 1/3 hrs b)7 hrs c)8 hrs d)14 hrs


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50

A large tanker can be filled by two pipes A and B in 60 min and 40 min resp.How many min will

it take to fill the tanker from empty state if B is used for half the time and A and B fill

it together for another half?

a)15 min b)20 min c)27.5 min d)30 min

Two taps A and B can fill a tank in 5 hrs and 20 hrs resp.If both the taps are open then due

to leakage,it took 30 min more to fill the tank.If the tank is full,how long will it take

for the leakage alone to empty the tank?

a)4 1/2hrs b)9 hrs c)18 hrs d)36 hrs

Two pipes A and B can fill a tank in 15 hrs and 20 hrs resp while a third pipe C can empty

the full tank in 25 hrs.All the three pipes are opened in the beginning.After 10 hours,C is

closed.In how much time will the tank be full?

a)12 hrs b)13 hrs c)16 hrs d)18 hrs

Two pipes A and B together can fill a cistern in 4 hours.Had they been opened separately,then

B would have taken 6 hours more than A to fill the cistern.How much time will be taken by A

to fill the cistern separately?a)1 hr b)2 hrs c)6 hrs d)9 hrs


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Two pipes A and B can fill a tank in 15 min and 20 min resp.Both the pipes are opened together

but after 4 min,pipe A is turned off.What is the total time required to fill the tank?

a)10 min 20 sec b)11 min 45 sec c)12 min 30 sec d)14 min 40 sec

One pipe can fill a tank three times as fast as another pipe.If together the two pipes canfill the tank in 36 min,then the slower pipe alone will be able to fill the tank in



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52

Numbers

Real complex

Rational irrational

Integers non terminating,

Decimal numbers non recurring

Terminating numbers

Recurring decimals

Integers

Negative non negative positive

(-1, -2, .. -) (0,1,2,..) (1,2,3,.)

Whole numbers Natural numbers


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Natural numbers

Even(E) Odd(O) Prime Composite

E+E = E E*E = E E/E = E

E+O = O E*O = E E/O = E

O+O = E O*O= O O/O = O

O/E = Decimal

Prime factorization:

N = P1a * P2b * P3c

Number of factors of N = (a+1)(b+1)(c+1)

Ex: 150 = 3*5*5*2

= 31

* 52* 2

1

Number of factors = (1+1)(2+1)(1+1) = 12.


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54

Highest common factor (greatest common multiple)

Method 1: long division method

2 80 140

2 40 70 HCF = 2*2*5 = 20

5 20 35 N1 = HCF*C1 = 20*4 = 80

4 7 N2 = HCF*C2 = 20*7 = 140

C1 C2

Method 2: prime factorization method

80 = 2*2*2*2*5

= 24

* 5 TAKE THE COMMON FACTORS

140 = 2*2*5*7 WITH THEIR LOWEST POWERS

= 22* 5 * 7 AND MULTIPLY THEM

HCF = 22* 5 = 20

d f d


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Method 3: simple division method

80 140 1

-80

60 80 1

-60

20 60 3

-60

00

HCF = 20

KPs method:

Difference= 60 *80/60= 80

80 140

To get back80/60=

4/3 HCF=

60/3= 20

10 d f i d


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56

Lowest common multiple:

Method 1: long division method

2 80 140

2 40 70 LCM = 2*2*5*4*7 = 560

5 20 35 N1 = LCM/C2 = 560/7=80

4 7 N2 = LCM/C1 =560/4 =140

C1 C2

Method 2: prime factorization method

80 = 2*2*2*2*5

= 24

* 5 TAKE ALL THE FACTORS

140 = 2*2*5*7 WITH THEIR HIGHEST POWERS

= 22* 5 * 7 AND MULTIPLY THEM

LCM = 24* 5 * 7 = 560

10 d f tit d


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KPs method:

Difference= 60 *80/60= 80

80 140

To get back80/60= 4/3 HCF=

60/3= 20

LCM = 140*4= 560

Sum of numbers N1+N2 = HCF*(C1+C2)

Difference N1-N2 = HCF*(C1-C2)

Product N1*N2 = HCF*LCM

For fractions:

LCM =LCM of numerators

/HCF of denominators

HCF =HCF of numerators/LCM of denominators



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58

Remainders

If N/a gives remainder R

And N/b gives the same remainder R

N = k * ( LCM(a, b) ) + R

k = 1,2,3, least value is obtained when k=1

Divisibility rules:

To be divisible by

2 => last digit should be even

3 => sum of digits should be divisible by 3

4 => last 2 digits divisible by 4 or 00

5 => last digit should be 0 or 5

6 => it should be divisible by 2 and 3



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7 => method 1

Ex:- 343 3 4 3

- 6 * 2

2 8

28 isdivisible by 7 so 343 is also divisible.

