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Calculations : Many of us have the same problem in exams, the calculations, tough and lengthy calculations.

Sometimes we are stuck in a situation where we face problems like 27 × 64 or 342 and we really have no idea whether to do it manually or leave it because of a shortage of time.  So here are some good tricks which make your calculation fast. But as you know, these are not shortcuts, these are few tricks to save your manual time by doing 80% of calculation in your mind in an easy way. So, it requires practice, a lot of practice and nothing but practice.

So, let’s check whether they are interesting and helpful or not.

MULTIPLICATION

We all can do a single digit by single-digit multiplication, a single digit by double-digit multiplication and many more.

But the problem comes when we get two two-digit numbers. Sometimes it becomes very lengthy and confusing when the numbers are like 87 and 63. So, let’s take some easy numbers like 21 & 47 and try to multiply them.

e.g. 1. 21 × 47 ?

Ans- We know that 21 multiply by 47 means we add 47 up to 21 times or we add 21 up to 47 times. Since 21 is very close to the number ending with 0 (i.e. 20), we can simply multiply 47 with 20 (or 2 and then add 0 to the right of rightmost digit) and then add 47 to it. i.e. 47 × 20 = 940, and then,

940 + 47 = 987. So, 987 is the answer.

Yeah, I know what you are thinking, that if we have to do the same ‘first multiplication and then addition’ work then we can do it simply. Also, it has limitations that the number should be near the multiples of 10.

So, here is the trick by which you can multiply two-digit numbers in one line by doing most of the calculation in mind. There are three simple steps.

(I-step) multiply the unit digits and write the unit digit of answer (say d) (carry the tens digit (if any))

(II-step) Cross multiply the digits (i.e. unit digit of first no. with tens digit of second number and vice versa) and add them (add the carried quantity too (if any)) and write the unit digit of the answer (say c) and carry the remaining number (if any)

[Note: Since these are just two-digit numbers to add, it can be done in our mind, there is no need to write them]

(III-step) Multiply the tens digits and add the carried quantity (if any) and write the answer (say ab)

Let’s take the above example and see how it works.

(Explanation)

(I) unit digits, 1 × 7 = 7

(II) cross multiplication, 7 × 2 = 14 & 4 × 1 = 4 and 14 + 4 = 18 (so 8, and 1 carry)

(III) tens digits, 4 × 2 = 8 and 8 + 1 = 9

So, 987 is the ans.

We can do it for any number, let’s take 87 and 63

e.g. 2. 87 × 63 ?

Ans-  ____?

(Explanation)

(I) unit digits, 7 × 3 = 21 (so 1, and 2 carry)

(II) cross multiplication, 3 × 8 = 24 & 6 × 7 = 42 and 24 + 42 = 66 and 66 + 2 = 68 (so 8, and 6 carry)

(III) tens digits, 8 × 6 = 48 and 48 + 6 = 54

So, 5481 is the ans.

Like that, we can multiply any number with any number without any lengthy manual work.

SQUARES AND SQUARE ROOTS

There are many ways to find out the square of any number. And the most common way is multiplication. But I will give you some short tricks to find the square of numbers.

Trick I- We can find square of some numbers by knowing the squares of other numbers.

Here is the general formula,

(n + 1)2 = n2 + 2n + 1

We can write it as,

(n + 1)2 = n2 + n + (n+1)

But how can we use this? Here, we will mostly talk about square of numbers from 1 to 100.

Also, I am just assuming that we all know the squares of 10, 20, 30, …, 100.

Now, firstly let’s just learn how to find the square of the numbers ended with 5.

It’s just simple. We take the tens digit number (or whatever on the left side of 5. It may be two digits or 3 digits or many more) and multiply it by its next consecutive integer and then write 25 to the rightmost of the answer. And that will be your final answer.

e.g. 3.  352 ?

Ans- Here the tens digit number is 3. So, 3 × 4 = 12. So the answer is, 1225.

e.g. 4.  652 ?

Ans- Here the number is 6. So, 6 × 7 = 42. So the answer is, 4225.

e.g. 5.  1252  ?

Ans- Here the number is 12. So, 12 × 13 = 156. So the answer is 15625.

e.g. 6.  1652  ?

Ans- Here the number is 16. So, 16 × 17 = 272. So the answer is 27225.

So by this, we can find the square of any number ended with 5.

Now, we can apply the general formula for any square. Let’s take some examples.

e.g. 7.  362  ?

Ans- According to the formula,

362 = 352 + 35 + 36 = 1225 + 71 = 1296.

e.g. 8.  412  ?

Ans- 412 = 402 + 40 + 41 = 1600 + 81 = 1681.

e.g. 9.  772  ?

Ans- 772 = 752 + (75 + 76) + (76 + 77) = 5929.

e.g. 10.  792  ?

Ans- Here, we can use another general formula of algebra,

a2 – b2 = (a + b) (a – b)

For a and b consecutive integers, a – b is always 1.

So, we can say that, the difference between the squares of two consecutive integers is the sum of the two integers.

So, 802 – 792 = 80 + 79

Which means, 792 = 802 – 80 – 79 = 6400 – 159 = 6241.

e.g. 11.  442  ?

Ans- 442 = 452 – 45 – 44 = 1936.

Trick 2- Let’s just write the squares of numbers from 1 to 50 and find if there is something common.

Here, we can see some interesting patterns. After 252, the last two digits of squares are repeating correspondingly (i.e. last two digits of (25 + n)2 = last two digits of (25 – n)2, where n ≤ 25)

It means, last two digits of 442 = last two digits of 62,

Last two digits of 332 = last two digits of 172 (and the rest of the number from 172 will be carried for further calculation), and so on.

So to find the square of any number from 1 to 125, we have to remember the squares from 1 to 25.

Now, let’s split this into two parts. In the first part, we will talk about squares from 25 to 75. In the second part we will talk about squares from 75 to 125.

[Note: The pattern of last two digits for the next fifty numbers and for further next fifty numbers and so on will be same as the first fifty numbers]

Part (1)- Square of numbers from 25 to 75.

For now, we know the last two digits of the number. So, for the first two digits (or three digits), we subtract the main number from 50 and then subtract the outcome from 25 (and also add the remaining number which was carried from the square). It seems a little bit complicated. Let’s see some examples.

e.g. 12.  342  ?

Ans- The last two digits will be the same as the last two digits of 162. i.e. 56. (∵ 34 – 25 = 25 – 16)

But 162 = 256, so 2 will be carry from here

For the first two digits, 50 – 34 = 16. So, 25 – 16 = 9 and then we add that carried number, i.e. 2

So, 9 + 2 = 11.

e.g. 13.  442  ?

Ans- The last two digits will be the same as the last two digits of 62. i.e. 36. (∵ 44 – 25 = 25 – 6)

[Note: Since 62 = 36 which is two-digit number, so here is nothing to carry]

For the first two digits, 50 – 44 = 6. So, 25 – 6 = 19.

e.g. 14.  572  ?

Ans- The last two digits will be the same as the last two digits of 72. i.e. 49 (∵ after 50, the pattern of last two digits will be same as first 50 numbers)

For the first two digits, 50 – 57 = -7. So, 25 – (-7) = 32.

e.g. 15.  732  ?

Ans- The last two digits will be same as the last two digits of 232. i.e. 29

But 232 = 529, so 5 will be carry.

For the first two digits, 50 – 73 = -23. So, 25 – (-23) = 48,

Now, 48 + 5 = 53.

Try some examples by yourself and you’ll find it easy too.

Part (2)- Square of numbers from 75 to 125.

For now, we know the last two digits of the number. So, for the first two digits (or three digits), we subtract the main number from 100 and then subtract the outcome from the main number (and also add the remaining number which was carried from the square). In other words, we subtract the main number from 100 and then subtract double of the outcome from 100 (and add the carried quantity).

Let’s see some examples.

e.g. 16.  922  ?

Ans- The last two digits will be the same as the last two digits of 422. i.e. last two digits of 82. (∵ 42 – 25 = 25 – 8). i.e. 64

For the first two digits, 100 – 92 = 8. So, 92 – 8 or 100 – (8 × 2) = 84.

e.g. 17.  882  ?

Ans- The last two digits will be same as the last two digits of 382. i.e. the last two digits of 122. i.e. 44.

But 122 = 144. So 1 will be carry,

For first two digits, 100 – 88 = 12. So, 88 – 12 = 76. Also, 76 + 1 = 77.

e.g. 18.  1032  ?

Ans- The last two digits will be same as the last two digits of 32. i.e. 09.

For the first two/three digits, 100 – 103 = -3. So, 103 – (-3) = 106.

e.g. 19.  1172  ?

Ans- The last two digits will be the same as the last two digits of 172. i.e. 89.

But 172 = 289, so 2 will be carry from here,

For first two/three digits, 100 – 117 = -17, So, 117 – (-17) = 134. Also, 134 + 2 = 136.

As I told you earlier, it seems too complicated but it requires practice. After some time, you will also find them easily.

Now, after squares, let’s come to the “Square Root” topic.

Though there are some methods (like factorization, square root method, etc.), but these methods are too lengthy and complex.

Here, I can tell you how to find the “SQUARE ROOT OF A PERFECT SQUARE” in a very short time. Remember, this trick will only work when the number is a perfect square. So, before applying this trick, be sure that the number is really a perfect square.

So, let’s take some examples and find out the way.

But before that, just remember the pattern. The last digit of square of 1 and 9 is 1.

The last digit of square of 2 and 8 is 4.

The last digit of square of 3 and 7 is 9.

The last digit of square of 4 and 6 is 6.

The last digit of square of 5 is 5.

The last digit of square of 0 is 00.

So, any number ended with the digit 2, 3, 7, 8 and an odd number of 0s will never be a perfect square.

e.g. 20.  Find the square root of 9409.

Ans- (Remember again, we are finding the square root of those numbers, which are perfect squares. Don’t even try this trick where you are not sure whether this is a perfect square or not. We will discuss the properties of perfect squares but in “Number System”)

We can roughly go through a square of any two numbers, one is greater than this and one is less than this. Let’s take 50 and 100. Now, 1002 = 10000. And 9409 seems much closer to 10000 than 2500 (i.e. 502). So let’s take 90 (we have to minimize the domain to minimize the range). 902 = 8100.

Here, 8100 < 9409 < 10000. But it is closer to 10000 than 8100. So our answer lies between 95 and 100. And there is only one number 97, whose square’s the last digit will be 9.

e.g. 21. Find the square root of 17689.

Ans- Square greater than 18000 will be 19600 which is 1402. And 1302 = 16900. So the answer lies between 130 and 140. But 17689 seems close to 16900. So our answer lies between 130 and 135. And 133 is the only number which can give us 9 as the last digit by squaring itself.

e.g. 22. Find the square root of 11025.

Ans- It lies between 10000 (i.e. 1002) and 12100 (i.e. 1102). Since it has the last digit 5.

Here is another trick. How to find “SQUARE ROOTS OF NON-PERFECT SQUARES” in a very short time. Though, this trick will not give you the exact answer square roots of some numbers are irrational (e.g. √2), so we can’t write them and hence, no trick can give you the exact answer.

But, this trick will give you the approximate answer which is too close to the actual answer. And, also, you can find the square root of any number in just 4 or 5 seconds. Hence, learning this trick is worth it.

For this trick, we have to learn a small formula.

i.e.

So, the trick is you have to break the number into a sum of two numbers or difference of two numbers in which a bigger number should be a perfect square. And then apply this formula.

e.g. 23. Find the approx. value of √17.

Ans- We can write 17 = 16 + 1. We have done it because 16 is a perfect square. Now, just put 16 in place of x and 1 in place of y.

e.g. 24. Find the approx. value of √32.

Ans- We can write 32 = 36 – 4. Now, again apply the formula,

We can verify this formula by both sides.

e.g. 25. Find the approx. value of √111.

Ans- We can write 111 = 100 + 11. Now, apply the formula,

Also,  We can write 111 = 121 – 10. Now, apply the formula,

So, as you can see, there is a very small change in the approx. value and actual value. This change is not big enough to provide some difficulty in your final answer or give you a whole another answer. Also, this is a time-saving trick.

More on MULTIPLICATION-

Let’s talk about the multiplication of two three-digit numbers. See, the reason to put this calculation in the expert section is that it is a little bit difficult to digest and also, there are very rare chances that the multiplication of two three-digit numbers will come in the exam.

Now, when we see numbers like 367 and 824, the first thing that comes to mind is that can we multiply these numbers in one line? So the answer is, yes we can. Let’s check how.

It requires 5 simple (not really) steps.

The explanation of these 5 steps is the same as the explanation of two-digit multiplication. Let’s take an example and try to understand.

e.g. 23.  367 × 824 ?

Ans-

(Explanation)

(I) 7 × 4 = 28 (so 8 and 2 carry)

(II) 6 × 4 = 24 & 7 × 2 = 14 and 24 + 14 = 38 and 38 + 2 = 40 (so 0, and 4 carry)

(III) 3 × 4 = 12 & 7 × 8 = 56 & 6 × 2 = 12. So, 12 + 56 + 12 = 80 and 80 + 4 = 84 (so 4, and 8 carry)

(IV) 3 × 2 = 6 & 6 × 8 = 48 and 6 + 48 = 54 and 54 + 8 = 62 (so 2, and 6 carry)

(V) 3 × 8 = 24 and 24 + 6 = 30