Easy Vocab : The fun way to learn words! (popular and helpful app for students)

Easy Vocab is a learner’s wish coming true. Whether you’re studying for a competitive exam like SAT, GRE, ACT or LSAT or a just casual learner looking to improve on vocabulary, look no further. Seriously! It’s simple in design, packed with awesome features and has a brilliant UI that is not only attractive (the hand made icons and buttons give a pleasant and calming feel) but also crisp, concise and easily navigable.

The app consists of three main features – Learning (through the flashcards option), Testing (through the quiz option) and Analysis (through the progress option). The Flashcards have a funky layout and when touched reveal the meaning, synonyms, antonyms and the usage of the word. ‘Okay, what makes it special then?’ might be your question. Aha! That is where Easy Vocab surprises everyone. The Flashcards have a button console made up of three buttons – red, green and yellow. They are to help the user set the ‘confidence level’ for each individual word. As you revise and learn new words you can give each a confidence level, unique to your learning ability and speed:

·         Red- if you think the word is really tough and you can’t master it

·         Yellow – if you think the word is moderately tough and you might require some help to learn it.

·         Green- if you know the word and are confident about its meaning and usage.  

Once, you’ve set a confidence level for a word, Easy Vocab uses its ‘magic algorithm’ to provide you more practice for the words you are finding tough to master. In short, it is focused on ensuring you REALLY learn (which app does that, right?). Truly a feature you must see (and use) to believe!

But, wait! There’s more. You can use the confidence levels in the quiz option too. In fact the quizzes are completely customizable – You make them! You can choose the number of questions you want, the duration of the quiz and what words you wish to test yourself on based on the confidence level (pick Red, Yellow or Green or a combination of them).

Once you’ve taken a few quizzes and you want to track your progress you select the Progress tab.  Easy Vocab provides an in-depth analysis of your learning progress. A graph plots the scores of all quizzes you have taken till date against the respective quiz number. You can also view data about how many flashcards you have learned and the percentage of words marked according to each confidence level.  There is an additional section providing details about your highest, lowest and average quiz scores. For the more meticulous learner, you can get information about each individual quiz too (as a pie chart).

Apart from these stand out features there are other cool features too like:

·         The Star button – to mark words for later review

·         Word search based on confidence level and starred words

·         The awesome Video Wall – where you can view some entertaining animated YouTube videos of the words (great way to take a break and still get some work done!)

Overall, this app is extremely user centric. It makes a great effort to understand the ability of the learner and focuses on helping them learn and improve. The simplistic design and power packed features makes this a must get for any ‘linguaphile’. The fact that you get so much for absolutely FREE has helped it find its own spot on my phone.

At present, Easy Vocab is available only on Android but there are plans to release it on other platforms as well. You can download the Easy Vocab app at: https://play.google.com/store/apps/details?id=com.last_bench.easyvocab

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Something's brewing at the LB! Secret Labs!

Something's brewing at the LB! Secret Labs! Watch out for something big coming out this week. Here's a sneak peek! Any guesses? Leave us a comment!  

(www.last-bench.com)

www.last-bench.com

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Accenture Sample Problems On Speed

Below are three problems dealing with speed and time calculations.

Question 1
Two buses leaving from two stations 20 km away from each other travel with constant speed of 45km/hr towards each other.Find the time taken to cross each other if the length of each bus is 100m.
a)10sec b)4sec c)15sec d)20sec
Answer : b)4sec
Solution :
The time taken to cross each other = (a + b) / (u + v) sec.
Here, a = b = length of the two buses = 100m = 1/10 km.
and u = v = speed of the two buses = 45km/hr.
Time taken to cross each other = (a + b) / (u + v) sec = (1/10 + 1/10) / (45 + 45) hours
1 / (10 x 90) = 1/900 hours.
since 1 hour = 3600 sec, 1/900 hour = 3600/900 = 4 sec
Hence the answer is 4 sec.

Question 2
A bus starts from A at 7 a.m and reaches the destination B at 7.30 a.m while a cyclist starts from B at 7 a.m and reaches A at 8.30 a.m. At what time the bus and the cyclist will cross each other?
a)7.34 a.m b)7.49 a.m c)7.23 a.m d)8.01 a.m
Answer : c)7.23a.m
Solution :
Let the distance between A and B is X km
Time taken by the bus to cover X km = 1/2 hour .(i.e., 7 a.m to 7.30 a.m)
Time taken by the cyclist to cover X km = 3/2 hour (i.e., 7 a.m to 8.30 a.m)
Speed = distance / time
Speed of the bus = X / 1/2 = 2X km/hr
And the speed of the cyclist = X / 3/2 = 2X / 3 km/hr.
Let they meet at point C at Y hours after 7 a.m.
Distance covered by the bus in Y hours = Speed X Time = 2XY km = AC
And the distance covered by the cyclist in Y hours = Speed x Time = 2XY/3 km = BC
Since C is the point where Bus and Cyclist cross each other, AC + BC = AB
In other words, 2XY + (2/3)XY = X
Y(2 + 2/3) = 1
Y = 3/8 hours.
Expressing in minutes, Y = 3/8 x 60 minutes = 22.5 minutes
Y = 23 minutes (approximately)
Therefore they meet at 7 a.m + 23 minutes = 7.23 a.m
Hence the answer is 7.23 a.m.

Question 3
A man rides a bike with a constant speed of 20 km/hr from A at 4 p.m and travels towards a destination B. Another rider with a speed of 25 km/hr starts from B at 5 p.m and travels towards A. If the straight distance between A and B is 110km then at what time they will meet?
a)7 p.m b)8 p.m c)7.30 p.m d)6 p.m
Answer : a)7 p.m
Solution :
Let the two riders meet at X hours after 4 p.m.
Speed of the first rider is 20km/hr and the distance covered by him in X hours = 20X km
Speed of the 2nd rider is 25 Km/hr and he starts at 5 p.m.
Distance covered by him in (X - 1) hours = 25(X - 1) km
Distance between A and B is 110km.
Therefore 20X + 25(X - 1) = 110
45X = 135
X = 135/45 = 3
So they meet at 3 hours after 4 p.m.
Hence the answer is 7 p.m.

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Wipro Sample Algebraic Problems

Below are three model problems dealing with the formulae (x+y)² = x² + y² + 2xy and (x-y)² = x² + y² - 2xy
Question 1
Find X when X - Y = 3 and X² + Y² = 89 where X and Y are integers.
a)10 b)-5 c)-10 d)-3
Answer : b)-5.
Solution :
We know that (x - y)² = x² + y² - 2xy
Sub. the given values,
3² = 89 - 2XY
9 - 89 = -2XY
80 = 2XY
XY = 40
Since X and Y are integers and XY = 40,the possibilities of X and Y are as follows:
(1,40), (2,20), (4,10), (5,8), (-1,-40), (-2,-20), (-4,-10) and (-5,-8)
X - Y is a positive integer(3), so X will be greater than Y.
By checking the given condition X - Y = 3, we have X = 8, Y = 5 or X = -5, Y = -8.
i.e., X is either 8 or -5.
From the given options we can conclude that X = -5.
Question 2
If the summation and multiplication of two integer is 24 and 143 respectively then the difference of them is:
a)2 b)1 c)12 d)4
Answer : a)2
Solution :
Let A and B be two integers.
Then A + B = 24 and AB = 143.
We know that (x+y)² = x² + y² + 2xy
Here, 24² = A² + B² + 2(143)
576 - 286 = A² + B²
A² + B² = 290.
i.e., Summation of the square of two integers is 290.
i.e., A² or B² is < or = 290.
Since A and B are integers, the possibilities are {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289}.
From the above set of square numbers, the values may be
A² + B² = 1 + 289 or 121 + 169 = 290
A² + B² = 1² + 17² or 11² + 13² = 290
The condition A + B = 24 and AB = 143, is satisfied by A = 11, B = 13 or A = 13, B = 11
We have to find the difference of A and B.
Hence the difference is 2.
Question 3
If X - Y = 9, X² + Y² = 257 and X,Y are integers then what will be the value of X and Y ?
a)21,12 b)15,6 c)none of these d)cannot be determined
Answer : d)cannot be determined
Solution :
Here we use the formula, (x-y)² = x² + y² - 2xy
9² = 257 - 2XY
XY = (257-81)/2 = 176/2 = 88
XY = 88
Since X and Y are integers and XY = 88 then the possibilities of X and Y are (1,88), (2,44), (4,22), (8,11), (-1,-88), (-2,-44), (-4,-22) and (-8,-11).
By observing, none of the above possibility of X and Y satisfies X - Y = 9.
Hence, the values of X and Y cannot be determined by given conditions.

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Moderate Problems on Mixture

Below are three problems based on the concept of mixtures in some ratio.
Question 1
A certain box of health drink powder is obtained by mixing the makers as 3 parts of milk powder with 5 parts of energy powder. If 2 kg of the mixture is needed and the individual makers can be purchased only in one kg or half kg box. What is the least amount of makers, in kg's, that must be purchased in order to measure out the portions needed for the mixture?
a)2kg b)3kg c)4.5kg d)2.5kg
Answer : d)2.5kg
Solution :
Step 1) "A certain box of health drink powder is obtained by mixing the makers as 3 parts of milk powder with 5 parts of energy powder. "
You should think: 3 parts of milk powder, 5 parts of energy powder sums to a total of 8 parts.
More specifically, 3/8 of the mixture is milk powder and 5/8 of the mixture is energy powder.
Step 2) "If 2 kgs of the mixture is needed and the individual makers can be purchased only in one kg or half kg cans"
So we know the mixture is 2 kg. How much of these 2 kg is milk powder and how much of these 2 kg is energy powder?
We know 3/8 of the mixture is milk powder. And the entire mixture is 2 kg.
So 3/8 of the 2 kg = 3/8 x 2 = 6/8 = 3/4 = .75 of a kg is milk powder
Likewise 5/8 of the 2 kg or 5/8 x 2 = 10/8 = 5/4 = 1.25 of the kg is energy powder
Step 3) " individual makers can purchase only in one-kg or half- kg cans"
That means the .75 needs to be rounded up to 1 kg
And it means that the 1.25 kg needs to be rounded up to 1.5 kgs (a 1 kg box and a half kg box)
Combine the 1 kg of milk powder + 1.5 kg of energy powder = 2.5 total kgs.
Hence the answer is option d.
Question 2
A certain shade of orange color is obtained by mixing 1 part of white color with 2 parts of red color. If 3.6 kgs of the mixture is needed and the white and red colors can be purchased only in 1 kg, what is the least amount of color, in kg's, that must be purchased for the mixture?
a)5kg b)4kg c)6kg d)4.5kg
Answer : a)5kg
Solution :
Since, the orange color is obtained by mixing 1 part of white color with 2 parts of red color.
i.e., 1 part white color, 2 part red color, that means totally there are 3 parts.
Then, 1/3 of the mixture is white color and 2/3 is red color.
Given that, 3.6 kg of the mixture is needed.
We know 1/3 of the mixture is white color and the entire mixture is 3.6 kg
So 1/3 of 3.6 kg = 3.6/3 = 1.2 kg is white color.
Likewise 2/3 of the 3.6 kg = 2 x 3.6/3 = 2 x 1.2 = 2.4 kg is red color.
Since the individual colors (white and red) can be purchased only in one-kg
i.e., 1.2 kg needs to be rounded up to 2 kg
And 2.4 kg needs to be rounded up to 3 kg.
Combine the white and red color = 2kg + 3kg = 5 kg.
Hence the least amount of colors to be purchased is 5 kg.
Question 3
A certain shade of violet color is obtained by mixing 2 parts of white color with 3 parts of blue color. If 8 kg of the mixture(violet color) is needed and the individual colors can be purchased only in 2 kg cans worth Rs.80, then what is the least amount of color in kg's that must be purchased for the mixture? And how much rupees spend to purchase the blue color cans?
a)6kg & Rs.480 b)10kg & Rs.400 c)10kg & Rs.800 d)6kg & Rs.320
Answer : b)10kg & Rs.400
Solution :
Since the violet color is obtained by mixing 2 parts of white color with 3 parts of blue color.
i.e., 2 parts of white color and 3 parts of blue color, that means totally there are 5 parts.
Then, 2/5 of the violet color is white and 3/5 is blue color.
Given that, 8kg of the mixture is needed.
So 2/5 of 8kg = 2 x 8/5 = 3.2 kg is white color
Likewise 3/5 of 8kg = 3 x 8/5 = 4.8 kg is blue color.
Since the individual colors (white and red) can be purchased only in 2-kg cans.
i.e., 3.2 kg needs to be rounded up to 4 kg.
And 4.8 kg needs to be rounded up to 6 kg.
Then the required least amount of colors = 4kg + 6kg = 10kg or 5 cans (2 cans of white + 3 cans of blue)
Since each can of 2 kg color worth Rs.80. Total cost for 5 cans = 5 x 80 = Rs. 400
Hence the answer is 10kg & Rs.400.

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Mobile Screen and their Resolution for android developers

Compatible devices MUST support dynamic orientation by applications to either portrait or landscape screen orientation. That is, the device must respect the application's request for a specific screen orientation. Device implementations MAY select either portrait or landscape orientation as the default.

There are some screen resolutions already defined in Android. They are: QVGA (240×320, low density, small screen) WQVGA (240×400, low density, normal screen) FWQVGA (240×432, low density, normal screen) HVGA (320×480, medium density, normal screen) WVGA800 (480×800, high density, normal screen) WVGA854 (480×854 high density, normal screen)

QVGA (Quarter Video Graphics Array)

QVGA (Quarter Video Graphics Array or Quarter VGA) is a type of resolution whose dimensions are 320×240 pixels. The QVGA display is normally found in devices such as mobile phones and PDAs. The display is also common on other hand-held devices such as game consoles.

Resolution

Most of the time, the display is in the portrait layout as opposed to the landscape layout. They are also known as 240×320 displays because they are taller than they are wide. So why is it known as Quarter Video Graphics Array? It has this name because the display is 1/4th of the 640×480 resolution of the original IBM machine that had the VGA display. The original display was almost a de facto standard in the 1980s.

WQVGA (400×240)

Wide QVGA or WQVGA is any display resolution having the same height in pixels as QVGA, but wider. This definition is consistent with other 'wide' versions of computer displays.

Since QVGA is 320 pixels wide and 240 pixels high (aspect ratio of 4:3), the resolution of a WQVGA screen might be 384×240 (16:10 aspect ratio), 400×240 (5:3—such as the Nintendo 3DS screen or the maximum resolution in YouTube at 240p), 428×240 or 432×240 (~16:9 ratio).

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