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) 2012] |
A) 8
B) 6
C) 4
D) 5
E) None of these
Correct Answer: E
Solution :
\[\frac{9-x}{15-x}=\frac{15-x}{27-x}\] |
\[\Rightarrow \]\[243-9x-27x+{{x}^{2}}=225-15x-15x+{{x}^{2}}\] |
\[\Rightarrow \]\[6x=243-225=18\]\[\Rightarrow \]\[x=3\] |
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