3
Q:

# If a, b, c be the $\inline p^{th},q^{th},r^{th}$terms of a GP then the value of (q-r) log a + (r-p) log b + (p-q) log c is :

 A) 0 B) 1 C) -1 D) pqr

Explanation:

Let m be the first term and k be the common ratio of GP, then

$\inline&space;a=mk^{p-1}$

$\inline&space;b=mk^{q-1}$

$\inline&space;c=mk^{r-1}$

$\therefore$ $\inline&space;(q-r)\log&space;a&space;+&space;(r-p)&space;\log&space;b&space;+&space;(p-q)&space;\log&space;c&space;=&space;\log&space;\left&space;[&space;mk^{(p-1)}&space;\right&space;]^{q-r}+\log&space;\left&space;[&space;mk^{(q-1)}&space;\right&space;]^{r-p}+\log&space;\left&space;[&space;mk^{(r-1)}&space;\right&space;]^{p-q}$

= $\inline&space;\log&space;(m^{q-r+r-p+p-q})(k^{p-1})^{q-r}(k^{q-1})^{r-p}(k^{r-1})^{p-q}$

= $\inline&space;\log&space;m^{0}k^{0}=\log&space;1=0$

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

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

6 585
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

Explanation:

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

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

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

10 800
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

Explanation:

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

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

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

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

8 811
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

Explanation:

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

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

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

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

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

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

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

15 2279
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

$\inline \fn_jvn Time = \frac{sum\: of\: length\: of\: the\: two\: train}{Difference\: in\: speeds}$
$\inline \fn_jvn 20 = \frac{(100+x)}{25/2}$
$\inline \fn_jvn \Rightarrow x= 150\: m$