
The percentage error in the ${11^{th}}$ root of the number 28 is approximately ___________times the percentage error in 28
a.$\dfrac{1}{{28}}$
b.$\dfrac{1}{{11}}$
c.$11$
d.$28$
Answer
578.1k+ views
Hint: First let's take $y = {x^{11}}$ and differentiate it with respect to x and dividing the equation by y . We need to replace $dy$ by $\vartriangle y$ and $dx$ by $\vartriangle x$. And now taking $y = 28$ and $x = \sqrt[{11}]{{28}}$ we get the required solution
Complete step-by-step answer:
Now at first let's suppose $y = {x^{11}}$ ………………(1)
Now let's divide the above equation with respect to x
$
\Rightarrow \dfrac{{dy}}{{dx}} = 11{x^{10}} \\
\Rightarrow dy = 11{x^{10}}dx \\
$
Now we can replace $dy$ by $\vartriangle y$ and $dx$ by $\vartriangle x$ we get
$ \Rightarrow \vartriangle y = 11{x^{10}}\vartriangle x$………..(2)
Let's divide equation (2) by y
$ \Rightarrow \dfrac{{\vartriangle y}}{y} = \dfrac{{11{x^{10}}\vartriangle x}}{y}$
Now substituting $y = {x^{11}}$ from equation (1)
$ \Rightarrow \dfrac{{\vartriangle y}}{y} = \dfrac{{11{x^{10}}\vartriangle x}}{{{x^{11}}}}$
Simplifying we get,
$ \Rightarrow \dfrac{{\vartriangle y}}{y} = \dfrac{{11\vartriangle x}}{x}$…………(3)
Now here we can see that $y = 28$
From this we get $x = \sqrt[{11}]{{28}}$
Substituting this in the (3) equation , we get
$
\Rightarrow \dfrac{{\vartriangle (28)}}{{28}} = \dfrac{{11\vartriangle \left( {\sqrt[{11}]{{28}}} \right)}}{{\sqrt[{11}]{{28}}}} \\
\Rightarrow \dfrac{1}{{11}}\left( {\dfrac{{\vartriangle (28)}}{{28}}} \right) = \dfrac{{\vartriangle \left( {\sqrt[{11}]{{28}}} \right)}}{{\sqrt[{11}]{{28}}}} \\
$
We can see that the percentage error in the ${11^{th}}$root of 28 would approximately $\dfrac{1}{{11}}$ times the error in 28
Therefore the correct option is b.
Note: Alternative method also we can solve:-
We need to find the percentage error in ${11^{th}}$ root with respect to 28.
We know that for a number ${x^n}$ then the percentage error in ${x^n}$ is equal to n times the percentage error in x
From this we can say that , the percentage error of ${n^{th}}$of a number is approximately $\dfrac{1}{n}$times the percentage error in number .
Therefore , ${11^{th}}$ root of 28 = ${28^{\dfrac{1}{{11}}}}$
Hence , here $x = 28,n = \dfrac{1}{{11}}$
Therefore the solution is $\dfrac{1}{{11}}$
Complete step-by-step answer:
Now at first let's suppose $y = {x^{11}}$ ………………(1)
Now let's divide the above equation with respect to x
$
\Rightarrow \dfrac{{dy}}{{dx}} = 11{x^{10}} \\
\Rightarrow dy = 11{x^{10}}dx \\
$
Now we can replace $dy$ by $\vartriangle y$ and $dx$ by $\vartriangle x$ we get
$ \Rightarrow \vartriangle y = 11{x^{10}}\vartriangle x$………..(2)
Let's divide equation (2) by y
$ \Rightarrow \dfrac{{\vartriangle y}}{y} = \dfrac{{11{x^{10}}\vartriangle x}}{y}$
Now substituting $y = {x^{11}}$ from equation (1)
$ \Rightarrow \dfrac{{\vartriangle y}}{y} = \dfrac{{11{x^{10}}\vartriangle x}}{{{x^{11}}}}$
Simplifying we get,
$ \Rightarrow \dfrac{{\vartriangle y}}{y} = \dfrac{{11\vartriangle x}}{x}$…………(3)
Now here we can see that $y = 28$
From this we get $x = \sqrt[{11}]{{28}}$
Substituting this in the (3) equation , we get
$
\Rightarrow \dfrac{{\vartriangle (28)}}{{28}} = \dfrac{{11\vartriangle \left( {\sqrt[{11}]{{28}}} \right)}}{{\sqrt[{11}]{{28}}}} \\
\Rightarrow \dfrac{1}{{11}}\left( {\dfrac{{\vartriangle (28)}}{{28}}} \right) = \dfrac{{\vartriangle \left( {\sqrt[{11}]{{28}}} \right)}}{{\sqrt[{11}]{{28}}}} \\
$
We can see that the percentage error in the ${11^{th}}$root of 28 would approximately $\dfrac{1}{{11}}$ times the error in 28
Therefore the correct option is b.
Note: Alternative method also we can solve:-
We need to find the percentage error in ${11^{th}}$ root with respect to 28.
We know that for a number ${x^n}$ then the percentage error in ${x^n}$ is equal to n times the percentage error in x
From this we can say that , the percentage error of ${n^{th}}$of a number is approximately $\dfrac{1}{n}$times the percentage error in number .
Therefore , ${11^{th}}$ root of 28 = ${28^{\dfrac{1}{{11}}}}$
Hence , here $x = 28,n = \dfrac{1}{{11}}$
Therefore the solution is $\dfrac{1}{{11}}$
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