Method 2

(1/7)r= 1, (10/7)r= 3, (100/7)r = 2, (1000/7)r = -1

(10000/7)r= -3 the same sequence follows that is

2 3 1 -2 -3 -1 2 3 1.

Ex:- 563829

The numbers along with place values are written.

5 6 3 8 2 9 result should be

*-2 *-3 *-1 *2 *3 *1 0 or 7

-10 -18 -3 +16` +6 +9 = 0



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60

=>8 last 3 digits should be divisible by 8 or 000

=> 9 sum of digits should be divisible by 9

=> 10 last digit 0

=> 11 sum of digits at odd places sum of digits at even places should be 0 or multiple of 11.

Ex:- 353, -3 +5 -3 = 1 it is not divisible

352, -3 +5 -2 = 0 it is divisible

=>12 should be divisible by both 4 and 3

For any number, its factors who are a prime composite pair should be taken

N is divisible by X, if X is divisible by both prime composite factors

Prime composite factors: pair of factors of the number who have only one common factor 1.

Power cycles:

Find units place of 22222435

* 893456

?

Sol: only last digit of answer is written in table below



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X1

X2

X3

X4

X5

frequency

1 1 1 1 1 1

2 4 8 6 2 4

3 9 7 1 3 4

4 6 4 6 4 25 5 5 5 5 1

6 6 6 6 6 1

7 9 3 1 7 4

8 4 2 6 8 4

9 1 9 1 9 2

We can observe from the table above that units place of the result repeats again.

This happens in a cyclic manner

The number of powers after which the number x occurs in units place is the frequency

For any number, to find the power cycle you need not write the whole table

Cycle of that particular number is enough.

If the 3rd

power of 8 ends with 2, 8*2 = 16 ends with 6 so 4th

power ends with 6, finding the units place

value of power is very easy



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62

22224to find units place for this, from cycle of 2, that is 2 4 8 6 we know that 4

thplace

corresponds to 6. Hence the result will end with 6.

222256

to find units place for this, from cycle of 2 we know that its frequency is 4. Find remainder of the

division (56/4)r= 0 if the remainder is 0 when divided by 4 it is same as remainder 4.

So 4th

place in cycle is 6, the result will end with 6

The question asks us to find units place of 22222435

* 893456

22222435

will end with?

Cycle of 2 has frequency 4,

(2435/4)r= from divisibility rules we know that for a number to be divisible by 4, its last two digits should

be divisible by 4. Remainder obtained by dividing last 2 digits is the same as remainder of whole division.

(35/4)r=3 , (2435/4)r=3

So third place in the cycle of 2 = 8.

893456

will end with?

Units place is 3, cycle of 3 has frequency 4

(456/4)r=0. Hence the 4th

place in cycle of 3 = 1.



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8 * 1 will end with 8.

Hence the above calculation ends with 8.

Finding remainders for numbers with powers

Find the remainder when 375

is divided by 4?

Sol:

express the numerator in terms of the denominator.

3 = 4-1 hence (375

/4)r= ((4-1)75

/4)r

From binomial theorem, we know that in the result of the expansion (4-1)75all the terms will have 4 except

the last term. So, all the terms except the last one is a multiple of 4 so, obviously they will not leave any

remainder when divided by 4. Proceed with the remaining

((-1)75

/4)r= -1 remainder is -1. It is the same as 4-1 =3.



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64

Progressions

Arithmetic progression:

a, a+d, a+2d, a+3d. a+(n-1)d (nth

term)

a = 1st

term, d = common difference, n = number of terms

nth

term , Tn= a + (n-1)d.

summation, Sn=n/2(1

stterm + last term)

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

Arithmetic mean = (a+b+c)/3.

Geometric progression:

a, ar, ar2, ar

3,a

4, . ar

n-1

a = 1st

term, r = common ratio, n = number of terms, Tn= arn-1

.

Sn= a(rn- 1)/(r-1) for r>1

Sn= a(1-rn)/1-r for r


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f p

Geometric mean = 2 Ex:- A ball is dropped from a height of 10 meter. Every time it bounces back to 3/5

thof its previous height.

What is the distance travelled by the ball before coming to rest.

