Edu Services » Blog » edubuzz360 blog
By: S@$h!k!r@n  September 12, 2017
Er.Palanivelraj, Mechanical Designer
is doing a great Service in Designing AutoCAD 2D drawings and Solid work 3D models at the Best Cost !
Students and Business people can approach him for converting your ideas into Designs !
Make use of this wonderful Service !
Category: Uncategorized
Tags:
By: S@$h!k!r@n  April 04, 2017
Tamil Nadu College  IIT, madras takes the First place in College Ranking of 2017
22 colleges in Tamil Nadu among the top 100 College List !
Category: education
Tags: College Ranking 2017
By: S@$h!k!r@n  February 02, 2017
Once one of the most sought after courses in the nation, engineering still holds it charm in the eyes of lakhs of applicants that apply every year for the entrance examinations. Engineering for some is their passion and there are those too who interested or not, consider it a rite of passage. Whichever category you fall in, this article aims to introduce you to some of the major national level examination for entrance into engineering courses. We hope that this post gives you some clarification regarding the programs offered under each of these examinations, along with the basic theme on which they function. So without wasting much time, let’s start with it.
JEE Main 2017
Conducted by the Central Board of Secondary Education (CBSE), ...
Category: education
Tags: Engineering Entrance Examinations for Academic Year 2017
By: S@$h!k!r@n  July 22, 2016
Ex. 1. A bill for Rs. 6000 is drawn on July 14 at 5 months. It is discounted on 5th October at 10%. Find the banker's discount, true discount, banker's gain and the money that the holder of the bill receives.
Sol.
Face value of the bill = Rs. 6000.
Date on which the bill was drawn = July 14 at 5 months. Nominally due date = December 14.
Legally due date = December 17.
Date on which the bill was discounted = October 5.
Unexpired time : Oct. Nov. Dec.
26 + 30 + 17 = 73 days =1/ 5Years
B.D. = S.I. on Rs. 6000 for 1/5 year
= Rs. (6000 x 10 x1/5 x1/100)= Rs. 120.
T.D. = Rs.[(6000 x 10 x1/5)/(100+(10*1/5))]
=Rs.(12000/102)=Rs. 117.64.
B.G. = (B.D.)  (T.D.) = Rs. (120  117.64) = Rs. 2.36.
Money received by the holde...
Category: Uncategorized
Tags: Aptitude, RS Aggarwal, bankers discount, problems in bankers discount
By: S@$h!k!r@n  July 22, 2016
IMPORTANT CONCEPTS
Banker's Discount : Suppose a merchant A buys goods worth, say Rs. 10,000 from another merchant B at a credit of say 5 months. Then, B prepares a bill, called the bill of exchange. A signs this bill and allows B to withdraw the amount from his bank account after exactly 5 months.
The date exactly after 5 months is called nominally due date. Three days (known as grace days) are added to it to get a date, known as legally due date.
Suppose B wants to have the money before the legally due date. Then he can have the money from the banker or a broker, who deducts S.I. on the face value (i.e., Rs. 10,000 in this case) for the period from the date on which the bill was discounted (i.e., paid by the banker) and the legally ...
Category: Quantitative Aptitude
By: S@$h!k!r@n  July 22, 2016
Ex. 1. Find the present worth of Rs. 930 due 3 years hence at 8% per annum. Also find the discount.
Sol.
P.W=100 x Amount /[100 + (R x T)]
=Rs.100 x 930/100+ (8x3)
= (100x930)/124
= Rs. 750,
T.D. = (Amount)  (P.W.) = Rs. (930  750) = Rs. 180.
Ex. 2. The true discount on a bill due 9 months hence at 12% per annum is Rs. Find the amount of the bill and its present worth.
Sol. Let amount be Rs. x. Then,
x*R*T/100 + (R x T)
=T.D.
=>x * 12*3/ 4/[100+[12*3/4]]
=540
x= 540x109 = Rs.6540
Amount  Rs. 6540. P.W. = Rs. (6540  540)  Rs. 6000.
Ex. 3. The true discount on a certain sum of money due 3 years hence is Rb. 250 and the simple interest on the same sum for the same time and at the same rate is Rs. 375. F...
By: S@$h!k!r@n  July 22, 2016
IMPORTANT CONCEPTS
Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to Rs. 156 in 4 years. So, the payment of Rs. 100 now will clear off the debt of Rs. 156 due 4 years hence. We say that:
Sum due = Rs. 156 due 4 years hence;
Present Worth (P.W.) = Rs. 100;
True Discount (T.D.) = Rs. (156  100) = Rs. 56
(Sum due)  (P.W.).
