A) 3 and 6
B) 6 and 12
C) 6 and 6
D) 3 and 12
Correct Answer: D
Solution :
Numbers \[=x\]and y (let) Mean proportional \[=\sqrt{xy}\] Third proportional \[=\frac{{{y}^{2}}}{x}\] \[\therefore \]\[\sqrt{xy}=6\Rightarrow xy=36\] \[\Rightarrow \]\[x=\frac{36}{y}\] ? (i) \[\therefore \] \[\frac{{{y}^{2}}}{x}=48\] \[\Rightarrow \] \[\frac{\frac{{{y}^{2}}36}{36}}{y}=48\] \[\Rightarrow \] \[{{y}^{2}}=36\times 48\] \[\Rightarrow \] \[y=\sqrt[3]{48\times 36}\] \[=\sqrt[3]{8\times 6\times 6\times 6}=2\times 6=12\] \[\therefore \]\[x=\frac{36}{y}=\frac{36}{12}=3\]
You need to login to perform this action.
You will be redirected in
3 sec
