A) 8, 12, 16
B) 16, 24, 32
C) 12, 18, 24
D) None of these
Correct Answer: B
Solution :
| [b] Let three numbers are 2x, 3x and 4x, respectively. Then, \[4{{x}^{2}}+9{{x}^{2}}+16{{x}^{2}}=1856\] \[\Rightarrow \]\[29{{x}^{2}}=1856\] \[\Rightarrow \] \[{{x}^{2}}=64\]\[\Rightarrow \]\[x=8\] Thus, required numbers are 16, 24 and 32. |
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