For the first time, nearly 30,000 degree students of Vinoba Bhave University (VBU) would be using computerized answer sheets at their degree Part III examination commencing from Tuesday.
Although college authorities fear that the students from rural background may not be able to fill the required information on the computerized first page of the answer books, the university has made elaborate arrangement to ensure that everything passes off peacefully.
The university has distributed dummy copies of the first page of the computerized answer sheet along with admit card to the students well in advance to avoid any last-minute trouble at the examination hall.
Dravyansh Sharma of Delhi Public School, Meerut Road Ghaziabad with an overall aggregate of 98.6 per cent stood first in the overall merit list of students who appeared in class 12th examinations conducted by Central Board of Secondary Education (CBSE) from Uttar Pradesh and Uttarakhand and the results for which were declared on Monday.
Sharma was closely followed in the second spot by Sachin Kumar of DAV Centenary Public School, Haridwar with an aggregate of 98 per cent followed by Kalyani Chaudhary of Army Public School, NOIDA, Gautam Buddh Nagar with 97.6 per cent.The fourth spot went to Sanchit Gangwar of Ben Hur Public School, Pilibhit with 97.6 percent while Paridhi Gupta of DPS Shastripuram , Agra with 97.4 percent bagged the fifth spot.
Releasing the Class XII results on Monday, CBSE said 91.32% of students who took the exam in the region had passed, well above the national average of 81.71%.
The region, which scored over the national average on almost all parameters, however, saw a dip when compared to the performance in 2010. The pass percentage saw a dip from last year as did the individual pass percentages of boys and girls.
CBSE does not release the list of top rankers nor the top marks. In Chennai, at least five students notched aggregate scores of 487 and above. Among the high scorers were S Divya of Modern Senior Secondary School in Nanganallur with 98.4% and K A Sumana of PS Senior Secondary School in Mylapore with 98.2%.
School heads said that while there didn't seem to be much change in the number of centums in maths from last year, the number of centums in physics had gone down. Students had said earlier that physics exam was tough.
The reform is likely to benefit more than 10 lakh students every year, who at present have to take multiple examinations to get into engineering courses.
In a meeting, which was attending by the directors of IIT Delhi, Kanpur and other engineering institutions, the government broadly agreed on “one nation one examination” policy. Dr T Ramaswamy, secretary, department of science and technology was entrusted the job to re-look at the test methodology of selecting students and have a common system for admission. According to sources, NAT will be similar to the single medical entrance test that the Medical Council of India (MCI) will conduct from next year.
The state governments may not approve the idea of NAT and the IITs do not want the JEE system to be destroyed. “You already have a good system working, why destroy it,” an IIT director said.
During the last meeting of the IITs council, there was no consensus on having a single entrance examination for admissions. Seven out of 15 IITs favoured having a separate “add-on” examination in addition to the NAT. However, HRD minister Kapil Sibal is of the view that if IITs have an additional exam then nothing stops other institutions like planning and architecture colleges etc to have their own exams.
The Science Advisory Council to the Prime Minister in its recommendations has also favoured a single entrance examination for engineering.
RAJKOT: A city boy has become the youngest student to equal the world
record of obtaining full 1,000 marks in an online computer exam
conducted by US-based multinational Microsoft Corp.
Akshit Jayesh
Dhruv, a Class V student, appeared in the exam on March 16 and obtained
full marks, said Chirag Kothari, Centre Head of Inside Tomorrow
Computer Academy, where the 11-year-old studied.
Jammu University, located in Jammu & Kashmir,
on Tuesday i.e. March 1, 2011, has declared the results of Master of
Education (M Ed) I Semester (DDE/Private) exam. The exams were held in
2010. To know more about the results, candidates can visit the
university's official website: http://www.jammuuniversity.in/.
Jammu University is located in Jammu & Kashmir. It is one of
the renowned universities of the region accredited with ‘A’ grade by the
National Assessment and Accreditation Council (NAAC), Bangalore. The
university offers education at different degree level in a wide variety
of disciplines - Life Sciences, Arts / Oriental Languages, Sciences,
Education, Commerce, Management Studies, Behavioral Science, Law, Social
Science, Medicine, Engineering and Music & Fine Arts.
Date of Results Declaration
March 1, 2011
Category of Exam
Education - Master of Education (M Ed) I Semester (DDE/Private) Exam
Rajasthan Technical University (RTU), Kota,
Rajasthan, on Tuesday i.e. March 1, 2011, has declared the results of
Master of Technology (M Tech) I Semester Main exam. To know more about
the results, candidates can visit the university's official website:
http://www.rtu.ac.in/.
