Tag: errors and approximations

Questions Related to errors and approximations

The approximate value of $\sqrt[10]{0.999}$ is 

maths the trapezium rule approximation errors and approximations the need for approximation

  1. 0.0998

  2. 0.9998

  3. 0.0999

  4. 0.9999

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Correct Option: A

The positive root of ${x}^{2}-98.8=0$ after first approximation by Newton Raphson method assuming initial approximation to the root is $14$ is

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $9.821$

  2. $9.814$

  3. $9.715$

  4. $9.915$

Show answer
Correct Option: B
Explanation:

Here ${x} _{0}=14, f(x)={x}^{2}-78.8$
and $f'(x)=2x$
$\therefore$ ${x} _{1}={x} _{0}-\cfrac{f({x} _{0}}{f'({x} _{0})}$
$=14-\cfrac { { \left( 14 \right)  }^{ 2 }-\left( 78.8 \right)  }{ 2\times 14 } =9.814$

State the following statement is True or False

According to the Newton-Raphson's method the approximate root of the equation $f(x) = 0$ is $x _{n}$ then be $x _{n} = x _{n + 1} - \dfrac {f(x)}{f'(x _{n})}$.

maths the trapezium rule approximation errors and approximations the need for approximation

  1. True

  2. False

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Correct Option: B
Explanation:
If $x _n$ is the approximate root of $f(x)=0$ then the value of $x _n$ given by Iterative eqn. for Newton-Raphson method is 
    ${ x } _{ n+1 }= { x } _{ n } - \dfrac { f({ x } _{ n }) }{ f' ({ x } _{ n }) } $, where $f'(x _n)$ is derivative of $f$ at $x _n$

$\therefore $ The given statement is FALSE.

The value of $\cdot23454\ E\ 06 +\cdot31063\ E06.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $ \cdot54517\ E\ 06$

  2. $\cdot12057\ E\ 09$

  3. $\cdot12057\ E\ 05$

  4. $\cdot64045\ E\ 09$

Show answer
Correct Option: A
Explanation:
$E06$ can be taken out common 
$= (.23454+.31036)E06$
$ = (.54517)E06 $

The value of $\cdot6235\ E\ 05 +\cdot5781\ E05.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot13056\ E\ 09$

  2. $\cdot12057\ E\ 09$

  3. $\cdot64045\ E\ 05$

  4. $\cdot12057\ E\ 05$

Show answer
Correct Option: C
Explanation:
Taking $E 05$ common, we get
$ = (.06235+0.5781)E05 $
$ = (.64045)E05 $

The value of $\cdot4136\ E\ 05 +\cdot5132\ E07.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot517336\ E\ 07$

  2. $\cdot164045\ E\ 09$

  3. $\cdot12057\ E\ 07$

  4. $\cdot33715\ E\ 01$

Show answer
Correct Option: A
Explanation:
Take out $E07$ common to get 
$= (.004136+.5132)E07$
$ = .517336E 07$

The value of $\cdot3656\ E\ 06 -\cdot7326\ E05.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot12057\ E\ 09$

  2. $\cdot6423\ E\ 05$

  3. $\cdot12057\ E\ 08$

  4. $\cdot29234\ E\ 06$

Show answer
Correct Option: D
Explanation:
Take out $E06$ common to get 
$= (.3456-.07326)E06 $
$ = .29234 E 06$

The value of $\cdot2642\ E\ 05 +\cdot3781\ E05.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot12057\ E\ 05$

  2. $\cdot54517\ E\ 05$

  3. $\cdot6423\ E\ 05$

  4. $\cdot64045\ E\ 05$

Show answer
Correct Option: C
Explanation:

$E05$ can be taken out common. 

Therefore we have 
$= (.2642+.3781)E05 $

$ = .6423E05 $

The value of $\cdot6321\ E\ 08 +\cdot5736\ E08.$ is?

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $\cdot12057\ E\ 05$

  2. $\cdot12057\ E\ 09$

  3. $\cdot64045\ E\ 09$

  4. $\cdot54517\ E\ 09$

Show answer
Correct Option: B
Explanation:

$E08$ can be taken common 

$ = (.6321+.5736)E08 $
$ = (1.2057)E08$
$ = (0.12057)E09$

Find the approximate error in the volume of a cube with edge $x$ cm, when the edge is increased by $2\%$

maths the trapezium rule approximation errors and approximations the need for approximation

  1. $4\%$

  2. $2\%$

  3. $6\%$

  4. $8\%$

Show answer
Correct Option: A
Explanation:

in a mutiplication while multiplying we have to add the percentage errors 


so if the edge of the cube is $x$cm then the volume will be ${ x }^{ 3 }$

but for errors we have to add the percentage 

therefore the total error  percentage now becomes $6\%$

therefore the error is increased by $4\%$