Question

divide 243 into three parts such that half of the first part, one-third of the second part and one-fpurth of the third part all are equal

Answer 1

Let the three parts be x, y and z,

x+y+z = 243

x/2 = y/3 = z/4= k, say.

So x = 2k, y = 3k, z = 4k.

Now 2k+3k+4k=243, or

9k = 243, or

k = 243/9 = 27.
x = 2k = 54
y = 3k = 81
z = 4k = 108
Thus
54 + 81 + 108 = 243