Located in Kota, Rajasthan Technical University (RTU) is one of the
premier technical universities of the state. It was set up in 2005 by
Govt. of Rajasthan to enhance technical education in Rajasthan.
Presently, 60 Engineering Colleges, 08 M Tech Colleges, 21 MCA Colleges,
69 MBA Colleges and three Hotel Management and Catering Institute are
affiliated to the university.
Date of Results Declaration
March 1, 2011
Category of Exam
Science/Engineering - Master of Technology (M Tech) I Semester Main Exam
The results of the sixth teachers' registration examination test-2010 have been published.
A total of 42,641 candidates passed the test, showing a pass percentage of 19.34.
Education minister Nurul Islam Nahid announced the results at a press conference on Wednesday.
The
results are available on the websites of Non Government Teachers'
Registration and Certification Authority (NTRCA) and government-run
mobile company Teletalk.
The number of successful candidates for colleges is 15,280 or 19.81 percent, while that for schools is 27,361 or 19.08 percent.
The exam was held on Dec 10 and 11 last year, with 220,517 candidates taking part in it.
The
minister said that NTRCA would gradually be developed to recruit
teachers. It will be established like Public Service Commission (PSC).
"It helps make the list of meritorious candidates," Nahid added.
On February 28, 2011 results of UG Semester I
(Lateral Entry) examination was declared by Sri Krishnadevaraya
University (SKU), Anantapur. For knowing more about the results one can
visit the University’s official website: www.skuniversity.org
Sri Krishnadevaraya University (SKU) is a public university in Anantapur
in the state of Andhra Pradesh. It has earned its name as a centre of
academic excellence and it is also accorded with a four star status by
the National Assessment and Accreditation Council (NAAC). The University
was founded in the year 1968, however, in 1987 it was conferred the
status of an affiliating University. Presently, it provides course at
various faculties via its numerous affiliated colleges / institutes.
Date of Results Declaration
February 28, 2011
Where to Go for Result www.skuniversity.org
Visvesvaraya Technological University (VTU),
Belgaum, Karnataka, on March 1, 2011, has declared the results of
Bachelor of Engineering (BE)/Bachelor of Technology (B Tech) III
Semester exam. The results have been announced for Belgaum, Gulbarga and
Mysore region. To know more about the results, candidates can visit the
university's official website: http://www.vtu.ac.in/.
Located in Belgaum, Karnataka, Visvesvaraya Technological
University (VTU) is one of the largest technical universities in India.
The university offers engineering and technology courses at various
levels. Currently, the university has more than 200 departments across
the affiliated colleges recognized as Research Centres.
Date of Results Declaration
March 1, 2011
Category of Exam
Science/Engineering - Bachelor of Engineering (BE)/Bachelor of Technology (B Tech) III Semester Exam
The results have been uploaded on the official website of the University.
The students will have to upload their registration number/Roll number to download the results.
We are providing below the links for the official site.
You can connect to the official website after clicking on the following links. http://www.ptu.ac.in/ http://ptuexam.com/
The question paper of an all-India
examination for the post of assistant administrative officer of Life
Insurance Corporation of India (LIC) was leaked here on Sunday ahead of
the test.
Delhi Police has taken into its custody a youth, who allegedly sold the
papers for Rs five lakh each and is believed to be the main conspirator.
The police swung into action after the crime branch officials got hold
of the copies of the question papers for both morning and afternoon
sessions from some persons much ahead of the scheduled time of the
examination.
The test is being conducted in two sessions in 160 centres across India. There are 16 centres for the test in New Delhi.
An LIC spokesperson said Educational Consultants India Limited (EDCIL), a
public sector company, is conducting the entire examination process.
He further said that there is no report of leakage of question papers from any other part of India.
Rashtrasant Tukadoji Maharaj Nagpur University
(RTMNU), formerly known as Nagpur University, located in Nagpur
(Maharashtra), on Sunday i.e. February 27, 2011, has declared the
results of Bachelor of Engineering (BE) III Semester Winter 2010 exam.
To know more about the results, candidates can visit the university's
official website: http://www.nagpuruniversity.org/.
Rashtrasant Tukadoji Maharaj Nagpur University (RTMNU), formerly
known as Nagpur University, which is located in Nagpur (Maharashtra), is
considered as one of the prominent universities in the state of the
Maharashtra. It conducts several undergraduate and postgraduate level
programmes.
