A) 8 and 16
B) 16 and 32
C) 8 and 32
D) 16 and 16
Correct Answer: C
Solution :
Mean proportional \[=\sqrt{xy}=16\] \[\Rightarrow xy=16\times 16\] ? (i) Third proportional \[=\frac{{{y}^{2}}}{x}=128\] \[\Rightarrow \]\[\frac{{{y}^{2}}}{\frac{16\times 16}{y}}=128\] \[\Rightarrow \]\[\frac{{{y}^{3}}}{16\times 16}=128\] \[\Rightarrow \]\[{{y}^{3}}=16\times 16\times 128\] \[\therefore \] \[y=\sqrt[3]{16\times 16\times 16\times 8}\] \[=16\times 2=32\] \[\therefore \]\[xy=16\times 16\] \[\Rightarrow x=\frac{16\times 16}{y}\] \[=\frac{16\times 16}{32}=8\]
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