Answer
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Hint: We first assume the salary of Ricky Ponting as a variable. We have the salary of Sachin Tendulkar is $20\%$ more than that of Ricky Ponting. Using the formula, we find the salary of Sachin Tendulkar. Then we find the difference between their salaries. In the end, we find the percentage of Ricky’s salary less than that of Sachin’s.
Complete step-by-step solution
The salary of Sachin Tendulkar is $20\%$ more than that of Ricky Ponting.
Let’s assume the salary of Ricky Ponting is x.
We can find the salary of Sachin Tendulkar which is $20\%$ more than x.
The salary of Sachin Tendulkar is $x\left( 1+\dfrac{20}{100} \right)=\dfrac{6x}{5}$.
So, the difference between their salaries is $\dfrac{6x}{5}-x=\dfrac{6x-5x}{5}=\dfrac{x}{5}$.
Now we need to find the percentage of Ricky’s salary less than that of Sachin’s.
So, we need to find the percentage of $\dfrac{x}{5}$ on $\dfrac{6x}{5}$.
So, the percentage is $\dfrac{\dfrac{x}{5}}{\dfrac{6x}{5}}\times 100$.
We solve the fraction and get $\dfrac{\dfrac{x}{5}}{\dfrac{6x}{5}}\times 100=\dfrac{x\times 5\times 100}{5\times 6x}=\dfrac{50}{3}=16.67$.
The nearest integer is $17\%.$
Therefore, Ricky’s salary is $17\%$ less than that of Sachin’s salary.
Note: We also could have used backward calculation where we have taken two variables for two types of salaries. From the $20\%$ relation, we find the variable dependent on each other. We express them in the form of Sachin's salary. We take the incomes as x and y for Ricky Ponting and Sachin Tendulkar respectively. Now $y=x\left( 1+\dfrac{20}{100} \right)=\dfrac{6x}{5}$. We expressed x as a function of y. So, $x=\dfrac{5y}{6}$. Now we find the percentage of x and how much it’s less than that of y. So, x is \[\dfrac{y-x}{y}\times 100=\dfrac{y-\dfrac{5y}{6}}{y}\times 100=\dfrac{100}{6}=16.67\].
Complete step-by-step solution
The salary of Sachin Tendulkar is $20\%$ more than that of Ricky Ponting.
Let’s assume the salary of Ricky Ponting is x.
We can find the salary of Sachin Tendulkar which is $20\%$ more than x.
The salary of Sachin Tendulkar is $x\left( 1+\dfrac{20}{100} \right)=\dfrac{6x}{5}$.
So, the difference between their salaries is $\dfrac{6x}{5}-x=\dfrac{6x-5x}{5}=\dfrac{x}{5}$.
Now we need to find the percentage of Ricky’s salary less than that of Sachin’s.
So, we need to find the percentage of $\dfrac{x}{5}$ on $\dfrac{6x}{5}$.
So, the percentage is $\dfrac{\dfrac{x}{5}}{\dfrac{6x}{5}}\times 100$.
We solve the fraction and get $\dfrac{\dfrac{x}{5}}{\dfrac{6x}{5}}\times 100=\dfrac{x\times 5\times 100}{5\times 6x}=\dfrac{50}{3}=16.67$.
The nearest integer is $17\%.$
Therefore, Ricky’s salary is $17\%$ less than that of Sachin’s salary.
Note: We also could have used backward calculation where we have taken two variables for two types of salaries. From the $20\%$ relation, we find the variable dependent on each other. We express them in the form of Sachin's salary. We take the incomes as x and y for Ricky Ponting and Sachin Tendulkar respectively. Now $y=x\left( 1+\dfrac{20}{100} \right)=\dfrac{6x}{5}$. We expressed x as a function of y. So, $x=\dfrac{5y}{6}$. Now we find the percentage of x and how much it’s less than that of y. So, x is \[\dfrac{y-x}{y}\times 100=\dfrac{y-\dfrac{5y}{6}}{y}\times 100=\dfrac{100}{6}=16.67\].
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