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SAMPLE NOTES

i) CURRENT AFFAIRS

ii) CSAT - QUANT


The Right to Fair Compensation and Transparency in Land Acquisition, Rehabilitation and

Resettlement Act 2013

1. Need for new law

A. In sync with Art.300A of the constitution.

B. Public Concern

Heightened public concern on Land Acquisition issues

Absence of national law to provide for resettlement & rehabilitation and compensation for loss of

livelihood.

C. Outdated law

The original act is Law Acquisition Act of 1894 enacted by British

D. Need for Balance

Addressing concerns of farmers and others whose livelihood depended on the land being acquired.

Facilitating land acquisition for industrialization, infrastructure and urbanization

2. Multi- crop irrigated land will not be acquired except as a last resort measure. If acquired , an equivalent

area of culturable wasteland will be developed for agricultural purpose.( Will not be applicable in linear

projects – railway, highways, canals)

3. If land acquired for urbanization – 20 % of the developed land will be reserved and offered to land owning

project affected family OR the Company for whom land is acquired may offer shares limited to 25% of the

Compensation amount.

4. R & R (Rehabilitation and Resettlement)

Subsistence allowance of Rs.3000 per family per month for 12 months

If jobs created – mandatory employment for one member of family OR Rs.5 lakh per family OR Rs.

2000 per month per family for 20 years with appropriate index of inflation.

5. Enhanced Role of PRI’s

SIA(Social Impact Assessment) has to be carried out in consultation with representatives of PRI’s

Representation of members of PRI’s in Expert group which has power to reject a project.

Consent of Gram Sabha is mandatory for acquisition in Scheduled areas under Schedule V under

the constitution.

Representation of Panchayat’s Chairman in R&R Committee at project level.

6. Provision for Farmers

All farmers in rural areas will get up to 4 times the highest sales price in the given area.

Consent of 70% of farmers is required for both acquisition and compensation amount in case of

acquisition for PPP project and that of 80 % if acquired for private company.

If the acquired land is sold to a third party at higher value, 40% of the profit amount to be

distributed amongst farmers.

7. Role of Collector

1894 Act Present Act 1. “Public Purpose “ to be decided by Collector 1.“Public Purpose “ clearly defined in Act. No

discretion of Collector. 2. Collector could decide the quantum of compensation

2. It has to be decided now as per the method described

3. Collector could decide when to take possession, even immediate possession.

3. Possession can only be taken after payment of compensation and R&R has been completed.

4. Collector had sweeping power to invoke urgency clause

Urgency limited to only two situations- natural disasters and national defence.


8. This law will have retrospective effect in cases where no award has been given or compensation not given or

possession not taken and proceedings pending for more than 5 years.

9. Transparency

SIA – Gram Sabha included, report to be made public

R&R Scheme – Made available to public

Public Disclosure – All documents mandatorily made available in the public domain and on the website.

Fixed timeline for Monetary R&R and Infrastructural R&R.

10. The new law will not apply to activities under 13 Acts of Central Government presently related to Land

acquisition like National Defence, Railways, National Highways etc. for initial one year after which it can apply

by notification by Central Government.

Kirit Parikh Committee report

The government had appointed an Expert Group on ‘Viable and Sustainable System of Pricing of Petroleum Products’. This

Committee, chaired by Shri Kirit S. Parikh, recently submitted its report.

Key observations include:

1. The reduced cash surplus of upstream public sector oil companies restricts their ability for exploration of domestic

fields and acquisition of overseas assets. Keeping oil firms viable is in the interest of self-sufficiency in domestic oil

production.

2. Price control, subsidies and taxes can introduce distortions which may not be desirable. For example:

A. The higher excise duty on petrol compared to diesel encourages use of diesel cars

B. Lower diesel prices lessen the incentive to shift freight movement from trucks to railways, which consume 4

times less diesel for every tonne km of freight

3. Control on pricing also restricts competition. Several oil marketing companies, viz. Reliance Industries, Essar Oil

and Shell India, that were not part of the subsidy sharing arrangement, closed down their retail marketing

businesses across the country.

