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
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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