Wednesday, July 17, 2013

Blogger Read More Option

Create 'After the jump' summaries

After the Jump is a feature which lets you create expandable post summaries in your blog posts, so longer posts appear as an intro with a link to Read More.
Creating jump breaks in your blog posts can be easily done right from the Post Editor, without the need for any HTML changes. First, decide where in the post you want to create the jump break, and place your cursor in that position:

The text where you want to jump
Once your mouse cursor is placed at the jump point, simply click the Insert Jump Breaktoolbar icon:

Jump icon in the toolbar
Clicking the icon will insert a grey bar at the cursor point, illustrating where in the post your break will appear. The bar can be dragged though, so you can always re-position it after insertion.

Gray jump line
Once you decide on the jump break's location within the post, you're ready to publish your post. After publishing, you'll notice that the Read More link is placed where you set the jump break. Clicking on the Read More link will then display the full text of the post.
If you feel like changing the Read More text to your own custom phrase, you can easily do this from the Layout tab. Click Edit on the Blog Post widget, and then change the Post page link text to whatever you'd like.
The Jump Break feature doesn't change how your post appears in your feed. You can configure post feed options by going to Settings | Basic | Other, and editing Allow Blog Feed.

HCL Time & Work Problems

Solved Time & Work Problems For HCL Aptitude Tests

Below are three problems on time and work which you can expect in HCL and other companies placement tests.
Question 1
In a factory, there are two machines A and B. A machine can produce 1000 watts power in 5 hours and by running both the machines together can produce same watts power in 3 hours. Find the time taken by B machine alone to produce same watts of power.
a) 5 hrs b) 7 hrs & 30 mins c) 6 hrs d) 6 hrs & 30 mins
Answer : b) 7 hrs & 30 mins
Solution :
A type machine can produce 1000 watts power in 5 hours.
Then A's 1 hour work = 1/5.
Both A and B can produce in 3 hours.
And 1 hour work of (A + B)'s = 1/3.
B's 1 hour work = 1/3 = 1/5 = 2/15.
Therefore, machine B can produce 1000 watts power in 15/2 hours.
i.e., 15/2 hours = 7.5 hours = 7 hours and 30 minutes.
Question 2
There is a building with 4 floors and each floor has equal area. There are three sweepers P, Q and R. P can clean 1 floor in 10 hours, Q can clean 8/5 of the floors in 40 hours and R can clean 4/3 of the floors in 13 hours. Then who will clean all the 4 floors at first?
a) R b) R & P c) Q d) Q & P
Answer : a) R
Solution :
P can sweep 1 floor in 10 hours.
Then 4 floors will be cleaned by P in = 10 x 4 = 40 hours.
Q can sweep 8/5 part in 40 hours.
Then 1 floor will be done by Q in 40/(8/5) = 25 hours.
And 4 floors will be done by Q in 25 x 4 = 100 hours.
R can clean 4/3 part of the floors in 13 hours.
Then 1 floor will be cleaned by R in 13/(4/3) = 13 x 3/4 hours.
And 4 floors will be done by R in 13 x 3/4 x 4 = 13 x 3 = 39 hours.
Thus P, Q and R takes 40, 100 and 39 hours respectively.
Hence the sweeper R will complete the work at first.
Question 3
A man and his son planned to paint their own house. The man alone can do in 20 days. He paints for 4 days and then his son completed in 8 days. Find how many days will they take working together?
a) 12 days b) 11 days c) 7 days d) 8 days
Answer : c) 7 days
Solution :
Man's 1 day work = 1/20
Then 4 days work = 4/20 = 1/5
Remaining work = 1 - 1/5 = 4/5
Now, 4/5 work is done by the son in 8 days.
Then, whole work will be done by him in (8 x 5/4) days = 10 days.
Then his 1 day work = 1/10
The man and his son's 1 day work = 1/10 + 1/10 = 3/10
They will complete the work together in 20/3 days = 6 2/3 days = 7 days(nearly).
Hence the answer is 7 days.

