23
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:   B) 1.2040



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

Q:

Solve the equation 122x+1 = 1 ?

A) -1/2 B) 1/2
C) 1 D) -1
 
Answer & Explanation Answer: A) -1/2

Explanation:

Rewrite equation as 122x+1 = 120

 

Leads to 2x + 1 = 0 

 

Solve for x : x = -1/2

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12 1658
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
 
Answer & Explanation Answer: B) (1+2m)/2

Explanation:

log4928 = 12log77×4

 

= 12+122log72
= 12+log72
12 + m
1+2m2.

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18 2405
Q:

If a2+b2 = c2 , then 1logc+ab + 1logc-ab = ?

A) 1 B) 2
C) 4 D) 8
 
Answer & Explanation Answer: B) 2

Explanation:

Given a2 + b2 = c2

 

Now  1logc+ab  + 1logc-ab 

 

logbc+a + logbc-a

 

logbc2-a2

2logbb = 2

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13 1797
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:

Time=Sum of length of the two trainsDifference in speeds

20=(100+x)252 

 X=150m

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24 2754
Q:

For xN, x>1,  and  p=logxx+1q=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
 
Answer & Explanation Answer: C) p > q

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)

 

 

 

logxx+1 >logx+1x+2

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16 1996
Q:

The Value of  logtan10+logtan20++logtan890 is

A) -1 B) 0
C) 1/2 D) 1
 
Answer & Explanation Answer: B) 0

Explanation:

= log tan10+log tan890 + log tan20+ log tan880++log tan450  

 

= log [tan10 × tan890] + log [tan20 × tan880 ] ++log1  

 

 tan(90-θ)=cotθ and tan 450=1  

 

= log 1 + log 1 +.....+log 1 

 

= 0.

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16 1921
Q:

What is the number of digits in 333? Given that log3 = 0.47712?

A) 12 B) 13
C) 14 D) 15
 
Answer & Explanation Answer: B) 13

Explanation:

 Let   Let x=333 333

 

 Then, logx = 33 log3  

 

= 27 x 0.47712 = 12.88224 

 

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

 

The number of digits in x is (12 + 1) = 13 

 

Hence the required number of digits in 333is 13.

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16 2231
Q:

Find value of log27 +log 8 +log1000log 120

A) 1/2 B) 3/2
C) 2 D) 2/3
 
Answer & Explanation Answer: B) 3/2

Explanation:

 = log 33 + log 23+ log 103log10×3×22 

 

 

 

=log33 12+log 23+log 10312log(10×3×22)  

 

 

 

 

 

 

 

=12log 33+3 log 2+12 log103log10+log3+log22  

 

 

 

 

 

 

 

=32log 3 + 2 log 2 + log 10log 3 + 2 log 2 + log 10 = 32  

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13 1846