28
Q:

# if

 A) 1 B) 3 C) 5 D) 10

Explanation:

$log105+log5x+1=log10x+5+1$

$log1055x+1=log1010x+5$

$55x+1=10x+5$

5x+1=2x+10

3x=9

x=3

Q:

It is given that log10 2= 0.301 and log10 3 = 0.477. How many digits are there in $(108)10$?

 A) 19 B) 20 C) 21 D) 22

Explanation:

38 3998
Q:

If 5x - 17 = -x + 7, then x = ?

 A) 4 B) 5 C) 7 D) 9

Explanation:

Given equation  5x-17 = -x+7

Add 1x to each side of the equation

5x-17+x = -x+7+x

6x=24

x=4

17 1994
Q:

Solve the equation  ?

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

Explanation:

Rewrite equation as

Leads to 2x + 1 = 0

Solve for x : x = -1/2

37 5486
Q:

If $log72$ = m, then $log4928$ is equal to ?

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

Explanation:

= $12+122log72$
= $12+log72$

$1+2m2$.

59 10615
Q:

If  , then

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

Explanation:

Given

Now

$logbc2-a2$

29 7573
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, log 64 = 1.8061

i.e $log43=1.8061$

--> 3 log 4 = 1.8061

--> log 4 = 0.6020

--> 2 log 4 = 1.2040

$⇒log42=1.2040$

Therefore, log 16 = 1.2040

49 15058
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:

$20=(100+x)252$

$⇒$ X=150m

43 6694
Q:

For ,  and  $p=logxx+1$$q=logx+1x+2$ then which one of the following is correct?

 A) p < q B) p = q C) p > q D) can't be determined

Explanation:

$kl>k+1l+1$ for (k,l) > 0 and  k > l

Let     k = x+1    and   l = x

Therefore, $x+1x>(x+1)+1(x)+1$

(x + 1) > x

Therefore, $log(x+1)log(x)>log(x+2)log(x+1)$