Sol: Distances travelled are 10m, 103/5m, 10(

3/5)

2m.. and so on till 0

a = 10, r =3/5, n = 1,2,3...

Sn as n

Sn = a/(1-r) = 10/(1-3/5) = 25m downwards,

Upwards, it is 25-10 = 15m upwards,

Total distance travelled = 25+15 = 40m.



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f p

66

Ex:-find total length of lines in the figure.

20 cm

Sol:

10 cm

10 cm

Length of hypotenuse from Pythagoras theorem = = 10*21/2 = 20/Perimeter = 4( 20 + 20/+ 20/)2+.)

= 4 Sn = 4 * ( 20/(1-1/) ) = 80

.

1.Find the number of factors that the number 2025 has.

a)12 b)13 c)15 d)14



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2.In how many ways can 14630 be writen as the product of two factors

a)14 b)15 c)23 d)16

3.Find the largest 3 digit multiple of 28.

a)980 b)987 c)986 d)578

4.Find the smallest number which when divided by 4 or 7 leaves a

remainder of 3 in each case and the number being greater than the

two divisors

a)30 b)76 C)21 d)31

5.Find the largest four digit number which when divided by 5 and 9

gives remainders of 3 and 7 respectively

a)8899 b)9988 c)9898 d)89896.Find the smallest number which when divided by 7 leaves a remainder

of 6 and when divided by 11 leaves a remainder of 8

a)40 b)20 c)59 d)41

7.Find the remainder of the division 3^42/4

a)0 b)1 c)2 d)3

8.Which of the following is divisible by 19?

a)614120 b)860472 c)921194 d)998777

9.What should be added to 475935 so that it becomes a multiple of 11?

a)2 b)3 c)7 d)8



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68

10.What should be subracted from 478185 so that it becomes a multiple

of 19?

a)7 b)8 c)10 d)12

11.There are some sweets with me.If i distribute them equally among 10,

16 or 20 children,i would be left with one sweet in each case.If i

distribute the sweets equally among 20 children,I do not have any

sweets left with me.How many sweets do i have?

a)92 b)115 c)150 d)161

12.What is the largest number which,divides 288,528 and 708, and leave

the same remainder in each case?

a)30 b)60 c)75 d)15013.What is the smallest five digit number which when divided by 7,11 and

21 leaves a reminder of 5 in each case?

a)10164 b)10169 c)10118 d)10123

14.From a certain city,buses start for four different places every 15,20,

25 and 30 minutes starting from 8 a.m.At what time,for the first time

after 8 a.m. would all the buses start together again?

a)10 a.m b)12 noon c)1 p.m d)2 p.m

15.A number when divided by 928 leaves a reminder 244.What would be the

reminder when the number is divided by 58?



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a)8 b)12 c)17 d)23

16.When a three digit number is divided by 64484 and 62767,the remainder

is the same in both the cases.What is the remainder?

a)101 b)458 c)757 d)cannot be determined

17.if x+1/x = 4 , find the value of x^4-1/x^4(x>1)

a)112 b)112sqrt3 c)224 d)none of these

18.Which of the following is a perfect square?

a)20,734 b)61,504 c)71,295 d)77,286

19.How many digits are there in the smallest number containing all 9's

which is divisible by 13?

a)4 b)5 c)6 d)820.If the number 7448x24y is divisible by 72,find the value of x-y given

x not equal to y


21.The difference of a four digit number and any number formed by permutin

its digits would always be divisible by

a)18 b)11 c)10 d)9

22.Find a number such that it exceeds 18 by three times the number by which

it is less than 86

a)32 b)69 c)54 d)67



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70

23.Find the smallest number,which when divided by 31 leaves as remainder 7

and when divided by 25 leaves a remainder 6

a)106 b)81 c)131 d)162

24.Find the greatest number with which when 565,847 and 1551 are divided,

the respective remainders are 5,7 and 11

a)70 b)140 c)170 d)210

25.Find the remainder when 3^50 is divided by 11


26.Find the smallest number which when increased by 5 is divisible by 36,

108,126 and 198

a)4422 b)8311 c)7764 d)686727.What is the minimum number of identical square tiles required to cover

a floor of dimensions 3 m 78 cm by 4 m 80 cm

a)3200 b)5040 c)7600 d)8100

28.The HCF and LCM of a pair of numbers are 11 and 1001 resp.Find the smaller

of the two numbers given that their sum is 220

a)11 b)33 c)55 d)77

29.Find the greatest power of 20,which can divide 200!

a)10 b)49 c)98 d)104

30.Find the number of factors of the number 46200



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a)64 b)78 c)96 d)106

31.Find the highest power of 5 contained in 250!

a)50 b)62 c)78 d)84

32.Find the largest five digit number which when divided by 8 leaves remainder 3

and when divided by 7 leaves remainder 1?

a)99948 b)99953 c)99960 d)99947

33.When the numbers 5,7 and 11 divide a multiple of 17,the remainders left are

,respectively,4,6 and 10.Which multiple of 17 gives the least number that

satisfies the given condition?


