How to calculate double time percent increase?

Question

Suppose that I have initial value 88. Now we increase 88 a certain amount of percentage and then also we increase another certain amount of percentage. And the final value is 112. What percent I have to increase in first time and second time to achieve 112. How could I solve this question easily.

Answer 1

Recall that to increase a quantity $Q$ by a percentage $p$ you multiply by $(1+p)$ because to increase $Q_0$ to $Q_1$ by a percentage $p$:

$$Q_1=Q_0+\Delta Q=Q_0+pQ_0=Q_0(1+p).$$

My guess is that the two percentages are the same, $x$, so you want:

$$88\longrightarrow 88(1+x)\longrightarrow 88(1+x)(1+x)=88(1+x)^2\overset{!}{=}112,$$

so you just solve

$$88(1+x)^2=112.$$

If you want really want the two percentages to be different you have an infinite number of solutions. Let $p_1$ and $p_2$ be the two percentages so that:

$$88(1+p_1)(1+p_2)=112.$$

You can solve for $p_2$ in terms of $p_1$ for all $p_1\neq -1$.

Answer 2

At the end of the first increase be by $x \%$: the result is $(88+\frac{x}{100}\times 88)$

Let the second increase be by $y \%$: $(88+\frac{x}{100}\times 88)\times \frac{y}{100}$

At the end of the second increase the result is

$(88+\frac{x}{100}\times 88)+\frac{y}{100}\times (88+\frac{x}{100}\times 88)=112$

For example: if $x=10 \%$ then $96.8+\frac{y}{100}\times 96.8=112$

$1+\frac{y}{100}=\frac{112}{96.8}\Rightarrow y=15.70$