Date of Results Declaration
February 27, 2011
Category of Exam
Science/Engineering - Bachelor of Engineering (BE) III Semester Winter 2010 Exam
Gulbarga University, Karnataka, has announced the
results of UG (undergraduate), held in November 2010. Students can check
the results are made available in the University’s official website:
http://www.gulbargauniversity.kar.nic.in/
Gulbarga University Results are available for B.Com, BA, B.Sc and BCA.
To view the results one must enter Register No. and Select Course &
Year/Semester to get results in the given link:
University of Rajasthan (UOR), popularly known as
Rajasthan University, Jaipur, is one of the oldest universities of the
state. The University has been accredited with ‘A+’ grade by National
Accreditation and Assessment Council (NAAC), Bangalore, for its academic
excellence. The University is an excellent quality higher education in
the state. Presently, the University has 36 post graduate departments,
15 recognized research centers, 6 constituent colleges and 500
constituent colleges, spread across 6 districts of Rajasthan.
Date of Results Declaration
February 25, 2011
Category of Exam
IT - BCA (Bachelor of Computer Applications) Part III (Supplementary) Exam
Pune University, formerly known as University of Poona, is situated in
Pune, Maharashtra. It is one of the renowned universities offering
quality education in the region. The university imparts education in
wide range of disciplines via its various affiliated colleges /
institutes.
Date of Results Declaration
February 24, 2011
Category of Exam
Science - Master of Science (M Sc) Computer Science Exam
The Institute of Company Secretaries of India (ICSI) was constituted
under an Act of Parliament i.e. the Company Secretaries Act, 1980 (Act
No. 56 of 1980). It is the only recognized professional body in India
to develop and regulate the profession of Company Secretaries in India.
The Institute of Company Secretaries of India awards the certificate of
bestowing the designation of Company Secretary (CS) to a candidate
qualifying for the membership of the Institute.
Institute of Company Secretaries of India (ICSI), New Delhi, declared the results of CS December 2010 examination today i.e. on February 25, 2011 at 12.00 noon. The results will be made available in the institute’s official website: www.icsi.edu/ Candidates will be able to check their CS December 2010 examination results in the aforementioned official website, once declared.
One can also get the results through email by visiting the following link: http://www.icsi.edu/icsiweb/email/resemail.asp.
For that candidate must select :
Syllabus / Course (From the Given Options),
enter their Email and
Roll No in the space provided and
click in the Submit Button in the aforementioned link.
Contact address:
Institute of Company Secretaries of India New Delhi,
Delhi
1) 2^2n-1 is always divisible by 3 2^2n-1 = (3-1)^2n -1 = 3M +1 -1 = 3M, thus divisible by 3
2) What is the sum of the divisors of 2^5.3^7.5^3.7^2? ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6 Funda : if a number 'n' is represented as a^x * b^y * c^z .... where, {a,b,c,.. } are prime numbers then
Quote:
(a) the total number of factors is
(x+1)(y+1)(z+1) ....
(b) the total number of relatively prime numbers less than the number is n *
(1-1/a) * (1-1/b) * (1-1/c)....
(c) the sum of relatively prime numbers less than the number is n/2 * n *
(1-1/a) * (1-1/b) * (1-1/c)....
(d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} *
...../(x*y*...)
http://hallosushant.blogspot.com/ 3) what is the highest power of 10 in 203!
ANS : express 10 as product of primes; 10 = 2*5
divide 203 with 2 and 5 individually 203/2 = 101 101/2 = 50 50/2 = 25 25/2 = 12 12/2 = 6 6/2 = 3 3/2 = 1 thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1
= 198
divide 203 with 5 203/5 = 40 40/5 = 8 8/5 = 1
thus power of 5 in 203! is, 49
so the power of 10 in 203! factorial is 49
4) x + y + z = 7 and xy + yz + zx = 10, then what is
the maximum value of x? ( CAT 2002 has similar question ) ANS: 49-20 = 29, now if one of the y,z is zero, then the
sum of other 2 squares shud be equal to 29, which means, x can take a max value
at 5
5) In how many ways can 2310 be expressed as a product
of 3 factors? ANS: 2310 = 2*3*5*7*11 When a number can be expressed as a product of n distinct
primes, then it can be expressed as a product of 3 numbers in
(3^(n+1) + 1)/2 ways
6) In how many ways, 729 can be expressed as a
difference of 2 squares? ANS: 729 = a^2 - b^2 = (a-b)(a+b), since 729 = 3^5, total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27. So 4 ways Funda is that, all four ways of expressing can be used to
findout distinct a,b values, for example take 9*81 now since 9*81 = (a-b)(a+b) by solving the system a-b = 9
and a+b = 81 we can have 45,36 as soln.
