Tag: basic mathematical concepts

Questions Related to basic mathematical concepts

If $x=\log _{ a }{ bc } ,y=\log _{ b }{ ca } ,z=\log _{ c }{ ab } $, then the value of $\dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $ will be

  1. $x+y+z$

  2. $1$

  3. $ab+bc+ca$

  4. $abc$


Correct Option: B
Explanation:

Now, $1+x=\log _{ a }{ a } +\log _{ a }{ bc } =\log _{ a }{ abc } $
$\Rightarrow \dfrac { 1 }{ 1+x } =\log _{ abc }{ a } $
Similarly, $\dfrac { 1 }{ 1+y } =\log _{ abc }{ b } $ and $\dfrac { 1 }{ 1+z } =\log _{ abc }{ c } $
$\therefore \dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $
                    $=\log _{ abc }{ a } +\log _{ abc }{ b } +\log _{ abc }{ c } $
                    $=\log _{ abc }{ abc } =1$

The characteristic of a number having $m$ $(m>1)$ digits is given by,

  1. $m-1$

  2. $m+1$

  3. $m$

  4. None of the above


Correct Option: A
Explanation:

If a number is $N>0$, then $\log _{10}N$ will have two parts, the integral part is known the characteristic and the decimal part is known as mantissa.

$2$ digit belongs from $[10,100]$ where $\log _{10}10=1$ and $\log _{10}100=2$
Similarly, $m$ digit number belongs from $[10^{m-1},10^m]$, where $\log _{10}10^{m-1}=m-1$ and $\log _{10}10^m=m$.
Thus any number between $[10^{m-1},10^m]$ will have $m-1$ as the integral part.
Thus the characteristic of a number having $m$ digits is given by $m-1$.

Calculate $x$, to the nearest tenth: $\log _{12} 640 = x$

  1. $1.7$

  2. $2.6$

  3. $2.8$

  4. $53.3$

  5. $7,680$


Correct Option: B
Explanation:

Given, $x= \log _{ 12 }{ 640 } $

$=\log _{ 12 }{ 144\times 4.44 } $
$=\log _{ 12 }{ 144 } +\log _{ 12 }{ 4.44 } $
$=2+\log _{ 12 }{ 4.44 }$
$= 2.6$

If $\displaystyle { log } _{ 5 }{ log } _{ 5 }{ log } _{ 2 }x=0$, then the value of $x$ is

  1. $32$

  2. $125$

  3. $625$

  4. $25$


Correct Option: A
Explanation:

Given, $\displaystyle { log } _{ 5 }{ log } _{ 5 }{ log } _{ 2 }x=0$
$\displaystyle \Rightarrow \quad { log } _{ 5 }{ log } _{ 2 }x=50$
$\displaystyle \Rightarrow \quad { log } _{ 5 }{ log } _{ 2 }x=1\Rightarrow { log } _{ 2 }x=5$
$\displaystyle \Rightarrow x={ 2 }^{ 5 }=32$

The value of $\log _{10} 0.0006024$ is equal to

  1. $\overline {3}.7979$

  2. $\overline {1}.9779$

  3. $\overline {4}.7799$

  4. $0.7279$


Correct Option: C
Explanation:

$\log _{10}0.0006024$

Characteristics$=-4$
For mantissa ,read from the table $6024.$
Mantissa$=7799$ 
Thus, $\log _{10}0.0006024=$ characteristics of $0.0006024+$ mantissa of $0.0006024$
$=-4+0.7799$
$=\bar4.7799$
Hence, C is the correct option.

Using logarithm table, determine the value of $\log _{10}0.5432$.

  1. $\overline {1}.7350$

  2. $\overline {2}.7350$

  3. $0.7350$

  4. $0.07350$


Correct Option: A
Explanation:

$\log _{10}0.5432$

Characteristics$=-1$
For mantissa ,read from the table $5432.$
Mantissa$=7350$ 
Thus, $\log _{10}0.5432=$ characteristics of $0.5432+$ mantissa of $0.5432$
$=-1+0.7350$
$=\bar1.7350$
Hence, A is the correct option.

Antilog of the number $( -8.654)$ is equal to

  1. $2.18\times 10^{-8}$

  2. $2.18\times 10^{-9}$

  3. $2.218\times 10^{-9}$

  4. $2.218\times 10^{-8}$


Correct Option: C
Explanation:
$(-8.654)=-8+(-0.654)$
$=-8+(-0.654)+1-1$
$=-9+(1-0.654)=-9+0.346$

Actual digits of 0.346 from the log table $=2.218$

$\therefore$ $ anti \log(-8.654)=2.218\times 10^{-9}$

The antilog of $\overline {1}.8840$is equal to

  1. $76.56$

  2. $\overline {1}.7656$

  3. $0.7656$

  4. $7.656$


Correct Option: C
Explanation:

Antilog $\bar1.8840$

Characteristics$=-1$
Value of $0.8840$ from antilog table $=7656$

Now number of zeroes after decimal will be $0.$
Antilog $\bar1.8840=0.7656$
Hence, C is the correct option.

The antilog of the number $2.9586$ is equal to

  1. $909.1$

  2. $90.91$

  3. $9.091$

  4. $0.9091$


Correct Option: A
Explanation:

The number before decimal point is $2,$ so decimal point will be after $3$ digits.

Value of $0.9586$ from antilog table $=9078+12=9090$
Now place a decimal point after $3$ digits of the number from left we get, $909.0$
Antilog $2.9586=909.0$
Hence, A is the correct option.

The antilog of $(1.32)$ is equal to

  1. $20$

  2. $20.98$

  3. $20.89$

  4. $20.79$


Correct Option: C
Explanation:

The number before decimal point is $1,$ so decimal point will be after $2$ digits.

Value of $1.32$ from antilog table $=2089$
Now place a decimal point after $2$ digits of the number from left we get, $20.89$
Antilog $1.32=20.89$
Hence, C is the correct option.