Question 82

x varies inversely as $$(2y^2 - 5)$$ and is equal to 25 when y = 15. Find the value of x when y = 10 (approximate value).

Solution

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Since x is inversely proportional to  $$(2y^2 - 5)$$, we get the equation as x= $$\frac{k\ }{2y^2-5}$$ where k is the proportionality constant.

Now, x= 25 when y=15. Putting this in the above equation, we get the value of k as 11,125.

When y= 10, x = $$\frac{11125\ }{200-5}$$, which is approximately equal to 57.

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