Finding the original value of a quantity when final value and the ratio in which it is altered is given.

Question

Question: A liquid when heated expands $0.00236$ of it's original volume. If it is finally $624.7 cm^3$ after expanding, what was it before heating?

Answer 1

@Moo's comment gives the correct answer. I would like to explain the rationale underlying how we arrive at the equation.

"A liquid when heated expands 0.00236 (times) of its original volume".

When we say that its volume expands, we mean that the volume is larger than it was previously; therefore we must have the ratio of the new volume to the old volume is greater than 1: $$\frac{\text{new volume}}{\text{old volume}} > 1$$

Now obviously we have 0.00236 < 1, so this number cannot give the ratio of the new volume to the old volume. Instead, it must give the percentage increase in the volume.

$$\text{new volume} = \text{old volume} + \text{(percentage increase written as a decimal)}\times\text{(old volume)}$$ $$ = (1 + \text{percentage increase})\times(\text{old volume})$$ $$\implies \text{new volume} = (1 + \text{percentage increase})\times(\text{old volume})$$

Now this might not seem to help, since we don't know the old volume. However, we do know the new volume, and we do know the percentage increase. $$\text{new volume}=624.7$$ $$\text{percentage increase}=0.00236$$

Therefore, the old volume is the only unknown, and we can use the above equation to solve for it. We represent/symbolize the old volume with $x$ as a shorthand.

$$624.7 = (1+0.00236)x = (1.00236)x$$ $$\implies x = \frac{624.7}{1.00236}=623.2$$ $$\text{old volume} = 623.2$$