MCQ
If ${\log _{12}}27 = a,$ then ${\log _6}16 = $
- A$2.{{3 - a} \over {3 + a}}$
- B$3.{{3 - a} \over {3 + a}}$
- ✓$4.{{3 - a} \over {3 + a}}$
- DNone of these
${\log _6}16 = {{\log 16} \over {\log 6}} = {{4\log 2} \over {\log 2 + \log 3}}$
$ = {{4\log 2} \over {\log 2 + {{2a\log 2} \over {3 - a}}}} = {{4(3 - a)} \over {3 - a + 2a}} = 4.{{3 - a} \over {3 + a}}$.
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$1.$ The probability of the drawn ball from $U_2$ being white is
$(A)$ $\frac{13}{30}$ $(B)$ $\frac{23}{30}$ $(C)$ $\frac{19}{30}$ $(D)$ $\frac{11}{30}$
$2.$ Given that the drawn ball from $U_2$ is white, the probability that head appeared on the coin is
$(A)$ $\frac{17}{23}$ $(B)$ $\frac{11}{23}$ $(C)$ $\frac{15}{23}$ $(D)$ $\frac{12}{23}$
Give the answer question $1$ and $2.$