Infosys Sample Clock Problems

Clock Problems for infosys placement

Below are three problems based on clocks, which you can expect in infosys and other companies placement test.
Question 1
Three analog clocks were set to correct time. First one run with the exact time. Second one loss one hour per day. Third one gains one hour per day. After how many days will they show the same time?
a)180 days & 10 hours b)287 days & 12 hours c)212 days d)199 days
Answer : b) 287 days & 12 hours
Solution :
Let a day be a cycle of 24 hours.
First one completes a cycle within a day
Second one completes a cycle in 25 hours
And third one completes a cycle in 23 hours.
Now, the LCM of 23,24,25 is 23 x 24 x 25 = 13800.
Therefore, after 13,800 hours (least period) they will show the correct time together. (for 24 hours cycle)
i.e., 13,800/24 = 575
On 575th day they will show the correct time.
Since the clocks are analog (12 hours cycle), after 287 days and 12 hours (575/2 = 287.5)they will show the same time.
Question 2
Three clocks were set to true time and starts to run together, makes a beep sound at intervals of 8,10 and 12 minutes respectively. In 2 days, how many times do they will make the sound together?
a)12 b)15 c)25 d)30
Answer : c)25
The clocks make sound at intervals of 8, 10 and 12 minutes respectively.
The LCM of 8,10 and 12 is 120.
So the clocks will make sound together after every 120 minutes.
i.e., after every 2 hours.
In 2 days(48 hours), they will make the sound together (48/2) = 24 times.
At the end of 2 day,they will make one more beep together.
Therefore, totally (24+1) 25 times beep sound is made together.
Hence the answer is 25 times.
Question 3
Three analog clocks were set to the exact time at 12 am and they run together and gain 3, 6 and 9 minutes per hour respectively. At what time, they will show the same time again?
a)2.42am b)12.02am c)1.58pm d)12.02pm
Answer : a)2.42am
Solution :
Let an hour(60 minutes) be a cycle.
The clocks complete a cycle early by 3, 6 and 9 minutes respectively.
i.e., a cycle is completed in 57, 54 and 51 minutes respectively.
LCM of 57,54 and 51 is 17442 minutes.
The clocks show the same time after 17442 minutes
To convert 17442 minutes, we have to divide by 60 :
17442/60 = 17400/60 + 42/60 = 290 hours + 42/60 hours = 290 hours and 42 minutes
(42/60 hours = 42/60 x 60 minutes = 42 minutes)
After 290 hours and 42 minutes from 12 am, an analog clock shows 2.42am. (since 290 = 24 x 12 + 2).
Hence, they show the exact time again at 2.42 am .

