Learn to Solve Simultaneous Equations Mentally

Do you struggle with solving simultaneous equations?

By simultaneous I mean equations with multiple unknown variables. Generally the number of equations given will be equal to the number of equations.

Let’s take an example,
3x + 4y = 18
5x + 7y = 31

Methods Taught at Schools

In our schools, we are taught to solve for x by equating the co-efficient of y by multiplying both the equations by some constants in such a manner that you get the same resultant value and then subtracting one equation from the other.

For instance in this case, to find the value of x, we will multiply first equation by 7 and second equation by 4 and get 28 in both the cases. This is done so that we get a zero on subtracting.
3x + 4y = 18 …………………(i) x 7
5x + 7y = 31 …………………(ii) x 4

We get 2 equations, where co-efficient of y is same
21x + 28y = 126
20x + 28y = 124

Subtracting second equation obtained above from the first one we get
(21x – 20x) + (28y – 28y) = 2
Hence, x = 2

Plugging the value of x in equation (i) we get y = 3

Problems with the Above Methods

  • This method can become quite laborious, especially when the co-efficients of the unknowns are such that they have to be multiplied by large numbers to make them equal to eliminate one of the unknown by adding or subtracting as the case may be.
  • The above calculations become cumbersome, when the co-efficient(s) are large prime numbers.
  • This method involves multiple steps where we need to do multiplication and addition/subtraction.
  • Also, there is no chance of using this method to solve the problem mentally as one has to keep track of the equations and various computations.

Solving Simultaneous Equations the Smarter Way

Our new method will give us the final answer in fractions, i.e. in Numerator and Denominator for both the variables: x and y.

First, we need to find the numerator of the value of x in the above case, take the simple following steps:
Step #1: Cross-multilply the coefficient of y in the first equation by the constant term (RHS) of the second equation
Step #2: Subtract from it the cross-product of the y coefficient in the second equation and the constant term (RHS) of the first equation.

Solving Simultaneous equation

So the numerator is 4×31 – 18×7 = 124-126 = -2.

Second, we need to find the denominator of the value of x:
Step #1: Cross-multiply the coefficient of y in the first equation by the coefficient of x in the second equation
Step #2: Subtract from it the cross-product of the y coefficient in the second equation and x coefficient in the first equation.

Solving linear equations

Hence the denominator is 4×5 – 7×3 = -1
Hence, the value of x = -2/-1 = 2

Now, let’s try with a simpler example,
x+2y = 8
3x + y = 9

Using the above method, in a single line calculation you can say,
x = (2×9 – 1×8)/(2×3 – 1×1)
x = 10/5 = 2
Therefore, y = 3

Isn’t this amazingly simpler? With some practice you can comfortably apply this technique to solve simultaneous equation mentally.

If you liked this method, you must explore another Vedic Mathematics trick of solving a special class of simultaneous equations in seconds.

Do you find simultaneous linear equations difficult to solve? Do you think you can start using above method in solving equations?

Understanding Exponential Function – “e” from Layman’s Perspective

In our journey of learning mathematics, we must have across a mathematical constant called “e”. It is approximately equal to 2.71828.

Unfortunately, most of us memorized the value and its usage without bothering about the concept behind it. That’s because mostly mathematics is taught in a way, where we try to explain concepts by their technical features without really explaining what it is and why it is used in the first place. Let us explore very basic insights about this so called “e”.

The natural exponential “e” explained in simple terms

In simple terms, “e” is manifestation of continuous growth in any naturally and continuously growing phenomenon.

Now what is this continuous growth and how do we deal it mathematically? To answer this question, we need to start with the other type of growth which we understand more easily, i.e. discrete growth.

Discrete Growth

Discrete growth can be visualized as step by step growth at the end of each period. If something doubles after every period; that is, at 100% growth per period

  • 1 becomes 2 at the end of 1st period;
  • 2 become 4 at the end of 2nd period;
  • 4 become 8 at the end of 3rd period; and so on…

This is basically 2^n where n is the number of periods. Remember, in the case of discrete growth nothing happens till we don’t reach the end of the period, then suddenly it doubles. It’s more like growth in step by step fashion depicted in the diagram below.  If you’re just interested in knowing how much time it takes to double your money at certain rate of interest, you will love the Rule of 72.

Discrete Growth vs Continuous Growth

In real life, the growth of anything like money, population, bacteria, plants, etc. is more smooth i.e. it is continuous process rather than discrete. 

Given the following formula for compounding, we get discrete results –

A = P (1 + r/n) ^ nt

  • A = Is the future value of investment including interest
  • P = Principal initial investment or the present value of the investment
  • r = yearly rate
  • n = number of compounding periods
  • t = no. of years

Assuming, the initial amount is Re.1, what happens when the yearly rate of growth is 100% and t is 1 year?

The above compounding formula gives us,

compunding computation

Calculation credit: Purplemath

Continuous Growth

Now, think how we can, mathematically shift from discrete stairs like growth curve to continuous smooth growth curve?


From the above table and graph, you know intuitively that higher the value of n, smoother the curve will become. So continuous growth is possible when the growth happens over infinitely small periods of time.  Thus to make it continuous n is assumed to be “x” which tends to infinity.


Thus, you have got a new number “e” which denotes continuous growth over 1 period in which growth is taken as 100%.

Different Growth Rates and Time Periods

Now the question in your mind must be what if the growth rate is not 100% per period but something else. Fair question.

Remember, “e” by itself expresses the continuous growth in something at the rate of 100% over 1 time period, say 1 year. At 100% growth r becomes 1 (as you know 100% = 1) and for single time period t = 1.

Essentially, continuous compounding is expressed as e^x or it can be further expanded as ert where r is the rate of interest and t is time period.

The number “e” is about continuous growth. For all practical purposes, we can use e^x, where x allows us to merge time and rate. 5 years at 100% growth is the same as 1 year at 500% growth, when continuously compounded. Intuitively, e^x means, how much growth do I get after after x units of time and 100% continuous growth.

So if rate is 20% and time is 3 years, the resultant growth at continuous compounding can be captured using e^rt = e^(0.2*3) = e^0.6. That’s like saying 100% growth rate for 0.6 time years

Hope now “e” has started making more sense to you. Yes or No, please let me know in comments below.

simultaneous equations

Solution of Linear Equations in 2 Seconds

Solving Simultaneous Equation using Vedic Mathematics

Let’s learn a very simple trick using which we can quickly solve linear algebraic equations with 2 variables using a Vedic Maths sutra named as Anurupye Shunyamanyat.  This is the 6th Sutra of Vedic Mathematics (flipkart affiliate link).

ANURUPYE SHUNYAMANYAT – which means, if one is in ratio, then the other one is zero.

simultaneous equations

This is used in solving simple simultaneous equations in which one of the variables are in the same ratio to each other as the independent terms. This is useful in certain circumstances only but saves time when applicable.

Let us take a simple example

3x + 8y = 42

6x + 18y = 84

Co-efficient of “x” (which is a variable or unknown) are in the ratio 3:6 = 1:2 and the independent terms are also in the ratio 42:84 = 1:2

Wherever you see this happening, simply use the above sutra which says in such cases

y = 0

Therefore, x = 42/3 = 14

Hence problem solved. Simple isn’t it.

Let’s take another example –

11x + 7y = 28

23x + 21y = 84

Here, co-efficient of “y” are in the ratio 7:21 = 1:3 and the independent terms are also in the ratio 28:84 = 1:3

That means the above mentioned vedic maths sutra is applicable in this situation too.


y = 0

y = 28/7 = 4

If you’ve any question related to simultaneous equations, please post them as comment below. Share your views / feedback also.

2 + 2 = 5? Two Plus Two Equals Five?

Two plus Two Equals Five, is it possible?

Two plus two equals five” (“2 + 2 = 5“) is a famous phrase used in George Orwell’s book Nineteen Eighty-Four; therein, it is used as an example of an obviously false dogma one may be required to believe because of political or religious pressures. Orwell’s protagonist, Winston Smith, uses the phrase to wonder if the State might declare “two plus two equals five” as a fact; he ponders whether, if everybody believes it, does that make it true?

Here in India we’ve a famous movie starring Amitabh Bachchan and Shashi Kapoor in the name “Do Aur Do Paanch”, the literal translation of which is two plus two equals five. So what’s the fallacy behind this simple calculation?

Authorities can make you believe something which may be completely baseless.  However, is it mathematically possible to proof: 2 + 2 = 5

Let us give it a shot.

Required to prove, 2+2 = 5?

We all know, 2+2 = 4

= 4 – 9/2 + 9/2
= √(4 – 9/2)2 + 9/2
= √(16 – 2*4*9/2 + (9/2)2) + 9/2
= √(16 – 36 + (9/2)2) + 9/2
= √(-20 + (9/2)2) + 9/2
= √(25 – 45 + (9/2)2) + 9/2
= √(52) – 2*5*9/2 + (9/2)2) + 9/2
= √(5 – 9/2)2 + 9/2
= 5 – 9/2 + 9/2
= 5
Therefore, 2+2 = 5 (proved)

Oh boy, how’s that possible? Amusing, isn’t it? Let me know if you like this post. If yes, share it on social media with your friends.

Use your grey cells and find out the fallacy. Where’s the mistake in above proof? Put your answers in the comment below.

How to Select Right Math Tutor for Your Child

right math tutor for your childThe most common way to help your child improve in his or her learning is by hiring a tutor. Hiring one does not necessarily mean that a child is a slow learner. Mathematics is one of the academic subjects in which many students need help for improvement. To ensure that your child can benefit the best help, you should look for a tutor that can address the learning style and temperament of your child. This way you can expect for the best possible results.

Though most of the suggestions below are generic in nature, however, this article is written keeping in mind audience from the USA

1. Tutor’s college degree & major

When hiring a tutor, the first thing to consider is the college degree completed. If you’re hiring a tutor through an agency, you should take time verifying the degree and the major of discipline. Since you want your child to improve in mathematics, obviously it is best to hire a tutor having college major in this subject.

2. Teaching credentials and Background

As much as possible, it’s a smart idea to hire a tutor with good teaching credentials and background. It’s not enough that a tutor has the capability to impart learning and knowledge. Thus, before hiring a tutor make sure that he or she has passed the licensure examination given by the state. Keep in mind that you will entrust your child to the tutor for some period of time that’s why it’s essential to implement background check and verify the employment background. Once you know that the tutor you’ve hired is legitimate and a member of a qualified organization, you’ll have peace of mind that the teaching process will flow smoothly.

3. Rapport

The connection between the student and a tutor plays a big role to the success of the process. This makes sense of hiring a tutor who can establish a good rapport with your child particularly if there’s some behavior issues.

4. Ask your kid’s teacher

If it’s your first time to look for a tutor, you can consider recommendation from people who have been doing the same. Your child teacher at school or the school counsellor can be the appropriate individuals who can recommend a qualified tutor.

5. Check community center bulletin boards

Another way to find a qualified tutor is through community boards or library wherein tutors ate posting information.

6. Select a tutor willing to work interactively on concepts

Prior to hiring a tutor you should check first his or her capability and the concepts of teaching. Hire a tutor who is willing to utilize interactive concepts. The tutor should not only provide lectures but also interacts with the student. Likewise, as a teacher, your child’s tutor should be willing and give time in addressing the questions of your child. The tutor must be always ready to assist your child in solving math problems.

7.  Try to search a tutor who is open to feedback

Select a tutor who welcomes feedback in appositive manner. This means that before hiring one, parents should discuss with the tutor a sort of guidelines along the teaching process. He or she should also be open to discussion and suggestions from the parents in order to rule out which concept will work or doesn’t to the child.

It is very challenging to find the right math tutor to your child knowing that parents should be the first and personal teacher of a child. But, there are times when parents don’t have enough time helping a child to improve in academic subjects, thus hiring a perfect math tutor can be the last option. There’s nothing wrong hiring a tutor as long as he or she can help to improve the ability of the student and not just to earn income.

Author Bio: Andy Bell is an everyday learner and editor, working at Tutoring help. He is passionate about helping online learning businesses to achieve their goals. He loves tutoring and his favourite subjects are math and English, but he has also taken honours classes in science and history. Outside the classroom, he enjoys tennis and is an active member of the Rancho Crandon Park.

Permutations and Combinations – Part 5- Grouping & Distribution

Grouping and Distribution

This is a very important concept of permutation and combination where some higher order fundamentals of permutation and combination is involved – the reason for reserving this topic for the end.

Here is the list of articles in this series of permutation-combination for quick reference. Ideally you should check these articles in order to gain better understanding of the whole concept –

  1. Principle of Counting
  2. Permutations Introduced
  3. Permutations – Special cases
  4. Combinations – Introduction
  5. Combinations – Grouping & Distribution

What is the difference between grouping and distribution?

To distribute something, first grouping is done. Only after  you have made groups of some objects, you might want to distribute these groups in various places. For example, after you made groups of some toys, you might want to distribute these groups among some children. Or, after dividing some number of toffees into groups, you might want to distribute these groups into boxes.

Just as the objects that we group can be similar or dissimilar, so can the places that we assign these groups to be similar or dissimilar.

While distributing groups, we need to keep one rule in mind: We permute the groups only if these places for distribution are dissimilar, otherwise not.

Say, we have 2 items- X and Y and I have to split them into two groups. There is only one way of doing it – X goes in one group and Y in the other. However, if I have to distribute among 2 people A and B, then these 2 groups can be permuted in 2! ways.

Division of dissimilar items into groups of EQUAL SIZE

Let’s take a very simple example. In how many ways can you divide 4 different things (say A, B, C and D) into two groups having two things each?

You would like to say that we select two things out of the four and two would be left behind, i.e. 4C2 = 6 ways. But are there really 6 ways?

Take a look. We can divide four things, A, B, C and D into two groups of two in the following ways:




You can keep trying but there is no fourth way to do it. So where have the remaining 3 ways calculated through 4C2 = 6 disappear? If you look carefully, there was an overlap. When we select 2 things out of 4, we can do it in 6 ways – AB, AC, AD, CD, BD and BC but when we select the first three groups, the last three get automatically selected without having to select them separately and vice-versa. So when we select AB, CD is automatically selected and vice-versa.

This overlap will manifold itself if we increase the number of items further.

So be very careful not to apply the usual combinations formula whenever we have to divide into groups of equal size. However, if the groups contain unequal number of things, then our earlier method of using combinations formula for selection will be valid as will be discussed in the next section.


Let me now increase the objects to 5. How would you divide these 5 distinct (dissimilar) objects into groups of 2, 2 and 1?

The single object can be chosen in 5C1 = 5 ways. The rest of the 4 objects can be divided into two equal groups in 3 ways as explained above. Therefore, total number of ways = 5 x 3 = 15.

  • The number of ways in which mn different things can be DIVIDED equally into m groups, each group containing n things = (mn)!/(n!)^m x 1/m!

  • The number of ways in which mn different things can be DISTRIBUTED equally into m groups, each group containing n things = (mn)! / (n!)^m

Note: In the distribution, order is important hence the divisible things can be arranged in m! ways since things are divided into m groups.

Division of dissimilar items into groups of UNEQUAL SIZE

Say we have k things and we have to divide them into 2 groups containing m and n things respectively such that m+n =k, then this can be done in k!/m!.n! ways. This is because m things can be selected out of k things in kCm ways and when m things are taken, n things are left to form the other group of n things which can only be done in nCn =1 way. Hence the required number of ways is kCm = m+nCm = (m+n)!/(m+n-n)!.n! = k!/m!.n!.

We can extend the same concept for increased number of groups as long as the number of items in all the groups add up to the total i.e. n.

The number of ways in which n distinct things can be DIVIDED into R unequal groups containing a1, a2, a3, ……, aR things (different number of things in each group and the groups are not distinct)

= nCa1 × (n-a1)Ca2 × … × (n-a1-a2-….-a(r-1))CaR

=n! / a1! a2! a3!…….aR!   (here a1 + a2 + a3 + … + aR = n)


What if there are n distinct things and we have to find out the number of ways in which they can be distributed among r persons such that some person get a1 things, another person get a2 things, . . . . and similarly someone gets aR things (each person gets different number of things)?

Number of ways in which n distinct things can be divided into R unequal groups containing a1, a2, a3, ……, aR things (different number of objects in each group and the groups are distinct)

=n! / a1! a2! a3!…….aR!x R!  (here a1 + a2 + a3 + … + aR = n)

Division of IDENTICAL / SIMILAR ITEMS into Groups

Number of ways in which n identical things can be divided into r groups, if blank groups are allowed i.e. each can receive zero or more things (here groups are numbered, i.e., distinct), where 0≤r≤n = (n+r-1)C(r-1)


Number of ways in which n identical things can be divided into r groups, if blank groups are not allowed i.e. each receives at least one item (here groups are numbered, i.e., distinct), where 1≤r≤n = (n-1)C(r-1)


Number of ways in which n identical things can be divided into r groups so that no group contains less than m items and more than k (where m<k) is coefficient of xn  in the expansion of (xm + xm+1 +…..xk)r


Some Important Results:

  • Number of squares in a square having ‘n’ columns and ‘n’ rows = 12 + 22 + 32 +….+ ∑n2 = n2 = n(n+1)(2n+1)/6

  • Number of rectangles in a square having ‘n’ columns and ‘n’ rows= 13 + 23 + 33 +….+ n3 = ∑n3

  • =[n(n+1)/2]2

  • Number of quadrilaterals if m parallel lines intersect n parallel lines = mC2 x nC2


With this discussion on Grouping and Distribution we come to an end on the topic Permutations and Combinations.

Please post in your doubts, queries and interesting problems in the comments box.

Hope you enjoyed reading these 5 articles as much as I enjoyed writing them.

induction Puzzle

Age of Three Sons – Interesting Induction Puzzle

Today I’m giving you a very interesting induction puzzle. I’m sure you’ll enjoy solving it. Put your answers in the comment below. I don’t have an option to hide the answers given, hence I would request you to avoid checking the answers given by others before trying it yourself. If you like puzzles you would surely like these mathematical puzzles also.

Induction Puzzle Definition

It might help you to know what’s induction puzzle. I’m giving you the Wikipedia definition here for you –

Induction Puzzles are logic puzzles which are solved via the application of the principle of induction. In most cases, the puzzle’s scenario will involve several participants with reasoning capability and the solution to the puzzle will be based on identifying what would happen in an obvious case, and then repeating the reasoning that: “as soon as one of the participants realises that the obvious case has not happened, they can eliminate it from their reasoning, so creating a new obvious case”.

Here’s the link to the Wikipedia page which has more examples of induction puzzle – https://en.wikipedia.org/wiki/Induction_puzzles

Solve this Puzzle – If you can!

A man bumps into his mathematician friend on the street that he hasn’t seen in 5 years. The man asks the mathematician how old his children are. The mathematician, who always replies in riddles said, “I now have three children. The sum of their ages is equal to the number of windows on the building in front of you and the product of their ages equals 36.” The friend then says “I need one more piece of information.” The mathematician then replies “My youngest child has blue eyes.” What are the ages of the mathematicians three children?

Waiting for your answers.

Permutations and Combinations – Part 4- Combinations

This article is the fourth in a series of 5 articles. Here we will be discussing a pet topic of all competitive examinations – Combinations. The list of 5 articles of the permutation combination series is –

  1. Principle of Counting
  2. Permutations Introduced
  3. Permutations – Special cases
  4. Combinations – Introduction
  5. Combinations – Grouping & Distribution

I would advise you to go ahead with this article only after you are clear on first 3 articles in this list above.

You will be better equipped to move on with Combinations after you are thorough with the above topics. We will discuss various cases of Grouping & Distribution in the article following this.

All right, let us get started!

Difference between Permutations and Combinations

Suppose there are 3 bags (A,B and C) in my home and I want to select any 2 out of them to take with me on my holiday. In how many ways can I make the selection?

Clearly I select either AB or BC or AC i.e. 3 ways.

An important point to note is that we are talking about selection and not order here. Obviously whether I select AB or BA makes no difference.

Let us take one more example.

Suppose from a class of 10 students I have to select 3 students for a play, it is a case of Combinations. But, if I have to arrange 3 students in a line from a class of 10 students, it is a case of Permutations.

I hope I have made it clear that in permutations (rearrangement) order matters but in combinations (selections) order does not matter.

We are now in a position to define Combinations.

 Definition of Combinations

Combinations is the selection of some or all of a total of n number of things.

If out of n things we have to select r things (1≤r≤n), then the number of combinations is denoted by  nCr =  n!/r!(n-r)!

Combinations does not deal with the arrangements of the selected things. This explains division by r! which denotes the arrangement of the selected r things.

Important relation between Permutations and Combinations

‘r’ selected things can be arranged in r! ways.

So, r! x  nCr  = nPr

or,  nCr = nPr / r!

or     nCr = n! / r! (n-r)!

Example: In a class there are 6 boys and 5 girls. In how many ways can a committee of 2 boys and 2 girls be formed?

Solution: 2 boys can be selected out of 6 in 6C2 ways.

2 girls can be selected out of 5 in 5C2 ways.

So the selection can be made in 6C2 x 5C2 ways. (Product Rule: ‘and’ stands for multiplication)

Example: In a class there are 6 boys and 5 girls. A committee of 4 is to be selected such that it contains at least 1 boy and 1 girl.

Solution: There are 3 different possibilities now-

i. 1 boy and 3 girls

ii. 2 boys and 2 girls

iii. 3 boys and 1 girl

In the 1st possibility, total number of combinations =  6C1 x 5C3

In the 2nd possibility, total number of combinations =  6C2 x 5C2

In the 3rd possibility, total number of combinations =  6C3 x 5C1

But only one of the above possibilities will occur; 1st OR 2nd OR 3rd.

So the total number of required combinations is  6C1 x 5C3 +  6C2 x 5C2 + 6C3 x 5C1

Some Important Results on Combinations

  • nCr = nCn-r                                           (0≤r ≤n)
  • nC0 = nCn = 1
  • nCr + nCr-1 = n+1Cr                  
  • If nCp = nCq , then p = q or p + q = n (p,q € W)

Restricted Combination

The number of combinations of ‘n’ different things taken ‘r’ at a time  subject to restriction that p particular things –

i) will never occur = n-pCr

ii) will always occur = n-pCr-p

Number of ways of selecting one or more things from a group of n distinct things = nC1 + nC2 + nC3 + …… +  nCn  = 2n – 1 .

Number of ways of selecting zero or more things from a group of n distinct things = nC0 + nC1 + nC2 + nC3 + …… +  nCn  = 2n

With this we conclude the topic Combinations. Feel free to post interesting questions which we all can share and solve together. In the following article we will discuss the rules on Grouping & Distribution- a very important concept of permutation and combination.

financial calculator

Which Financial Calculator to Buy?

Are you taking any financial certification exams like CFA, CFP or CPA?

Are you having a tough time calculating your loan payments?

Are you facing difficulties figuring out interest amounts and payment schedule?

Are you interested in calculating cash expenses and making other financial projections for your business?

If the answer to any one of the above mentioned questions is a yes, you might consider buying a financial calculator for yourself.

A financial calculator is a highly powerful calculating machine with numerous additional features and calculation capabilities as compared to the simple add-subtract-multiply-divide functions of a standard calculator. It essentially works like a mini computer. Some of the higher-end models of financial calculators also include graphing capabilities that enables you to calculate future growth rates based on current mathematical estimates or to store formulas inside the built-in data drives.

Based on functionality, price considerations and exam acceptance, I am presenting here a list of the best financial calculators to choose from.


The HP10BII is one of the most affordable financial calculators available and thankfully, you do not need to compromise on quality on this one. It is easy to use and learn and a true value for money. Of course there are more powerful financial calculators out there but not one is easier to use with all the features that you’ll need as a beginner. It is capable of completing over 100 different math functions and it has special programs for bond valuations and bond yield. Unless you’re a financial expert with specific needs, you can definitely look at this powerful equipment from HP.

The HP 10BII+ takes everything that is great about the HP 10BII and then simply extends on the functionality. Buy Online

CFP Standards: The Certified Financial Planning Board (CFP) has strict regulations regarding acceptable calculator models for students taking the exam. It does not accept any calculator with alphabetical keys. HP 10BII is one of the few calculators allowed for the exam. Therefore, I would strongly recommend this particular model to students who are planning to take this certification test in the future.

Texas Instruments BA II Plus Financial

There are two versions of this particular calculator. The Texas Instruments BAII Plus Financial version is the scaled-down model of the professional one. It is best for students who have just started with preparations for financial certification examinations. The better versions can always be bought later as you progress in your career path. Its price is probably one of the best reasons to select this calculator, especially for those on strict budgets.

The calculator features finance functions such as amortization, bonds, cash flows (up to 24 uneven), depreciation, and more.

CFA & CFP Standards: Only two calculator models are authorized by the CFA Program exams and one of these is the BAII Plus. It is also acceptable for the CFP certification exam.  Buy Online

HP 12C

HP 12 C is the ideal purchase for students and professionals alike. Price wise it strikes just the right balance. It is more expensive than the budget financial calculators and cheaper than the high-end graphic calculators.

The large layout of buttons and 10-digit display screen is perhaps its biggest draw making it extremely easy to use. In terms of functionality, it has over 120 business and finance functions such as, cash flow, depreciation, bonds, loan payments, amortization, and more. Overall, a great calculator except its price. There are less expensive financial calculators that perform the same functions at a lower price but for banking, real estate, and insurance professionals it may be worth the extra money.

CFA Standards: This is the other calculator acceptable for use in CFA certification apart from BAII Plus discussed above. It is also acceptable for use in CFP certification examinations. Buy Online

Texas Instruments BAII Plus Professional Financial

Slightly more expensive than the entry level Plus II, the BAII Plus Professional is surely a power performer. In addition to the basic features, it comes with many advanced functions for trignometric and logarithmic (including natural log) calculations . It is also an ideal buy for students who need to incorporate advanced calculus.

Though it is pricier than the standard BAII Plus, it definitely is an intelligent buy for students looking for an accounting or economic career as it will last you long.

CFA & CFP Standards: It is acceptable for use in both CFA and CFP certification examinations. Buy Online


The HP 17BII can be easily compared to the HP 12C both in terms of performance and price. The main difference is the display screen. While 12C has a single line 10-digit large display, 17BII has a split screen display that allows users to see two different numerical lines. That is not all. The 17BII has additional 250 finance functions as compared to the HP 12c. One of these is the dual variable forecasting feature, which is surely a big plus point as compared to other budget and moderately-priced calculators. It is helpful for statistics classes, but obviously if your pocket permits, a graphic calculator will serve a better purpose there.

CFP Standards: This particular calculator is acceptable for the CFP exam. Buy Online

 Texas Instruments TI-84 Plus

Essentially a graphic calculator, TI 84 is an improved version of the TI 83. It incorporates graphical and statistical programs in addition to all the regular functions of a financial calculator. Fact is fact, it is a very expensive calculator. So it suits only those people who actually need the added graphical and statistical capabilities of this power machine. It takes a while to get accustomed to this one because of the many many features.

Like most graphic calculators, it has a display screen large enough to show an entire sequence of numbers as well as the display of graphs.

Certification Exam Standards: It is forbidden for use in any financial certification exam! Buy Online

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Which calculator do you use and recommend?

manhole puzzle

Manholes Cover Puzzle

Why is it better for manhole cover to be round rather than square?

Any problem can be solved in many creative ways. The more street smart you’re, the more logical you’re you can think of many ways of solving a given problem, specially when you get an unusual question to answer. There can be many reasons for the manholes being circular rather than squarish or rectangular. Don your thinking hats and come up with some cool reasons. You can explore many such puzzles at Quickermaths.com

Leave your answers below.