Play With Numbers


  • Being MBAtious!


    Calculation speed is not a chapter to learn but a skill to achieve. One among the most hyped element in aptitude test preparation is about achieving that impeccable calculation speed where we need not even touch a pen to get our answer. Sounds great! But at exam room, chances are more that we need our pen! Aspirants get overwhelmed with the idea of calculation speed and start learning methods that can help in quick solving.

    Some of these techniques are really helpful, but most of them are helpful only when we are practicing it for a period of time and got habituated with the same.Here the trap is if we decide to use some method just because it is considered as a “Shortcut”, we probably end up rechecking answers with our natural methods! Shortcuts don’t always save time. Methods we know for more than a decade are more likely to help us under exam pressure. Still, some techniques are easy to learn and handy to use. Some such concepts are shared here. Keep practicing and tweaking these methods so that they becomes the natural way of calculation. Two major sections in aptitude exams (Quant & DI) are calculation intensive. Even if we save 15 seconds from a question through smart approach we should be able to attend more questions or improve our accuracy. Considering each mark scored (or lost) has a huge impact on the result, you do the math. Literally! :-)

    The main differentiator between people who consistently score good marks in quant and others who doesn’t is the way they approach their basic math like multiplication and division. Those who are comfortable with quant save a lot of time in calculation and they don’t mess up with their accuracy. If you think you are not getting enough time in quant section, call your math loving friends for a coffee and "steal" their calculation methods :)

    Most of us require lesser time for addition and subtraction when compared to multiplication and division… when life teach us to use our strength over weakness, why not here!

    Multiplication and division can be represented as addition and subtraction operations. 39 * 46 = 1794 says that  the number 39 when added 46 times yields 1794 or when 46 is subtracted 39 times from 1794, result is zero. Similarly when we say 819/13 = 63 in math, English will say 63 should be added 13 times to get 819 or 13 should be subtracted 63 times from 819 to reach zero.

    We are assuming you have good speed and accuracy in addition/subtraction of 2 digits and 3 digits numbers. If you feel a little bit practice will not hurt then try using the time you spent sitting in a bus or auto… start looking at the number plates around and do your addition/subtraction practice sessions, where else you will get so many numbers flashing before you! Not convinced...? Numbers which are the smallest that can be expressed as the sum of two cubes in n distinct ways have been dubbed as "taxicab numbers"!

    Here goes the famous anecdote of the British mathematician G. H. Hardy regarding a visit to the hospital to see our great mathematician Srinivasa Ramanujan.

    “I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."

    The quotation is sometimes expressed using the term "positive cubes", since allowing negative perfect cubes (the cube of a negative integer) gives the smallest solution as 91.This ambiguity is eliminated by the term "positive cubes". 91 = 63+ (-5)3= 43+ 33

    As a first step, learn squares and square roots of natural numbers till 30 and also be comfortable with some common fractions …

    Knowing square roots also will help to save time in many scenarios.

    Some useful fractional values are given below

    Calculation of squares of numbers ending with 5

    If the number is of the form X5 then (X5)2= X * (X + 1) and append 25 at the end.
    652= 6 * (6+1) and append 25 at the end = 4225
    (135)2= 13 * (13+1) and append 25 at the end = 18225

    This comes handy in scenarios like 85 * 87

    = 85 * 85 + 85 * 2 = 7225 + 170 = 7395

    This method is actually derived from another useful Vedic math concept. It gives an easy way to find the product of two numbers if their unit digits add up to 10 and rest of the digits are same.

    Say, 72 * 78

    Units digits adds up to 10 (2 + 8 ) and other digits are same (7).
    Here product can be found out as 7 * (7 + 1)   (2 * 8 ) = 5616.

    Similarly 144 * 146 = 14 * 15     4 * 6 = 21024

    Many more neat tricks are there to easily find the square of numbers…

    To calculate squares of two digit numbers

    Let the number be xy then (xy)2 = [x2][y2] + 20 * x * y

    Take 58, (58)2 = [52][82] + 20 * 5 * 8 = 2564 + 800 = 3364

    Take 81, (81)2 = [82][12] + 20 * 8 * 1 = 6401+ 160 = 6561 ( note: we wrote 12 as 01 ( 2 digits) )

    To calculate squares for numbers closer to 100

    Take 94, 94 is 6 less than 100, so 942 = [94 – 6][62] = 8836 (note: we subtracted the difference as the given number is lesser than 100)

    Take 108, 108 is 8 more than 100, so 1082 = [108 + 8][82] = 11664 (note: we added the difference as the given number is greater than 100)

    Take 119, 119 is 19 more than 100, so 1192 = [119 + 19] [192] = [138][361] as the second part is more than 2 digit take the most significant number (here 3) and add it to the first part (138 ).
    So 1192 =  [138 + 3][61] = 14161

    Take 125, 125 is 25 more than 100, so 1252 = [125 + 25]... we dont need to do that... we have a better trick, right? 1252 = [12 * 13][25] = 15625  :)

    To calculate squares of numbers closer to 50

    58 = 50 + 8
    582 = [25 + 8] [82] = 3364 [Note: Add the difference to 25 to get the first 2 digits as the given number is greater than 50]

    63 = 50 + 13
    632 = [25 + 13] [132] = [38] [169], as the second part is more than 2 digit take the most significant number (here 1) and add it to the first part (38 ).
    632 = [38 + 1][69] = 3969

    51 = 50 + 1
    512 = [25 + 1][12] = 2601 [Note: we wrote 01 as we need 2 digits in each bracket]

    47 = 50 – 3
    472 = [25 – 3][32] = 2209 [Note: Subtract the difference to 25 to get the first 2 digits as the given number is lesser than 50]

    33 = 50 – 17
    332 = [25 – 17][172] = [8][289], as the second part is more than 2 digit take the most significant number (here 2) and add it to the first part (8 ).
    332 = [8+2][89] = 1089

    So now we know how to find squares faster. If just knowing was enough, life would have been much easier… right? We should be able to apply our strengths to our advantage in solving a situation…  our confidence in calculating squares can help us in some quick multiplication… How?

    We all know the basic formula (a + b) (a - b) = a2 – b2

    87 * 93 = ( 90 – 3 ) * (90 + 3) = 902 – 32 = 8091

    similarly, 43 * 47 = 452 – 4 = 2025 – 4 = 2021

    Ok.. Now solve 35 * 45 in 5 seconds  :)

    Multiplication

    9 * 12 = ____ Ok, you got 108; question is how you did it?
    Most of you might have done it using the logic 9 * 10 + 9 * 2 = 108.

    27 * 23 = ___
    27 * 20 + 27 * 3 = 540 + 81 = 621

    634 * 126 = ___
    634 * 126 = 634 * 100 + 634 * 26 = 63400 + 634 * 20 + 634 * 6
    = 63400 + 12680 + 3804 = 79884
    To find 634 * 6 we used 634 * 5 + 634, an easy way to multiply by 5 is multiply by 10, i.e. put a zero at the end, and divide by 2;   636 * 6 = 3170 + 634 = 3804)

    Try solving the below questions using the above method.

    1. 85 * 11
    2. 69 * 71
    3. 45 * 43
    4. 112 * 120
    5. 852– 652
    6. 814 * 629

    Solutions

    1. 850 + 85 = 935
    2. (70-1) * (70+1) = 4900 – 1 = 4899 (a2- b2= (a + b) (a - b))
    or 71 * 69 = 71 * 70 - 71 = 70 * 70 + 70 - 71 = 4900 - 1 = 4899
    3. 45 * (45 - 2) = 452– 2 * 45 = 2025 – 90 = 1935
    4. 120 (100 + 12) = 12000 + 1440 = 13440
    5. 7225 – 4225 = 3000
    6. (800+14) * (630-1) = 800 * 630 * 14*630 - 800 - 14,
    800 * 630 = 800 * (600+30) = 480000 + 24000 = 504000
    14 * 630 = 10 * 630 + 4 * 630 = 6300 + 2520 = 8820
    Our answer is 504000 + 8820 – 800 – 14 = 512006

    Division

    Easiest way is to represent numerator as sum/ difference of terms related to denominator.

    183 / 16 = ___
    183 = 160 (16 * 10) + 16 (16 * 1) + 8 (16 * 0.5) – 1
    183 / 16 = 10 + 1 + 0.5 - x (some small value) ≈ little less than 11.5 (approx)

    3840 / 23 = ____
    3840 = 2300 + 1150 + 230 + 115 + 45
    3840 / 23 = 100 + 50 + 10 + 5 + (little less than 2) ≈ little less than 167

    There will be a lot of calculations based on fractions waiting for you, mostly in data interpretation section. It’s very important to have a high comfort level with such numbers. Keep below fractions handy.

    Let’s play!

    Which is greater, 10/9 or 7/6?
    10/9 = 10 * 0.111 = 1.11
    7/6 = 7 * 1/2 * 1/3 = 7 * 0.5 * 0.333 = 3.5 * 0.333 = 1.1655
    7/6 > 10/9

    Another method is 10 = 9 + 0.9 + 0.09 + …
    10/9 = 1 + 0.1 + 0.01 + … = 1.111

    7 = 6 + 0.6 + 0.36 + 0.03 + …
    7/6 = 1 + 0.1 + 0.06 + 0.005 +... = 1.165 (approx)

    Personally I feel second method is better as it’s a generic way of tackling fractions and percentages. You just need to brush up your addition skills on two and three digit numbers. Represent numerator in terms of denominator. That’s all it takes.

    321/562 = ____
    321 = 281 (562/2) + 28.1(562/20) + 11.24(562/50) + …
    321/562 = 0.5 + 0.05 + 0.02 + … = 0.57(approx)

    562/321 = ___
    562 = 321 + 160.5 (321/2) + 80.25 (321/4) + …
    562/321 = 1 + 0.5 + 0.25 = 1.75 (approx)

    You will find these methods very useful while dealing with calculation intensive DI problems or number system questions in quant.If you feel this method is helping (or can help) to save some time in calculation, practice well so that these stuffs will come automatically when you solve a question. Reduce your pen usage just to write the sub values of an expression and not to write the steps.

    Here we are not using any speed math techniques, but we are trying to do smart math. Forget the “magic trick” of finding the square of 4 digit numbers by drawing circles or finding the square root by connecting dots. Stick to the basics and make it strong which will be good enough for kind of calculations expected in our aptitude exams. Everything we need to win an aptitude exam is taught to us when we were in high school... We just need to brush it up.

    There will be lot more techniques which comes under the criteria Simple to learn & easy to use. Please do share your secret tricks... You can publish your gyan notes or can post your tricks as comments here.

    Happy Learning! :)


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