Question

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

Answer 1

Step-by-step explanation:

a+b+c = 243

Now, according to the question ;

(1/2)a = (1/3)b = (1/4)c

-------------(multiply each by LCM of 2,3,4 = 12)

(12/2)a = (12/3)b = (12/4)c

6a = 4b = 3c

_______________________

a:b = 4/6 = 2:3

b:c = 3:4

a:b:c = 2:3:4

_______________________

Let the common ratio be x, then

2x + 3x + 4x = 243

9x = 243

x = 27

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Hence the three parts are ;

a = 2x = 2×27 = 54

b = 3x = 3×27 = 81

c = 4x = 4×27 = 108

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Answer 2

X + Y + Z = 243

1/2X=1/3Y=1/4Z

So, Y=1/2X*3

And Z=1/2X*4

Substituting those in, we have

X+1/2X*3+1/2X*4=243

X+X+1/2X+2X=243

9/2X=243

X=2/9(243)

X=2(27)=54

Substituting in X to find Y from the equation above

Y=1/2(54)*3=27*3=81

Substituting in X to find Z from the equation above

Z=1/2X*4=27*4=108

54+81+108=243