We define : T.D. = Interest on P.W.
Amount = (P.W.) + (T.D.).
Interest is reckoned on P.W. and true discount is reckoned on the amount.
Let rate = R% per annum and Time = T years. Then,
1. P.W.=[100 x Amount /100 + (R x T)
=100 x T.D./ RxT
2. T.D.=[(P.W.) x R x T /100]
= [ Amount x RxT/100 + (R x T)]
3.(S...
Category: Quantitative Aptitude
By: S@$h!k!r@n  July 22, 2016
SOLVED EXAMPLES
Ex 1. In a throw of a coin ,find the probability of getting a head.
sol. Here s="{h,t} and e={h}.
P(E)=n(E)/n(S)=1/2
Ex2.Two unbiased coin are tossed .what is the probability of getting atmost one head?
sol.Here s={hh,ht,th,tt}
Let Ee="event" of getting one head
e={tt,ht,th}
p(e)=n(e)/n(s)=3/4
Ex3.An unbiased die is tossed .find the probability of getting a multiple of 3
sol. Here s={1,2,3,4,5,6}
Let e be the event of getting the multiple of 3
then ,e={3,6}
p(e)=n(e)/n(s)=2/6=1/3
ex4. in a simultaneous throw of pair of dice .find the probability of getting the total more than 7
sol. Here n(s)=(6*6)=36
let e=event of getting a total more than 7
={(2,...
Category: Quantitative Aptitude
Tags: Aptitude, RS Aggarwal, probability, problems in probability
By: S@$h!k!r@n  July 22, 2016
important facts and formula
1.Experiment :An operation which can produce some welldefined outcome is called an experiment
2.Random experiment: An experiment in which all possible outcome are known and the exact out put cannot be predicted in advance is called an random experiment
Eg of performing random experiment:
(i)rolling an unbiased dice
(ii)tossing a fair coin
(iii)drawing a card from a pack of well shuffled card
(iv)picking up a ball of certain color from a bag containing ball of different colors
Details:
(i)when we throw a coin. Then either a head(h) or a tail (t) appears.
(ii)a dice is a solid cube, having 6 faces ,marked 1,2,3,4,5,6 respectively when we throw a die , the outcome is the number that appear on its top face .
(iii)a pack of cards has 52 cards it has 13 cards of each suit ,namely spades, clubs ,hearts and diamonds
Cards of spades and clubs are black cards
Cards of hearts and diamonds are red cards
There are 4 honors of each suit
These are aces ,king ,queen and jack
These are called face cards
3.Sample space :When we perform an experiment ,then the set S of all possible outcome is called the sample space
eg of sample space:
(i)in tossing a coin ,s={h,t}
(ii)if two coin are tossed ,then s="{hh,tt,ht,th}.
(iii)in rolling a die we have,s={1,2,3,4,5,6}.
4.event:Any subset of a sample space.
5.Probability of occurrence of an event.
let S be the sample space and E be the event .
then,EÍS.
P(E)=n(E)/n(S).
6.Results on Probability:
(i)P(S) = 1 (ii)0<P(E)<1 (iii)P(f)=0
(iv)For any event a and b, we have:
P(aÈb)=P(a)+P(b)P(aÈb)
(v)If A denotes (nota),then P(A)=1P(A).
Category: Quantitative Aptitude
Tags: Aptitude, RS Aggarwal, probability
By: S@$h!k!r@n  July 22, 2016
Ex. 1. Evaluate: 30!/28!
Sol. We have, 30!/28! = 30x29x(28!)/28! = (30x29) = 870.
Ex. 2. Find the value of (i) ^{60}p_{3} (ii) ^{4}p_{4}
Sol. (i) ^{60}p_{3} = 60!/(603)! = 60!/57! = 60x59x58x(57!)/57! = (60x59x58) = 205320.
(ii) ^{4}p_{4} = 4! = (4x3x2x1) = 24.
Ex. 3. Find the vale of (i) ^{10}c_{3 }(ii) ^{100}c_{98} (iii) ^{50}c_{50}
Sol. (i) ^{10}c_{3 }= 10x9x8/3! = 120.
(ii) ^{100}c_{98} = ^{100}c_{(10098)} = 100x99/2! = 4950.
(iii) ^{50}c_{50} = 1. [^{n}c_{n} = 1]
Ex. 4. How many words can be formed by using all letters of the word “BIHAR”
Sol. The word BIHAR contains 5 different letters.
Required number of words = ^{5}p_{5} = 5! = (5x4x3x2x1) = 120.
