**Blog-25**

Dear Students,

###
I delivered the lecture on Mathematical Fun. It was delivered in Indian Language “Marathi”. If you understand Marathi, Please click here to see my Lecture on youTube.

One Student of 10

^{th}Standard / 10^{th}grade asked me one question. I would like to share the method of solving the problem & its presentation with you. Please go through it & make the habit of applying the same technique to solve other problems & please share this technique with your friends & relatives. (Please note one thing that “By distributing our knowledge, it will boost & by keeping it only with us, it will reduce”).
The Problem is very simple. Please see the following Problem.

The Volume of a sphere is 3052.08 cubic centimeters, find its radius. (π = 3.14) (Note: Don’t use calculator)

Actually this problem is very simple. But I would like to tell you the simple method of calculations & the proper method of presentation of this problem on paper.

First I would like to discuss about presentation, which will be useful for all the problems.

I would like to give numbers to each step. These numbers are to be written to the left of each step. Secondly the equation numbers are to be given only to the equations & to be written to the right of the equations. You can see the above problem which is solved by using the same technique.

Now we will discuss the basics of the problem.

Here we know that V = 4/3 π r

^{3}
Here we want to find radius so r

^{3 }= 3V/4π. So many students will solve this problem by taking the product of numerator & the product of Denominator & then take the division. I had done this problem with different method of calculations which are very simple and also authentic.
(Note: The description written in the box bracket in BLACK is only to understand the steps & not to be written in the solution of the problem)

4 π r

^{3}
1) We know that, V = ----------

3

3V

So r

^{3}^{ }= ---------- ---------------- (1)
4 π

3 x 3052.08

2) So r

^{3}^{ }= ---------------------
4 x 3.14

[Multiply Numerator & denominator by 100]

3 x 305208

3) r

^{3}^{ }= ---------------------
4 x 314

[Divide Numerator & denominator by 4]

3 x 76302

4) r

^{3}^{ }= ------------------
314

[Here we need to get Right Hand Side as the perfect cube so try to get 9 from 76302 as one 3 is already available in the numerator]

3 x 9 x 8478

5) r

^{3}^{ }= ---------------------
314

[The Number 8478 is even which is divisible by 2. But as tenth Placed digit of 8478 is 7 which is odd & when divided by 2 gives reminder as 1 and the next number for division will be 18 which when divided by 2 gives 9 which is odd & so we don’t have two more 2’s for getting required three 2’s for perfect cube. So divide numerator & denominator by 2]

3 x 9 x 4239

6) r

^{3}^{ }= ---------------------
157

[Here 4239 is odd. The sum of the digits (4+2+3+9 = 18) which is divisible by 3 so take 3 out from 4239]

3 x 9 x 3 x 1413

7) r

^{3}^{ }= --------------------------
157

[Here the sum of the digits (1+4+1+3 = 9) which is divisible by 9 so take 9 out from 1413]

3 x 9 x 3 x 9 x 157

8) r

^{3}^{ }= -----------------------------
157

[Here dividing Numerator & denominator by 157 directly, we get]

9) r

^{3}^{ }= 3 x 9 x 3 x 9
[So we get the direct Answer as]

10) r

^{3}^{ }= 3 x (3 x 3) x 3 x (3 x 3)
r

^{3}^{ }= (3 x 3)^{ 3}^{ }r = 3 x 3

r = 9

^{ }
So the radius of the sphere is 9.

^{ }
So dear Students,

Like this you can also find so many ways of simple calculation methods and apply the same to other problems.