Question

Divide 243 into three 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 1

let the parts be x,y,z
x/2=y/3=z/4
x:y=2:3,y:z=3:4
x:y:z=2:3:4
x=2×243/9=54
y=3×27=81
z=4×27=108

Answer 2

let the numbers be x, y , z

So 1/2 x = 1/3 y = 1/4 z = k

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

Also,

x+y+z = 243

2k+3k+4k = 243

9k = 243

k = 27

So numbers are x = 54 , y = 81 , z = 108

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or

Let x be the number in the first part.

Let y be the number in the second part.

Let z be the number in the third part.

We know that

x + y + z = 243

Then “half of the first part, a third of the second part, and a fourth of the third part are all equal” means

x/2 = y/3 = z/4

Let's look at just one variable. We know that

x = (2/3)y

And

z = (4/3)y

So

(2/3)y + y + (4/3)y = 243

3y = 243

y = 81

Then

x = (2/3)(81)

x = 54

And

z = (4/3)(81)

z = 108

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or

Hence, 243 is divided into 54, 81 and 108 as the three parts respectively. Therefore, 243 is divided into three parts such that the half of the first part which is 54, one-third of the second part which is 81 and one-fourth of the third part which is 108 are all equal.

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