Ex. 5. How many words can be formed by using all letters of the word ‘DAUGHTER’ so that the vow...
By: S@$h!k!r@n  July 22, 2016
Factorial Notation: Let n be a positive integer. Then, factorial n, denoted by n! is defined as:
n! = n(n1)(n2)........3.2.1.
Examples: (i) 5! = (5x 4 x 3 x 2 x 1) = 120; (ii) 4! = (4x3x2x1) = 24 etc.
We define, 0! = 1.
Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Ex. 1.All permutations (or arrangements) made with the letters a, b, c by taking two at a time are: (ab, ba, ac, bc, cb).
Ex. 2.All permutations made with the letters a,b,c, taking all at a time are:
(abc, acb, bca, cab, cba).
Number of Permutations: Number of all permutations of n things, taken r at a time, given by:
^{n}P_{r }= n(n1)(n2)......
By: S@$h!k!r@n  July 22, 2016
SOLVED EXAMPLES
Ex. 1. Find the cost of:
(i) Rs. 7200, 8% stock at 90;
(ii) Rs. 4500, 8.5% stock at 4 premium;
(iii) Rs. 6400, 10% stock at 15 discount.
Sol. (i) Cost of Rs. 100 stock = Rs. 90
Cost of Rs. 7200 stock = Rs. (90/100 * 7200 ) = Rs. 6480.
(ii) Cost of Rs. 100 stock = Rs. (100+4)
Cost of Rs. 4500 stock = Rs. (104/100 * 4500 ) = Rs. 4680
(iii) Cost of Rs. 100 stock = Rs. (10015)
Cost of Rs. 6400 stock = Rs. (85/100 * 6400 ) = Rs. 5440.
Ex. 2. Find the cash required to purchase Rs. 3200, 7(1/2) % stock at 107 (brokerage (1/2) %)
Sol. Cash required to purchase Rs. 100 stock = Rs (107+(1/2)) = Rs. (215/2).
Cash required to purchase Rs. 100 stock = Rs [(215/2)*(1/100)*3200] =...
By: S@$h!k!r@n  July 22, 2016
To start a big business or an industry, a large amount of money is needed. It is beyond the capacity of one or two persons to arrange such a huge amount. However, some persons associate together to form a company. They, then, draft a proposal, issue a prospectus(in the name of company), explaining the plan of the project and invite the public to invest money in this project. They, thus, pool up the funds from the public, by assigning them shares of the company.
IMPORTANT FACTS AND FORMULAE
1. Stockcapital: The total amount needed to run the company is called the stockcapital
2. Shares or stock: The whole capital is divided into small units, called shares or stock.
For each investment, the company issues a sharecertificate,...
Category: Quantitative Aptitude
Tags: Aptitude, RS Aggarwal, stocks, shares, stocks and shares
By: S@$h!k!r@n  July 22, 2016
The Face or dial of a watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.
A clock has two hands, the smaller one is called the hour hand or short hand while the larger one is called the minute hand or long hand..
i) In 60 minutes, the minute hand gains 55 minutes on the hour hand.
ii) In every hour, both the hands coincide once.
iii) The hands are in the same straight line when they are coincident or opposite to each other.
iv) When the two hands are at right angles, they are 15 minute spaces apart.
v)When the hand's are in opposite directions, they are 30 minute spaces apart.
vi)Angle traced by hour hand in 12 hrs = 360°.
vii)Angle traced by minute hand in 60 min. = 360°.
To...
Category: Quantitative Aptitude
By: S@$h!k!r@n  July 22, 2016
Ex: 1.Wbat was the day of the week on, 16th July, 1776?
Sol: 16th July, 1776 = (1775 years + Period from 1st Jan., 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day. 100 years have 5 odd days.
75 years = (18 leap years + 57 ordinary years)
= [(18 x 2) + (57 x 1)] odd days = 93 odd days
= (13 weeks + 2 days) = 2 odd days.
.. 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.
Jan. Feb. March April May June July
31 + 29 + 31 + 30 + 31 + 30 +16 = 198days
= (28 weeks + 2 days) =2days
:. . Total number of odd days = (0 + 2) = 2. Required day was 'Tuesday'.
Ex. 2. What was the day of the week on 16th August, 1947?
Sol. 15th August, 1947 = (1946 years + Period from 1s...
Category: Quantitative Aptitude
By: S@$h!k!r@n  July 22, 2016
Under this heading we mainly deal with finding the day of the week on a particular given date the process of finding it lies on obtaining the number of odd days.
