9
Q:

# Find the logarithm of 144 to the base $\inline 2\sqrt{3}$ :

 A) 2 B) 4 C) 8 D) None of these

Explanation:

$\inline&space;\log_{2\sqrt{3}}144=&space;\log_{2\sqrt{3}}\left&space;(&space;2\sqrt{3}&space;\right&space;)^4$ = 4

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

Explanation:

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

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

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

7 503
Q:

For $\inline \fn_jvn x\epsilon N, x>1$$\inline \fn_jvn P = \log_{x}(X+1)$ and  $\inline \fn_jvn Q = \log_{(X+1)}(X+2)$, then which one of the following is correct?

 A) P < Q B) P = Q C) P > Q D) can't be determined

Explanation:

$\inline \fn_jvn \frac{k}{l}>\frac{k+1}{l+1}$ for (k,l) > 0 and  k > l

Let     k = x+1    and   l = x

$\inline \fn_jvn \therefore \frac{x+1}{x}>\frac{(x+1)+1}{(x)+1}$

$\inline \fn_jvn \because (x+1)>x$

$\inline \fn_jvn \therefore \frac{log(x+1)}{logx}>\frac{log(x+2)}{log(x+1)}$

$\inline \fn_jvn \Rightarrow \log_{x}(x+1)>\log_{x+1}(x+2)$

6 473
Q:

The Value of  $\inline \fn_jvn (log\: tan \: 1^{o}+log\: tan\: 2^{o}+....+log\: tan \: 89^{o})$ is

 A) -1 B) 0 C) 1/2 D) 1

Explanation:

$\inline \fn_jvn log\: tan\: 1^{o} + log\: tan \: 2^{o}+.....+log\: tan\: 89^{o}$$\inline \fn_jvn =(log\: tan\: 1^{o}+log\: tan\: 89^{o})+(log\: tan\: 2^{o}+log\: tan\: 88^{o})+.....+ log \: tan\: 45^{o}$

$\inline \fn_jvn =log(tan\: 1^{o}.tan\: 89^{o})+log(tan\: 2^{o}.tan\: 88^{o})+....log\: 1$

$\inline \fn_jvn =log\: 1+log\: 1+....+log\: 1$

$\inline \fn_jvn \left ( \because \: \: tan(90-\Theta )=cot\Theta\: \: and \: \: tan \: 45^{o} =1\right )$

$\inline \fn_jvn =0$

6 515
Q:

What is the number of digits in $\inline 3^{3^{3}}$? Given that log3 = 0.47712?

 A) 12 B) 13 C) 14 D) 15

Explanation:

Let   $\inline x = 3^{3^3}=(3)^{3^3}$

Then ,    $\inline log x = 3^{3} log 3$

= 27 x 0.47712 = 12.88224

Since the characteristic in the resultant value of log x is 12

$\inline \therefore$ The number of digits in x is (12 + 1) = 13

Hence the required number of digits in $\inline 3^{3^3}$ is 13.

4 545
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

Explanation:

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

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

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

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

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