34.How many times the HCF , of the fractions 3/14,6/35 and 16/21 is their LCM?a)1440 b)643144 c)210 d)48

35.What is the maximum number of ways in which the number 11025 can be expressed

as the product of a pair of co prime factors

a)1 b)2 c)4 d)8

36.What is the number of different ways in which the number 784 can be expressed

as a product of the two different factors?


37.What is the sum of all the coprimes of 24,which are less than 24?

a)96 b)72 c)52 d)144



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72

38.How many co prime of 53 are there between 29 and 41,including the two numbers?


39.What is the HCF of the numbers represented by n(n^2 + 20), n being an even

number?

a)16 b)12 c)8 d)48

40.When a three digit number is divided by 64484 and 62767,the remainder

is the same in both the cases.What is the remainder?


41) find the units place of the calculation 3334223455

*789678

+5686

42) find the remainder of the calculation 42

123

/8.



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Calendars

This is a very interesting and puzzling topic for most students. The method discussed here is abstract

and very simple. Once you finish reading this section you will be able to find out day of the week for any

date from start of the calendar, be it Mahatma Gandhis day of birth or say 25 thDecember of year 300028

within 30 seconds(average) and for the years that we regularly come across youll take 5 to 10 seconds.

Building the platform:

Before entering to the method and an example certain names and concepts should be made familiar.

These follow.

The Remainder Function:

The remainder of any division is written as the result, its called remainder function. (23/7)r=2

The subscript r indicates remainder function. (28/7) r=0, (31/4)r=3 . this should be clear before we proceed

further.

The Quotient function:

Only the whole number result is taken as result of the quotient function. (24/7) q=3 . The subscript q

indicates the quotient function. (18/4)q=4, (14/4)q=2, (94/4)q=23, (21/7)q=3.



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Concept of leap year:

Check if the last 2 digits of the year are divisible by 4, if they are it is a leap year. In case the last 2 digits are

00 then, check if the digits prior to them are divisible by 4, if they are it is a leap year (century year should

be a multiple of 400 to be a leap year).

1236 is a leap year (36 divisible by 4)

1300 is not (13 not divisible by 4)

1384 is a leap year (84 is divisible by 4)

1933 is not (33 not divisible by 4)

13200 is a leap year (132 is divisible by 4)



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Principle of counting:

Start counting from the bottom of your little finger, each joint corresponds to a number. The figure better

illustrates counting.

6

5 3

7

4 1

2

Principle of counting

Count from 1 to 7 as shown to count 8 come back to the position that holds 1 , 11 would be at the position

of 4.

Similarly the days of the week also correspond to these positions 1mon, 2tue, 3wed. 7

sun. be very clear and practice at least 4 to 5 times before proceeding.



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KPs Rhyme:

This is the most important part without which the method doesnt work. Read it again and again till you

can remember it exactly in the same sequence.

my maths teacher taught me 3+ 3is 6,

In my exam I wrote 1+ 4as 6,

The exam was for 25marks and I got 0,

That is why I couldnt score 35% in 1stattempt.

Observe the numbers underlined, there are 12 numbers corresponding to 12 months of the year.

-} A date 12/5/1950 means that from the start of the calendar 1949 years, 4 months are completely over

and it is the 12th

day.

The method: it is explained with 3 examples stage by stage.

Ex1:- to find day of the week of 21/09/2011

2011 is the present year, 2010years are completely over

9

thmonth is going on, 8are over

It is the 21st

day.



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2010 8 21

(2010/400)r=

10

(10/4)q=22+10 = 12

(12/7)r=5

Start counting

on your fingers

as described in

the principle of

counting, it will

Stop at fifth bit

leave it there.

Do not remove

the thumb

placed on fifth

bit.

Remember the

KPs Rhyme,

the 8th

number

in sequence is 55

Continue

counting with

the help of

thumb on your

fingers

6,7,8,9,10 it

will end on the

3rd

bit keep

your thumb

there.

(21/7)r = 0

0

Add nothing to

your fingers.The bit that

you lie your

thumb on is the

last bit bit of

little finger and

that

corresponds to

Wednesday

based on

principle of

counting.

