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Question Number 57284 by Jmasanja last updated on 01/Apr/19

y is varies directly as the square of x and  inversely as z.  if x is inceased by 10% and z  is   decreased by 20%, find the percentage  change in y.

$${y}\:{is}\:{varies}\:{directly}\:{as}\:{the}\:{square}\:{of}\:{x}\:{and} \\ $$$${inversely}\:{as}\:{z}. \\ $$$${if}\:{x}\:{is}\:{inceased}\:{by}\:\mathrm{10\%}\:{and}\:{z}\:\:{is}\: \\ $$$${decreased}\:{by}\:\mathrm{20\%},\:{find}\:{the}\:{percentage} \\ $$$${change}\:{in}\:{y}. \\ $$

Commented by mr W last updated on 01/Apr/19

x_2 =1.1x_1   z_2 =0.8z_1   (y_2 /y_1 )=((x_2 /x_1 ))^2 ((z_1 /z_2 ))=1.1^2 ×(1/(0.8))=1.5125  ⇒y is increased by 51.25%.

$${x}_{\mathrm{2}} =\mathrm{1}.\mathrm{1}{x}_{\mathrm{1}} \\ $$$${z}_{\mathrm{2}} =\mathrm{0}.\mathrm{8}{z}_{\mathrm{1}} \\ $$$$\frac{{y}_{\mathrm{2}} }{{y}_{\mathrm{1}} }=\left(\frac{{x}_{\mathrm{2}} }{{x}_{\mathrm{1}} }\right)^{\mathrm{2}} \left(\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\right)=\mathrm{1}.\mathrm{1}^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{0}.\mathrm{8}}=\mathrm{1}.\mathrm{5125} \\ $$$$\Rightarrow{y}\:{is}\:{increased}\:{by}\:\mathrm{51}.\mathrm{25\%}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19

y∝x^2   y∝(1/z)  combining  y∝(x^2 /z)→y=k×(x^2 /z)  let x changes from 100→110  z changes from 100→80  y_1 =((kx^2 )/z)  y_1 =((k×100^2 )/(100))  so y_1 =100k  now y_2 =((k×110^2 )/(80))  y_2 =((k×1210)/8)  △y=y_2 −y_1 =((1210k)/8)−((100k)/1)=((1210k−800k)/8)  △y=((410k)/8)  % increase of y is  ((△y)/y_1 )×100  =((410k)/(8×100k))×100  =((410)/8)=51.25% increase

$${y}\propto{x}^{\mathrm{2}} \\ $$$${y}\propto\frac{\mathrm{1}}{{z}} \\ $$$${combining} \\ $$$${y}\propto\frac{{x}^{\mathrm{2}} }{{z}}\rightarrow{y}={k}×\frac{{x}^{\mathrm{2}} }{{z}} \\ $$$${let}\:{x}\:{changes}\:{from}\:\mathrm{100}\rightarrow\mathrm{110} \\ $$$${z}\:{changes}\:{from}\:\mathrm{100}\rightarrow\mathrm{80} \\ $$$${y}_{\mathrm{1}} =\frac{{kx}^{\mathrm{2}} }{{z}} \\ $$$${y}_{\mathrm{1}} =\frac{{k}×\mathrm{100}^{\mathrm{2}} }{\mathrm{100}}\:\:{so}\:{y}_{\mathrm{1}} =\mathrm{100}{k} \\ $$$${now}\:{y}_{\mathrm{2}} =\frac{{k}×\mathrm{110}^{\mathrm{2}} }{\mathrm{80}} \\ $$$${y}_{\mathrm{2}} =\frac{{k}×\mathrm{1210}}{\mathrm{8}} \\ $$$$\bigtriangleup{y}={y}_{\mathrm{2}} −{y}_{\mathrm{1}} =\frac{\mathrm{1210}{k}}{\mathrm{8}}−\frac{\mathrm{100}{k}}{\mathrm{1}}=\frac{\mathrm{1210}{k}−\mathrm{800}{k}}{\mathrm{8}} \\ $$$$\bigtriangleup{y}=\frac{\mathrm{410}{k}}{\mathrm{8}} \\ $$$$\%\:{increase}\:{of}\:{y}\:{is}\:\:\frac{\bigtriangleup{y}}{{y}_{\mathrm{1}} }×\mathrm{100} \\ $$$$=\frac{\mathrm{410}{k}}{\mathrm{8}×\mathrm{100}{k}}×\mathrm{100} \\ $$$$=\frac{\mathrm{410}}{\mathrm{8}}=\mathrm{51}.\mathrm{25\%}\:{increase} \\ $$

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