Question

successive discounts of 10%and 20% is equivalent to a single discount of
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Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

Let , the cost price of a product is = X Rs

Now after giving a 10%discount ,

it's selling price will become

$\implies \: selling_{1st} =( x - 10\% \: of \: x) \\ \implies \:selling_{1st} =x - \frac{10x}{100} \\ \implies \:selling_{1st} =x(1 - \frac{10}{100} ) \\ \implies \:selling_{1st} =x( \frac{100 - 10}{100}) \\ \implies \:selling_{1st} =x( \frac{90}{100} \\ \implies \boxed{selling_{1st} = \frac{9x}{10} }$

Now again giving a 20%discount on the price of

(9x/10) Rs

Selling_{2nd}=

$\implies \:Selling_{2nd}= \frac{9x}{10} - 20\% \: of \: \frac{9x}{10} \\ \implies \:Selling_{2nd}= \frac{9x}{10} (1 - \frac{20}{100} ) \\ \implies \:Selling_{2nd}= \frac{9x}{10} ( \frac{80}{100} ) \\ \implies \boxed{Selling_{2nd}= \frac{72x}{100} }$

therefore total discount

$\implies \: total \: discount =cost \: price - final \: selling \: price \\ \implies \: total \: discount =x - \frac{72x}{100} \\ \implies \: total \: discount = \frac{28x}{100}$

therefore in a single discount the discount percentage will be

$\implies \: dicount\% = \frac{dicount \: amount}{c.p} \times 100\% \\ \implies \: dicount\% = \frac{ \frac{28x}{100} }{x} \times 100\% \\ \implies \: dicount\% = \frac{28 \cancel x}{ \cancel{100} \cancel x} \times \cancel{100}\% \\ \implies \boxed{ \red{ dicount\% = 28\%}}$

$\therefore successive \:discounts\: of\: 10\%\:and\: 20\%\: \\is\: equivalent \:to \:a\: single\: discount \:of \:28\%$

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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