4. Petrol is largely an item of final consumption. Its price, therefore, has a very small impact on inflation due to

forward linkages.

5. Trucks accounted for 37% and buses 12% of total diesel consumption in 2008-09. Agriculture’s share was 12%.

6. The cost of diesel in agriculture can be accounted for by the Government while fixing the Minimum Support Price

(MSP) for major crops and hence an increase should not adversely affect farmers.

7. The inflationary impact of increase in diesel prices (due to increased costs in transport and industrial usage) would

be comparable with the inflationary impact of subsidies.

Recommendations:

1. Petrol and diesel prices should be market determined both at the refinery gate and at the retail level.

2. An additional excise duty of Rs 80,000 should be levied on diesel cars.

3. The price of PDS kerosene should be increased by at least Rs 6/litre. Thereafter, price of PDS kerosene should be

raised every year in step with the growth in per capital agricultural GDP at nominal prices.

4. Prices of domestic LPG should be increased by at least Rs 100 per cylinder. Thereafter, the price of domestic LPG

should be periodically revised based the rising per capita income.

5. The subsidy on domestic LPG should be discontinued for all others except the BPL households once an effective

targeting system is in place.


Direct Cash Transfer Introduction Poverty elimination and inclusive growth are the top most priority for the welfare of every state. To meet these socio-developmental objectives, a number of Government sponsored programs and schemes have been introduced. However, there have been issues associated with the efficiency and effectiveness of the same. Rampant leakages and corruption have made many of the schemes and programs dysfunctional. Direct Cash Transfer to the poor has been aimed to mitigate these malaises. Need for Direct Cash Transfer Recent studies by the Planning Commission have shown that the Public Distribution System has become so inefficient that 58% of the subsidized grains do not reach the targeted group and almost a third of it is siphoned off the supply chain. According to the Finance Ministry the inefficiencies of the PDS ensure that the Government is forced to spend Rs.3.65 for transferring of Rs. 1 to the poor. The idea behind the Direct Cash Transfer is to cut down wastage, duplication and leakages and also to enhance efficiency. The idea is to move to a completely electronic cash transfer system for the entire population. Launch of the programme The programme is now called direct benefit transfer (DBT). On January 01, 2013, the government of India rolled out the DBT covering seven welfare schemes in 20 districts in 16 states. The programme covers schemes like educational scholarship for the Scheduled Castes and the Scheduled Tribes and pensions to widows. Food, fertilisers, LPG, diesel and kerosene have been kept out for the present. Among other objectives like better delivery, more accurate targeting, giving broader choice to the beneficiaries, reducing pilferage and corruption, the programme is also aimed at cutting the massive subsidy bill of Rs 1,64,000 crore .

India’s welfare spending is a major contributor to its fragile public finances. The country’s budget deficit was 5.8 percent of gross domestic product in the financial year ended March 2012. The government is of the opinion that the programme can generate much-needed budget savings by eliminating corruption. The Prime Minister has set up a three-tier architecture for monitoring the scheme. This includes a national ministerial committee, a national executive committee and implementation committees. The seven schemes that will now employ direct cash transfers to beneficiaries’ accounts are mostly related to student scholarships and stipends, the Indira Matrutva Yojna and the Dhanalakshmi schemes. It is estimated that at least two lakh beneficiaries will receive cash benefits from Jan 1. Cash benefits in the remaining 19 schemes will be available from February and March when the government will cover 23 other districts across the country. The government had originally identified 51 districts across 16 states to be covered by the programme under which cash subsidy benefits will directly go to the bank accounts of beneficiaries with mandatory requirement of Aadhaar number. Subsequently, four districts each of Himachal Pradesh and Gujarat were exempted from the roll-out because of the assembly elections. Later, the programme was scaled down as gaps in infrastructure, like beneficiary lists and bank accounts were noticed and are yet to be fixed. Scope of DBT: The DBT program aims that entitlements and benefits are transferred directly to the beneficiaries. The beneficiaries could include widows, students and pension takers. This would be done through biometric-based Aadhaar-linked bank accounts. This would reduce several layers of intermediaries and delays in the system. Advantages of DBT system:

The use of Aadhaar or other biometric based systems would dissolve problems like duplicates or ghosts. Duplicates are when the name of the beneficiary is repeated and Ghosts is when the name of a nonexistent beneficiary is mentioned.

