Question

Divide 243 into 3 parts such that half of the first par, one-third of the second part and one-fourth of the third part are all equal.

Answer 1

Given that half of the first par, one-third of the second part and one-fourth of the third part are all equal.

Let it be k

So first part = 2k
second part = 3k
third part = 4k

according to the question,

2k + 3k + 4k = 243

=> 9k = 243

=> k = 243/9 = 27

All parts are:

first part = 2k = 2×27 = 54

second part = 3k = 3×27 = 81

third part = 4k = 4×27 = 108

Answer 2

$\bf\huge\boxed{\boxed{\boxed{Cybary\:Radhe\:Radhe}}}$

According to Question :-

Half of the first par, one-third of the second part and one-fourth of the third part are all equal.

Let it be m

Here

First = 2m

Second = 3m

Third = 4m

Here

2m + 3m + 4m = 243

=> 9m = 243

=> m = $\bf\huge\frac{243}{9}$

=> m = 27

All parts are:

First part = 2m

= 2 × 27 = 54

Second part = 3m

=> 3 × 27 = 81

Third part = 4m

= 4 × 27

= 108

$\bf\huge\boxed{\boxed{\boxed{Radhe\:Radhe}}}$

$\bf\huge\boxed{\boxed{\boxed{Together\:we\:go\:far}}}$