Tuesday, July 5, 2011

Application Form for Karnataka Post Graduate Common Entrance Test 2011

1. The candidates have to procure the PGCET 2011 MCA / MBA brochure cum application form by Speed Post only by sending a self addressed cloth-lined envelope of size 25 cm X 30 cm along with the prescribed fee of 600/- in the form of demand draft drawn in favour of “The Executive Director, Karnataka Examinations Authority”, payable at Bangalore. ( 300/- in case of Karnataka SC, ST and Category-I Candidates )

2. If a candidate likes to apply for more than one course ( eg. for both MBA and MCA ) then he she is required to apply separately by sending two different covers with two different requisite demand drafts each, as the case may be.

3. The contents on the self addressed envelope should be in English or Kannada language only. The cover in which the demand draft and self addressed envelope are sent should be boldly super scribed with the words “Request for PGCET 2011 Brochure for course” ( MBA / MCA / M.E. / M. Tech. / M. Arch / MBA ( Infrastructure Management etc. ) and should be sent to the following address By Speed Post only.

The Executive Director,
Karnataka Examinations Authority,
18th Cross, Sampige Road,
Bangalore – 560 012.
4. KEA will not take responsibility for the envelopes sent through courier or given in person or through any other private arrangement and also for incomplete and insufficient address on the envelope.

5. The Last Date for PGCET Entrance Exam 2011 Application submission of Demand Drafts to KEA to obtain brochure cum application form is 25-06-2011 5.00 P.M.

PGCET 2011 Karnataka Important Dates :

1) The PGCET Karnataka brochure cum application form will be sent by speed post to the home address of the candidates furnished on the self addressed cloth lined cover on or after 02-07-2011.

2) Last date for submission of filled PGCET Application form by Speed Post to KEA is 15-07-2011 5.00 P.M.

3) Karnataka Examination will be conducted only in the state of Karnataka at various places as specified in the PGCET 2011 Brochure.

Sunday, May 1, 2011


What are prepositions?

Prepositions are words that specify place, direction, and time. There are three types of prepositions.

Time Prepositions - Time prepositions define time. Time prepositions - In - At - On For - During - While

Place Propositions - Place prepositions clarify the place someone or somebody. Place Prepositions - In - On -At

Direction Prepositions - Direction prepositions are used to clarify the direction of someone or something. Examples: under, over, right, left etc.

Sunday, April 17, 2011

Problems on Clocks

Problems on Clocks

General Concepts
The face or dail 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 hand is called the minute hand or long hand.

In 60 minutes , the minute hand gains 55 minutes on the hour hand.
In every hour , both the hands coincide once.
The hands are in the same straight line when they are coincident or opposite to each other
When the two hands are at right angles, they are 15 minute spaces apart.
When the hands are in opposite directions, they are 30 minute spaces apart.
Too Fast and Too Slow :If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15 minutes too fast

On the other hand , if it indicates 7.45, when the correct time is 8, it is said to be too slow.

Solved Problems

1. Find the angle between the minute hand and hour hand of a clock when the time is 7.20

Sol : Angle traced by hour hand in 12 hours = 360 Degrees

Angle traced by it in 7 hrs 20 min i.e 22/3 hrs = (360/12) x (22/3)= 2200

Angle traced by minute hand in 60 min = 3600

Angle traced by it in 20 minutes = (360/60) x 20 = 1200

Required angle = (2200 - 1200) = 1000

2. At what time between 2 and 3 o' clock will the hands of a clock together ?

Sol : At o' clock, the hour hand is at 2 and minute hand at 12, i.e they are 10 minute spaces apart.

To be together , the minute hand must gain 10 minutes over the hour hand.

Now, 55 minutes are gaines by it in 60 minutes

10 minutes will be gained in {(60/55) x 10} min =10 10/11 min

The hands will coincide at 10 10/11 min past 2

3. At what time between 4 and 5 o' clock will the hands of a clock be at right angle ?

Sol : At 4 o' clock, the minute hand will be 20 min spaces behind the hour hand.

