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Three-variables-u-v-and-w-are-related-such-that-u-varies-directly-as-v-and-inversely-as-the-square-of-w-If-v-increases-by-15-and-w-decreased-by-10-find-the-percentage-change-in-u-




Question Number 38636 by pieroo last updated on 27/Jun/18
Three variables u,v and w are related such that  u varies directly as v and inversely as the square of   w. If v increases by 15% and w decreased by 10%,  find the percentage change in u.
$$\mathrm{Three}\:\mathrm{variables}\:\mathrm{u},\mathrm{v}\:\mathrm{and}\:\mathrm{w}\:\mathrm{are}\:\mathrm{related}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{u}\:\mathrm{varies}\:\mathrm{directly}\:\mathrm{as}\:\mathrm{v}\:\mathrm{and}\:\mathrm{inversely}\:\mathrm{as}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\: \\ $$$$\mathrm{w}.\:\mathrm{If}\:\mathrm{v}\:\mathrm{increases}\:\mathrm{by}\:\mathrm{15\%}\:\mathrm{and}\:\mathrm{w}\:\mathrm{decreased}\:\mathrm{by}\:\mathrm{10\%}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{percentage}\:\mathrm{change}\:\mathrm{in}\:\mathrm{u}. \\ $$
Answered by MrW3 last updated on 28/Jun/18
u=k(v/w^2 )  v_2 =1.15v_1  (increase by 15%)  w_2 =0.90w_1  (decrease by 10%)  u_2 =k((1.15v_1 )/(0.90^2 w_1 ^2 ))=((1.15)/(0.90^2 ))×k(v_1 /w_1 ^2 )=((1.15)/(0.90^2 ))u_1 =1.42u_1   ⇒u increases by 42%.
$${u}={k}\frac{{v}}{{w}^{\mathrm{2}} } \\ $$$${v}_{\mathrm{2}} =\mathrm{1}.\mathrm{15}{v}_{\mathrm{1}} \:\left({increase}\:{by}\:\mathrm{15\%}\right) \\ $$$${w}_{\mathrm{2}} =\mathrm{0}.\mathrm{90}{w}_{\mathrm{1}} \:\left({decrease}\:{by}\:\mathrm{10\%}\right) \\ $$$${u}_{\mathrm{2}} ={k}\frac{\mathrm{1}.\mathrm{15}{v}_{\mathrm{1}} }{\mathrm{0}.\mathrm{90}^{\mathrm{2}} {w}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{\mathrm{1}.\mathrm{15}}{\mathrm{0}.\mathrm{90}^{\mathrm{2}} }×{k}\frac{{v}_{\mathrm{1}} }{{w}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{\mathrm{1}.\mathrm{15}}{\mathrm{0}.\mathrm{90}^{\mathrm{2}} }{u}_{\mathrm{1}} =\mathrm{1}.\mathrm{42}{u}_{\mathrm{1}} \\ $$$$\Rightarrow{u}\:{increases}\:{by}\:\mathrm{42\%}. \\ $$
Commented by pieroo last updated on 28/Jun/18
Thanks Sir.
$$\mathrm{Thanks}\:\mathrm{Sir}. \\ $$

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