MathByVemuri: May 2012

Thursday, 24 May 2012

SSC Results 2012 - A new Grading System: Math concept on number ranges

The SSC results are out today. This time there is a huge difference in the way of results announcement. A new "Grade Point System" surprised one and all. It's the talk of the day.

A scale of 10:

A student getting 90 in Hindi and all 92's in the remaining 5 subjects gets a 10 out of 10 scale, but do you believe what his total score is? 550 out of 600.

But a student getting all 100's in five subjects and 91 in the remaining one subject gets a scale of 9.8. you get surprised at his total.It is 591. This is the surprise part of the scaling system. "A student with a total marks of 591 has a GPA of 9.8 where as a student with a total of 550 can get 10 out of 10".

Rules of the game: The Grade point system in a nut shell

There are six subjects: 1st Language, 2nd Language, 3rd Language, Maths, Science and Social Studies. The grade points are awarded for these subjects individually based on the actual marks obtained in those subjects and the key table followed for this process is:

1st Language, 3rd Language, Non-language subjects	2nd Language	Grade	Grade Point


92-100	90-100	A1	10
83-91	80-89	A2	9
75-82	70-79	B1	8
67-74	60-69	B2	7
59-66	50-59	C1	6
51-58	40-49	C2	5
43-50	30-39	D1	4
35-42	20-29	D2	3
34-0	19-0	E	Fail

A student with marks 92,90,92,92,92,92 for 1st Language, 2nd Language, 3rd language, Maths, Science and Social Studies in that order (A total of 550)gets grade points 10,10,10,10,10 and 10. The Grade Point Average becomes (10+10+10+10+10+10)/6 = 60/6 = 10

A student with marks 100,100,91,100,100,100 for 1st Language, 2nd Language, 3rd language, Maths, Sciencs and Social Studies in that order (A total of 591) gets grade points 10,10,9,10,10 and 10. The Grade Point Average becomes (10+10+9+10+10+10)/6 = 59/6 = 9.8

GPA workout:

These GPAs are rounded to one decimal point.

Some possible GPAs:

(10+10+10+10+10+10)/6 = 60/6 = 10

(9+10+10+10+10+10)/6 = 59/6 = 9.8

(9+9+10+10+10+10)/6 = 58/6 = 9.7

(9+9+9+10+10+10)/6 = 57/6 = 9.5

56/6 = 9.3; 55/6 = 9.2; 54/6 = 9.0......

The Total of Grade Points can be decremented by a minimum of 1.

Let us see how the Range of Total marks varies for different GPAs:

GPA	Minimum Total	Maximum Total
10	92+92+90+92+92+92 = 550	100+100+100+100+100+100 = 600
9.8	92+92+80+92+92+92 = 540	100+100+100+100+100+91 = 591

Try to fill out the table for some other possible GPAs. It's not an easy game. It's number crunching all along...

check out the results here:

GPA	Min. Total	Max. Total
10	550	600
9.8	540	591
9.7	530	582
9.5	520	574
9.3	510	566
9.2	500	558
9	490	550
8.8	480	542

Monday, 21 May 2012

CSAT 2012 solved paper (paper-II)

click here for useful concepts click here for puzzles
Q1)
Mr Kumar drives to work at an average speed of 48 km per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. How far is his office?

(a) 30 km (b) 40 km (c) 45 km (d) 48 km
Solution:

Let the distance be x.

First 60% distance is covered in 10 more minutes more time than that of the rest 40% distance

=> 0.6*x distance is covered in 10 more minutes more than that of the rest 0.4*x distance

=> 0.2*x is covered in 10 minutes ie., 1/6 hr

=> average speed = distance/time = 0.2*x/(1/6) = 48

=> x = 48/6*0.2 =40 km

Answer-(b)

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.

Saturday, 12 May 2012

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’…..

Friday, 11 May 2012

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?

Monday, 7 May 2012

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

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.