Now , when the two hands are at right angles , they are 15 min. spaces apart.

So, they are at right angles in following two cases

Case I : When minute hand is 15 minute spaces behind the hour hand

In this case min hand will have to gain (20 -15) = 5 minute spaces

55 min. spaces will be gained by it in (60 x 5)/55 min = 5 5/11 min

They are at right angles at 5 5/1 min past 4

Case II : When the minute hand is 15 minute spaces ahead of the hour hand

They are at right angles at 38 2/11 min past 4

Pipes and Cisterns

Important Facts:

1.INLET:A pipe connected with a tank or cistern or a reservoir,
that fills it, it is known as Inlet.

OUTLET:A pipe connected with a tank or a cistern or a reservoir,
emptying it, is known as Outlet.

2. i) If a pipe can fill a tank in x hours, then :
part filled in 1 hour=1/x.

ii)If a pipe can empty a tank in y hours, then :
part emptied in 1 hour=1/y.

iii)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where y>x), then on
opening both the pipes, the net part filled in
1 hour=(1/x -1/y).

iv)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where x>y), then on opening
both the pipes, the net part filled in 1 hour=(1/y -1/x).

v) If two pipes A and B can fill a tank in x hours and y hours
respectively. If both the pipes are opened simultaneously, part
filled by A+B in 1 hour= 1/x +1/y.

Simple Problems

1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.
If both the pipes are opened simultaneously, how much time will be
taken to fill the tank?

Sol: Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45;
Part filled by (A+B)’s in 1 hour=1/36 +1/45= 9/180 =1/20
Hence, both the pipes together will fill the tank in 20 hours.

2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While
3rd pipe empties the full tank n 20 hours. If all the three pipes
operate simultaneously,in how much time will the tank be filled?

Sol: Net part filled in 1 hour=1/10 +1/12 -1/20
The tank be filled in 15/2hours= 7 hrs 30 min

3)A cistern can be filled by a tap in 4 hours while it can be emptied
by another tap in 9 hours. If both the taps are opened simultaneously,
then after how much time will the cistern get filled?

Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours.

4)If two pipes function simultaneously, the reservoir will be filled in
12 days.One pipe fills the reservoir 10 hours faster than the other.
How many hours does it take the second pipe to fill the reservoir.

Sol: Let the reservoir be filled by the 1st pipe in x hours.
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x=20
So, the second pipe will take 30 hours to fill the reservoir.

5)A cistern has two taps which fill it in 12 min and 15 min respectively.
There is also a waste pipe in the cistern. When all the three are opened,
the empty cistern is full in 20 min. How long will the waste pipe take to
empty the full cistern?

Sol: Work done by a waste pipe in 1 min
=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)

6)A tap can fill a tank in 6 hours. After half the tank is filled, three
more similar taps are opened. What is the total time taken to fill the
tank completely?

Sol: Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2
Therefore, 2/3:1/2::1:x
or x=(1/2)*1*(3/2)=3/4 hours.
i.e 45 min
So, total time taken= 3hrs 45min.

7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B
can empty it in 6 min. If both pipes are open, how long will it take to
empty or fill the tank completely ?

Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.
Part to be emptied=2/5.
Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15
Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.
So, the tank be emptied in 6 min.

8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for
Bucket P to fill the empty drum. How many turns it will take for both the
buckets P&Q, having each turn together to fill the empty drum?

Sol: Let the capacity of P be x lit.
Then capacity of Q=x/3 lit
Capacity of the drum=60x lit
Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45

Complex Problems

1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The
pipes are opened simultaneously and it is found that due to leakage in the
bottom it took 32min more to fill the cistern. When the cistern is full,
in what time will the leak empty it?

Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours.

2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After
7 min, C also opened. In how much time, the tank s full.

Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20
Remaining part= 1-7/20=13/20
Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60
Now, 1/60 part is filled in 1 min.
13/20 part is filled n (60*13/20)=39 min
Total time taken to fill the tank=39+7=46 min

3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If
both the pipes are opened simultaneously, after how much time B should
be closed so that the tank is full in 18 min.

Sol: Let B be closed after x min, then part filled by (A+B) in x min+
part filled by A in (18-x) min=1
x(1/24+1/32) +(18-x)1/24 =1
=> x=8
Hence B must be closed after 8 min.

4)Two pipes A& B together can fill a cistern in 4 hours. Had they been
opened separately, then B would have taken 6 hours more than A to fill
the cistern. How much time will be taken by A to fill the cistern

Sol: Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x+6 hours
1/x + 1/x+6=1/4
Solving this we get x=6.
Hence, A takes 6 hours to fill the cistern separately.

5)A tank is filled by 3 pipes with uniform flow. The first two pipes
operating simultaneously fill the tan in the same time during which
the tank is filled by the third pipe alone. The 2nd pipe fills the tank
5 hours faster than first pipe and 4 hours slower than third pipe. The
time required by first pipe is :

Sol: Suppose, first pipe take x hours to fill the tank then
B & C will take (x-5) and (x-9) hours respectively.
Therefore, 1/x +1/(x-5) =1/(x-9)
On solving, x=15
Hence, time required by first pipe is 15 hours.

6)A large tanker can be filled by two pipes A& B in 60min and 40 min
respectively. How many minutes will it take to fill the tanker from
empty state if B is used for half the time & A and B fill it together for
the other half?

Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24+1/40)=1
=> (x/2)*(1/15)=1
=> x=30 min.

7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.
If they are opened on alternate hours and if pipe A s opened first, in
how many hours, the tank shall be full.

Sol: (A+B)’s 2 hours work when opened alternatively =1/6+1/4 =5/12
(A+B)’s 4 hours work when opened alternatively=10/12=5/6
Remaining part=1 -5/6=1/6.
Now, it is A’s turn and 1/6 part is filled by A in 1 hour.
So, total time taken to fill the tank=(4+1)= 5 hours.

8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.
If A is open all the time and B and C are open for one hour each
alternatively, the tank will be full in.

Sol: (A+B)’s 1 hour’s work=1/12+1/15=9/60=3/20
(A+C)’s 1 hour’s work=1/20+1/12=8/60=2/15
Part filled in 2 hours=3/20+2/15=17/60
Part filled in 2 hours=3/20+2/15= 17/60
Part filled in 6 hours=3*17/60 =17/20
Remaining part=1 -17/20=3/20
Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.
Therefore, total time taken to fill the tank=6+1=7 hours

9)A Booster pump can be used for filling as well as for emptying a tank.
The capacity of the tank is 2400 m3. The emptying capacity of the tank is
10 m3 per minute higher than its filling capacity and the pump needs 8
minutes lesser to empty the tank than it needs to fill it. What is the
filling capacity of the pump?

Sol: Let, the filling capacity of the pump be x m3/min
Then, emptying capacity of the pump=(x+10) m3/min.
So,2400/x – 2400/(x+10) = 8
on solving x=50.


A Bird In The Hand Is Worth Two In The Bush:
Having something that is certain is much better than taking a risk for more, because chances are you might lose everything.
A Blessing In Disguise: Something good that isn't recognized at first.
A Chip On Your Shoulder: Being upset for something that happened in the past.
A Dime A Dozen: Anything that is common and easy to get.

