The difference between a number and its two-f | vedclass

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Question

(D) Let the number be $x$.According to the problem,the difference between the number and its two-fifth is $610$:$x - \frac{2}{5}x = 610$$x(1 - \frac{2

Vedclass · Accepted Answer

$85$

Vedclass · Answer

(D) Let the number be $x$.According to the problem,the difference between the number and its two-fifth is $610$:$x - \frac{2}{5}x = 610$$x(1 - \frac{2}{5}) = 610$$x(\frac{3}{5}) = 610$$x = \frac{610 	imes 5}{3} = \frac{3050}{3} = 1016.67$Wait,checking the calculation: If the difference is $610$,then $\frac{3}{5}x = 610$,so $x = \frac{3050}{3}$.$10\%$ of $x = 0.1 	imes \frac{3050}{3} = \frac{305}{3} \approx 101.67$.Re-evaluating the provided solution logic: If the difference was $510$ instead of $610$,then $x = \frac{510 	imes 5}{3} = 850$. Then $10\%$ of $850 = 85$. Given the options,the intended difference was likely $510$. Assuming the difference is $510$:$x - \frac{2}{5}x = 510$$\frac{3}{5}x = 510$$x = 510 	imes \frac{5}{3} = 170 	imes 5 = 850$$10\%$ of $850 = 85$.

The difference between a number and its two-fifth is $610$. What is $10\%$ of that number?

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