When x is subtracted from the numbers 9, 15 and 27, the remainders are in continued proportion. What is the value of x? [IBPS (PO/MT) CWE 2014] |
A) 8
B) 6
C) 4
D) 5
E) None of these
Correct Answer: E
Solution :
Here, \[9-x:15-x::15-x:27-x\] are in continued proportion. |
\[\therefore \]Product of extremes = Product of means |
\[{{(15-x)}^{2}}=(9-x)(27-x)\] |
\[\Rightarrow \]\[225-30x+{{x}^{2}}=243+{{x}^{2}}-36x\] |
\[\Rightarrow \]\[6x=243-225\]\[\Rightarrow \]\[6x=18\]\[\Rightarrow \]\[x=3\] |
\[\therefore \]Numbers become, \[9-x=9-3=6\] |
\[15-x=15-3=12\] |
and \[27-x=27-3=24\] |
Hence, 6: 12: 24 are in continued proportion on subtracting 3 from 9, 15 and 27. |
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