For More Idioms Visit http://www.idiomsite.com/

Monday, April 4, 2011

Compound Interest

Compound Interest

Compound Interest: sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half-yearly or quarterly to settle the previous amount.
            In such case, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on.
            After a specified period, the difference between the amount and the money borrowed is called the compound Interest (abbreviated as C.I.) for that period.
 Let Principal =p, Rate = R% per annum, Time=n years
I.                   When interest is compounded annually:
Amount = P[ 1 + (R/100)]n 
II.                When the interest is compounded half-yearly :
                             Amount = p[ 1 + (R/2)/100]2n
III.             When the interest is compounded quarterly :
                            Amount = P[ 1 + (R/4)/100 ]4n 
IV.              When the interest is compounded annually but time is in                    fraction, say3 2/5 Years.   
                 Amount = P{ [1 + (R/100)]3  x [ 1 + (2R/5)/100] }
V.                 When rates are different for different years, say R1%, R2%,               R3% for 1st,   2nd  and 3rd year respectively.
             Then, Amount= P [ 1 + R1/100][ 1 + R2/100][ 1+ R3/100]
VI.              Present worth of Rs. X Due n years hence is given by :
                        Present worth = X/ (1 + R/100)n
                                                            Solved Problem
Ex.1. Find compound interest on Rs.6250 at 16% per annum for 2 years, compounded annually.
 Sol.  Amount = Rs. [ 6250 x ( 1 + 16/100)2 ]
                      =  Rs.[ 6250 x 29/25 x 29/25 ] = Rs.8410.
          .. C.I = Rs.(8410 – 6250) = Rs. 2160.
 Ex.2. Find compound interest on Rs. 5000 at 12% per annum for 1 year,  compounded  half-yearly.
Sol.   Principal = Rs.5000, Rate=6% per half year
          Time = 1 year   = 2 half-years
    Amount = Rs. {5000 x ( 1 + 6/100)2] = Rs. [ 5000 x 53/50 x 53/50 ]
                           = Rs. 5618
          .. C.I. = Rs. (5618 – 5000) = Rs. 618
 Ex.3. Find compound interest on Rs. 16000 at 20% per annum for 9 months,compounded quarterly. 
Sol.  Principal = Rs. 16000,  Time = 9 months = 3 quarters
         Rate = 20% per annum = 5% per quarter.
         .. Amount = Rs. [ 16000 x { 1 + 5/100 )3 ]
                          = Rs. [ 16000 x 21/20 x 21/20 x 21/20 ] = Rs. 18522.
            .. C.I. = Rs. (18522 – 16000) = Rs. 2522.
 Ex.4.  The difference between the compounded interest and simple interest on a certain sum at 10% per annum for 2 years is Rs.631. Find the sum.
Sol. Let the sum be Rs. X .Then,
        C.I.. = x[ 1 + 10/100 ]2  - x ] = 21x/100
         S.I.  = [ (Xx10x2)/100] = X/5
         .. (C.I.) – (S.I.) = [ (21x)/100 – (x/5) ] = x/100
         .. x/100 = 631     x= 63100
          Hence, the sum is Rs. 63100
 Ex.5. If the compound interest on a certain sum for 2 years at 12% per annum is  Rs. 1590, what would be the simple interest?
Sol. Let the sum be Rs. X. Then
       X{ 1 + 12/100 }2 – X = 1590  or  (784X/625) – X = 1590
       Or 159X/625 = 1590 or x = {1590 x 625}/159 =6250
       .. Sum = Rs. 6250
       .. S.I. = Rs. [ (6250 x 12 x 2)/100] = Rs. 1500
 Ex.6. A sun of money amounts to Rs.6690 after 3 years and to Rs 10035 after 6  years on compound interest. Find the sum.
   P{ 1 + R/100 }3 = 6690 ….(i)  and  P{ 1 + R/100 }6  = 10035  … (ii)
          On Dividing, we get [ 1 + R/100 ]3 = 10035/6690 = 3/2
          Substituting this value in (i), we get;
          P x 3/2 =6690  or  P=[6690 x 2/3] = 4460
          Hence, the sum is Rs. 4460
Ex.7. A sum of money doubles itself at compound interest in 15 years. In how many  years will it become eight times?
Sol.       P { 1 + R/100 }15 = 2P   ie 2P= [ 1 + R/100]15 = 2    … (i
 Let  P[ 1 + R/100]n = 8P      [ 1 + R/100]n =8 = 23 = { [1 + R/100]15 }3
                   [ 1 + R/100]n = [ 1 + R/100]45
                   i.e      n=45
              Thus required time = 45 years
Ex.8. A certain sum amounts to Rs.7350 in 2 years and to Rs.8575 in 3 years . Find  the sum and rate per cent.
Sol.  S.I. on Rs.7350 for 1 year = Rs.(8575 – 7350) = Rs.1225.
         .. Rate = [(100 x 1225)/(7350 x 1)]% = 16 2/3%
          Let the sum be Rs. X .Then
          X[1 + 50/(3 x100)]2 = 7350   
          i.e X x 7/6 x 7/6 =7350
              X= {7350 x 36/49]
           .. Sum = Rs. 5400