It helps in the quick and direct cash transfer to the intended beneficiary. The cash transfer happens through a dense Business Correspondent system on the ground with micro ATM’s. This

ensures that the poor get the same level of service that the rich and the middle classes in the society receive. The financial inclusion offered by the DBT infrastructure can also be used by internal migrants to send their

remittances. Approximately, Rs.75,000 crore worth of within-country remittances are made in India every year. However, seventy per cent of these remittances are channeled through informal (and often illegal) channels which impose high costs on them. The Aadhaar-based microATM network could ensure that remittances take place instantly and at much lower cost to migrants.


Areas of concern

1. Inflation could easily erode the purchasing power of cash transfers. Concern about indexing of cash transfers to the price levels.Even if some indexation does happen, small delays or gaps in price information could cause significant hardship for poor people.

2. Biometric-based unique identity or Aadhaar is leading to huge problems for people working for the rural employment guarantee scheme and for others receiving welfare benefits. Not only have enrolments been done shoddily but the experience of the pilot projects shows that it is almost impossible to authenticate the work-hardened fingerprints of the poor.

2. There is also the overwhelming issue of deficient online connectivity. 3. Some states such as Tripura have opposed the central government’s direct cash transfer scheme saying banking was not

available in most villages. 4. Many fear that the direct cash transfer in PDS will affect the system and gives scope to corruption. Rights groups argue that

the decision to stop food grains and introduce cash transfer in its place is a unilateral decision and unscientific. Instead of targeted rationing system, they want the government to universalise the PDS.

5. There are increased chances of money that is transferred into a female member’s account being misused by others in the family.

6. Many NGOs have pointed out the problem of the Government’s decision of ending the public distribution system of food grain and give money directly to the people without proper identification of the poor in the country. About 40 per cent of the poor are still not officially recognised in India according to many welfare economists.

Kotkasim Case Study

Kotkasim town in Alwar district of Rajasthan saw the rollout of pilot experiment with “ direct cash transfer “ of kerosene subsidies.

Positives –

1. According to district administration, the scheme led to net savings of 79 percent in kerosene subsidies by weeding out ‘fake

users”

2. Instead of getting kerosene from the local FPS at a subsidised price of Rs.15 a litre, as they used to do, households now pay the

full market price . The subsidy , that is , the difference between the market price and the subsidised rate of Rs.15 a litre, is

deposited into their bank accounts. Reduce leakages – If FPS dealers get the same price from their legitimate customers as

from the black market, there is no incentive to cheat.

Problems –

1. Erratic payments of subsidies, due to lack of coordination between the banks. Also, since the subsidy transfer requires Core

Banking Solutions (CBS) enabled bank branches, the post office accounts of MGNREGA workers were not considered.

2. Because of erratic and cumbersome transfer of subsidies, the effective price of kerosene actually shot up. Amartya Sen – “ It is not the modality of cash transfer that is the only issue, but also how much, and for whom, and also, instead of

what.”

“If the cash transfer is not additional to food subsidies, and is given “instead of” food subsidies, it would be important to make sure

that the money given would be used for nutritional purposes and, equally importantly, that it would be divided within the family in a

way that addresses the manifest problems of undernourishment and deprivation of young girls.”


CSAT - QUANT TECHNIQUES FROM INDIA’s No.1 APTITUDE TRAINER BYJU

INDEX

1) CONSTANT PRODUCT RULE

2) POWER CYCLE

3) “MINIMUM OF ALL” REGIONS IN VENN DIAGRAMS

4) LAST 2 DIGITS TECHNIQUE

5) SIMILAR TO DIFFERENT GROUPING (P&C)

6) APPLICATION OF FACTORIALS

7) GRAPHICAL DIVISION FOR GEOMETRY

8) ASSUMPTION METHOD

1) CONSTANT PRODUCT RULE &

This rule can be applied when we have two parameters whose product is constant, or in other words, when they are inversely proportional to each other. eg)Time x Speed = Distance, Price x consumption = Expenditure, Length x breadth= Area

A increase in one of the parameters will result in a decrease in the other parameter if the parameters are

inversely proportional. Let’s see the application with the following examples 1) A man travels from his home to office at 4km/hr and reaches his office 20 min late. If the speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to office?