HCL Distance Puzzles

Solved Distance Puzzles For HCL Placement Tests

Below are three problems based on distance and time calculations.
Question 1
A boy drops a bouncing toy from 1st floor to ground floor. Each time, it bounces vertically to a height of 1/2 of the preceding bounce's height. Find the distance the toy will cross if it totally takes 6 bounces and the distance between each floor is 10m.
a)39 m b)30 m c)40 m d)15 m
Answer : b)30 m
Solution :
The distance between two floors is 10m.
The boy dropped the bouncing toy from a height of 10 m.
As the toy is initially dropped, the toy falls 10 meters. Then, the toy rebounds to 10 x 1/2 = 5 meters and falls back down 5 meters. This process will repeat for 5 more times.
Distance that the toy travels down = 10 + 10x1/2 + 10x 1/2 x 1/2 + ...+ 10x(1/2)6
= 10 + 10x(1/2)1 + 10(1/2)2 + 10(1/2)3 + 10x(1/2)4 + 10(1/2)5 + 10(1/2)6 which is a g.p series with 7 terms and first term = 10 common term = 1/2.
we know that, sum of n terms in a g.p series is
sum = a(1 - rn)/1 - r when r is less than one
Here, distance = 10 x [1 - (1/2)7]/(1-1/2) = 10 x 127/128 / 1/2
= 20x127/128 = 19.84 meters.
Distance that the toy travels up = 10x(1/2)1 + 10(1/2)2 + 10(1/2)3 + 10x(1/2)4 + 10(1/2)5 + 10(1/2)6
= 19.84 - 10 = 9.84 meters.
Thus, the total distance is 19.84 + 9.84 = 29.68 meters = 30 meters (approximately)
Question 2
A spider starts jumping up from the bottom of a ladder with 32 steps. Each minute, it jumps 4 steps and slips back 2 steps. How much time would the spider take to reach the top of the ladder?
a)15 minutes b)14 minutes c)16 minutes d)18 minutes
Answer : a)15minutes.
Solution :
Think of it like a number line: In the first minute, the spider goes 4 steps up, and slips back 2 steps.
(Imagine that's going up instead of horizontal.)
The spider effectively climbs 4 steps - 2 steps = 2 steps higher after the first minute.
In the second minute, the spider goes up 4 more steps, and slips back 2 steps. So he's at 2 steps + 4 steps - 2 steps = 4 steps.
Every minute, it gains 4 steps and loses 2 steps, for a total gain of 2 steps. So after the 14th minute, it is at 28 steps height.
Now, here's the twist: in the 15th minute, it jumps up 4 steps. Since 28 steps + 4 steps = 32 steps, it is at the top of the ladder.
Hence the answer is 15 minutes.
Question 3
A man throws a ball from certain height to the ground and it covers a distance of X meters on the ground. Each time it's bounce covers 2/3 of previous distance. The ball stops bouncing after 4 bounces. Then the total distance it covers on the ground is:
a)191X/43 meters b)250X/21 meters c)172X/75 meters d)211X/81 meters
Answer : d)211X/81 meters
Solution :
This problem is slightly different from the 1st one.
Here, we have to find the distance which covered by the ball on the ground.
Given that, it covers X meters by the through.
And it takes four bounce, each bounce covers 2/3 of previous distance.
Then the distance covered in 1st bounce = (2/3)X
The distance covered in 2nd bounce = (2/3) of (2/3)X = X(2/3)2
The distance covered in 3rd bounce = (2/3) of X(2/3)2 = X(2/3)3
And the distance covered in 4th bounce = (2/3) of X(2/3)3 = X(2/3)4
Required distance = X + (2/3)X + X(2/3)2 + X(2/3)3 + X(2/3)4
The above series is a G.P series of 5 terms with a = X and r = 2/3
Then sum = a(1 - rn)/(1 - r)
= X[1 - (2/3)5]/(1 - 2/3)
= X[1 - 32/243]/(1/3)
= 3X(211/243) = 211X/81
Hence the required answer is 211X/81 meters.

