 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} Verified
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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 $20\%$.

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 $p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100$.
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$p=\left( 1-\prod\limits_{i=1}^{n}{\left( \dfrac{100-{{x}_{i}}}{100} \right)} \right)\times 100$.
\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 $28\%$.
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.