Question #1cb8f

2 Answers
Ratnaker Mehta
Oct 1, 2017

#"Decrease by "10.4%.#

Explanation:

We are given that,

#z prop x*y^2. :. z=kxy^2...(1); k" is const., "kne0.#

Let, the initial values of #x=x_0=100, and, y=y_0=100.#

Therefore, initially, #z=z_0=k*100*100^2=10^6k...[because, (1)].#

Now, #x_0 uarr" by "40%, and, y darr" by "20%.#

If the corresponding values are, #x_1,y_1, and, z_1,# then,

#x_1=140, y_1=80. :. (1) rArr z_1=k140*80^2=896xx10^3k.#

#:. z_1-z_0=896xx10^3k-10^3xx10^3k=-104xx10^3k.#

Since, #z_1-z_0 < 0,# there is decrease in #z.#

The Percentage Decrease, in #z,# is,

#((z_1-z_0)/z_0)xx100=((104xx10^3k)/(10^3xx10^3k))xx100=10.4%.#

Enjoy Maths.!

Jim G.
Oct 1, 2017

#"z decreases by "10.4%#

Explanation:

#"the initial statement is "zpropxy^2#

#"to convert to an equation multiply by k the constant of"#
#"variation"#

#rArrz=kxy^2#

#"now "xto1.4" and "yto0.8#

#rArrz=kxx1.4xx(0.8)^2#

#color(white)(rArrz)=0.896k#

#"change in z "=1-0.896=0.104=10.4%#

#"that is a decrease of "10.4%#

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