So, 21/09/2011 is a Wednesday.

Ex2:- a small addition when the year to which the date belongs is a leap year.

15/08/2012



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2011 years are over

7 months are over

It is the 15th

day.

2011 7 15(2011/400)r=11

(11/4)q+11 = 13

(13/7)r =6

3

3

6

1

4

6

2

5

03

5

1

Nos are in order as

in the rhyme from

the second number

all the numbers are

one more than in

the rhyme for a leap

year.7

thno is 2 but since

it is a leap year, 3 is

taken.

3

(15/7)r=1

1

Count 6,3 and 1

on your fingers

it ends at the

last bit of little

finger, henceWednesday



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Initially we should always check if the year is a leap year or not. For leap year theres a slight change in the

months code everything else is the same.

Ex3:- 15/08/1947

100 200 300

5 3 1

1946 years are over

7 months are over

It is the 15th

day.

1946 7 15

(1946/400)r=346Additional step

when the

reminder

exceeds 2 digits,

346

300 + 46From table above 300

=> 1

The 7th

numberin the series is

2

(15/7)r=1finally adding

this too you see

that thumb

indicates that it

is the 5th

position,

Friday.



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Add 1 to your finger ,

continue as usual.

(46/4)q+46 = 57

(57/7)r = 1

Different questions on calendars along with simple methods to solve them

:-} If day before yesterday was Sunday 90 days from today will be?

Sol: since there are 7 days in a week, the same day will repeat after 7 days. Here, we use the remainder

function that gives output corresponding to number of days ahead from today. If day before yesterday wasSunday, today is Tuesday. (90/7)r=6. Hence, the day will correspond to 6 days from today(Tuesday) that is

Monday.

-} Find the day of the week corresponding to following dates

a) 10/10/2011

b) 7/5/1947

c) 8/1/2044

d) 21/9/1994



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Sol: a) 10/10/2011

2010 years are over

9 months are over

It is the 10th

day.

2010 9 10

(2010/400)r=10

10 + (10/4)q=12

(12/7)r= 5

9th

number in

code is 0

(10/7)r=3

It ends on first

bit of little

finger Monday

b) 7/5/1947

=> 1946 years are over

=> 4 months are over

=> its the 7th

day

1946 4 7

(1946/400)r=346

300 + 46

300 => 1

Remaining 46,

4th

number is

1

(7/7)r=0

Last bit of

little finger

that is



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(46/4)q+46= 57

(57/7)r=1

Wednesday

c) 8/1/2044, 2044 is a leap year



=> it is 8th

day

2043 0 8

(2043/400)r=43

(43/4)q+43=53

(53/7)r=4

0so, nothing

added.

(8/7)r=1it

corresponds

to a Friday.

d) 21/9/1994, 1994 is not a leap year



=> It is the 21st

day



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1993 8 21

(1993/400)r=393

300+93

300 => 1

(93/4)q+93=116(116/7)r=4

8th

number in

series is 5

(21/7)r=0

Thumb

indicates last

bit of middlefinger,

Wednesday.

Profit and Loss

We purchase various commodities from shops in our daily life. But, do we ever try to find the

percentage profit owner makes. Why sales of a shop are better than the others? How is it that a

shopkeeper is so rich, earns more than a techie?

Ram once visits Shyams shop and buys jeans and T shirt. The price tagged on it is 3000/-. Shyam has

put up an attractive offer that 20% discount is given on tagged price. Ram calculates 20% of 3000 = 600/-,

deducts it from the tagged price and buys it at 2400/-. Shyam had bought this pair at 2000/-. He calculates

his percentage profit. His profit is 24002000 = 400/-, percentage profit indicates what part of the



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investment is the profit. For an investment of 2000/-, he gets a profit of 400/-. Profit percent = 400/2000

*100 = 20% .

Amount Formulae

MP 3000/-

D% 20%

D 600/- D%/100*MP

SP 2400/- MP-D

CP 2000/-

P 400/- SP-CP

P% 20% P/CP*100The tagged price is called marked price.(MP)

Discount percentage is applicable to MP.

Discount value (D) is based on MP.

Selling price (SP) is the price for which customer buys the article.

Cost price (CP) is the price at which shopkeeper buys it.

Profit is SPCP.

Profit and loss percentage is always calculated based on investment made (CP).



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-} A man sells a pen at a profit of 25%. Had he bought it at 30% less and sold it for Rs 4 less, he would have

gained 20%. Find cost price of the pen?

Sol:

Case1: take x as the CP of the pen.

P = 25/100*x (25% of CP)

SP = CP + P

SP = x + 0.25x = 1.25x

Case2 : brings it for 30% less

CP = x30/100*x = 0.7x

SP = 1.25x4 (sells at 4 Rs less)

Given, P% = 20% =( SP-CP)/CP*100.

Substitute and solve to find CP.

CP = 16.32/-.



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-} The selling price of 20 articles is same as the cost price of 21. Find the gain%.

Sol: assume CP of each article is 1Re.

CP of 21 articles = 21/-

SP of 20 articles = 21/-

SP of 1 article =21

/20/- P =21

/201 per article.

P% = (21

/201)/1*100

= 1/20*100 = 5%.

-} A man loses 12% by selling a book for 246 Rs. What should be his selling price to gain 15%?

Sol:

CP SP

X 0.88x = 246

X 1.15x = ?

? = 1.15x/0.88x*246 = 312.42/- (**10 seconds**)

Note: if a man has a gain or loss of x% and he wants a gain or loss of y% then,



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New SP = (1+/- y/100) * old SP/(1

+/- x/100)

+ for gain, - for loss.

-} A man loses 8% by selling an article for Rs 46/-. If he wants to gain 8% what should his SP be?

Sol: new SP = (1.08)*(46/0.92) =54/-.

-} A shopkeeper gains 20% on an article by selling it at Rs 132. Another shopkeeper who had bought it at

same cost loses 10%. What is the SP of another shopkeeper?

Sol: SP 2= 0.9*(132/1.2) =99/-.

-} A man sells 2 articles for Rs 300 each. On one he gains 10% and on the other he loses 10%. Find his net %

gain or loss.

Sol: when we have equal percentage gain and loss, multiply them to find the net loss. There is always a loss

regardless of the SP.

Loss% = 10/100 * 10/100 = 1/100 = 1% loss.

-} A shop keeper sells 2 pens at Rs 100 each one with a gain of 25% and other with a loss of 25 percent.

Find the total cost price.

Sol: net loss = 25/100 * 25/100 = 6.25% loss



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Total SP = 200/-

Total CP = SP/(1-0.0625) = 200/93.75 = 213.33/-.

-} A businessman sells 25% stock on 15% profit, 2/3rd

of the remaining on 10% loss. At what percent should

he sell the remaining to get a 5% overall gain?

Sol: Assume that he has 100 grams stock that costs 100/- he should sell it at 105 for overall gain.

Quantity SP

available sold

100 25 28.75 (15% P)

75 50 45 (10% L)

25 25 ? = 31.250

Sum of SP = 105

? = 10528.7545 = 31.25/-

P% = (31.25-25)/25*100 = 25%.



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Simple interest:

The simple interest(SI) is the extra amount more than the principle a person has to pay when he has

borrowed a sum of money.

The interest he pays depends on the rate of interest(r), principal he has borrowed(p), and time after which

he returns(t).

Time should always be taken in years.

Rate in percentage per annum

A = SI+P = amount

Compound interest:

After a certain period, if the interest for the period till then is taken into account, interest is calculated and

added to the principal to obtain the new principal, it is called compounding.



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n = number of compoundings per year.

CI = AP.

Recurring deposits:

A sum of money is invested every month in the post office or in banks for a certain fixed time called

the maturity period. When this period is over, the investor gets the total principal he has paid along with

interest.

Consider the case where a person has deposited Rs 100 every month for year at the rate of 6% per

annum.

1st

months deposit should be given interest for 12 months, 2nd

months deposit should be given interest

for 11 months and so on till the last month. Let number of months be n. here, n=12.



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.

Sum of all these interests,

(

)

(NA)

Normalized average is that value for which interest for a period of 1 month is equal to the interest of the

recurring deposit.



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Partnership:

Simple partnership:

A group of people together invest in a partnership where the amount of investment remains the

same for the entire period. At the end of this period profit is divided amongst them in the ratio of their

investment.

Compound Partnership:

When the amount invested or the period of time varies in the group, the profits are divided based on

time value of money.

The sum of money is multiplied by period of investment for each partner. The end values are in a particular

ratio. In this ratio the money is divided.

Ex:- Rahul, Priya and David enter into a partnership. Rahul invests 3000Rs for 8 months and reduces it to

2000 till end of the year. Priya invests 2500 for the whole year. Dvid invests 2000in the beginning but at the

end of 3 mnths he increases it to 4000. Find the ratio in which profit should be divided.

