buy selling a flower pot for 528 a woman gains 20% at what price should selling to gain 25%

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1. ➤ Given :-

   Selling price of the pot :- ₹528

   Gain percentage :- 20%

   [tex]\:[/tex]

   ➤ To Find :-

   The selling price if it’s sold at 25%

   [tex]\:[/tex]

   ★ How to do :-

   Here, we are given with the selling price and the gain percentage obtained for that in the first statement. In the other statement, we are given with an other gain percentage and we are asked to find the selling price if it’s sold at a different gain percentage. So, first we should find the cost price of the flower pot by using the information in first statement. Later, we can find the cost price by applying the different profit percentage. The particular formulas needed to be applied are given while solving. So, let’s solve!!

   [tex]\:[/tex]

   ➤ Solution :-

   Cost price of the pot :–

   [tex]{\tt \leadsto \underline{\boxed{\tt \dfrac{100}{(100 + Gain \bf\%)} \times SP}}}[/tex]

   Substitute thr given values.

   [tex]{\tt \leadsto \dfrac{100}{(100 + 20)} \times 528}[/tex]

   First, solve the bracket given in denominator.

   [tex]{\tt \leadsto \dfrac{100}{120} \times 528}[/tex]

   Write the fraction in lowest form by cancellation method.

   [tex]{\tt \leadsto \cancel \dfrac{100}{120} \times 528 = \dfrac{5}{6} \times 528}[/tex]

   Write the denominator and the whole numbers in lowest form by cancellation method.

   [tex]{\tt \leadsto \dfrac{5}{\cancel{6}} \times \cancel{528} = \dfrac{5}{1} \times 88}[/tex]

   Now, multiply the remaining fractions.

   [tex]{\tt \leadsto \dfrac{5 \times 88}{1}}[/tex]

   Multiply to get the cost price.

   [tex]{\tt \leadsto \cancel \dfrac{440}{1} = \underline{440}}[/tex]

   [tex]\:[/tex]

   Now, let’s find the selling price of the second statement applied.

   Selling price of the pot :–

   [tex]{\tt \leadsto \underline{\boxed{\tt \dfrac{(100 + Gain \bf\%)}{100} \times CP}}}[/tex]

   Substitute the values.

   [tex]{\tt \leadsto \dfrac{(100 + 25)}{100} \times 440}[/tex]

   First, solve the bracket given in the numerator.

   [tex]{\tt \leadsto \dfrac{125}{100} \times 440}[/tex]

   Write the fraction in lowest form by cancellation method.

   [tex]{\tt \leadsto \cancel \dfrac{125}{100} \times 440 = \dfrac{5}{4} \times 440}[/tex]

   Write the denominator and the whole number in lowest form by cancellation method.

   [tex]{\tt \leadsto \dfrac{5}{\cancel{4}} \times \cancel{440} = \dfrac{5}{1} \times 110}[/tex]

   Now, multiply the remaining fractions.

   [tex]{\tt \leadsto \dfrac{5 \times 110}{1}}[/tex]

   Multiply to get the final answer.

   [tex]{\tt \leadsto \cancel \dfrac{550}{1} = \pink{\underline{\boxed{\tt Rs \: \: 550}}}}[/tex]

   [tex]\Huge\therefore[/tex] The selling price of the pot when sold at 25% gain is ₹550.

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   [tex]\dashrightarrow [/tex] Some related formulas :-

   [tex]\small\boxed{\begin{array}{cc}\large\sf\dag \: {\underline{More \: Formulae}} \\ \\ \bigstar \: \sf{Gain = S.P – C.P} \\ \\ \bigstar \:\sf{Loss = C.P – S.P} \\ \\ \bigstar \: \sf{Gain \: \% = \Bigg( \dfrac{Gain}{C.P} \times 100 \Bigg)\%} \\ \\ \bigstar \: \sf{loss \: \% = \Bigg( \dfrac{loss}{C.P} \times 100 \Bigg)\%} \\ \\ \bigstar \: \sf{S.P = \dfrac{100-loss\%}{100} \times C.P} \\ \\ \bigstar \: \sf{C.P =\dfrac{100}{100-loss\%} \times S.P}\end{array}}[/tex]

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