Question Number 68349 by peter frank last updated on 09/Sep/19
![](https://www.tinkutara.com/question/9228.png)
Answered by mr W last updated on 09/Sep/19
![w_(new) =1.03w (3% more=factor 1.03) d_(new) =0.975d (2.5% less=factor 0.975) t_(new) =1.04t (4% more=factor 1.04) y=((Kwd^4 )/t^3 ) y_(new) =((Kw_(new) d_(new) ^4 )/t_(new) ^3 ) =((K1.03w0.975^4 d^4 )/(1.04^3 t^3 ))=0.827((Kwd^4 )/t^3 ) =0.827y=82.7% of y 100%−82.7%=17.3% ⇒y is decreased by 17.3%](https://www.tinkutara.com/question/Q68358.png)
$${w}_{{new}} =\mathrm{1}.\mathrm{03}{w}\:\:\:\left(\mathrm{3\%}\:{more}={factor}\:\mathrm{1}.\mathrm{03}\right) \\ $$$${d}_{{new}} =\mathrm{0}.\mathrm{975}{d}\:\:\:\left(\mathrm{2}.\mathrm{5\%}\:{less}={factor}\:\mathrm{0}.\mathrm{975}\right) \\ $$$${t}_{{new}} =\mathrm{1}.\mathrm{04}{t}\:\:\left(\mathrm{4\%}\:{more}={factor}\:\mathrm{1}.\mathrm{04}\right) \\ $$$${y}=\frac{{Kwd}^{\mathrm{4}} }{{t}^{\mathrm{3}} } \\ $$$${y}_{{new}} =\frac{{Kw}_{{new}} {d}_{{new}} ^{\mathrm{4}} }{{t}_{{new}} ^{\mathrm{3}} } \\ $$$$=\frac{{K}\mathrm{1}.\mathrm{03}{w}\mathrm{0}.\mathrm{975}^{\mathrm{4}} {d}^{\mathrm{4}} }{\mathrm{1}.\mathrm{04}^{\mathrm{3}} {t}^{\mathrm{3}} }=\mathrm{0}.\mathrm{827}\frac{{Kwd}^{\mathrm{4}} }{{t}^{\mathrm{3}} } \\ $$$$=\mathrm{0}.\mathrm{827}{y}=\mathrm{82}.\mathrm{7\%}\:{of}\:{y} \\ $$$$\mathrm{100\%}−\mathrm{82}.\mathrm{7\%}=\mathrm{17}.\mathrm{3\%} \\ $$$$\Rightarrow{y}\:{is}\:{decreased}\:{by}\:\mathrm{17}.\mathrm{3\%} \\ $$
Commented by peter frank last updated on 09/Sep/19
![please sir elaborate](https://www.tinkutara.com/question/Q68364.png)
$${please}\:{sir}\:{elaborate} \\ $$
Commented by peter frank last updated on 21/Sep/19
![thank you](https://www.tinkutara.com/question/Q69216.png)
$${thank}\:{you} \\ $$