Sol: time value of money ratio

Rahul : Priya : David



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32 : 30 : 36

16 : 15 : 18 is the ratio in which profits should be divided.

Problems to be solved:

Braun invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years,Braun got back 4 times his original investment. What is the value of n?

Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest beingcompounded annually, for the same 2 years at the same rate of interest and received $605 as interest.What was the value of his total savings before investing in these two bonds?

Ann invested a certain sum of money in a bank that paid simple interest. The amount grew to $240 atthe end of 2 years. She waited for another 3 years and got a final amount of $300. What was theprincipal amount that she invested at the beginning?

Peter invested a certain sum of money in a simple interest bond whose value grew to $300 at the end of

3 years and to $ 400 at the end of another 5 years. What was the rate of interest in which he invested

his sum?



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A father left a will of Rs.35 lakhs between his two daughters aged 8.5 and 16 such that they may getequal amounts when each of them reach the age of 21 years. The original amount of Rs.35 lakhs hasbeen instructed to be invested at 10% p.a. simple interest. How much did the elder daughter get at thetime of the will?

What will Rs.1500 amount to in three years if it is invested in 20% p.a. compound interest, interestbeing compounded annually?

If a sum of money grows to 144/121 times when invested for two years in a scheme where interest iscompounded annually, how long will the same sum of money take to treble if invested at the same rateof interest in a scheme where interest is computed using simple interest method?

The population of a town was 3600 three years back. It is 4800 right now. What will be the populationthree years down the line, if the rate of growth of population has been constant over the years and hasbeen compounding annually?

A man invests Rs.5000 for 3 years at 5% p.a. compound interest reckoned yearly. Income tax at therate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of thethird year.

The difference between the compound interest and the simple interest on a certain sum at 12% p.a. fortwo years is Rs.90. What will be the value of the amount at the end of 3 years?

A stairway 10ft high is such that each step accounts for half a foot upward and one-foot forward. Whatdistance will an ant travel if it starts from ground level to reach the top of the stairway?

A sum of money invested for a certain number of years at 8% p.a. simple interest grows to Rs.180. The

same sum of money invested for the same number of years at 4% p.a. simple interest grows to Rs.120only. For how many years was the sum invested?



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How long will it take for a sum of money to grow from Rs.1250 to Rs.10,000, if it is invested at 12.5%p.a simple interest?

Rs. 5887 is divided between Shyam and Ram, such that Shyam's share at the end of 9 years is equal toRam's share at the end of 11 years, compounded annually at the rate of 5%. Find the share of Shyam

The question for the day is from the topic simple and compound interest. Shawn invested one half ofhis savings in a bond that paid simple interest for 2 years and received Rs.550 as interest. He investedthe remaining in a bond that paid compound interest, interest being compounded annually, for the same2 years at the same rate of interest and received Rs.605 as interest. What was the value of his totalsavings before investing in these two bonds?

Rs.100 doubled in 5 years when compounded annually. How many more years will it take to get

another Rs.200 compound interest?

How long does it take a principal of $25,000 at a simple interest rate of 5% to become $30,000? $45,000 is deposited into a savings account. After one year, 4 months and 20 days it totals $52,500.

Calculate the simple interest rate for this account.

Determine the simple interest rate applied to a principal over 20 years if the total interest paid equalsthe borrowed principal.

How long does it take a principal payment to triple at a simple interst rate of 6%?

Find the total amount of simple interest that is paid over a perod of five years on a principal of $30,000 at a simple interest rate of 6%.

Calculate the total worth of an investment after six months with a principal of $10,000 at a simpleinterest rate of 3.5%.

A sum of money deposited at C.I. amounts to $2420 in 2 years and to $2662 in 3 years. Find the ratepercent



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Find the sum lend at C.I. at 5 p.c per annum will amount to $441 in 2 years

A property decreases in value every year at the rate of % of its value at the beginning of the year. Itsvalue at the end of 3 years was $21093.95. Find its value at the beginning of the first year.

Find the least number of complete years in which a sum of money put out at 25% compound interest

will be more than double of itself? A property decreases in value every year at the rate of % of its value at the beginning of the year. Its

value at the end of 3 years was $21093.95. Find its value at the beginning of the first year.

The difference between simple interest and C.I. at the same rate for $5000 for 2 years is $72. The rateof interest is?



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Permutation and combinations

Principle of counting: considering a very simple example, we understand the principle of counting.

Ex: there are 4 cities A, B, C and D connected by many roads as shown in figure.

