If-a-seller-could-purchase-goods-at-an-8-lower-price-by-keeping-the-selling-price-fixed-his-profit-over-the-cost-price-would-increase-from-the-current-x-to-x-10-So-what-is-the-value-of-x-

Question Number 71439 by Mr. K last updated on 15/Oct/19

$${If}\:{a}\:{seller}\:{could}\:{purchase}\:{goods}\:{at}\:{an} \\ $$$$\mathrm{8\%}\:{lower}\:{price}\:{by}\:{keeping}\:{the}\:{selling} \\ $$$${price}\:{fixed},\:{his}\:{profit}\:{over}\:{the}\:{cost} \\ $$$${price}\:{would}\:{increase}\:{from}\:{the}\: \\ $$$${current}\:{x\%}\:{to}\:\left({x}+\mathrm{10}\right)\%.\:{So},\:{what}\:{is} \\ $$$${the}\:{value}\:{of}\:{x}? \\ $$

Commented by ajfour last updated on 16/Oct/19

his profit over the cost price would increase from the current x% to (x+10)%. (its given so in question) c_(new) =(100%−8%)c=0.92c

$${his}\:{profit}\:{over}\:{the}\:{cost} \\ $$$${price}\:{would}\:{increase}\:{from}\:{the}\: \\ $$$${current}\:{x\%}\:{to}\:\left({x}+\mathrm{10}\right)\%. \\ $$$$\left({its}\:{given}\:{so}\:{in}\:{question}\right) \\ $$$${c}_{{new}} =\left(\mathrm{100\%}−\mathrm{8\%}\right){c}=\mathrm{0}.\mathrm{92}{c} \\ $$

Answered by ajfour last updated on 15/Oct/19

((s−0.92c)/(0.92c))−((s−c)/c)=0.1 where x=(((s−c)/c))×100 ⇒ (s/c)=(x/(100))+1 ⇒ (x/(100))+1−0.92−((0.92x)/(100))=0.092 ⇒ ((0.08x)/(100))=0.012 ⇒ x= 15 .

$$\:\:\frac{{s}−\mathrm{0}.\mathrm{92}{c}}{\mathrm{0}.\mathrm{92}{c}}−\frac{{s}−{c}}{{c}}=\mathrm{0}.\mathrm{1} \\ $$$${where}\:\:\:{x}=\left(\frac{{s}−{c}}{{c}}\right)×\mathrm{100} \\ $$$$\Rightarrow\:\:\frac{{s}}{{c}}=\frac{{x}}{\mathrm{100}}+\mathrm{1} \\ $$$$\Rightarrow\:\frac{{x}}{\mathrm{100}}+\mathrm{1}−\mathrm{0}.\mathrm{92}−\frac{\mathrm{0}.\mathrm{92}{x}}{\mathrm{100}}=\mathrm{0}.\mathrm{092} \\ $$$$\Rightarrow\:\frac{\mathrm{0}.\mathrm{08}{x}}{\mathrm{100}}=\mathrm{0}.\mathrm{012}\:\:\Rightarrow\:\:{x}=\:\mathrm{15}\:. \\ $$

Commented by Mr. K last updated on 15/Oct/19

$${How}\:{did}\:{you}\:{develop}\:{the}\:{first}\: \\ $$$${expression}? \\ $$

Commented by ajfour last updated on 15/Oct/19

p%=((s−c)/c)×100 new cost price = ((92)/(100))×c = 0.92c new p%−old p%=10 ⇒ (((s−0.92c)/(0.92c)))×100−(((s−c)/c))×100=10 ⇒ ((s−0.92c)/(0.92c))−((s−c)/c)= 0.1

$${p\%}=\frac{{s}−{c}}{{c}}×\mathrm{100} \\ $$$${new}\:{cost}\:{price}\:=\:\frac{\mathrm{92}}{\mathrm{100}}×{c}\:=\:\mathrm{0}.\mathrm{92}{c} \\ $$$${new}\:{p\%}−{old}\:{p\%}=\mathrm{10} \\ $$$$\Rightarrow\:\:\left(\frac{{s}−\mathrm{0}.\mathrm{92}{c}}{\mathrm{0}.\mathrm{92}{c}}\right)×\mathrm{100}−\left(\frac{{s}−{c}}{{c}}\right)×\mathrm{100}=\mathrm{10} \\ $$$$\Rightarrow\:\frac{{s}−\mathrm{0}.\mathrm{92}{c}}{\mathrm{0}.\mathrm{92}{c}}−\frac{{s}−{c}}{{c}}=\:\mathrm{0}.\mathrm{1}\: \\ $$

Commented by Mr. K last updated on 15/Oct/19