Calculating Faster

Speed in solving questions is of crucial importance if one wants to crack the MBA entrance exams. In fact, the only two skills tested in the Data Interpretation section are those of understanding data or interpreting information from raw data and calculating fast.

In this article we will look at the basic groundwork you must do before you can even think of doing calculations involving complex divisions within 20 seconds.

One needs to thoroughly learn:

1. Tables up to 30 × 30

2. Squares up to 30

3. Cubes up to 15

4. Square roots up to 10

5. Cube roots up to 5

6. Reciprocal percentage equivalents up to 30

It seems like a very tedious and time consuming task. However, it is not as tough as it seems. Try this -- what is 7 × 8? I bet anyone would have answered 56. Now, what is 14 × 8? Even if I don't know the tables, I can understand it would be twice of 7 × 8, i.e. twice of 56, i.e. 112. Even though one did not know table of 14, one could have arrived at the answer within couple of seconds. Thus, except tables of prime numbers, i.e. 13, 17, 19, 23 and 29 all other tables till 30 can be done in this way if one knows the tables till 12.

Reciprocal Percentage Equivalents

Reciprocal percentage equivalents are the reciprocals of numbers 1 to 30 in percentages, e.g. the reciprocal of 3 is 0.3333 or 33.33%. Reciprocal percentage equivalent of 5 is 20%, of 6 is 16.66% and so on. Reciprocal percentage equivalents are an absolute must for one to crack quantitative section. Not only do they immensely help in division but also in many quant questions. So be sure to learn them by heart. You can also make and use flashcards to help you in memorizing them.

Memorising Reciprocal Percentage Equivalents

Let's see how reciprocals can be memorized. Almost everyone knows that reciprocal of 2 is 50%, of 3 is 33.33% and of 5 is 20%. If reciprocal of 2 is 50%, the reciprocal of 4 is half of 50%…25%? The reciprocal of 8 will be half of 25%...12.5%. Similarly, reciprocal of 16 will be 6.25%. Also if I know reciprocal of 3 as 33.33%, I can also conclude reciprocal of 6, 9 will be 16.66% and 11.11% respectively.

Thus, from 1 to 10, one has to only mug up reciprocal of 7 which is 14.28% (simple two times 7 is 14 and two times 14 is 28…thus 14.28).

If reciprocal of 9 is 11.11, reciprocal of 11 is 09.090909. Reciprocal of 9 is composed of 11s and reciprocal of 11 is composed of 09s.

Reciprocal of 12 will be half of reciprocal of 6, i.e. half of 16.66%, i.e. 8.33%.

Thus, we see that except for prime numbers, we can very easily remember the reciprocals of all others. Thus, effectively we need to mug up reciprocals of only 7, 13, 17, 19, 23 and 29.

Some other numbers that can be remembered easily and the methods are:

Reciprocal of 20 is 5%. Reciprocal of 21 is 4.76% and of 19 is 5.26%. Thus, we can easily remember the reciprocals of 19, 20, 21 as 5.25%, 5, 4.75% respectively.

Reciprocal of 29 is 3.45% (i.e. 345 in order) and reciprocal of 23 is 4.35% (same digits but order is different)

Reciprocal of 22 is half of 09.0909%, i.e. 4.545454%, i.e. consists of 4s and 5s.

Reciprocal of 18 is half of 11.1111%, i.e. 5.55555%, i.e. consists of only 5s.

Thus, the work may seem to be a huge task, but if we use a smart approach, it is hardly anything. And compare it with the time it can save and the confidence it leads to. . . if any calculation has 9 in the denominator, I know for sure the decimal part will be only 0909. . . or 1818… or 2727… or 3636…, e.g. 84/9 will be 9.272727, and can be found out in a jiffy

One can also calculate any fraction of the type (n-1)/n (n <= 30) within two seconds if one knows the reciprocal percentage equivalent. e.g. 11/12 is nothing but 1 – 1/12, i.e. the complement of 0.08333 which is 0.91666. Similarly, if I know 1/23 is 0.0435, 22/23 will be 0.9565.