Solution: Assume original speed= 4km/hr. Percentage increase in speed from 4 to 6= 50% or ½. increase in

speed will result in decrease in original time=30 minutes.(from given data). Original time= 90 minutes= 1.5

hours Answer is Distance= speed x time = 4x1.5=6 km 2) There is a 20% increase in price of sugar. Find the percentage decrease in consumption a family should adopt so that the expenditure remains constant. Solution: Here price x consumption= expenditure (constant)

Using the technique, 20% ( ) increase results in 16.66%( ) decrease in consumption. Answer=16.66%


2)Power Cycle The last digit of a number of the form ab falls in a particular sequence or order depending on the unit digit of the number (a) and the power the number is raised to (b). The power cycle of a number thus depends on its’ unit digit. Consider the power cycle of 2 21=2, 25=32 22=4 26=64 23=8 27=128 24=16 28=256 As it can be observed, the unit digit gets repeated after every 4th power of 2. Hence, we can say that 2 has a power cycle of 2,4,8,6 with frequency 4. This means that, a number of the form 24k+1 will have the last digit as 2 24k+2 will have the last digit as 4 24k+3 will have the last digit as 8 24k+4 will have the last digit as 6 (where k=0, 1, 2, 3…) This is applicable not just for 2, but for all numbers ending in 2. Therefore to find the last digit of a number raised to any power, we just need to know the power cycle of digits from 0 to 9, which are given below

For example 3) Find the remainder when 375 is divided by 5.

1) Express the power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3. 2) Take the power cycle of 3 which is 3,9,7,1. Since the form is 4k+3, take the third digit in the cycle, which is 7 Any number divided by 5, the remainder will be that of the unit digit divided by 5. Hence the remainder is 2. Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in the fastest way possible. For example,

4) Find the unit digit of Put n=1, the problem reduces to 73^4, which is 781. Since 81=4k+1, take the first digit in the power cycle of 7, which is 7. 5) What is the first non zero integer from the right in 83301957 + 83701982? a) 3 b) 1 c) 9 d) none of these 83701982 will end with more number of zeroes so we need to consider only the first part. Rightmost non-zero integer of the expression will be = unit digit of 8331957

Unit digit Power cycle Frequency

0 0 1

1 1 1

2 2,4,8,6 4

3 3,9,7,1 4

4 4,6 2

5 5 1

6 6 1

7 7,9,3,1 4

8 8,4,2,6 4

9 9,1 2


= unit digit of 31957. Since 1957=4k+1, take the first digit in the power cycle of 3, which is 3.

6) If N = (13)1! + 2! + 3! + ..+ 13! + (28)1! + 2! + 3!..+ 28! + (32)1! + 2! + 3! + ...+ 32!+ (67) 1! + 2! + 3! + ......+ 67!, then the unit digit of N is (a) 4 (b) 8 (c) 2 (d) none of these Based on Power Cycle After 4! Every number is of the form 4k+4 , here we need to check the nature of the power till 4! Every term’s power is of the form 4k+1. So taking the first digit from the power cycles of 3,8,2, and 7 we will get the unit digit as (3+8+2+7 = ..0). Ans = 0

3) Useful technique to find the last 2 digits of any expression of the form ab

Depending on the last digit of the number in question, we can find the last two digits of that number. We can classify the technique to be applied into 4 categories. TYPE METHOD EXAMPLES

1) Numbers ending in 1 The last digit is always 1. The 2nd last digit = product of tens digit of base * unit digit of the power. In 2167; 2 is the tens digit of base and 7 is the unit digit of power

1) 2167 =__41(2 * 7=__4_) 2) 4187 =__ 81( 4 *7= __8_) 3) 1261167 =__21 ( 6 * 7= __2_) 4) 31124 =__ 21( 3 * 4=__2_)

2) Numbers ending with 5 Last two digits: always 25 or 75

e.g.) 155534 = __ 25

3) Numbers ending in 3, 7, 9 Change the power so that the base ends with 1 and then use the same technique as for those numbers ending with 1. eg) 34, 74&92 all will end in1.

e.g.) 17288 =(174)72 (taking the power 4 as 74 will end in 1. (172* 172) 72 =( _89*_89)72(as last 2 digits of 172=89)

= ( _21)72 (as last 2 digits of 89*89=21) Answer =__41( as 2*2=4)

4) For even numbers (2,4,6,8) Use the pattern of the number 1024 =210 i.e. *210 raised to even power ends with 76 and * 210raised to odd power ends with 24.

e.g.) 2788 = (210) 78 * 28 = __76 * __56 = __56.

It is also important to note that,


1. 76 multiplied by the last 2 digits of any power of 2 (>21)will end in the same last 2 digits as of that power of 2 E.g. 76*04 = 04, 76*08 = 08, 76*16 = 16, 76*32 = 32 2. The last two digits of x2, (50-x)2 , (50+x)2 , (100-x)2 will always be the same. For example last 2 digits of 122,382,622,882,1122…. will all be the same (..44). Also, last two digits of 112=392=612=892 =1112=1392=1612=1892 and so on 3.To find the squares of numbers from 30-70 we can use the following method 7) To find 412 Step1 : Difference from 25 will be first 2 digits = 16 Step 2 : Square of the difference from 50 will be last 2 digits = 81 Answer = 1681. 8): To find 432 Step1 : Difference from 25 will be first 2 digits = 18 Step 2 :Square of the difference from 50 will be last 2 digits = 49 Answer = 1849 4. Combining all these techniques we can find the last 2 digits for any number because every even number can be written as 2* an odd number

4) “Minimum of all” regions in Venn Diagrams

9) In a survey conducted among 100 men in a company, 100 men use brand A, 75 use brand B, 80 use

brand C, 90 use brand D & 60 use brand E of the same product. What is the minimum possible number of

men using all the 5 brands, if all the 100 men use at least one of these brands?

Sum of the difference from 100 = (100-100) + (100-75)+(100-80)+(100-90)+(100-60) = 95 Again take the difference from 100 = 5 (answer)

5) Similar to different Grouping in Permutation & Combination All questions in Permutation and Combination fall into 4 categories, and if you master these 4 categories, you

can understand all concepts in P&C easily.

1) Similar to Different

2) Different to Similar

3) Similar to Similar

4) Different to Different

In this booklet, we will look at the first category; i.e. Similar to Different, where I will give a unique approach to

the number of ways of dividing ‘n’ identical (similar) things into ‘r’ distinct (different) groups


a) NO LIMIT QUESTIONS

Let me explain this with an example. Suppose I have 10 identical chocolates to be divided among 3 people. The 10 chocolates need to be distributed into 3 parts where a part can have zero or more chocolates. So let us represent chocolates by zeroes. The straight red lines (Call then “ones”) are used to divide them into parts. So you can see that for dividing into 3 parts, you need only two lines. Suppose you want to give 1st person 1 chocolate, 2nd 3 chocolates and 3rd 6 chocolates. Then you can show it as:

Suppose you want to give one person 1 chocolate, another person 6 chocolates and another one 3, then it can

be represented as:

Now if first person gets 0, second gets 1 and third gets 9 chocolates then it can be represented as:

Now suppose you want to give first person 0, second also 0 and third all of 10 then you can show it like:

So, for dividing 10 identical chocolates among 3 persons you can assume to have 12 (10 zeroes +2 ones)

things among which ten are identical and rest 2 are same and of one kind. So the number of ways in which

you can distribute ten chocolates among 3 people is the same in which you can arrange 12 things, among

which 10 are identical and of one kind while 2 are identical and of one kind which can be done in

The above situation is same as finding the number of positive integral solutions of a + b + c = 10. a, b, c is the

number of chocolates given to different persons.

b) LOWER LIMIT QUESTIONS

Now suppose we have a restriction that the groups cannot be empty i.e. in the above example all 3

persons should get at least 1.