Wednesday, March 30, 2011

Number Word Problems

Number Word Problems
Number" word problems are fairly contrived, but they're also fairly standard, so you should learn how to handle them. After all, the point of these problems isn't their relation to "real life", but your ability to extract the math from the English.
  • The sum of two consecutive integers is 15. Find the numbers.
    They've given me two pieces of information here. First, I know that I am adding two numbers, and their sum is fifteen. Second, I know that the numbers are nice neat round numbers (like –3 or 6), not messy ones (like –4.628 or 17/32), and that the second number is one more than the first. This last piece of information comes from the fact that "consecutive integers" (or "consecutive whole numbers", if they're restricting the possibilities to only positive numbers) are one unit apart. Examples of "consecutive integers" would be –12 and –11, 1 and 2, and 99 and 100. Using these facts, I can set up the translation. I will represent the first number by "n". Then the second number has to be "n + 1". Their sum is then:
      n + (n + 1) = 15 2n + 1 = 15 2n = 14 n = 7   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
    The exercise did not ask me for the value of the variable n; it asked for the identity of two numbers. So my answer is not "n = 7"; the actual answer is:
      "The numbers are 7 and 8."
  • The product of two consecutive negative even integers is 24. Find the numbers.
    They have told me quite a bit about these two numbers: the numbers are even and they are negative. (The fact that they are negative may help if I come up with two solutions — a positive and a negative — so I'll know which one to pick.) Since even numbers are two apart (for example, –4 and –2 or 10 and 12), then I also know that the second number is two greater than the first. I also know that, when I multiply the two numbers, I will get 24. In other words, letting the first number be "n" and the second number be "n + 2", I have:
      (n)(n + 2) = 24 n2 + 2n = 24 n2 + 2n – 24 = 0 (n + 6)(n – 4) = 0
    Then the solutions are n = –6 and n = 4. Since the numbers I am looking for are negative, I can ignore the "4" and take n = –6. Then the next number is n + 2 = –4, and the answer is
      The numbers are –6 and –4.
In the exercise above, one of the answers was one of the solutions to the equation; the other answer was the negative of the other solution to the equation. Warning: Do not assume that you can use both solutions if you just change the signs to be whatever you feel like. While this often "works", it does not always work, and it's sure to annoy your teacher. Throw out invalid results, and solve properly for valid ones.
  • Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71. What are the numbers?
The point of exercises like this is to give you practice in unwrapping and unwinding these words, and turning the words into algebraic equations. The point is in the solving, not in the relative "reality" of the problem. That said, how do you solve this? The best first step is to start labelling:
      the larger number:  x the smaller number:  y twice the larger:  2x three more than five times the smaller:  5y + 3 relationship between ("is"):  2x = 5y + 3 four times the larger:  4x three times the smaller:  3y relationship between ("sum of"):  4x + 3y = 71
    Now I have two equations in two variables:
      2x = 5y + 3 4x + 3y = 71
    I will solve, say, the first equation for x:
      x = (5/2)y + (3/2)
    Then I'll plug the right-hand side of this into the second equation in place of the "x":
      4[ (5/2)y + (3/2) ] + 3y = 71 10y + 6 + 3y = 71 13y + 6 = 71 13y = 65 y = 65/13 = 5
    Now that I have the value for y, I can solve for x:
      x = (5/2)y + (3/2) x = (5/2)(5) + (3/2) x = (25/2) + (3/2) x = 28/2 = 14
    As always, I need to remember to answer the question that was actually asked. The solution here is not "x = 14", but is the following sentence:
      The larger number is 14, and the smaller number is 5.
The trick to doing this type of problem is to label everything very explicitly. Until you become used to doing these, do not attempt to keep track of things in your head. Do as I did in this last example: clearly label every single step. When you do this, these problems generally work out rather easily.

