Lesson: Working with Formulas

Thanks for the heads up! I've fixed that link.

Your approach to the question is perfectly valid.
However, we need to recognize that we're squaring the r term, but not squaring the h term.

So, if the radius = r and the height = h, the volume = (1/3)πr²h = K

If the radius = r/2 and the height = 2h, the volume = (1/3)π(r/2)²(2h)
= (1/3)π(r²/4)(2h)
= (1/3)π(r²/2)h
= (1/3)πr²h/2 = K/2

Does that help?

Cheers,
Brent

Question link: https://gmatclub.com/forum/if-c-is-the-temperature-in-degrees-celsius-an...

Unfortunately, that approach won't work.
To help understand why this is the case, let's see what would happen if the question said the temperature extremes differed by 32 degrees on the Fahrenheit scale (instead of 45 degrees).

If we plugged in F = 32, the equation, 9C = 5(F – 32), would become: 9C = 5(32 – 32).
When we solve that equation for C we get C = 0
In other words, a temperature difference of 32° Fahrenheit would be the equivalent of a temperature difference of 0° Celsius.
This makes no sense.

In my solution (https://gmatclub.com/forum/if-c-is-the-temperature-in-degrees-celsius-an...), I take two temperatures in Fahrenheit that differ by 45°(32° and 77°), convert those temperatures to Celsius, and then determine the temperature difference in Celsius.

Notice that we can use any two temperatures (in Fahrenheit) that differ by 45°
If you convert them to Celsius and then calculate the difference, you will always get a difference of 25° Celsius