Odd Days : Number of days more than the complete number of weeks in a given
Period ., is the number of odd days during that period.
LeapYear: Every year which is divisible by 4 is called a leap year.
Thus each one of the years 1992, 1996, 2004, 2008, 2012, etc. is a leap year. Every 4th century is a leap year but no other century is a leap year.thus each one of 400, 800, 1200,' 1600, 2000, etc. is a leap year.
None of 1900, 2010, 2020, 2100, etc. is a leap year.
An year which is not a leap year is called an ordinary year.
(I )An ordinary year has 365 days. (II) A leap year has 366 days.
Counting of Odd Days:
i)1 ordinary year = 365 days = (52 weeks + 1 day).
:. An ordinary year has 1 odd day.
ii)1 leap year = 366 days = (52 weeks + 2 days).
:. A leap year has 2 odd days.
_ iii)100 years = 76 ordinary years + 24 leap years
= [(76 x 52) weeks + 76 days) + [(24 x 52) weeks + 48 days]
= 5200 weeks + 124 days = (5217 weeks + 5 days).
:. 100 years contain 5 odd days.
200 years contain 10 and therefore 3 odd days.
300 years contain 15 and therefore 1 odd day.
400 years contain (20 + 1) and therefore 0 odd day.
Similarly, each one of 800, 1200, 1600, 2000, etc. contains 0 odd days.
Remark: (7n + m) odd days, where m < 7 is equivalent to m odd days.
Thus, 8 odd days ≡ 1 odd day etc.
No of odd days  0  1  2  3  4  5  6 
Day  Sun.  Mon.  Tues.  Wed.  Thur.  Fri.  Sat. 
Category: Quantitative Aptitude
Tags: Aptitude, RS Aggarwal, calendar
By: S@$h!k!r@n  July 22, 2016
SOLVED EXAMPLES :
Ex. 1. In a km race, A beats B by 28 metres or 7 seconds. Find A's time over the course.
Sol. Clearly, B covers 28 m in 7 seconds.
:. B's time over the course = (278 x 1000) sec = 250 seconds.
:. A's time over the course = (250  7) sec = 243 sec = 4 min. 3 sec.
Ex. 2. A runs 1 ¾ times as fast as B. if A gives B a start of 84 m, bow far must
winning post be so that A and B might reach it at the same time?
Sol. Ratio of the rates of A and B = 7/4 : 1 = 7 : 4.
So, in a race of 7 m, A gains 3m over B.
:. 3 m are gained by A in a race of 7 m.
:. 84 m are gained by A in a race of (7/3 x 84) m = 196 m.
:. Winning post must be 196 m away from the starting point.
Ex. 3. A can run 1 km in 3 min. 10 sec. and ...
Category: Quantitative Aptitude
By: S@$h!k!r@n  July 22, 2016
RACES AND GAMES
IMPORTANT FACTS
Races: A contest of speed in running, riding, driving, sailing or rowing is called race
Course: The ground or path on which contests are made is called a race course.
Starting Point: The point from which a race begins is known as a starting point.
Winning Point or Goal: The point set to bound a race is called a winning paint or a goal.
Winner: The person who first reaches the winning point is called a winner.
Dead Heat Race: If all the persons contesting a race reach the goal exactly at the same time, then the race is said to be a dead heat race.
Start: Suppose A and B are two contestants in a race. If before the start of the race, A is at the starting point and B is ahead of A by 12...
Category: Quantitative Aptitude
By: S@$h!k!r@n  July 21, 2016
Let length = 1, breadth = b and height = h units. Then, 1. Volume = (1 x b x h) cubic units.
2. Surface area=" 2(lb + bh + lh) sq.units.
3. Diagonal.=Öl^{2} +b^{2} +h^{2} units
Let each edge of a cube be of length a. Then,
1. Volume = a^{3} cubic units.
2. Surface area = 6a^{2} sq. units.
3. Diagonal = Ö3 a units.
Let radius of base = r and Height (or length) = h. Then,
1. Volume = (P r^{2}h) cubic units.
2. Curved surface area = (2P rh). units.
3. Total surface area =2Pr (h+r) sq. units
Let radius of base = r and Height = h. Then,
1. Slant height, l =Ö h^{2}+r^{2}
2. Volume = (1/3) Pr^{2}h cubic units.
3. Curved surface area = (Prl) sq. units.
4. Total surface area = (Prl + Pr^{2 }) ...
Category: Quantitative Aptitude
2015 © All Rights Reserved  A Website owned by S.S.Sashikiran Our website aims Bringing Engineering closer to Technology 
Our Sponsors
 We Work With
