Question

Question:

In a shop advertising a cash discount of 10% over and above a Summer Discount of 15%, a sales clerk wrongly allowed a single discount of 25% to sell a frock for 562.50. At what price should the frock have been sold?


Please do the solution with full explanation and in detail.​

Answer 1

$\large\underline{\sf{Solution-}}$

First of all, we find the Marked Price of the frock, by using the data, that a sales clerk wrongly allowed a single discount of 25% to sell a frock for 562.50.

So, we have from this,

Selling Price of frock = 552.50

Discount % = 25 %

Let assume that Marked Price of the frock be x

So, We know

Marked Price, Selling Price and Discount % are connected by the relationship

$\boxed{\tt{ Marked \: Price \: = \: \frac{100 \: \times \: Selling \: Price}{100 \: - \: Discount\%} \: }}$

So, on substituting the values, we get

$\rm \: x = \dfrac{100 \times 562.50}{100 - 25}$

$\rm \: x = \dfrac{56250}{75}$

$\rm\implies \:x = 750$

So, Marked Price of the frock is 750.

Now, we have to find the Selling Price of the frock when two successive discounts of 10 % and 15 % is given.

We know,

Selling Price of an Article having Marked Price (MP) having two successive discounts of a % and b % is given by

$\boxed{\tt{ Selling \: Price = MP\bigg[1 - \dfrac{a}{100} \bigg]\bigg[1 - \dfrac{b}{100} \bigg] \: }}$

So, on substituting the values

Marked Price, MP = 750

a = 10

b = 15

$\rm \: Selling \: Price = 750\bigg(1 - \dfrac{10}{100} \bigg) \bigg(1 - \dfrac{15}{100} \bigg)$

$\rm \: Selling \: Price = 750\bigg( 1- \dfrac{1}{10} \bigg) \bigg(1 - \dfrac{3}{20} \bigg)$

$\rm \: Selling \: Price = 750\bigg( \dfrac{10 - 1}{10} \bigg) \bigg(\dfrac{20 - 3}{20} \bigg)$

$\rm \: Selling \: Price = 750\bigg( \dfrac{9}{10} \bigg) \bigg(\dfrac{17}{20} \bigg)$

$\rm\implies \:\rm \: Selling \: Price = 573.75$

So, Selling Price of the frock is 573.75

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$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: to \:know}}}} \\ \\ \bigstar \: \bf{Gain = \sf S.P. \: – \: C.P.} \\ \\ \bigstar \:\bf{Loss = \sf C.P. \: – \: S.P.} \\ \\ \bigstar \: \bf{Gain \: \% = \sf \Bigg( \dfrac{Gain}{C.P.} \times 100 \Bigg)\%} \\ \\ \bigstar \: \bf{Loss \: \% = \sf \Bigg( \dfrac{Loss}{C.P.} \times 100 \Bigg )\%} \\ \\ \\ \bigstar \: \bf{S.P. = \sf\dfrac{(100+Gain\%) or(100-Loss\%)}{100} \times C.P.} \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$