Question:
Divide 243 into 3 parts such that half of the first part,one-third of the second part and one-fourth of the third part are all equal
Answer:
Let the parts are x, y and z
Now, x + y + z = 243 ..............1
Again, given
x/2 = y/3 = z/4
Now, x/2 = y/3
=> 3x = 2y
=> y = 3x/2
Again, x/2 = z/4
=> 4x = 2z
=> z = 4x/2
=> z = 2x
From equation 1, we get
x + 3x/2 + 2x = 243
=> 3x + 3x/2 = 243
=> 9x/2 = 243
=> x/2 = 243/9
=> x/2 = 27
=> x = 54
y = 3x/2 = (3*54)/2 = 3*27 = 81
z = 2x = 2*54 = 108
So, the parts are: 54, 81 and 108
Not what you are looking for? Go ahead and submit the question, we will get back to you.