7) How many times the digit 0 will appear from 1 to
10000 ANS: In 2 digit numbers : 9, In 3 digit numbers : 18 + 162 = 180, In 4 digit numbers : 2187 + 486 + 27 = 2700, total = 9 + 180 + 2700 + 4 = 2893
8 ) What is the sum of all irreducible factors between
10 and 20 with denominator as 3? ANS : sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 +
13.33 + 13.66……. = 21 + 23 + …… = 300
9) if n = 1+x where x is the product of 4 consecutive
number then n is, 1) an odd number, 2) is a perfect square SOLN : (1) is clearly evident (2) let the 4 numbers be n-2,n-1,n and n+1 then by multing
the whole thing and adding 1 we will have a perfect square
10) When 987 and 643 are divided by same number 'n' the
reminder is also same, what is that number if the number is a odd prime number? ANS : since both leave the same reminder, let the reminder
be 'r', then, 987 = an + r and 643 = bn + r and thus 987 - 643 is divisible by 'r' and 987 - 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43
hence 'r' is 43
11) when a number is divided by 11,7,4 the reminders
are 5,6,3 respectively. what would be the reminders when the same number is
divided by 4,7,11 respectively? ANS : whenever such problem is given, we need to write the numbers in top row and rems in the
bottom row like this
11 7 4 | \ \ 5 6 3
( coudnt express here properly )
now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 +
6) + 5 that is 302 + LCM(11,7,4) and thus the rems when the same
number is divided by 4,7,11 respectively are,
302 mod 4 = 2 75 mod 7 = 5 10 mod 11 = 10
12) a^n - b^n is always divisible by a-b
13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc EXAMPLE: 40^3-17^3-23^3 is divisble by since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus,
the number is divisible by 3,5,8,17,23 etc.
http://hallosushant.blogspot.com/ 14) There is a seller of cigerette and match boxes who
sits in the narrow lanes of cochin. He prices the cigerattes at 85 p, but found
that there are no takers. So he reduced the price of cigarette and managed to
sell all the cigerattes, realising Rs. 77.28 in all. What is the number of
cigerattes?
a) 49 b) 81 c) 84 d) 92
ANS : (d) since 77.28 = 92 * 84, and since price of cigarette is
less than 85, we have (d) as answer
Quote:
i have given this question to make the funda clear
15) What does 100 stand for if 5 X 6 = 33 ANS : 81 SOLN : this is a number system question, 30 in decimal system is 33 in some base 'n', by solving we
will have n as 9 and thus, 100 will be 9^2 = 81
16) In any number system 121 is a perfect square, SOLN: let the base be 'n' then 121 can be written as n^2 + 2*n + 1 = (n+1)^2 hence proved
17) Most of you ppl know these, anyways, just in case
Quote:
(a) sum of first 'n' natural numbers -
n*(n+1)/2
(b) sum of the squares of first 'n' natural numbers - n*(n+1)*(2n+1)/6
(c) sum of the cubes of first 'n' natural numbers - n^2*(n+1)^2/4
(d) total number of primes between 1 and 100 - 25
18 ) See Attachment to
know how to find LCM, GCF of Fractions
Quote:
CAT 2002 has 2 questions on the above
simple concept
19) Converting Recurring Decimals to Fractions
let the number x be 0.23434343434........
thus 1000 x = 234.3434343434...... and 10 x = 2.3434343434......... thus, 990 x = 232 and hence, x = 232/990
20) Reminder Funda
(a) (a + b + c) % n = (a%n + b%n + c%n) %n EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are
divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder
when you divide 3993 with
9? ( never seen such question though ) the reminder would be (6 + 8 + 1) % 9 = 6
(b) (a*b*c) % n = (a%n * b%n * c%n) %n EXAMPLE: What is the remainder left when 1073 * 1079 *
1087 is divided by 119 ? ( seen this kinda questions alot ) 1073 % 119 = ? since 1190 is divisible by 119, 1073 mod 119 is 2 and thus, "the remainder left when 1073 * 1079 * 1087
is divided by 119 " is 2*8*16 mod 119 and that is 256 mod 119 and that is
(238 + 18 ) mod 119 and that is 18
Glossary : % stands for reminder operation
find the number of
zeroes in 1^1* 2^2* 3^3* 4^4.............. 98^98* 99^99* 100^100
the expresion can be
rewritten as (100!)^100 / 0!* 1!* 2!* 3!....99!
