Indian Economy, a CR point

I was reading through a write-up on Indian Economy-2012. I wanted to relate one point from here to the critical reasoning stuff.

The following are the facts found in a recent survey:

Fact-1: Indian economy may be broadly divided in to three sectors- Industry, Agriculture and Services. The contributions of these sectors to the GDP are as follows:

Agriculture-15%, Industry-28%, Services-57%

Fact-2: For over a considerable period, there is exodus observed in a significant extent. People are moving from rural to urban areas. Most of the people who are moving out of rural areas are not finding agriculture as a lucrative sector and trying to find jobs in industries in urban areas.

A Concept on Days and Dates

Problems are built on simple know-how things. For example, we can consider problems on Days and Dates. Here, the concept is simple and is based on simple remainder logic. Let us consider some known-facts and probable questions based on those known-facts.

What we know: “A week consists of 7-days”

What can be asked: 

Question: what is the date of next immediate Monday?

Statement-A: Today is Monday

Statement-B: Today is 3^rd of Feb

The way approach is like this:

How many days are there in a week? Answer “7”. That means, the week gets repeated on every eighth day. If Day-1 is X, one cycle of the week gets completed on Day-7 and the second cycle gets started on Day-8. We get Day-8 by adding 7 days to Day-1. So, continue the logic, we get the next X-day on 8+7 = Day-15. A simple addition of 7 gives us the next date of the same-day.

If we consider the two given statements, the input is, today is Monday and 3^rd Feb. Let us add a 7 to the date. 10^th Feb is Monday and 17^th Feb is also a Monday so on and so forth.

Obviously, the next immediate Monday is 10^th Feb and the statements together are sufficient to get the answer.

Let us introduce a small twist in to this logic.

Question: what is the date of next immediate Monday?

Statement-A: Today is Monday

Statement-B: Today is 23^rd Feb

How to answer this? Can we simply add 7 to 23 and tell the answer 30^th Feb? No, it’s a blunder. No year can have 30 days in its February month. It’s given that, 23^rd Feb of the year is Monday. It’s not given which year is that. It may be a leap-year or non-leap year. Is the given information enough to answer the question? No, definitely not.

What is this Leap-Year?

A normal year consists of 365 days and a leap-year consists of 366 days. Where from this extra day in a leap-year comes from? A normal year has 28 days in its Feb’ month where as a leap-year’s Feb’ month consists of 29 days. There it is.

What we know: ”A normal year consists of 365 days and a leap-year consists of 366 days”

What can be asked: 

Question: Is that a leap year?

Statement-A: That year starts with Tuesday

Statement-B: That year ends with Wednesday

How to find? To find whether it is a leap-year or not, it requires to find the number of days in that year. Is it possible to find it from the given conclusions? Yes, it’s possible as there are only two possibilities, 365 or 366 days.

Case for 365-days: 365/7 gives a remainder of 1 => 365 = 364+1 = (a multiple of 7) + 1

=> A period of 365 days contains one day in excess of some exact number of weeks

=> If the start-day is Tuesday, the 364^th day becomes Monday such that it completes a set of weeks and the 365^th day becomes a Tuesday again.

Case for 366-days: Following the logic mentioned above, for a Leap-year, if start-day is Tuesday, then the end-day becomes Tuesday+1 = Wednesday.

A Yahoo moment, just vanished...

x/y > 1

when I saw this expression, my thought went in this way:

Multiply ‘y’ on both sides of the inequality.

But can we do so? 
Yes, but if y is a negative number, then the inequality-sign changes.

x/y > 1 => x < y, for all y < 0

hold, for all the values of y on negative side of the real nunmber line, x is less than y. That means if y lies to the left of zero on the real number line, then x lies to the left of y. That means, x must be negative. I can say in other words: As y is less than zero and x is less than y, x must be less than zero.

In this case, we can conclude that, both x and y are negative.

Next we proceed to the other case, where y is positive. As y is a positive number, even if we multiply it on both sides of the inequality, the inequality-sign won't change.

x/y > 1 => x > y, for all y > 0

Here it goes, as y is greater than zero and x is greater than y, x must be greater than zero. So it concludes that, in this case, both x and y are positive.

Putting everything at one place,

If x/y > 1, either both x and y are positive are both x and y are negative.

I thought ‘Yahoo’…

But the bottom line is here, which always haunts me…

If x*y is positive, both x and y have same sign. Either both are positive or both are negative. This holds good for the clause “if x/y is positive” as well. This is a great old concept having nothing new to find out now in that. But, small variations come to our mind now and then and make us feel ‘Yahoooo’…..

Data Sufficiency: Quadratic Equations Problem Set

Q1) What is the value of x?

(1)  x²-6x+9 = 0

(2) x > 0

Q2) What is the value of ‘x’?

(1)  x²-8x+15 = 0

(2) x > 0

Fun with Numbers: some styles of manipulation

There are several ways of manipulations while dealing with number-problems. Trying out these ways is fun and helps in building math-analytical skills.

Consider a product of consecutive odd integers,
X = 3*5*7*9....47.

We observe a regular pattern of numbers spread over with a common difference between every two consecutive numbers. Yes, this is from a topic called 'Series and Progressions' of 'Algebra' and this is an arithmetic progression.

The number of terms = {(Final term-Initial term)/(common difference)}+1 = (47-3)/2 + 1 = 23

One thought goes like this:

Pair up the first and last terms, second and last-but-one, etc..Then all the terms get paired up except the middle one '25'.

(3*47)(5*45)(7*43)...(23*27)(25)

If you observe it, sum of the two terms in each pair is constant and is 50.

3+47 = 50

5+45 = 50

....

23+27 = 50.

This is natural as the two terms in any pair are centred around the middle term 25 and hence have an arithmetic mean of 25.

Arithmetic Mean?, yes, now gears are changed to 'statistics'

(3+47)/2 = 25

"The two terms in any pair are centred around 25", this observation leads us to a little manipulation of each term in relation to the number '25'.

{(25-22)(25+22)}{(25-20)(25+20)}.......{(25-2)(25+2)}(25)

Again it's the turn of 'Algebra'. The idea that comes to the mind is

(a-b)(a+b) = a²-b², apply it..

{25²-22²}{25²-20²}{25²-18²}.....{25²-2²}(25)  

All the steps in a nut shell:

X = 3*5*7*9....47

  =(3*47)(5*45)(7*43)...(23*27)(25)

  ={(25-22)(25+22)}{(25-20)(25+20)}.......{(25-2)(25+2)}(25)

  ={25²-22²}{25²-20²}{25²-18²}.....{25²-2²}(25)

The other thought goes like this:

Pair up every two consecutive terms starting from the first one. As there are odd number of terms, the last term remains unpaired.

(3*5)(7*9)(11*13)...(43*45)(47)

The two numbers in every pair can be manipulated in terms of the mean of that pair. For example, take the pair (3,5). Its mean = (3+5)/2 = 8/2 = 4. The number 3 can be manipulated as '4-1' and 5 can be manipulated as '4+1'.

All the steps in a nut shell:

X = 3*5*7*9....47

    =  (3*5)(7*9)(11*13)...(43*45)(47)

    = {(4-1)(4+1)}{(8-1)(8+1)}.....{(44-1)(44+1)}(47)

    = (4²-1)(8²-1)(12²-1)....(44²-1)(47)

Let us see one problem here:

Find the sum : 2+4+6+...48

There is no need of remembering any formulae or typical math terms to solve this problem.

First let us find the number of terms. To find it, we can take out ‘2’ common from all the terms,

2(1+2+3+....24)

This shows that there are 24 terms.

As we discussed in the first approach, build pairs of numbers with first and last, second and last-but-one,...

(2+48)+(4+46)+(6+44)+....(24+26)

Each pair contributes to an equal sum of '50'. Now we can easily find the sum of all terms. Isn't it?

GMAT Preparation: Some traps in Data Sufficiency

Look at the following example:

-------------------------------

Is x+y even?

(1)x-y is odd

(2)x²-y² is odd
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Stop reading here and try to find your answer option. Got it?, then proceed..

GMAT Quant Preparation: Different Strategies

Quant Section:

The 37 questions in this section comprise two kinds of questions:
Problem Solving (PS) and Data Sufficiency (DS). The two kinds do not have a definite break-up, usually there are around 20 PS and 17 DS questions. The level of Math-skills required is of High-school level. All the stuff tested here is on "the basics". One should be thorough in basic formulae and concepts to crack this section.

DS-Strategy:
DS Problem is a bit difficult in the sense that it requires a bit more time to solve when compared to a PS problem. It requires a careful approach which deals a step by step elimination of the options. If one does not follow the process, there is a high chance of landing on a wrong choice. Weigh the two given statements independently to decide whether either one alone is enough to answer. If both the statements are not independently enough to solve the problem, then check whether both the statements together can do. Go through this link for more light on DS strategies.
Silly Mistakes:

Don't be too fast while solving the problems. Even if you are strong in Quant section, go with a balanced approach. Be careful with the traps. More balance is required especially in dealing with DS problems as we are more prone to traps there.

Prep-stuff:

After completing OG material, try out some advanced books like "Manhattan Advanced Quant Book" to get more varieties of flavour. Go through the challenge Q's, 700-800 level Q's on prominent sites and stuff. 

Time-Strategy:

Time would not be a big constraint while solving the 'Quantitative' section, provided enough practice is made judiciously.

For Quantitative section, the given 75 minutes of time to solve 37 Q's leaves us a solid 2 minutes (or 120 seconds) time for each Q' on average. But as an average-PS problem generally takes lesser time than that of an average-DS one does, there should not be any strict emphasis on limiting each problem to 2 minutes. One problem may take 150 seconds while the other may take 90 seconds only. Some lighter ones can be cracked even in less than a minute's time. So instead of keeping track on time taken for each problem, it is better to have an overall track on a bunch of problems (say 5 or 10 at a time). Timed Practice with keeping these things in mind builds one's confidence in tackling any issues with time.

Best method of calculation:

"Calculations performed in mind save a lot of time when compared to calculations put on paper".

"Performing calculations in mind" can be built by developing liking towards numbers and practising a variety of exercises on a regular basis in study and non-study hours. Whenever we come across a number, trying to find out its factors, trying for divisibility tests on it, checking whether it is a prime or not etc. in mind automatically make us develope a liking towards calculations and shed off fear of math.

GMAT DS Problem: Dilemma whether to solve it or not?

Don't get carried away with solving a DS problem, all the times. The main point here is, in most of the cases, we need not really solve the problem. “Checking that whether the given statements are enough to solve the problem” is only required than actually solving the problem.

For example, see this one:

Q1)What percent of 10 is 'x':

(1) x is 50 percent of 12

(2) 10 is 5 percent of x

Here note that each statement is individually sufficient to solve the problem. We can find the value of ‘x’ from each statement independently and then get the required answer ie., “what percent of 10 is x”. But don't get carried away with solving. Just by seeing each of the statements, we can say that we can solve the problem with the help of it. That’s all, that’s enough. The answer option is "D"

Let us see one more example here:

Q2)Find the value of y:

(1) 3x-7y = 20

(2) x+y = 2+y

Here the ace is statement(2). It directly gives the value of x (of course, it is 2). But is it sufficient to answer the Q'? It is asked to find the value of 'y', not 'x'. Hence by using the value of 'x' from statement(2) and substituting it in statement(1), we can get the value of 'y'. As both the statements are required to solve this one, the answer option is "C".

But suppose that it is asked for the value of 'x' in the main stem of the problem. In that case, irrespective of statement(1), statement(2) alone is sufficient to answer the Q’. And also, as statement(1) alone is not sufficient to answer the Q’, the answer option is "B".

But in some cases, it is required to solve the problem. This arises in cases where the given statements seem to be enough to answer the Q’ but if we really solve, it may result in "ambiguous" or "not defined" or no answers. This is where GMAC can really trick us. For a given DS problem, “the clarity on whether to go for solving or not” will come only through practicing different types of problems and a careful noting of the special cases there on.

For example, see this one:

Q3)What is the value of 'x":

(1) x-2y = 7

(2) 3x-5 = 6y

We know that two unknowns (in this case x and y) can be solved by using at least two equations. And hence we conclude that, both the given conditions are required together to find the value of x. That’s where we commit mistake.

See the second statement carefully. 3x-5 = 6y => 3x-6y = 5 => 3(x-2y) = 5 => “x-2y = 5/3”, which is inconsistent with the first statement "x-2y = 7". Hence, this Q’ cannot be answered even if we consider both the given statements together. To get this clarity, as just now done on the second statement, we need to shed a little on solving the given statements. I think we have no doubts to mark the answer option as "E" for the above Q’.

Finally, we will see one more example, where we need to really work on the given statements to get answer for the Q’.

Q4) Find the value of “5x-8y+z”:

(1) 2x+y-2z = 9

(2) 5x-y-3z = 6

A little brainstorming is required here whether we can find two numbers with which if we respectively multiply the two statements and do some manipulation to finally yield the value of “5x-8y+z”. In the case, where we can find those two numbers, we can select the answer option “C”. But, there is a possibility that we cannot find those two numbers ultimately and may land to the answer option “E” as well. Try this out…

Concept on Clocks

Speed of minutes hand = 360⁰ per hour = 6⁰ per minute

Speed of hours hand = 360⁰ per 12 hours = 0.5⁰ per minute

Relative speed of minutes hand over hours hand = 5.5⁰ per minute

If time is expressed as h:m, to find angle between minutes and hours hands:

At time h:0, angle between minutes and hours hands = 30h

After m minutes, ie., at h:m, minutes hand travels an angle of ‘5.5m’ relative to hours hand, and may or may not exceed the angle ‘30h’

=> the angle = 30h-5.5m (or) 5.5m-30h

If time is expressed as h:m, angle between minutes and hours hands 

= 30h-5.5m (or) 5.5m-30h

Numbers Concepts -1

1. For any positive integer n, product of ‘n’ or more than ‘n’ consecutive positive integers is divisible by n!.

For example, 63*64*65*.......*91 is divisible by 29!
2. If (n-1)! is not divisible by ‘n’, then n is a prime number. 4 is a special case here, which obeys this rule but not a prime.
3. From 1 & 2 above, we can conclude that,  for any positive integer 'n-1', if the product of ‘n-1’  consecutive positive integers is not divisible by n, then n is a prime. But if divisible, we can't say that n is not a prime.
4. If a > b ≥ 3, then b^a > a^b where a,b Є Z+
5. Product of two successive integers always ends in 2,6, or 0
6. The general perception about even numbers is even numbers start with 0 and go on 2,4,6,8,... and the odd numbers are 1,3,5,.... But negative integers are also to be categorized in to even and odd sets.

Hence the even number set is: ....-4,-2,0,2,4,6,.....
and the odd number set is: ....-5,-3,-1,3,5,7,.....
7. Why '1' is not a prime number?
A prime number can be defined as a positive integer that has exactly two different positive divisors, 1 and itself.  As the number '1' has only one positive divisor (ie., itself), it is not a prime number. 

Probability-11 (GMAT)

Ellen invites 5 people to dinner, 2 women and 3 men. If Ellen's dining table is circular and the guests choose their seats at random around her table, what is the probability that no female is seated next to another female?

(A) 1/90          (B) 1/72          (C) 5/72          (D) 5/36          (E) 1/10

Solution follows here:

Progressions-13 (NCERT)

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed?

(1)17   (2) 20   (3) 25   (4) 27   (5) 30

Solution follows here:

Algebra-4

You own a pet store. If you put in one canary per cage, you have one canary too many. If you put in two canaries per cage, you have one cage too many. How many canaries and cages do you have?
      (A) five canaries and four cages

(B)four canaries and five cages

(C)three canaries and four cages

(D)four canaries and three cages

(E) none of the above

Solution follows here:

Magic of Numbers-1 (Unit digit Patterns of number exponents)

Theory of numbers is wonderful. Here we see the unit digit patterns of exponents of numbers. As we go on incrementing the exponent of a number the unit digits of the results follow regular patterns.  Each number has its own song. Now its time to listen to those songs:

11 = 1	21 = 2	31 = 3	41 = 4	51 = 5	61 = 6
12 = 1	22 = 4	32 = 9	42 = 16	52 = 25	62 = 36
13 = 1	23 = 8	33 = 27	43 = 64	53 = 125	63 = 216
	24 = 16	34 = 81	44 = 256	54 = 625	64 = 1296
	25 = 32	35 = 243
	26 = 64	36 = 729
	27 = 128	37 = 2187
	28 = 256	38 = 6561

71 = 7	81 = 8	91 = 9	101 = 10
72 = 49	82 = 64	92 = 81	102 = 100
73 = 243	83 = 512	93 = 729	103 = 1000
74 = 1701	84 = 4096	94 = 6561
75 = 11907	85 = 32768
76 = 83349	86 = 262144
77 = 583443	87 = 2097152
78 = 4084101	88 = 16777216

From the above two tables we can observe the unique feature of patterns of unit

digits of the exponent results. (Observe the unit digits in bold font)

The unit digits of exponent-results of each number follow a unique pattern: