A) 144, 96, 64
B) 162, 108, 72
C) 180, 120, 80
D) 189, 126, 84
Correct Answer: B
Solution :
| [b] \[\frac{x}{y}=\frac{y}{z}=\frac{3}{2}\]\[\Rightarrow \]\[\frac{y}{z}=\frac{3}{2}\] \[\Rightarrow \] \[y=\frac{3z}{2}\] and \[\frac{x}{y}=\frac{3}{2}\] \[\Rightarrow \] \[x=\frac{3}{2}\,y\] \[=\frac{3}{2}(x)\frac{3}{2}z=\frac{9}{4}z\] According to the question, \[\frac{9}{4}z+\frac{3}{2}z+z=342\] \[\Rightarrow \] \[\frac{(9+6+4)}{4}z=342\] \[\Rightarrow \] \[z=\frac{342\times 4}{19}=72\] \[\therefore \] \[y=\frac{3}{2}z=\frac{3}{2}\times 72=108\] and \[x=\frac{3}{2}y=\frac{3}{2}\times 108=162\] |
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