Now the numerator has 2400 zeros
the formular for finding number of zeros in n! is
[n/5]+[n/5^2]...[n/5^r] where r is such that 5^r<=n<5^(r+1)
and [..] is the grestest integer function
for the numerator find the number of zeros using the above
formulae..
for 0!...4! number of zeros ..0 5!...9!.number os zeros ..1 9!...14!... 2 15!..19!..................3 20!..24!..................4! now at 25! the series makes a jump to 6 25!...29!.................6 30!...34!.................7 this goes on and again makes a jump at 50! and then at 75!
so the number of zeros is...
5(1+2....19) + 25+ 50+ 75
the last 3 terms 25 50 and 75 are because of the jumps..
this gives numerator has 1100 zeros
now total number of zeros in expression is no of zeros in
denominator - no of zeros in numerator 2400 - 1100
If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.
eg: x^4+3x^2+2x+6=0 has no positive roots . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .
Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
For a cubic equation ax^3+bx^2+cx+d=o
sum of the roots = - b/a sum of the product of the roots taken two at a time = c/a product of the roots = -d/a ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0
sum of the roots = - b/a sum of the product of the roots taken three at a time = c/a sum of the product of the roots taken two at a time = -d/a product of the roots = e/a +++++++++++++++++++++++++++++++++++++++++++++++++++++++++
If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if
x=y(=k/2). The maximum product is then (k^2)/4 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++
If for two numbers x*y=k(=constant), then their SUM is MINIMUM if
x=y(=root(k)). The minimum sum is then 2*root(k) . +++++++++++++++++++++++++++++++++++++++++++++++++++++++++
|x| + |y| >= |x+y| (|| stands for absolute value or modulus )
(Useful in solving some inequations) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Product of any two numbers = Product of their HCF and LCM .
Hence product of two numbers = LCM of the numbers if they are prime to each other
For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .
If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximum
if a/p = b/q = c/r = d/s . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Consider the two equations
a1x+b1y=c1 a2x+b2y=c2
Then , If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations. If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to ) If a1/a2 <> b1/b2 , then we have a unique solutions for these equations.. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is
0.5*d1*d2, where d1,d2 are the lenghts of the diagonals. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is ,
the minute hand describes 6 degrees /minute the hour hand describes 1/2 degrees /minute .
Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .
The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight. (This can be derived from the above) . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
If n is even , n(n+1)(n+2) is divisible by 24
If n is any integer , n^2 + 4 is not divisible by 4
Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2 =a*b*c/(4*R) where R is the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle .
In any triangle a=b*CosC + c*CosB b=c*CosA + a*CosC c=a*CosB + b*CosA ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
If a1/b1 = a2/b2 = a3/b3 = .............. , then each ratio is equal to
(k1*a1+ k2*a2+k3*a3+..............) / (k1*b1+ k2*b2+k3*b3+..............) , which is also equal to (a1+a2+a3+............./b1+b2+b3+..........) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
In any triangle
a/SinA = b/SinB =c/SinC=2R , where R is the circumradius ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 - 14^3)
If Q be the volume of a vessel q qty of a mixture of water and wine be removed each time from a mixture n be the number of times this operation be done and A be the final qty of wine in the mixture
then , A/Q = (1-q/Q)^n +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height
where median is the line joining the midpoints of the oblique sides. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
when a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .
Let a be the side of an equilateral triangle . then if three circles be drawn inside this triangle touching each other then each's radius = a/(2*(root(3)+1)) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Let 'x' be certain base in which the representation of a number is 'abcd' , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d
Many of u must b aware of this formula, but the ppl who don't know it must b useful for them.
a+b+(ab/100)
this is used for succesive discounts types of sums. like 1999 population increses by 10% and then in 2000 by 5% so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999
and if there is a decrease then it will be preceeded by a -ve sign and likeiwse
On February 28, 2011 results of UG Semester I
(Lateral Entry) examination was declared by Sri Krishnadevaraya
University (SKU), Anantapur. For knowing more about the results one can
visit the University’s official website: 
Gulbarga University, Karnataka, has announced the
results of UG (undergraduate), held in November 2010. Students can check
the results are made available in the University’s official website:
