12
Q:

# Find value of $\inline \frac{log\sqrt{27}+ log 8+log\sqrt{1000}}{log 120}$

 A) 1/2 B) 3/2 C) 2 D) 2/3

Answer:   B) 3/2

Explanation:

$\inline =\frac{log(27)^{\frac{1}{2}}+log2^{3}+log(10^3)^{\frac{1}{2}}}{log(3\times 2^{2}\times 10)}$

$\inline =\frac{\frac{1}{2}log3^{3}+3log2+\frac{1}{2}log(10)^{3}}{log3+log2^{2}+log10}$

$\inline =\frac{\frac{3}{2}log3+\frac{3}{2}.2log2+\frac{3}{2}log(10)}{log3+log2^{2}+log10}$

$\inline =\frac{\frac{3}{2}(log3+2log2+log10}{log3+2log2+log10}$

$\inline =\frac{3}{2}$

Q:

Solve the equation $\inline \fn_jvn \small \left ( \frac{1}{2} \right )^{2x+1} =1$  ?

 A) -1/2 B) 1/2 C) 1 D) -1

Answer & Explanation Answer: A) -1/2

Explanation:

Rewrite equation as $\inline \fn_jvn \small \left ( \frac{1}{2} \right )^{2x+1} = (\frac{1}{2})^{0}$

Leads to 2x + 1 = 0

Solve for x : x = -1/2

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6 808
Q:

If $\inline \fn_jvn \log _{7}2$ = m, then $\inline \fn_jvn \log _{49}28$ is equal to ?

 A) 1/(1+2m) B) (1+2m)/2 C) 2m/(2m+1) D) (2m+1)/2m

Answer & Explanation Answer: B) (1+2m)/2

Explanation:

$\inline \fn_jvn \log _{49}28 = \frac{1}{2\log _{7}(7x4)}$

= $\inline \fn_jvn \frac{1}{2}(1+\log _{7}4)$
= $\inline \fn_jvn \frac{1}{2}+\frac{1}{2}(2\log _{7}2)$
$\inline \fn_jvn \frac{1}{2}+\log _{7}2$
$\inline \fn_jvn \frac{1}{2}+m$

$\inline \fn_jvn \frac{(1+2m)}{2}$.

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11 1072
Q:

If $\fn_jvn \small a^{2}+b^{2}=c^{2}$ , then $\inline \fn_jvn \frac{1}{\log_{c+a}b} + \frac{1}{\log_{c-a}b}= ?$

 A) 1 B) 2 C) 4 D) 8

Answer & Explanation Answer: B) 2

Explanation:

Given $\fn_jvn \small a^{2}+b^{2}=c^{2}$

Now $\inline \fn_jvn \frac{1}{\log _{c+a}b}+\frac{1}{\log _{c-a}b}$ = $\inline \fn_jvn \log _{b}(c+a)+\log _{b}(c+a)$

$\inline \fn_jvn \log _{b}(c^{2}-a^{2})$

$\inline \fn_jvn \log _{b}b^{2}=2\log _{b}b = 1$

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9 1060
Q:

If log 64 = 1.8061, then the value of log 16 will be (approx)?

 A) 1.9048 B) 1.2040 C) 0.9840 D) 1.4521

Answer & Explanation Answer: B) 1.2040

Explanation:

Given that, $\inline \fn_jvn \small \log 64=1.8061$

$\inline \fn_jvn \small i.e. \log 4^{3}=1.8061$

$\inline \fn_jvn \small \Rightarrow 3\log 4=1.8061$

$\inline \fn_jvn \small \Rightarrow \log 4=0.6020$

$\inline \fn_jvn \small \Rightarrow 2\log 4=1.2040$

$\inline \fn_jvn \small \Rightarrow \log 4^{2}=1.2040$

$\inline \fn_jvn \small \Rightarrow \log 16=1.2040(approx)$

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17 2796
Q:

A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.

 A) 100 m B) 125 m C) 150 m D) 175 m

Answer & Explanation Answer: C) 150 m

Explanation:

$\inline \fn_jvn Time = \frac{sum\: of\: length\: of\: the\: two\: train}{Difference\: in\: speeds}$

$\inline \fn_jvn 20 = \frac{(100+x)}{25/2}$

$\inline \fn_jvn \Rightarrow x= 150\: m$

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19 1978