Question

Sixth-five percent of a number is 21 less than four-fifth of that number. What is the number?

Answer 1

Hint:
Here, we need to find the number such that \[65\% \] of it is 21 less than four-fifth of the number. We will assume the number to be \[x\]. We will apply the given operations on \[x\] to form a linear equation in terms of \[x\]. Then, we will solve this equation to find the value of \[x\] and hence, find the required number.

Complete step by step solution:
Let the number be \[x\].
We need to apply the given operations on the number \[x\] to obtain a linear equation in terms of \[x\].
First, we will find \[65\% \] of the number.
\[65\% {\rm{ of }}x = \dfrac{{65}}{{100}} \times x = \dfrac{{65x}}{{100}}\]
Simplifying the expression, we get
\[65\% {\rm{ of }}x = \dfrac{{13x}}{{20}}\]
Now, we will find the four-fifth of the number.
\[{\left( {\dfrac{4}{5}} \right)^{{\rm{th}}}}{\rm{of }}x = \dfrac{4}{5} \times x = \dfrac{{4x}}{5}\]
Now, it is given that \[65\% \] of the number is 21 less than \[{\left( {\dfrac{4}{5}} \right)^{{\rm{th}}}}\]of that number.
Thus, we get
\[65\% {\rm{ of }}x = \left[ {{{\left( {\dfrac{4}{5}} \right)}^{{\rm{th}}}}{\rm{of }}x} \right] - 21\]
Substituting \[65\% {\rm{ of }}x = \dfrac{{13x}}{{20}}\] and \[{\left( {\dfrac{4}{5}} \right)^{{\rm{th}}}}{\rm{of }}x = \dfrac{{4x}}{5}\], we get the equation
\[ \Rightarrow \dfrac{{13x}}{{20}} = \dfrac{{4x}}{5} - 21\]
We will solve this equation to get the value of \[x\] and hence, find the number.
Rewriting the equation, we get
\[ \Rightarrow \dfrac{{4x}}{5} - \dfrac{{13x}}{{20}} = 21\]
The L.C.M. of 5 and 20 is 20.
Taking the L.C.M., we get
\[ \Rightarrow \dfrac{{16x - 13x}}{{20}} = 21\]
Subtracting the terms, we get
\[ \Rightarrow \dfrac{{3x}}{{20}} = 21\]
Multiply both sides by 20, we get
\[\begin{array}{l} \Rightarrow \dfrac{{3x}}{{20}} \times 20 = 21 \times 20\\ \Rightarrow 3x = 420\end{array}\]
Dividing both sides by 3, we get
\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{420}}{3}\]
Thus, we get the value of \[x\] as
\[ \Rightarrow x = 140\]
\[\therefore \] We get the number as 140.
We can verify our answer by applying the stated operations on the number 140.
\[65\% {\rm{ of 140}} = \dfrac{{65}}{{100}} \times 140 = \dfrac{{9100}}{{100}} = 91\]
\[{\left( {\dfrac{4}{5}} \right)^{{\rm{th}}}}{\rm{of 140}} = \dfrac{4}{5} \times 140 = 4 \times 28 = 112\]
We can observe that \[91 = 112 - 21\].

Thus, the \[65\% \] of 140 is 21 less than \[{\left( {\dfrac{4}{5}} \right)^{{\rm{th}}}}\]of 140.
Hence, we have verified our answer.

Note:
We have used the given information to form a linear equation in one variable in terms of \[x\] in the solution. A linear equation in one variable is an equation that can be written in the form \[ax + b = 0\], where \[a\] is not equal to 0, and \[a\] and \[b\] are real numbers. A linear equation of the form \[ax + b = 0\] has only one solution.