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1. A box-and-whisker plot uses a number line to represent the data visually. I.E. It shows visually the GRE - Mean, Median, Mode, and Range

2. It displays a data set along a number line using medians. Quartiles divide the data set into four equal parts. The median (second quartile) divides the data set into two halves. The median of the lower half is the first quartile. The median of the upper half is the third quartile.

3. The box plot does not show individual data but it analyzes the spread of data. A measure of spread is a single number that describes how to spread out or clustered together the data are.

4. Another measure of spread is the interquartile range (IQR). The “Interquartile Range” is the difference between the smallest value and the largest value of the middle 50% of a set of data.

The "Interquartile Range" is from Q1 to Q3:




To find the interquartile range of a set of data:

  • First put the list of numbers in order;
  • The list must be divided into four equal parts;
  • These four equal parts (25% each, regardless of the numbers contained in each individual sector) are the quartiles;
  • The interquartile range is the distance between the two middle sets of data, or in other words: the third quartile (or chunk) less the second quartile or chunk.


How to construct a box plot?





More Examples


The first step in constructing a box-and-whisker plot is to first find the median (Q2), the lower quartile (Q1) and the upper quartile (Q3) of a given set of data.

Box plot with odd numbers in that data set

The first step in constructing a box-and-whisker plot is to first find the median (Q2), the lower quartile (Q1) and the upper quartile (Q3) of a given set of data.

Step 1: Find the median. The median is the value exactly in the middle of an ordered set of numbers.





Step 2: Consider the values to the left of the median: 18 27 34 52 54 59 61 Find the median of this set of numbers. The median is 52 and is called the lower quartile.

Step 3: Consider values to the right of the median: 78 82 85 87 91 93 100 Find the median of this set of numbers. The median 87 is therefore called the upper quartile.


Note that when there is an odd number of values, as in this example. we don't include the median in the set of numbers used to calculate the upper and lower quartiles.

We are able now to find the interquartile range (IQR). The interquartile range is the difference between the upper quartile and the lower quartile. In example 1, the IQR = Q3 — Q1 = 87 - 52 = 35.
The IQR is a very useful measurement. It is useful because it is less influenced by extreme values as it limits the range to the middle 50% of the values.


Box plot with even numbers in that data set


Step 1: Find the median. The median is the value exactly in the middle of an ordered set of numbers.




The median in this example is in-between 70 and 73, so the median is calculated by taking the mean of 70 and 73:

Median \(= \frac{70+73}{2} =71.5\)

Step 2: Consider the values to the left of the median: 42 63 64 64 70 Find the median of this set of numbers. The median is 64.

Step 3: Consider the values to the right of the median: 72 76 77 81 81


Note that when the number of values is even the median lies between the two middle values. As in this example, we include the data value just below the median in the set of numbers used to calculate the lower quartile. and the number just above the median in the set of numbers used to calculate the upper quartile.


The median is 77 and is called the upper quartile. We are able now to find the interquartile range (IQR). The interquartile range is the difference between the upper quartile and the lower quartile. The IQR = Q3 — Q1 = 77 - 64 = 13. The IQR is a very useful measurement. It is useful because it is less influenced by extreme values as it limits the range to the middle 50% of the values.


The Box Plots can be either vertical




Or horizontal




Sources used to compile the notions above:
Students Learning Centre at Flinders University
Flowingdata.com
bigideasmath.com
What do the minimum, first quartile, median, third quartile, and maximum of a box and whisker plot represent? by Kate M.
Introductory Statistics by OpenStaxCollege is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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­1) Sum of the first positive integers

\(1+2+3+4+5....................... n=\frac{n(n+1)}{2}\)


2) Sum of squares of the first positive integers

\(1^2+2^2+3^2+4^2+5^2+...........n^2=\frac{n(n+1)(2n+1)}{6}\)


3) Sum of cubes of the first positive integers

\(1^3+2^3+3^3+4^3+5^3+...........n^3=\left[ \begin{array}{cc|r} \frac{n(n+1)}{2} \end{array} \right]^2\)


4) Sum of odd integers

\(1+3+5......................(2n-1)=n^2\)

Note that the \(n^{th}\) term is \((2n-1)\)


5) Sum of even integers

\(2+4+6.......................2n=n(n+1)\)
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­Simple Interest


Principal or Sum : The money borrowed or lent out for a certain period is called the principal or the sum.

Interest : Extra money paid for using other’s moeny is called interest.

Simple Interest : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

Formulae: Let Principal = P, Rate = R% per annum and time = T years. Then, S.I. \(= \left( \begin{array}{cc} \frac{P \times R \times T}{100} \end{array} \right)\)



Compound Interest


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:\( Amount = P \left( \begin{array}{cc} 1+\frac{ R}{100} \end{array} \right)^n\)


II: When interest is compounded Half-yearly : \(Amount = P \left[ \begin{array}{cc|r}  1 + \frac{\frac{R}{2}}{100}  \end{array} \right]^{2n}\)


III: When interest is compounded Quarterly : \(Amount = P \left[ \begin{array}{cc|r}  1 + \frac{\frac{R}{4}}{100}  \end{array} \right]^{4n}\)


IV: When interest is compounded Annually but time is in fraction, say \(3 \frac{2}{5}\) years


\(Amount = P \left( \begin{array}{cc} 1+\frac{ R}{100} \end{array} \right)^3\) \( \times\) \( \left[ \begin{array}{cc|r}  1 + \frac{\frac{2}{5 }R}{100}  \end{array} \right]\)



Ex 1. A certain sum of money at C.I. amounts in 2 years to Rs 811.2 and in 3 years to Rs 843.65. Find the sum of money.
 
Sol. Since \(A = P (1 + \frac{R}{100})^n ⇒ 811.2 = P (1 + \frac{R}{100})^2\) .....(1) and \(843.65 = P (1 + \frac{R}{100})^3\) ....(2)

On dividing (2) by (1), we get : \(\frac{843.65}{811.2} = (1 + \frac{R}{100}) ⇒ 1.04 = (1 + \frac{R}{100}) ⇒ \frac{R}{100} = 0.04 ⇒ R = 4\)

Now, putting R = 4 into (1), we get \(811.2 = P (1 + \frac{4}{100})^2 ⇒ 811.2 = P (1.04)^2 ⇒ P = \frac{811.2 }{(1.04)^2} = 750 ⇒\) The sum of money is Rs 750. Ans.

Ex 2. Find the compound interest on Rs 4500 for 3 years at 6% p.a. interest being payable half yearly.
 
Sol. \(A = P (1 + \frac{R}{100})^{2n} = 4500 (1 + \frac{6}{200})^6 = 4500 (1.03)^6 = 5373 ⇒\) Compound interest = Rs (5373 – 4500) = Rs 873


Ex 3. If the population of a town increases at 6 % p.a., but decreases due to emigration by 1% p.a. , what is the next % increase in the population in 3 years?

Sol. ⇒ Say the population  before the increase  in \(P_0\) and after 3 years is \(P_3\)

So  \(P_n=P_0 \left( \begin{array}{cc} 1 \pm \frac{r}{100} \end{array} \right)^n\)

Net increase in population = r= Growth rate - Emigration rate =6-1=5%

\(P_3=100 \left( \begin{array}{cc} 1 + \frac{(6-1)}{100} \end{array} \right)^3 = 100 \left( \begin{array}{cc} 1 + \frac{1}{20} \end{array} \right)^3 = 100 \left( \begin{array}{cc} \frac{21}{20} \end{array} \right)^3 = 100 \times 2.1 \times 2.1\times 2.1=115.76\)%

Therefore, the net % increase \(=115.76-100=15.76 \)%


Ex 4. In a population of a town increases at 5% for the first 2 years, increases at 2% for the next 3 years , and then decreases at 2% for the next 2 years , what is the next % increase in the population in 7 years ?

Sol. ⇒ \(P_n=P_0 \left( \begin{array}{cc} 1 \pm \frac{r_1}{100} \end{array} \right)^{n_1} \left( \begin{array}{cc} 1 \pm \frac{r_2}{100} \end{array} \right)^{n_2} \left( \begin{array}{cc} 1 \pm \frac{r_3}{100} \end{array} \right)^{n_3}\)


\(P_7=100 \left( \begin{array}{cc} 1 + \frac{5}{100} \end{array} \right)^{2} \left( \begin{array}{cc} 1 + \frac{2}{100} \end{array} \right)^{3} \left( \begin{array}{cc} 1 - \frac{2}{100} \end{array} \right)^{2} = 112.37\)

Therefore, the net % increase \(= 112.37-100=12.37\) %
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­Many students believe that the GRE sections are easy, especially when they achieve a good mastering of the key concepts. However, it is the case that their own accuracy is limping. Why? Most students attribute this to “silly” or “careless” mistakes that they could have avoided. Even worse, they do not fully address this grey zone in future review and preparation. In other words, they do not fix this because they are confident in their math skills and that the only way to bypass this situation is to stay more attentive. As a result, they end up being stuck in mid 70 percentile scores at best with very mediocre accuracy in quant.

The real reason for the aforementioned scenario is that the students themselves believe first that the GRE is a test about math, concepts, solve questions per se. The truth is that the GRE is a strategy test. It is conceived and implemented to make the test-takers pick certain types of answers, which are wrong, even though they are proficient in that math area. In other words: I know everything about number properties and still get some questions wrong.

1) Read the question carefully

While you are reading the question stem, you mustn't miss any word. Even the most minuscule detail IS important: a word, a shift in the argument, a comma, a preposition. After all, if that element is there, then there is a specific and valid reason WHY is there. If you miss out on crucial details in the question stem, you will find yourself either wasting a lot of time or picking an incorrect answer.

Read everything carefully and methodically. Focus when you read. Make sure that you understand the main point of each sentence and the key concepts in each problem. if you do not fully grasp what you are reading - reread it !


Suppose I produce 100 hammers. And I buy a new machine that will permit me to produce the double amount in still one hour.

What is the increase in production of
What is the increased production

Notice how the first statement implies I can produce 100 more hammers in the same amount of time.
The second implies that OVERALL I can produce in one hour 200 hammers. OVERALL


2) Use your scratch paper

Whether the question requires you to do simple math calculations or use the calculator, make it a rule to use your scratch paper instead. Organize your work. Get used to performing a systematic job. This way, you will not lose track of the work you are doing, even though it looks straightforward to ask yourself: " it is the case to jot down how it does 2+2". The answer is NO, of course. However, the idea is to be organized. No room for errors or improvisation during the time we solve a question. More relevant if this question is tough.

The more a question or a single step appears simple, the more we tend to solve that step mentally. There is nothing wrong with that. But, for instance, we have the following two systems of equations.

4x + 2y = 23
–(3x+ 3y = 22)

The result is obviously x-y=1.

The case is that we tend to solve mentally: of x is positive in the first one and negative in the second, as such we can subtract the second from the first, and we do have x...now take the chance of y......and we spent 5 minutes only to do this process in our mind.

Jot down. Make the subtraction. Take the result. Move on


3) Deal with Conversions carefully

Many avoidable mistakes happen because unit conversions end up being treated carelessly. 

Quote:
Runner A ran \(\frac{4}{5}\) kilometer and Runner B ran 800 meters.


[quantity=The distance that A ran]The distance that B ran[/quantity]

Just make sure to follow these easy steps in sequence.

Quote:
Runner A = D= R*T. He run 4/5 kilometers which are = 800 meters
D= 800R
Runner B =D=800R


4) Ask always and constantly to yourself: WHY?


You need to understand what the question is actually testing you on

  • What is the question asking me to find?
  • What information do I have to help me solve this question?
  • What information do I need to solve this question?


5) Do not skip: solve!


It would be best if you were organized and follow a methodical process. If not that, you will make mistakes that will make your accuracy drop. Make sure you learn the correct steps to approaching different question formats or approach the same question with several approaches, i.e., to solve a question with more than one method, possible at least two.

Ensure that you follow each step for every question you practice.
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­For every five hours of studying combinatorics-type questions, the average GRE student increases their chances of being able to correctly answer a question type that is found only on the very difficult end of the GRE spectrum. Meanwhile, the same student will have to compute hundreds of basic computations without the aid of a calculator. For students who know how to quickly do these computations, they are rewarded with extra minutes that can be spent double-checking their work and critically thinking about whether their answers make sense. As I would say- a second saved is a second earned on the GRE, but it doesn’t matter if those extra seconds come from being faster at doing combinatorics questions or quicker at computations. So check out these five math tricks, learn the ones that you like, and practice them daily to give yourself some extra time to finish off that final quant question.

Note: like everything else on the GRE, being able to do something and being able to do something QUICKLY are two different tasks. If you like any of the following tricks, make sure you know it inside and out before you try using it during your test.

1. Add or Subtract 2 or 3 Digit Numbers


To add numbers that aren’t already a multiple of ten or one-hundred, round the number to the nearest tens or hundreds digit, add, and then add or subtract by the number you rounded off. Do the opposite when subtracting.

Examples:
144 + 48 = 144 + 50 – 2 = 192

1385 – 492 = 1385 – 500 + 8 = 893

Why?

This math trick comes down to the order of operations- adding and subtracting occur in the same step and can happen in either order. Like many other computation tricks, this one comes down to replacing one tricky computation with two simpler ones.

2. Multiply or Divide by 5


To multiply a number by 5, divide by 2 and then multiply by 10. To divide a number by 5, divide by 10 and then multiply by 2.

Example:

82 X 5 = 82/ 2 x 10 = 410

Why?

This math trick comes down to the order of operations- multiplying and dividing occur in the same step and can happen in either order. But instead of doing the (somewhat) difficult task of multiplying by 5, do the easier task of multiplying by the fraction 10/2. And since you can do this in either order, you can start by dividing a number by 2 or multiplying the number by 10. Starting with division is usually easier when you start with an even number (34 × 5 = 17 x 10 = 170) while starting with multiplication is easier when beginning with a non-integer (6.4 × 5 = 64 ÷ 2 = 32). And instead of thinking about dividing by 5, think about multiplying by 2/10 (455 ÷ 5 = 45.5 × 2 = 91).

3. Multiply Numbers Between 11 & 19


To multiply two numbers that are between 11 and 19, add the ones digit of one number to the other number, multiply by 10, and then add the product of the ones digits.

Example:

14 × 13 = (17 × 10) + (4 × 3) = 182
Why?

In the standard way that most American-students are taught to multiply numbers, you set up two numbers on top of one another like this:

    14
 ×  13
__________
     42
    14X
__________
    182
This leads to a problem where you multiply 3 by 14, then multiply 10 by 14 and add the two products together. But you can rearrange this problem further to say you want to multiply 3 by 4, 3 by 10, and 10 by 14. Because you are multiplying both 3 and 14 by the same factor of 10 (which only happens when both numbers are between 11-19), you can combine this into one step. So instead of doing one tricky computation (3 × 14) and two easy ones (10 × 14 and 42 + 140), you make four easy computations (14 + 3, 17 × 10, 4 × 3, and 170 + 12).

4. Square Any Number Between 11 & 99


To square any number n, first find the nearest multiple of 10 and find out how much you would have to add or subtract (k) to get to that number. Then do the opposite function (addition or subtraction) to get two numbers that average out to n (i.e. n + k and n – k). Multiply those two numbers and add the square of k.

Examples:

\(23^2 = (26 * 20) + 3^2 = 529\)
\(97^2 = (100 * 94) + 3^2 = 9409\)

Why?

\(a^2\) – \(b^2\) = \((a+b)(a-b)\)
\(a^2 = (a+b)(a-b) + b^2\)
If this special product doesn’t look familiar to you, write it down right now and memorize it because there are a plethora of GRE questions that test you on this very concept. But for this special trick, you are (once again) trading a difficult calculation (23 x 23) for a few simpler ones.

\(23^2\)– \(3^2\) = \((23+3)(23-3)\)
\(23^2\)– \(3^2\) = \((26)(20)\)
\(23^2 = (26)(20) + 3^2\)
\(23^2 = 529\)
Since multiplying by multiples of ten are usually easier than non-multiples of ten, you find the nearest multiple of ten. While this may be very confusing at first, it’s a neat trick if you can get quick with it and is especially helpful when squaring numbers ending in five, since you will always add 25 to the lower and higher multiple of 10:

\(45^2 = (40 * 50) + 5^2 = 2025\)
\(65^2 = (60 * 70) + 5^2 = 4225\)

Another easier version of the above trick is:
\((ab)^2 = _ _ _ _\)
When b is 5:
\((ab)^2\) = a x (a+1)  2 5 

Example:
\(55^2\) = (5 X 6) 2 5 = 3025
\(75^2\)= (7 X 8) 2 5 = 5625

5. Estimate Root 2 & Root 3


\(\sqrt{2}\)= 1.4(approx)
\(\sqrt{3}\)= 1.7(approx)
Example:

The length of some side of a figure is about equal to 13 and you are down to the following three options:

a) 9 √2

b) 9 / √2

c) 15 * (1 – 1/√2)

By estimating √2 ≈ 1.4 or even 3/2, you could quickly recognize that only answer (A) could be correct in this problem.

Why?

Because your calculator says so! But rather than trying to remember these two roots on test day, remember your two favorite holidays in the second and third months of the year- Valentine’s Day & St. Patrick’s Day, 2/14 and 3/17. Root of 2 is about 1.4 and Root of 3 is about 1.7. The above example is adapted from an old Official Guide problem and is a reminder of how the GRE’s wrong answer choices aren’t there because they are close to correct value, but because they are some incorrect computational jumble of the numbers given in the problem. If you can estimate in this problem, you can find four incorrect answer choices and that’s just as good as finding one correct answer.
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­



Currency:

• 1 Dollar = 100 cents
• Quarter = 25 cents

Length:
• 1 m(metre) = 100 cm(centimetres)
• 1 km(kilometre)=1000 m 
• 1 in(inch) = 2.54 cm 
• 1 ft(foot) = 12 in 
• 1 mile = 1.6 km
• 1 mile = 5280 ft 
• 1 yard = 3 feet

Weight:
• 1 Kg(Kilogram) = 1000 gm(gram) 
• 1 t(tonne) = 1000 Kg 
• 1 lb (Pound) = 0.45 Kg 
• 1 ounce = 28.3495 grams 

Volume:
• 1 L(Litre) = 1000 ml(millilitre) 
• 1 L = 1 dm 3 (decimetre) 
• 1 L = 0.001 m3U 
• 1 gallon = 3.78 L

Quantity
• 1 Dozen = 12 
• 1 gross = 12 Dozen 
• 1 great gross =12 gross 
• 1 million = 1000000 = 106
• 1 billion = 1000000000 = 109
• I trillion = 1000000000000 = 1012

Years:
• 1 Decade = 10 years
• 1 Century = 100 year

source image: https://www.open.edu/openlearn/­
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Really helpful explanation! The breakdown of PEMDAS and how to approach absolute value problems is super useful, especially when prepping for tests like the GRE. I also like how you clarified the relationship between square roots and absolute values—many people forget that √(x2) equals |x|, not just x. When I practice problems like this, I usually double-check large numbers using a rounding calculator just to make sure I'm not missing any small detail. Great resource overall!
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