If one number is 80 % of the other and the su | vedclass

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Question

(C) Let the two numbers be $x$ and $y$.According to the problem,one number is $80 \%$ of the other:$y = \frac{80}{100} x = \frac{4}{5} x$This implies

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$16, 20$

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(C) Let the two numbers be $x$ and $y$.According to the problem,one number is $80 \%$ of the other:$y = \frac{80}{100} x = \frac{4}{5} x$This implies $\frac{x}{y} = \frac{5}{4}$. Let $x = 5k$ and $y = 4k$ for some constant $k$.The sum of their squares is given as $656$:$x^2 + y^2 = 656$$(5k)^2 + (4k)^2 = 656$$25k^2 + 16k^2 = 656$$41k^2 = 656$$k^2 = \frac{656}{41} = 16$$k = 4$Substituting the value of $k$ back into the expressions for $x$ and $y$:$x = 5 	imes 4 = 20$$y = 4 	imes 4 = 16$Thus,the numbers are $16$ and $20$.

If one number is $80 \%$ of the other and the sum of their squares is $656$,then the numbers are:

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