TCS Measurement Problems

Solved Measurement Problems TCS Placement Tests

Below are three problems dealing with parameters of circles and squares.
Question 1
Let ABCD be a square. Alex wants to draw 5 circles of equal radius 'r' with their centres on BD such that the two extreme circles touch two sides AB, BC and AD, CD of the square respectively and each middle circle touches two circles on either side. Find the ratio of r to that of BD.
a) 2 : 11.828 b) 5 : 19.214 c) 1 : 10.828 d) 1 : 12.515
Answer : c) 1 : 10.828
Solution :
As per the statements in question, Alex would have drawn a diagram as shown below.
Let the radius of each circle be r.
Let the side of the square ABCD be a.
Then Diagonal BD = AC = a x sqrt 2 (You will get this formula by applying Pythagoras theorem)
We have to find this diagonal length as follows:
Given, he had drawn 5 circles on diagonal BD.
After sketching 5 circles according to the question, we see that some gap between corner of square and extreme circles. Now, let us calculate that space:
Draw perpendicular lines from sides AB and BC to extreme circle. Now you get a small square of side r and of diagonal r x sqrt2.
In the below diagram, Here, Y is the centre of left extreme circle, BXYZ is a small square with side r and diagonal r sqrt2.
Similar observation can be made on the other extreme circle as well.
Summing up our observations in the main diagram, we get,
Hence the diagonal of big square = r sqrt2 + r + 6r + r + r sqrt2 = 8r + 2r sqrt2
Then the required ratio = r : 8r + 2r sqrt2 = r : r(8 + 2sqrt2)
We can remove r on both sides as it is common. Therefore, we get,
= 1 : 8+2 sqrt2
= 1 : 8+2(1.414) = 1:10.828
Hence the required ratio is 1 : 10.828.
Question 2
In an exam, students were asked to find the perimeter of a square which contains three circles such that their centres on the diagonal of the square, the middle circle touches two extreme circles on either side and two extreme circles touch two sides of the square and the radius of the three are equal. And at what times the radius equals the perimeter?
a) 19.312 b) 12.312 c) 21.312 d) 17.312
Answer : a) 19.312
Let the radius of circle be r.
Let the side of the square be a.
Then Diagonal = a(sqrt2)
By applying a similar logic as that of problem 1, we will get a diagram as follows.
a(sqrt2) = diagonal = r sqrt2 + r + 2r + r + r sqrt2 = 4r + 2r sqrt2
a = (4r + 2r sqrt2) / sqrt2
a = 4r/sqrt2 + 2r sqrt2 / sqrt2
a = (2 x 2r) / sqrt2 + 2r
But we know, 2/sqrt2 = sqrt2. Therefore, above equation becomes,
a = 2r sqrt2 + 2r = 2r(sqrt2+1)
Then perimeter of the square = 4a = 4 x 2r(sqrt2+1) = 8r(sqrt2+1)
= 8r(1.414+1) = 8r(2.414) = r(19.312)
Hence 19.312 time radius equals the perimeter of the square.
Question 3
Find the length of diagonal BD of a square ABCD when 7 circles of radius 2 cm are located in ABCD such that their centres on BD, two extreme circles touch two sides of ABCD and each middle circle touches two circles on either side.
a)32.98cm b)29.65cm c)31.12cm d)28.56cm
Answer : b)29.65cm
Solution :
Given that, ABCD is a square where AB, BC, CD and AD are sides of square and BD, AC are diagonals of the square.
Applying similar logic to that of first and second questions, we will get the below diagram corresponding to this question :
The length of the diagonal BD = r sqrt2 + r + 10r + r + r sqrt2
= 2 rsqrt2 + 12r
Given that r = 2 cm
Then BD = 2(2)sqrt2 + 12(2) = 4 sqrt2 + 24 = 4(1.414) + 24 = 29.65 cm.
Hence the required answer is 29.65cm

Simplification Problems HCL Aptitude Tests

Simplification Problems To Prepare For HCL Aptitude Tests

Below are three fraction based simplification problems which you can expect in HCL and all other companies aptitude tests.
Question 1
A man spent 5/16 of his age plus two as student, 1/40 plus 1 as a husband, 1/4 as a good politician and 3/40 as a father. And the remaining 6 years as a good grand father. Then the living days of the man is:
a) 72 b) 89 c) 80 d) 69
Answer : c)80
Solution :
Lets say his age is X.
So years spent as student = 5X / 16 + 2,
as Husband = X/4 + 1,
as Politician = X/4,
and as Father = 3X/40.
Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + X/4 + 3X/40}
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40}
6 = X - [13X / 16 + 3 + 3X / 40]
6 = X - [71X / 80 + 3]
9X/80 = 9
X = 80.
Hence the age of the man is 80 years.
Question 2
In a village, every weekend, 1/4 th of men and 1/6 th of women participate in the social activity. If the total number of participants is 75, out of them 25 are men then the total number of men and women in the village is:
a) 100 b) 200 c) 300 d) 400
Answer : d)400
Solution :
Number of men who participated in social activities = 25
Therefore, 1/4 th of the men = 25
Or, total number of men in the village = 100
Number of women who participated in social activities = 75 - 25 = 50
Therefore, 1/6 th of the women = 50
Or, total number of women in the village = 300
Hence the required answer is 300 + 100 = 400.
Question 3
A father has divided his properties in such a way that one-half of total properties goes to A, two-third of the remaining shared equally to B, C and D and the rest to E, F, G and H such they equally gets Rs.30,000 by their sharing property. Then the amount will D get is:
a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000
Answer : a) Rs.80,000
Solution :
A's share = 1/2
Then remaining = 1 - 1/2 = 1/2 ...(A)
2/3 of this remaining 1/2 (as in eq A) goes to B,C and D.
Total share of B, C and D = 2/3 of 1/2 = 1/3
Therefore, now remaining property = Total share of E,F,G and H
= value of (A) - total share of B,C and D
= 1/2 - 1/3 = 1/6
Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 1/24
Given 1/24 = Rs.30,000
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000
Total share of B, C and D = 1/3 x 7,20,000
Then their individual share amount = (1/3 x 7,20,000) / 3 = Rs.80,000.
Hence D's share is Rs.80000.

Monday, July 15, 2013

Java The Complete Reference Seventh Edition

Java The Complete Reference - 7th Edition
Herbert Schildt | 2006 | 1024 Pages | ISBN: 0072263857 | PDF | 6.67 MB

Download :Java Complt Refe7thEd (6.67 MB)

Description: In this comprehensive resource, top-selling programming author Herbert Schildt shows you everything you need to develop, compile, debug, and run Java programs. This expert guide has been updated for Java Platform Standard Edition 6 (Java SE 6) and offers complete coverage of the Java language, its syntax, keywords, and fundamental programming principles. 

You'll also find information on Java's key API libraries, learn to create applets and servlets, and use JavaBeans. Herb has even included expanded coverage of Swing--the toolkit that defines the look and feel of the modern Java GUI. Essential for every Java programmer, this lasting resource features the clear, crisp, uncompromising style that has made Herb the choice of millions of programmers worldwide. 

Download Link : Java Complete Reference7 Edition(6.67 MB)

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how to change label names in blogger

Change the name of a label

Let's say you have a number of posts with the label Hawaii but you've decided that you'd rather call it Vacation instead. You can't edit the name of a label directly, but there's a simple workaround to accomplish your goal:
  1. Go to the Posts tab for your blog.
  2. Click on the "All labels" drop-down on top-right and select "Hawaii." This will filter the posts, so you'll only see those with the "Hawaii" label.
  3. Click the check box at the top of the list of posts to select all the posts.
  4. Then, click the Label icon and select "Hawaii" to remove the label "Hawaii" from all the posts that are selected.
  5. While the posts are still selected, click the Label icon again and select "Add new label." Then, type in "Vacation."
  6. You're done! Now, all the posts that used to have the label "Hawaii" have the label "Vacation" instead.

Saturday, July 13, 2013

IBPS Model Question Papers

IBPS-Dailythanthi part - 1 (Clerk Exam )

IBPS Clerk Exam Model Question Paper & Study Materials || IBPS Model Questions And Study Materials Free Download
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IBPS Clerk Exam Model Question Paper

IBPS-Dailythanthi part - 2 (Clerk Exam )

IBPS Clerk Exam Model Question Paper & Study Materials || IBPS Model Questions And Study Materials Free Download
* Answers in the bold Letters.

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Wednesday, July 10, 2013

VAO Model Question Paper 2011

Tamilnadu tnpsc study materials . This study materials based on the VAO exams. Also you can study all other group exams. It is very helpful for tnpsc exam prepare students.

 VAO Model Question Paper 2011.

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Tnpsc group-4 model question papers

Dinakaran news paper prepared TNPSC Group-IV 2013 study materials.
Download Link (6.12 MB)

This study materials are very useful to tnpsc group-IV VAO exams, Also it helps to other TNPSC group exams.
Download Link(6.12 MB)
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Tnpsc , TET Exam Tamil Study Materials free Download

TNPSC, TED exam Tamil study materials free download, this study materials are prepared from,
 6th standard to 12th standard, samacheerkalvi 
text books.

This study materials contains 264 pages.

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