A single discount is equal to discount series of \[10\%\] and \[20\%\] is
\[\begin{align}
  & \text{A}\text{. 25 }\!\!\%\!\!\text{ } \\
 & \text{B}\text{. 30 }\!\!\%\!\!\text{ } \\
 & \text{C}\text{. 35 }\!\!\%\!\!\text{ } \\
 & \text{D}\text{. 28 }\!\!\%\!\!\text{ } \\
\end{align}\]

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Hint: We know that if p is the single discount of discount series of \[{{x}_{1}},{{x}_{2}},......,{{x}_{n}}\]., then \[p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100\]. From the question, it is given to find the single discount of the discount series of \[10\%\] and \[20\%\]. By using the above formula, we can find the single discount of the discount series of \[10\%\] and \[20\%\].

Complete step-by-step answer:
Before solving the question, we should know that if p is the single discount of discount series of \[{{x}_{1}},{{x}_{2}},......,{{x}_{n}}\]., then \[p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100\].
Now by using this formula, we can find the single discount of a certain discount series.
From the question, it is clear that we have to find the discount series of \[10\%\] and \[20\%\].
We know that if p is the single discount of discount series of \[{{x}_{1}},{{x}_{2}},......,{{x}_{n}}\]., then\[p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100\].
So, we get
\[\begin{align}
  & \Rightarrow p=\left( 1-\left( \dfrac{100-10}{100} \right)\left( \dfrac{100-20}{100} \right) \right)\times 100 \\
 & \Rightarrow p=\left( 1-\left( \dfrac{90}{100} \right)\left( \dfrac{80}{100} \right) \right)\times 100 \\
 & \Rightarrow p=\left( 1-\dfrac{72}{100} \right)\times 100 \\
 & \Rightarrow p=\left( \dfrac{100-72}{100} \right)\times 100 \\
 & \Rightarrow p=\left( \dfrac{28}{100} \right)\times 100 \\
 & \Rightarrow p=28.....(1) \\
\end{align}\]
From equation (1), it is clear that the value of p is equal to 28. So, we can say that a single discount is equal to the discount series of \[10\%\] and \[20\%\] is \[28\%\].
So, the correct answer is “Option D”.

Note: Students may have a misconception that if p is the single discount of discount series of \[{{x}_{1}},{{x}_{2}},......,{{x}_{n}}\]., then \[p=\left( \prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100\]. If this misconception is followed, then we get
\[\begin{align}
  & \Rightarrow p=\left( \left( \dfrac{100-10}{100} \right)\left( \dfrac{100-20}{100} \right) \right)\times 100 \\
 & \Rightarrow p=\left( \left( \dfrac{90}{100} \right)\left( \dfrac{80}{100} \right) \right)\times 100 \\
 & \Rightarrow p=\left( \dfrac{72}{100} \right)\times 100 \\
 & \Rightarrow p=72.....(1) \\
\end{align}\]
From equation (1), it is clear that the value of p is equal to 28. So, we can say that a single discount is equal to the discount series of \[10\%\] and \[20\%\] is \[72\%\]. But we know that the single discount is equal to \[28\%\]. So, this misconception should get avoided.