You have to divide ten chocolates among 3 persons so that each gets at least one. Start by giving them one

each initially and take care of this condition. You can do this in just 1 way as all the chocolates are identical.

Now, you are left with 7 chocolates and you have to divide them among 3 people in such that way that each

gets 0 or more. You can do this easily as explained above using the zeroes and ones. Number of ways =

The above situation is same as finding the number of natural number solutions of a + b + c = 10. (a, b, c are the number of chocolates given to different persons) Now suppose I change the question and say that now you have to divide 10 chocolates among 3 persons in such a way that the first person gets at least 1, the second at least 2 and the third at least 3. It’s as simple as the last one. First fullfill the required condition. Give the 1st person “1”, second person “2” and the third person“3” chocolates and then divide the remaining 4 (10–1–2-3) chocolates among those 3 in such a way that each gets at least 1. This is same as arranging 4 zeroes and 2 ones which can be done in 6C2 ways.

The above situation is same as finding the number of positive integral solutions of a + b + c = 10 such that

a > 1, b > 2, c > 3. a, b, c is the number of chocolates given to different persons.In this case the answer is 9C2. 10) Rajesh went to the market to buy 18 fruits in all. If there were mangoes, bananas, apples and

oranges for sale then in how many ways can Rajesh buy at least one fruit of each kind? a) 17C3 b) 18C4 c) 21C3 d) 21C4 This is a Grouping type 1 Similar to Different question, with a lower limit condition. M+B+A+O=18 Remove one from each group, therefore 4 is subtracted from both sides. The problem changes toM+B+A+O=14. Using the logical shortcut you just learnt, the answer is based on the arrangement of 14 zeroes and 3 ones (i.e. 17C3) 11) There are four zeroes to be put in five boxes where each box can accommodate any number of

zeroes. In how many ways can one do this if: a) Zeroes are similar and boxes are different

a) 275 b) 70 c) 120 d) 19 When the zeroes are similar and the boxes are different, it’s a grouping type 1 question A+B+C+D+E=4, where A,B,C,D,E are the different boxes. The number of ways of selection and distribution=8C4=70 12) The number of non negative integral solutions of x1+x2+x3 ≤ 10 a) 84 b) 286 c) 220 d) none of these By non-negative integral solutions, the conditions imply that we can have 0 and natural number values for x1, x2, x3, and x4. To remove the sign ≤ add another dummy variable x4. The problem changes tox1+x2+x3+x4=10 This is an example of grouping type1 (Similar to Distinct). It is the arrangement of 10 zeroes and 3 ones. Using the shortcut of zeroes and ones, Therefore the answer is 13C3=286


6) APPLICATION OF FACTORIALS A thorough understanding of Factorials is important because they play a pivotal role not only in understanding concepts in Numbers but also other important topics like Permutation and Combination Definition of Factorial N! = 1x2x3x…(n-1)xn Eg 1) 5!= 1x2x3x4x5=120 eg 2) 3!=1x2x3=6 Let us now look at the application of Factorials

I) Highest power in a factorial or in a product

Questions based on highest power in a factorial are seen year after year in CAT. Questions based on this can be categorized based on the nature of the number (prime or composite) whose highest power we are finding in the factorial, i.e

a) Highest power of a prime number in a factorial:

To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

13) The highest power of 5 in 100!

=20; 4; Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24

ALTERNATIVE METHOD

=20+4=24 (We take upto 52 as it is the highest power of 5 which is less than 100)

b) Highest power of a composite number in factorial

Factorize the number into primes. Find the highest power of all the prime numbers in that factorial using the previous method. Take the least power.

14) To find the highest power of 10 in 100! Solution: Factorize 10=5*2. 1. Highest power of 5 in 100! =24 2. Highest power of 2 in 100! =97 Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available. So take the lesser number i.e. 24 & this is the answer.

15) Highest power of 12 in 100! Solution: 12=22 *3. Find the highest power of 22 and 3 in 100! First find out the highest power of 2.