A B C D

A person has to go to D from A.

He has 3 ways to reach B first from A. then, any one way taken, he has 5 more ways to reach C from B.from A to C he has 3*5 = 15 ways. Each of these 15 ways taken she has 2 more ways to reach D from C.

this gives rise to new possibilities = 15*2 = 30 ways to reach D from A. a person can travel from A to D in

30 ways(3*5*2).

If the whole event of reaching D from A has to be done then number of ways to reach D is given by

n1*n2*n3 where, n1 is the number of ways in which he can reach B from A. Number of ways from B to C is

n2 and for C to D is n3.



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Conclusion: if an event can happen in n1 ways and each of these can happen in n2 ways and further each

one can happen in n3 ways and so on. The number of ways in which the whole event can happen is

n1*n2*n3**nX ways.

Permutation (possible mutations):

Possible arrangements

To explain permutations, I consider this example:- we have 3 students, Amit(A), Prashant(P) and

Shashank(S). there are 2 chairs X and Y.

X Y

The number of ways in which they can be arranged in these chairs is called permutations.

Ways X Y

1 A B

2 A S

P P A

4 P S

5 S A

6 S PTotally we have 6 ways in which 2 people can be seated. If these three people have to be arranged in only

one chair, then we have 3 arrangements.



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Now we observe from the example above that one person is fixed to one position first next the possibilities

for arrangements further are done. The second person is fixed in first place again and all possible changes

are done. When all three seats are empty A, P or S can be put in the first place. So we have 3 people and 3

ways in which any one of them can be fixed in the first position. Two places are left out and second

position can be filled by any one of the remaining 2 people. (if we fix A in the first place, P or S can be put

in the second place). When 2 people are seated, theres one person and one place left out that can be

selected in one way. Hence from the principle of counting, we arrive at the conclusion that 3 people can be

arranged in 3 chairs in 3*2*1 = 6ways.

Here, we denote such arrangements by the factorial expression. i.e, 3!

4!=4*3*2*1 and so on.

n! = n*(n-1)*(n-2)*(n-3)**1 .

Further analysis: till now we had 3 people hence, 3 ways to put a person on the first chair. If we have 4

people A, P, S and T we have four ways to select a person to sit on the first chair. Consider the situation

when 4 people have to sit in 2 chairs.

Ways X Y

1 A P

S A S

3 A T



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4 P A

5 P S

6 P T

7 S A

8 S P9 S T

10 T A

11 T P

12 T S

Here first place can be selected in 4 ways and next place in 3 ways. Total no of possibilities is 4*3 = 12

ways.

For 4 places it is 4*3*2*1 ways, for 3 places it is 4*3*2 ways and for 2 places it is 4*3 ways.

4 people 4 places 4*3*2*1 ways

4 people 3 places 4*3*2 ways

4 people 2 places 4*3 ways

4 people 1 place 4 ways

Number of places is number of nos to be multiplied so to remove the other part from 4!.

It can be written as 4!/(4-4)!, 4!/(4-3)!, 4!/(4-2)!, 4!/(4-1)! .



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Conclusion:

If we have n people and r places, number of ways in which they can be arranged is given by n!/(n-r)!. this

expression is given a notation nPrand nPrle = n!/(n-r)!.

nPris the number of permutations of n different things taken r at a time.

I have 2 similar books, a pen and an eraser. I can arrange them such that 1st

place can be selected in 4

ways, second place in 3 ways and 3rd

and 4th

place are same since both are similar. Either of the book put in

this place gives the same result. So, number of possible mutations are 4!/2! Where, n objects out of which

x are similar. In general if out of n objects x are alike, y are alike and z are alike, then,n!

/x!y!z!is the number

of permutations.

The ghost theory:

We have 3 small safe chairs and 5 people A,B,C,D and E. 3 among the 5 can sit on the chairs. There is

no go but to let the left out ones be eaten by the ghost. Now, if I have to choose a person to sit on the first

chair, since there are 5 people, I have 5 options. Now one person is seated. To choose the next person to

sit for the second chair, there are 4 people left so, 4 options. For the last chair similarly I have 3 options.

The total number of ways in which I can choose 3 from 5 is 5*4*3. This includes different arrangements in

same group of people also. i.e if I interchange person sitting in the first chair and the one sitting in the

second chair, it is considered as a new possibility. But, whatever be the arrangement inside the selected 3



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they are safe from the ghost. To find no of different groups that I can make I have to divide all possible

ways by number of arrangements in a single group. The number of po

Documents

10 seconds part 2