FactorisationFactorisation is a process which goes a long way in reducing the calculations required. Factorisation in its basic sense has been used by many of us, e.g. if we want to find 17 × 21, we would do 17 × 20 + 17, i.e. 357. We have factorised 21 as 20 + 1.

Let's see how we can use this for even more tougher problems.

Let's say we want to find 14.25% of 3267.What will 10% of 3267 be… 326.7And 1%… obviously 32.67. Then what would 4% be… 128 for 32 and 2.68 for 0.67..., i.e. 130.6Thus, 14% will be 326.7 + 130.6, i.e. 457.3If I want an even more accurate answer, if 1% is 32.6, then 0.25% will be 1/4 of 32.6, i.e 8.1514.25% of 3267 will be 465.4

The best part of this method is that I have the liberty of deciding how accurate an answer do I require. Thus, if alternatives are wide apart, I may stop the process in-between.

Knowing reciprocal percentage equivalent, I should have thought of an even better factorisation as 14.28% – 0.03% and since 14.28% is nothing but 1/7, the answer can directly be found by dividing 3267 by 7, i.e. 466.7

Factorisation can also be used in division. If I have to find 1465/320, I would write it as (1280 + 185)/320 which is nothing but 4 + (160 + 25)/320, which is 4.5 + 25/320.So my answer will be slightly more than 4.5 and less than 4.6

Let's see another example where I can save a lot of calculations.

If I have to find 4835/7280 is of what percentage and the alternatives to choose from, area. 59.6%b. 63.8%c. 66.4%d. 71.4%

Just focus on denominator. 10% of the denominator would be 728. Thus, the answer will be definitely less than 70% (since 7* 10% i.e. 70% of the denominator will be more than 4900 i.e. more than the numerator) and also answer will be more than 60% (since 6*10% will be around 4360). 2/3 of 7280 will be 4860. Since the numerator is less than 4860, the answer choice has to be less than 2/3 or 66.66% and hence (c) is the obvious choice.

For making use of approximations, one must make sure that he treats the alternatives as also a

part of the questions. Thus, one must consciously use the process of elimination immediately after finishing reading the questions, this will force the person to have a look at the alternatives.

Vedic Mathematics

Multiplying by base method:This is applicable to multiplications where the numbers are close to a base like 10, 100, 1000 or so on. Let's take an example: 105 × 107

Here the base is 100 and the 'surplus' is 5 and 7 for the two numbers. The answer will be found in two parts — the right-hand should have only two digits (because base is 100) and will be the product of the surpluses. Thus, the right-hand part will be 5 × 7, i.e. 35. The left-hand part will be one multiplicand plus the surplus of the other multiplicand. The left part of the answer in this case will be 105 + 7 or for that matter 107 + 5 i.e. 112. The answer is 11235.

There can also be a carry-over from the right-hand part, e.g. 112 × 113. The right part will be 12 × 13, i.e. 156. But the right part should have only 2 digits. Thus, 1 will be carried over to the left part and the right part will be only 56. The left part will be 112 + 13 + 1 (the carry-over), i.e. 126. The answer will be 12656.

For 102 × 104 the answer will be 10608. Please note the right part will be 08 and not simply 8.

Can we use it for 92 × 97? Yes. In this case, the left part will be (–8) × (–3), i.e positive 24. The left part will be 92 + (–3) or 97 + (–8), i.e. 89. The answer is 8924.

How about 96 × 108? The right part will now be (–4) × 8, i.e. –32. To take care of the negative we will borrow 1 from the left part, which is equivalent to borrowing 100 (because we are borrowing from the hundred's digit of the answer). Thus, the right part will be 100 – 32 = 68. The left part will be 96 + 8 – 1 (the borrowed one) or 108 + (–4) – 1, i.e. 103. The answer will be 10368.

The same process can be done with other bases as well.

Corrolaries to base multiplication: Base multiplication can be used with other bases as well, like 20, 30 , 50 , 200, 500, etc.

24 × 28Using base as 20 (which is twice of 10) do calculations as if you would do with base 10. Only thing to keep in mind is double the left-hand part (because 20 is twice of 10) before adding the carry-over if any. The right part will be product of surpluses from 10, i.e 4 × 8 = 32. But as per calculations for base 10, right-hand part will have only one digit. So 3 will be carried over. The left part will be twice of 24 + 8, i.e. 64 add to which the carry over 3. Thus, the answer is 672.

Squaring of numbers ending with 5:Multiply the number formed after ignoring the 5 in unit's place with the next integer to it. This is the left part of answer. Append a 25 as the right part, e.g. 652 is 6 × 7 with 25 appended to it, i.e. 4225; 1152 is 11 × 12 with 25 appended at the end, i.e. 13225.

Squaring of other numbers: This method assumes that you are through with all squares of numbers up to 30.

A. If a number is in-between 30 to 80: Find out the 'surplus' (or for that matter the 'deficit') with 50 as base. The right part of the answer is the square of this surplus, just remember that the right part will have only 2 digits, all else will be carried over to left part. The left part is 25 + Surplus + Carry-over (if any)572 is 25 + 7 // 72, i.e. 3249382 is 25 + (–12) // (–12)2 i.e. 1444. In this case there is a carry-over of 1 from right part to left

part.

B. If a number is in-between 70 to 130: The right-hand part of the answer is same as above with the surplus being calculated with respect to 100. The left-hand part of the answer is the given number itself + Surplus1132 is 113 + 13 // 132 i.e. 12769.742 is 74 + (–26) // (–26)2 i.e. 48 // 676, i.e 5476

Cubing of numbers close to 100, 1000, etc.:In this case, our answer will be in three parts. The right-most and the next part will have two digits each if base is 100 and three digits if base is 1000, as usual, the rest will be carried over if needed. The answer in parts will be: The number + Twice the surplus // Thrice the surplus square // Surplus cube.Please note that surplus can also be a deficit, all we have to do is take care of the sign.

1053 is 105 + 10 // 3 × 25 // 53, i.e. 11576259983: This is a case with base 1000 and deficit 2. Thus, the answer is 998 + (–4) // 3 × (–2)2 // (–2)3 i.e. 994 // 012 // (–008) i.e. 994011992

Multiplication: a. When the sum of the digits of unit's place is 10

b. When the rest of the digits are same

In this case, the right part of the answer is the product of unit's place (no restriction on number of digits). The left part is the remaining number (after eliminating the unit's digit) multiplied to the next integer, e.g. 13 × 17 is 1 × 2 // 3 × 7, i.e. 221. Another example 124 × 126... The answer is 12 × 13 // 4 × 6, i.e. 15624.

Multiplication when one multiplicand consists of only 9s:Let's do this with an example 3425 × 999. Since there are three 9s in the multiplicand, ignore three rightmost digits and take the remaining integer, in this case 3. Add 1 to this integer (always 1) to get 4. Subtract this 4 from the whole multiplicand, i.e. 3425 to get 3421. This is the left part of the answer.To get the right part of the answer, take the complement of the three righmost digits of the multiplicand i.e. complement of 425 is 575. Thus, the answer is 3421575.Another example 3236437 × 99999. There are five 9s. Ignoring five digits from the right we get 32, adding 1 we get 33, subtracting this from the whole multiplicand we get 3236404. The right part is complement of 36437, i.e. 63563. So the answer is 323640463563.One more example, 456 × 999999. There are six 9s. So ignoring six digits from right we get 0, adding 1 we get 1, subtracting from multiplicand we get 455 as left part. Right part is the complement of 000456, i.e. 999544. So the answer is 455999544.