Listing out the quotients: ; ; ; ; ;

Highest power of 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97. So highest power of 22 = 48 (out of 97 2’s, only 48 are 22)

Now for the highest power of 3. ; ; ; ;

Highest power of 3 = 48. Thus, the highest power of 12 = 48


II) Number of zeros in the end of a factorial or a product Finding the number of zeroes forms the base concept for a number of application questions. In base 10, number of zeros in the end depends on the number of 10s; i.e. effectively, on the number of 5s In base N, number of zeroes in the end highest power of N in that product

16) Find the number of zeroes in 13! In base 10 Solution: We need to effectively find the highest power of 10 in 13! = Highest power of 5 in 13! As this power

will be lesser.

17) Find the number of zeroes at the end of 15! in base 12. Solution: Highest power of 12 in 15! =highest power of 22 *3 in 15! =Highest power of 3 in 15!= 5

III) Number of factors of any factorial

Let us look at an example to understand how to find the number of factors in a factorial

18) Find the factors of 12! STEP 1: Prime factorize 12! i.e. find out the highest power of all prime factors till 12 ( i.e. 2,3,5,7,11). 12! = 210*35*52*7*11 STEP2: Use the formula N=am*bn(a, b are the prime factors). Then number of factors= (m+1)(n+1) The number of factors= (10+1)(5+1)(2+1)(1+1)(1+1) =792. Answer=792

APPLICATION QUESTION BASED ON FACTORIAL

20) How many natural numbers are there such that their factorials are ending with 5 zeroes? 10! is 1*2*3*4*(5)*6*7*8*9*(2*5). From this we can see that highest power of 5 till 10! is 2. Continuing like this, 10!-14!, highest power of 5 will be 2. The next 5 will be obtained at 15 = (5*3). Therefore, from 15! To 19! - The highest power of 5 will be 3. 20!-24! – Highest Power = 4, In 25, we are getting one extra five, as 25=5*5. Therefore, 25! to 29!, we will get highest power of 5 as 6. The answer to the question is therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

7) USE OF GRAPHICAL DIVISION IN GEOMETRY Let’s look at a technique which will help you solve a geometry question in no time 21) ABCD is a square and E and F are the midpoints of AB and BC respectively. Find the ratio of Area( ABCD): Area(DEF)?


Lets divide the figure using dotted lines as shown in Figure B. Area of ABCD=100%. Area AEDG=50%. Then Area in shaded region 1(AED)= 25%. Similarly, DCFH=50%. Area in shaded region 2(DCF)=25%. Now Area of EOFB= 25%. .Area of shaded region 3(BEF)=12.5%. Total area outside triangle= 62.5%. Area inside triangle= 100-62.5=37.5%. required ratio = 100/37.5 = 8:3. To learn this directly from BYJU, refer the video in the CD given

8) ASSUMPTION method This involves assuming simple values for the variables in the questions, and substituting in answer options

based on those values. Assumption helps to tremendously speedup the process of evaluating the answer as

shown below.

22) k & 2k2 are the two roots of the equation x2 – px + q. Find q + 4q2 + 6pq = a) q2 b) p3 c) 0 d) 2p3

Solution: Assume an equation with roots 1&2 (k=1) =>p (sum of roots)= 3 and q(product of roots)=2. Substitute in q + 4q2 + 6pq = 54. Look in the answer options for 54 on substituting values of p=3 and q=2. we get 2p3 = 54.=>Ans = 2p3. 23) Consider the set S={2,3,4……2n+1), where n is a positive integer larger than 2007. Define X as the average of odd integers in S and Y as the average of the even integers in S. What is the value of X-Y?

a) 1 b) c) d) 2008 e) 0

The question is independent of n, which is shown below. Take n=2. Then S= {2,3,4,5). X= 4 and Y=3. X-Y =1, Take n=3. then S={2,3,4,5,6,7}. X=5 and Y=4. X-Y=1 Hence you can directly mark the answer option (a) .You can solve the question in less than 60 seconds.

There were more questions which could be solved using similar strategies. The methods given above clearly show that for someone with good conceptual knowledge and right strategies the quant section is a cakewalk.


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