Saturday, March 26, 2011

Types of Prepositions with example

What are the different types of Preposition?

Preposition of place
Preposition of time
Preposition of movement

Preposition of place - at, on, in, below, beside, under, behind, between
Prepositions of place are used to describe the place or position of nouns.

1.The dog is in the garden.
2.The box is behind the door
3. He threw the ball over the roof
4. She lives near the school.
5. The house is between 3rd Street and 4th Street.
6. London is on the river Thames.

Preposition of time-at, on, in, during,
Prepositions of time- used to show when something happened.

More examples: since, for, by, from -to, from-until, during, in.
1. I haven’t seen her since yesterday.
2. I'm leaving to Chennai in a week.
4. I waited for him from 8 a.m. until 10 a.m., and later went out
5. I watch TV during the evening.

Preposition of Movement
Preposition of Movement is used to show movement.

Example: to, towards, through, across, over, into, onto
To: signifies movement toward a specific destination.
Example: He went to the park.
Towards: suggests movement in a general direction,
Example: The helicopter flew towards south.
Into: Suji fell into the well.
Onto: Clara jumped onto the bench.

Articles usage in english

A, An or The?

When do we say "the dog" and when do we say "a dog"? (On this page we talk only about singular, countable nouns.)
The and a/an are called "articles". We divide them into "definite" and "indefinite" like this:

thea, an

We use "definite" to mean sure, certain. "Definite" is particular.
We use "indefinite" to mean not sure, not certain. "Indefinite" is general.
When we are talking about one thing in particular, we use the. When we are talking about one thing in general, we use a or an.
Think of the sky at night. In the sky we see 1 moon and millions of stars. So normally we would say:
  • I saw the moon last night.
  • I saw a star last night.
Look at these examples:

thea, an
  • The capital of France is Paris.
  • I have found the book that I lost.
  • Have you cleaned the car?
  • There are six eggs in the fridge.
  • Please switch off the TV when you finish.
  • I was born in a town.
  • John had an omelette for lunch.
  • James Bond ordered a drink.
  • We want to buy an umbrella.
  • Have you got a pen?

Of course, often we can use the or a/an for the same word. It depends on the situation, not the word. Look at these examples:
  • We want to buy an umbrella. (Any umbrella, not a particular umbrella.)
  • Where is the umbrella? (We already have an umbrella. We are looking for our umbrella, a particular umbrella.)
This little story should help you understand the difference between the and a, an:

A man and a woman were walking in Oxford Street. The woman saw a dress that she liked in a shop. She asked the man if he could buy the dress for her. He said: "Do you think the shop will accept a cheque? I don't have a credit card."

Monday, March 21, 2011

Number series

  1.Prime Number Series


2.Even Number Series

3.Odd Number Series:

4.Perfect Squares:


5.Perfect Cubes:


.Multiples of Number Series: 


           3,6,9,12,15,..............are multiples of 3

7.Numbers in Arthimetic Progression(A.P):


8.Numbers in G.P: