Quantitative Aptitude Techniques

If you are a bit slow in Quantitative Aptitude problems in Mathematics, which generally appears in many competitive exams like CAT, AFCAT, SNAP, XAT, IIFT etc… These tricks will only be helpful if you at least remember Tables up to 20 and can do addition of two numbers very fast. If you are not at that level, then practice first, come back here latter.

Various techniques which I use in problem solving are-

  • Digital Sum Technique
  • Divide and Rule Technique
  • Using Algebraic formulas
  • Percentages
  • Vedic Mathematics
  • Russian Peasant Mathematics
  • Base Method
  • Duplex Method

And there is no end to this list as you get deep into mysteries of mathematics. I will be briefing about these methods in this post, with the links to original source articles for further explanation. If you got any doubts, please let me know through comments section.

  1. Digital Sum Technique-
    As we work on decimal number system, it is inherent property of a number to retain its digital sum even when it is multiplied, divided, added or subtracted.
    Digital sum is sum of digits with which a number is made, till we get a single digit number, for example
    753729026 has digital sum = 7+5+3+7+2+9+0+2+6 = 41 = 4 + 1 = 5. Now if we do any operation with this number our basic mathematics operations can be easily verified with this technique, Example-
    753729026 X 2132 = 1609650283432
    Writing digital sum of each number-
    753729026 => 5
    2132 => 8
    1609650283432 => 4
    Multiplying digital sums – 8 X 5 = 40 (digital sum = 4)
    So this method might help you in quick verification of calculation or answer option elimination.
  2. Divide and Rule Technique-
    Suppose you have to perform a lot of calculation, you can arrive at an approximate answer using this technique-
    Example- Find 13.8 X 6.3 + 173% of 600
    Now, divide and rule by breaking them into parts
    This can be written as-
    13 X 6 + 0.8 X 6.3 + 0.3 X 13.8 + 175% of 600 – 2% of 600 = 72 + 5 + 4 + 7/4 of 600 – (2 x 1% of 600)
    = 83+ 1050 – (2×6) = 1121 (approximately)
    1124.94 (Exactly)
    Hence using this we got an error of 0.35% and calculation can be done mentally.
  3. Using algebraic formulas-
    Keep some of the algebraic formulas in your mind like-
    a2 – b2 = (a-b) (a+b)
    Example – 222 – 132 = 9 x 35 = 315
    There are lot more other formulas too, lookout for them.!
  4. Percentages
    Well, if you know all the percentages and corresponding fractions, a lot of hard work can be reduced!
    Try learning the common percentages, some of these are shown below-
    Percentage
  5. Vedic Mathematics
    I use this to do faster multiplication. See example below-
    Multiply and write the numbers as shown.
    Multiply
  6. Russian Multiplication
    To use this you need to be very fast in doubling the numbers and arriving at halves as well. Here is a quick demo how to perform such multiplication.
    Multiply
  7. Base Method
    This one is applicable to digits near same base, like here is an example-
    Multiply
  8. Duplex Method
    This one is used in calculating squares as shown below-
    Square

 

References-

  1. http://mathworld.wolfram.com/RussianMultiplication.html
  2. http://en.wikipedia.org/wiki/Digital_root
  3. http://www.sciencenews.org/view/generic/id/8203/title/Math_Trek__Divide-and-Conquer_Multiplication
  4. http://en.wikibooks.org/wiki/Vedic_Mathematics/Techniques/Multiplication
  5. Bonus – Japanese Multiplication

Mass Point Geometry

Well, as I am preparing for CAT 2012, so i thought it would be nice to share some out of the box things which i learn at my coaching classes. This might help you somewhere, and will be saved as an archive for my reference also. There is something very special kind of geometry I learned last week, and I am now exploring all the application part. The concept is very simple, and can be applied to any problem. Here is how I can explain it-

The concept-
It uses very simple principle of physics, which we have often observed in our daily life, see the figure below

Consider AB as line segment and C as some point in between, if the length of AC = 2 unit and length of BC = 8 unit, then the amount of weights need to balance will be 8 units on the side AC and 2 units on the side BC, as shown in figure.
Now this as obvious concept can be used in solving few problems relating to geometry, taking an example I can explain how..

An Example-
Question : – Consider the case below here are of small triangles contained in big triangle are known, now we have to find the area of whole triangle, I bet you can’t do this question in less than 10 minutes if you don’t know this method. Give it a try if you want to, before reading the solution.

Answer : – As we know base of triangle is in the same ratio as the areas are, when a line cuts triangle, hence the ratio of base will be same as ratio of areas.. Therefore, ratio of base-

 Now, using mass geometry i can complete this triangle as shown below-

Here, numbers written are nothing but ratios of sides, we know the area and ratio of sides. Therefore, we can find unknown area of quadrilateral. Picture below shows the actual areas in squares and ratios of sides. Using this we can say that, ratio of area of triangle ABE and triangle BCE = (4+x)/(8+13) = 7/1, where x is area of quadrilateral.

Therefore, 4+x = 7(8+13) = 21 x 7 = 147 => x = 143
Therefore, area of triangle ABC = 143 + 4 + 8 + 13 = 168
Please let me know through comments if you have any doubts. 🙂

Java & Prorgramming

So.. It was a long time I was not doing anything new in computers, so now I decided to make an Android application. But you need to know java and then you need to know How android SDK works.. Well now I’ve officially started learning java lets see how it goes. BTW, I like the coffee icon of java 😉

I am using the java help file which is used to create tutorials at official Sun Java website. If you are also intrested you can download the e-book from here [http://ge.tt/8F4ZhOB?c]